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Learning Objectives
• Understand what it means for a distribution to balance on a fulcrum
• Learn which measure of central tendency will balance a distribution
Instructions
• The box plot in the simulation is based on the data shown to the left of the box plot.
• Notice that various aspects of the box plot such as the mean and median are labeled.
• These labels can be hidden by unchecking the "show labels on box plot" button.
• Beneath the data is a pair of buttons that let you specify whether you want to be able to enter data or to see statistics based on the data.
• When you use the simulation, try to modify the data in various ways and see how it affects the box plot.
• Try putting in some extreme values and see if they get labeled as outliers.
• Outside values are shown as "\(o's\)" beyond the inner fence.
• Far out values are shown as \(*'s\) outside the outer fence.
• You can delete all the data (by pressing the "Clear All button") and enter your own data.
• Your data can be typed in or pasted in from another application.
• When pasting, use keyboard shortcut for pasting (Command-V for Mac, CTRL-V for Windows).
• When you change the data, the box plot will disappear.
After you have entered new data, click the "Draw box plot" button to redraw the box plot.
Illustrated Instructions
The screenshot below shows the box plot simulation with its default data on the left.
You can change some or all of the data on the left by editing individual numbers or pasting in your own data. When pasting, you must use the keyboard shortcut for pasting (Command-V for Mac, CTRL-V for Windows).
2.08: Bar Charts
Learning Objectives
• Create and interpret bar charts
• Judge whether a bar chart or another graph such as a box plot would be more appropriate
In the section on qualitative variables, we saw how bar charts could be used to illustrate the frequencies of different categories. For example, the bar chart shown in Figure $1$ shows how many purchasers of iMac computers were previous Macintosh users, previous Windows users, and new computer purchasers.
In this section, we show how bar charts can be used to present other kinds of quantitative information, not just frequency counts. The bar chart in Figure $2$ shows the percent increases in the Dow Jones, Standard and Poor 500 (S & P), and Nasdaq stock indexes from $\text{May 24 2000 to May 24 2001}$. Notice that both the S & P and the Nasdaq had “negative increases” which means that they decreased in value. In this bar chart, the $Y$-axis is not frequency but rather the signed quantity percentage increase.
Bar charts are particularly effective for showing change over time. Figure $3$, for example, shows the percent increase in the Consumer Price Index (CPI) over four three-month periods. The fluctuation in inflation is apparent in the graph.
Bar charts are often used to compare the means of different experimental conditions. Figure $4$ shows the mean time it took one of us (DL) to move the mouse to either a small target or a large target. On average, more time was required for small targets than for large ones.
Although bar charts can display means, we do not recommend them for this purpose. Box plots should be used instead since they provide more information than bar charts without taking up more space. For example, a box plot of the mouse-movement data is shown in Figure $5$. You can see that Figure $5$ reveals more about the distribution of movement times than does Figure $4$.
The section on qualitative variables presented earlier in this chapter discussed the use of bar charts for comparing distributions. Some common graphical mistakes were also noted. The earlier discussion applies equally well to the use of bar charts to display quantitative variables.
2.09: Line Graphs
Learning Objectives
• Create and interpret line graphs
• Judge whether a line graph would be appropriate for a given data set
A line graph is a bar graph with the tops of the bars represented by points joined by lines (the rest of the bar is suppressed). For example, Figure \(1\) was presented in the section on bar charts and shows changes in the Consumer Price Index (CPI) over time.
A line graph of these same data is shown in Figure \(2\). Although the figures are similar, the line graph emphasizes the change from period to period.
Line graphs are appropriate only when both the \(X\)- and \(Y\)-axes display ordered (rather than qualitative) variables. Although bar graphs can also be used in this situation, line graphs are generally better at comparing changes over time. Figure \(3\), for example, shows percent increases and decreases in five components of the Consumer Price Index (CPI). The figure makes it easy to see that medical costs had a steadier progression than the other components. Although you could create an analogous bar chart, its interpretation would not be as easy.
Let us stress that it is misleading to use a line graph when the \(X\)-axis contains merely qualitative variables. Figure \(4\) inappropriately shows a line graph of the card game data from Yahoo, discussed in the section on qualitative variables. The defect in Figure \(4\) is that it gives the false impression that the games are naturally ordered in a numerical way. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/02%3A_Graphing_Distributions/2.07%3A_Box_Plot_Demo.txt |
Learning Objectives
• Create and interpret dot plots
• Judge whether a dot plot would be appropriate for a given data set
Dot plots can be used to display various types of information. Figure \(1\) uses a dot plot to display the number of M & M's of each color found in a bag of M & M's. Each dot represents a single M & M. From the figure, you can see that there were \(3\) blue M & M's, \(19\) brown M & M's, etc.
The dot plot in Figure \(2\) shows the number of people playing various card games on the Yahoo website on a Wednesday. Unlike Figure \(1\), the location rather than the number of dots represents the frequency.
The dot plot in Figure \(3\) shows the number of people playing on a Sunday and on a Wednesday. This graph makes it easy to compare the popularity of the games separately for the two days, but does not make it easy to compare the popularity of a given game on the two days.
The dot plot in Figure \(4\) makes it easy to compare the days of the week for specific games while still portraying differences among games.
2.11: Statistical Literacy
Are Commercial Vehicles in Texas Unsafe?
Prerequisites
Graphing Distributions
A news report on the safety of commercial vehicles in Texas stated that one out of five commercial vehicles have been pulled off the road in \(2012\) because they were unsafe. In addition, \(12,301\) commercial drivers have been banned from the road for safety violations.
The author presents the bar chart below to provide information about the percentage of fatal crashes involving commercial vehicles in Texas since \(2006\). The author also quotes DPS director Steven McCraw:
Commercial vehicles are responsible for approximately \(15\) percent of the fatalities in Texas crashes. Those who choose to drive unsafe commercial vehicles or drive a commercial vehicle unsafely pose a serious threat to the motoring public.
Example \(1\)
Based on what you have learned in this chapter, does this bar chart below provide enough information to conclude that unsafe or unsafely driven commercial vehicles pose a serious threat to the motoring public? What might you conclude if \(30\) percent of all the vehicles on the roads of Texas in \(2010\) were commercial and accounted for \(16\) percent of fatal crashes?
Solution
This bar chart does not provide enough information to draw such a conclusion because we don’t know, on the average, in a given year what percentage of all vehicles on the road are commercial vehicles. For example, if \(30\) percent of all the vehicles on the roads of Texas in \(2010\) are commercial ones and only \(16\) percent of fatal crashes involved commercial vehicles, then commercial vehicles are safer than non-commercial ones. Note that in this case \(70\) percent of vehicles are non-commercial and they are responsible for \(84\) percent of the fatal crashes.
Linear By Design
Example \(2\)
Fox News aired the line graph below showing the number unemployed during four quarters between \(2007\) and \(2010\).
Does Fox News' line graph provide misleading information? Why or Why not?
Solution:
There are major flaws with the Fox News graph. First, the title of the graph is misleading. Although the data show the number unemployed, Fox News’ graph is titled "Job Loss by Quarter." Second, the intervals on the \(X\)-axis are misleading. Although there are \(6\) months between September \(2008\) and March \(2009\) and \(15\) months between March \(2009\) and June \(2010\), the intervals are represented in the graph by very similar lengths. This gives the false impression that unemployment increased steadily.
The graph presented below is corrected so that distances on the \(X\)-axis are proportional to the number of days between the dates. This graph shows clearly that the rate of increase in the number unemployed is greater between September \(2008\) and March \(2009\) than it is between March \(2009\) and June \(2010\).
Contributors and Attributions
• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.
• Seyd Ercan and David Lane | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/02%3A_Graphing_Distributions/2.10%3A_Dot_Plots.txt |
General Questions
Q1
Name some ways to graph quantitative variables and some ways to graph qualitative variables. (relevant section & relevant section)
Q2
Based on the frequency polygon displayed below, the most common test grade was around what score? Explain. (relevant section)
Q3
An experiment compared the ability of three groups of participants to remember briefly-presented chess positions. The data are shown below. The numbers represent the total number of pieces correctly remembered from three chess positions. Create side-by-side box plots for these three groups. What can you say about the differences between these groups from the box plots? (relevant section)
Non-players
Beginners
Tournament players
22.1
32.5
40.1
22.3
37.1
45.6
26.2
39.1
51.2
29.6
40.5
56.4
31.7
45.5
58.1
33.5
51.3
71.1
38.9
52.6
74.9
39.7
55.7
75.9
43.2
55.9
80.3
43.2
57.7
85.3
Q4
You have to decide between displaying your data with a histogram or with a stem and leaf display. What factor(s) would affect your choice? (relevant section & relevant section)
Q5
In a box plot, what percent of the scores are between the lower and upper hinges? (relevant section)
Q6
A student has decided to display the results of his project on the number of hours people in various countries slept per night. He compared the sleeping patterns of people from the US, Brazil, France, Turkey, China, Egypt, Canada, Norway, and Spain. He was planning on using a line graph to display this data. Is a line graph appropriate? What might be a better choice for a graph? (relevant section & relevant section)
Q7
For the data from the $1977$ $\text{Stat. and Biom. 200 }$ class for eye color, construct: (relevant section)
1. pie graph
2. horizontal bar graph
3. vertical bar graph
4. a frequency table with the relative frequency of each eye color
Eye Color
Number of students
Brown
11
Blue
10
Green
4
Gray
1
(Question submitted by J. Warren, UNH)
Q8
A graph appears below showing the number of adults and children who prefer each type of soda. There were $130$ adults and kids surveyed. Discuss some ways in which the graph below could be improved. (relevant section)
Q9
Which of the box plots below has a large positive skew? Which has a large negative skew? (relevant section & relevant section)
Questions from Case Studies
The following questions are from the Angry Moods (AM) case study.
Q10
(AM#6) Is there a difference in how much males and females use aggressive behavior to improve an angry mood? For the "Anger-Out" scores:
1. Create parallel box plots. (relevant section)
2. Create a back to back stem and leaf displays (You may have trouble finding a computer to do this so you may have to do it by hand.) (relevant section)
Q11
(AM#9) Create parallel box plots for the Anger-In scores by sports participation. (relevant section)
Q12
(AM#11) Plot a histogram of the distribution of the Control-Out scores. (relevant section)
Q13
(AM#14) Create a bar graph comparing the mean Control-In score for the athletes and the non-athletes. What would be a better way to display this data? (relevant section)
Q14
(AM#18) Plot parallel box plots of the Anger Expression Index by sports participation. Does it look like there are any outliers? Which group reported expressing more anger? (relevant section)
The following questions are from the Flatulence (F) case study.
Q15
(F#1) Plot a histogram of the variable "per day." (relevant section)
Q16
(F#7) Create parallel box plots of "how long" as a function gender. Why is the $25^{th}$ percentile not showing? What can you say about the results? (relevant section)
Q17
(F#9) Create a stem and leaf plot of the variable "how long" What can you say about the shape of the distribution? (relevant section.1)
The following questions are from the Physicians' Reactions (PR) case study.
Q18
(PR#1) Create box plots comparing the time expected to be spent with the average-weight and overweight patients. (relevant section)
Q19
(PR#4) Plot histograms of the time spent with the average-weight and overweight patients. (relevant section)
Q20
(PR#5) To which group does the patient with the highest expected time belong?
The following questions are from the Smiles and Leniency (SL) case study
Q21
(SL#1) Create parallel box plots for the four conditions. (relevant section)
Q22
(SL#3) Create back to back stem and leaf displays for the false smile and neutral conditions. (It may be hard to find a computer program to do this for you, so be prepared to do it by hand). (relevant section)
The following questions are from the ADHD Treatment (AT) case study.
Q23
(AT#3) Create a line graph of the data. Do certain dosages appear to be more effective than others? (relevant section)
Q24
(AT#5) Create a stem and leaf plot of the number of correct responses of the participants after taking the placebo ($d0$ variable). What can you say about the shape of the distribution? (relevant section)
Q25
Create box plots for the four conditions. You may have to rearrange the data to get a computer program to create the box plots.
The following question is from the SAT and College GPA case study.
Q26
Create histograms and stem and leaf displays of both high-school grade point average and university grade point average. In what way(s) do the distributions differ?
Q27
The $\text{April 10th}$ issue of the Journal of the American Medical Association reports a study on the effects of anti-depressants. The study involved $340$ subjects who were being treated for major depression. The subjects were randomly assigned to receive one of three treatments: St. John's wort (an herb), Zoloft (Pfizer's cousin of Lilly's Prozac) or placebo for an $8$-week period. The following are the mean scores (approximately) for the three groups of subjects over the eight-week experiment. The first column is the baseline. Lower scores mean less depression. Create a graph to display these means.
Placebo 22.5 19.1 17.9 17.1 16.2 15.1 12.1 12.3
Wort 23.0 20.2 18.2 18.0 16.5 16.1 14.2 13.0
Zoloft 22.4 19.2 16.6 15.5 14.2 13.1 11.8 10.5
The following questions are from . Visit the site
Q28
For the graph below, of heights of singers in a large chorus, please write a complete description of the histogram. Be sure to comment on all the important features.
Q29
Pretend you are constructing a histogram for describing the distribution of salaries for individuals who are $40$ years or older, but are not yet retired.
1. What is on the $Y$-axis? Explain.
2. What is on the $X$-axis?
3. What would be the probable shape of the salary distribution? Explain why.
Select Answers
S2
$85$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/02%3A_Graphing_Distributions/2.E%3A_Graphing_Distributions_%28Exercises%29.txt |
Descriptive statistics often involves using a few numbers to summarize a distribution. One important aspect of a distribution is where its center is located. Measures of central tendency are discussed first. A second aspect of a distribution is how spread out it is. In other words, how much the numbers in the distribution vary from one another. The second section describes measures of variability. Distributions can differ in shape. Some distributions are symmetric whereas others have long tails in just one direction. The third section describes measures of the shape of distributions. The final two sections concern (1) how transformations affect measures summarizing distributions and (2) the variance sum law, an important relationship involving a measure of variability.
03: Summarizing Distributions
Central tendency is a loosely defined concept that has to do with the location of the center of a distribution. The section "What is Central Tendency" presents three definitions of the center of a distribution. "Measures of Central Tendency" presents the three most common measures of the center of the distribution. The three simulations that follow relate the definitions of the center of a distribution to the commonly used measures of central tendency. The findings from these simulations are summarized in the section "Mean and Median." The "Mean and Median" allows you to explore how the relative size of the mean and the median depends on the skew of the distribution.
Less frequently used measures of central tendency can be valuable supplements to the more commonly used measures. Some of these measures are presented in "Additional Measures." Finally, the last section compares and summarizes differences among measures of central tendency.
3.02: What is Central Tendency
Learning Objectives
• Identify situations in which knowing the center of a distribution would be valuable
• Give three different ways the center of a distribution can be defined
• Describe how the balance is different for symmetric distributions than it is for asymmetric distributions
What is "central tendency," and why do we want to know the central tendency of a group of scores? Let us first try to answer these questions intuitively. Then we will proceed to a more formal discussion.
Imagine this situation: You are in a class with just four other students, and the five of you took a $5$-point pop quiz. Today your instructor is walking around the room, handing back the quizzes. She stops at your desk and hands you your paper. Written in bold black ink on the front is "$3/5$." How do you react? Are you happy with your score of $3$ or disappointed? How do you decide? You might calculate your percentage correct, realize it is $60\%$, and be appalled. But it is more likely that when deciding how to react to your performance, you will want additional information. What additional information would you like?
If you are like most students, you will immediately ask your neighbors, "Whad'ja get?" and then ask the instructor, "How did the class do?" In other words, the additional information you want is how your quiz score compares to other students' scores. You therefore understand the importance of comparing your score to the class distribution of scores. Should your score of $3$ turn out to be among the higher scores, then you'll be pleased after all. On the other hand, if $3$ is among the lower scores in the class, you won't be quite so happy.
This idea of comparing individual scores to a distribution of scores is fundamental to statistics. So let's explore it further, using the same example (the pop quiz you took with your four classmates). Three possible outcomes are shown in Table $1$. They are labeled "$\text{Dataset A}$," "$\text{Dataset B}$," and "$\text{Dataset C}$." Which of the three datasets would make you happiest? In other words, in comparing your score with your fellow students' scores, in which dataset would your score of $3$ be the most impressive?
In $\text{Dataset A}$, everyone's score is $3$. This puts your score at the exact center of the distribution. You can draw satisfaction from the fact that you did as well as everyone else. But of course it cuts both ways: everyone else did just as well as you.
Table $1$: Three possible datasets for the $5$-point make-up quiz
Student Dataset A Dataset B Dataset C
You 3 3 3
John's 3 4 2
Maria's 3 4 2
Shareecia's 3 4 2
Luther's 3 5 1
Now consider the possibility that the scores are described as in $\text{Dataset B}$. This is a depressing outcome even though your score is no different than the one in $\text{Dataset A}$. The problem is that the other four students had higher grades, putting yours below the center of the distribution.
Finally, let's look at $\text{Dataset C}$. This is more like it! All of your classmates score lower than you so your score is above the center of the distribution.
Now let's change the example in order to develop more insight into the center of a distribution. Figure $1$ shows the results of an experiment on memory for chess positions. Subjects were shown a chess position and then asked to reconstruct it on an empty chess board. The number of pieces correctly placed was recorded. This was repeated for two more chess positions. The scores represent the total number of chess pieces correctly placed for the three chess positions. The maximum possible score was $89$.
Two groups are compared. On the left are people who don't play chess. On the right are people who play a great deal (tournament players). It is clear that the location of the center of the distribution for the non-players is much lower than the center of the distribution for the tournament players.
We're sure you get the idea now about the center of a distribution. It is time to move beyond intuition. We need a formal definition of the center of a distribution. In fact, we'll offer you three definitions! This is not just generosity on our part. There turn out to be (at least) three different ways of thinking about the center of a distribution, all of them useful in various contexts. In the remainder of this section we attempt to communicate the idea behind each concept. In the succeeding sections we will give statistical measures for these concepts of central tendency.
Definitions of Center
Now we explain the three different ways of defining the center of a distribution. All three are called measures of central tendency.
Balance Scale
One definition of central tendency is the point at which the distribution is in balance. Figure $2$ shows the distribution of the five numbers $2, 3, 4, 9, 16$ placed upon a balance scale. If each number weighs one pound, and is placed at its position along the number line, then it would be possible to balance them by placing a fulcrum at $6.8$.
For another example, consider the distribution shown in Figure $3$. It is balanced by placing the fulcrum in the geometric middle.
Figure $4$ illustrates that the same distribution can't be balanced by placing the fulcrum to the left of center.
Figure $5$ shows an asymmetric distribution. To balance it, we cannot put the fulcrum halfway between the lowest and highest values (as we did in Figure $3$). Placing the fulcrum at the "half way" point would cause it to tip towards the left.
The balance point defines one sense of a distribution's center. The simulation in the next section "Balance Scale Simulation" shows how to find the point at which the distribution balances.
Smallest Absolute Deviation
Another way to define the center of a distribution is based on the concept of the sum of the absolute deviations (differences). Consider the distribution made up of the five numbers $2, 3, 4, 9, 16$. Let's see how far the distribution is from $10$ (picking a number arbitrarily). Table $2$ shows the sum of the absolute deviations of these numbers from the number $10$.
Table $2$: An example of the sum of absolute deviations
Values Absolute Deviations from 10
2 8
3 7
4 6
9 1
16 6
Sum 28
The first row of the table shows that the absolute value of the difference between $2$ and $10$ is $8$; the second row shows that the absolute difference between $3$ and $10$ is $7$, and similarly for the other rows. When we add up the five absolute deviations, we get $28$. So, the sum of the absolute deviations from $10$ is $28$. Likewise, the sum of the absolute deviations from $5$ equals $3 + 2 + 1 + 4 + 11 = 21$. So, the sum of the absolute deviations from $5$ is smaller than the sum of the absolute deviations from $10$. In this sense, $5$ is closer, overall, to the other numbers than is $10$.
We are now in a position to define a second measure of central tendency, this time in terms of absolute deviations. Specifically, according to our second definition, the center of a distribution is the number for which the sum of the absolute deviations is smallest. As we just saw, the sum of the absolute deviations from $10$ is $28$ and the sum of the absolute deviations from $5$ is $21$. Is there a value for which the sum of the absolute deviations is even smaller than $21$? Yes. For these data, there is a value for which the sum of absolute deviations is only $20$. See if you can find it. A general method for finding the center of a distribution in the sense of absolute deviations is provided in the simulation "Absolute Differences Simulation."
Smallest Squared Deviation
We shall discuss one more way to define the center of a distribution. It is based on the concept of the sum of squared deviations (differences). Again, consider the distribution of the five numbers $2, 3, 4, 9, 16$. Table $3$ shows the sum of the squared deviations of these numbers from the number $10$.
Table $3$: An example of the sum of squared deviations
Values Squared Deviations from 10
2 64
3 49
4 36
9 1
16 36
Sum 186
The first row in the table shows that the squared value of the difference between $2$ and $10$ is $64$; the second row shows that the squared difference between $3$ and $10$ is $49$, and so forth. When we add up all these squared deviations, we get $186$. Changing the target from $10$ to $5$, we calculate the sum of the squared deviations from $5$ as $9 + 4 + 1 + 16 + 121 = 151$. So, the sum of the squared deviations from $5$ is smaller than the sum of the squared deviations from $10$. Is there a value for which the sum of the squared deviations is even smaller than $151$? Yes, it is possible to reach $134.8$. Can you find the target number for which the sum of squared deviations is $134.8$?
The target that minimizes the sum of squared deviations provides another useful definition of central tendency (the last one to be discussed in this section). It can be challenging to find the value that minimizes this sum. You will see how you do it in the upcoming section "Squared Differences Simulation."
Contributor
• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.
• David M. Lane and Heidi Ziemer | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.01%3A_Central_Tendency.txt |
Learning Objectives
• Compute mode
In the previous section we saw that there are several ways to define central tendency. This section defines the three most common measures of central tendency: the mean, the median, and the mode. The relationships among these measures of central tendency and the definitions given in the previous section will probably not be obvious to you. Rather than just tell you these relationships, we will allow you to discover them in the simulations in the sections that follow. This section gives only the basic definitions of the mean, median and mode. A further discussion of the relative merits and proper applications of these statistics is presented in a later section.
Arithmetic Mean
The arithmetic mean is the most common measure of central tendency. It is simply the sum of the numbers divided by the number of numbers. The symbol "$\mu$" is used for the mean of a population. The symbol "$M$" is used for the mean of a sample. The formula for $\mu$ is shown below:
$\mu =\dfrac{\sum X}{N}$
where $\sum X$ is the sum of all the numbers in the population and $N$ is the number of numbers in the population.
The formula for $M$ is essentially identical:
$M = \dfrac{\sum X}{N}$
where $\sum X$ is the sum of all the numbers in the sample and $N$ is the number of numbers in the sample.
As an example, the mean of the numbers $1, 2, 3, 6, 8$ is $20/5 = 4$ regardless of whether the numbers constitute the entire population or just a sample from the population.
Table $1$ shows the number of touchdown (TD) passes thrown by each of the $31$ teams in the National Football League in the $2000$ season.
Table $1$: Number of touchdown passes
$\begin{matrix} 37 & 33 & 33 & 32 & 29 & 28 & 28 & 23 & 22 & 22 & 22 & 21\ 21 & 21 & 20 & 20 & 19 & 19 & 18 & 18 & 18 & 18 & 16 & 15\ 14 & 14 & 14 & 12 & 12 & 9 & 6 & & & & & \end{matrix}$
The mean number of touchdown passes thrown is $20.4516$ as shown below.
\begin{align*} \mu &= \sum X/N\ &= 634/31\ &= 20.4516 \end{align*}
Although the arithmetic mean is not the only "mean" (there is also a geometric mean), it is by far the most commonly used. Therefore, if the term "mean" is used without specifying whether it is the arithmetic mean, the geometric mean, or some other mean, it is assumed to refer to the arithmetic mean.
Median
The median is also a frequently used measure of central tendency. The median is the midpoint of a distribution: the same number of scores is above the median as below it. For the data in Table $1$, there are $31$ scores. The $16^{th}$ highest score (which equals $20$) is the median because there are $15$ scores below the $16^{th}$ score and $15$ scores above the $16^{th}$ score. The median can also be thought of as the $50^{th}$ percentile.
Computation of the Median
When there is an odd number of numbers, the median is simply the middle number. For example, the median of $2, 4$, and $7$ is $4$. When there is an even number of numbers, the median is the mean of the two middle numbers. Thus, the median of the numbers $2, 4, 7, 12$ is $(4+7)/2 = 5.5$. When there are numbers with the same values, then the formula for the third definition of the $50^{th}$ percentile should be used.
Mode
The mode is the most frequently occurring value. For the data in Table $2$, the mode is $18$ since more teams ($4$) had $18$ touchdown passes than any other number of touchdown passes. With continuous data such as response time measured to many decimals, the frequency of each value is one since no two scores will be exactly the same (see discussion of continuous variables). Therefore the mode of continuous data is normally computed from a grouped frequency distribution. Table $2$ shows a grouped frequency distribution for the target response time data. Since the interval with the highest frequency is $600-700$, the mode is the middle of that interval ($650$).
Table $2$: Grouped frequency distribution
Range Frequency
500-600 3
600-700 6
700-800 5
800-900 5
900-1000 0
1000-1100 1 | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.03%3A_Measures_of_Central_Tendency.txt |
Learning Objectives
• Learn which measure of central tendency will balance a distribution
Instructions
This demonstration allows you to change the shape of a distribution and see the point at which the distribution would balance.
The graph in the right panel is a histogram of \(600\) scores. The mean and median are equal to \(8\) and are indicated by small vertical bars on the \(X\)-axis The top portion of the bar is in blue and represents the mean. The bottom portion is in pink and represents the median. The mean and median are also shown to the left of the \(Y\)-axis.
You can see that the histogram is balanced on the tip of the triangle (the fulcrum). You can change the shape of the histogram by painting with the mouse. Notice that the triangle beneath the X-axis automatically moves to the point where the histogram is balanced. Experiment with different shapes and see if you can determine whether there is a relationship between the mean, median, and/or the mode and the location of the balance point.
Illustrated Instructions
Below is a screen shot of the simulaton's beginning screen. Note that the distribution is balanced on the fulcrum. The mean and median are shown to the left and also as small vertical bars below the \(X\)-axis. The mean is in blue and the median is in pink. The next figure illustrates this more clearly.
You can change the distribution by painting with the mouse when running the simulation.Below is an example of the distribution after it has been changed. Note that the mean and median are marked by vertical lines.
3.05: Absolute Differences Simulation
Learning Objectives
• Learn which measure of central tendency minimizes the sum of absolute deviations
Instructions
This demonstration allows you to examine the sum of absolute deviations from a given value. The graph to the right shows the numbers \(1, 2, 3, 4,\) and \(5\) and their deviations from an arbitrary starting value of \(0.254\) (the figure displays this rounded to \(0.25\)).
The first number, \(1\), is represented by a red dot. The deviation from \(0.254\) is represented by a red line from the red dot to the black line. The value of the black line is \(0.254\). Similarly, the number \(2\) is represented by a blue dot and its deviation from \(0.254\) is represented by the length of a blue line.
The portion of the graph with the colored rectangles shows the sum of the absolute deviations. The sum of the deviations is equal to \(0.746 + 1.746 + 2.746 + 3.746 + 4.746 = 13.73\) as shown by the height of the colored bar.
In this demonstration, you can move the black bar by clicking on it and dragging it up or down. To see how it works, move it up to \(1.0\). The deviation of the red point from the black bar is now \(0\) since they are both \(1\). The sum of the deviations is now \(10\).
As you move the bar up and down, the value of the sum of absolute deviations changes. See if you can find the placement of the black bar that produces the smallest value for the sum of the absolute deviations. To check and see if you found the smallest value, click the "OK" button at the bottom of the graph. It will move the bar to the location that produces the smallest sum of absolute deviations.
You can also move the individual points. Click on one of the points and move it up or down and note the effect. Your goal for this demonstration is to discover a rule for determining what value will give you the smallest sum of absolute deviations. When you have discovered the rule, go back and answer the questions again.
Illustrated Instructions
Below is a screen shot of the simulaton's beginning screen.
Below is an example after the vertical line has been changed. The distances to the line have been recalculated.
You can change the data by clicking on a data point and dragging. An example with changed data is shown below. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.04%3A_Balance_Scale_Simulation.txt |
Learning Objectives
• Learn which measure of central tendency minimizes the sum of squared deviations
Instructions
This demonstration allows you to examine the sum of squared deviations from a given value. The graph to the right shows the numbers $1, 2, 3, 4,$ and $5$ and their deviations from an arbitrary starting value of $0.254$ (the figure displays this rounded to $0.25$).
The first number, $1$, is represented by a red dot. The deviation from $0.254$ is represented by a red line from the red dot to the black line. The value of the black line is $0.254$.
Similarly, the number $2$ is represented by a blue dot and its deviation from $0.25$ is represented by a blue line.
The height of the colored rectangles represents the sum of the absolute deviations from the black line. The sum of the deviations of the numbers $1, 2, 3, 4,$ and $5$ from $0.25$ is $0.746 + 1.75 + 2.746 + 3.746 + 4.746 = 13.73$.
The area of each rectangle represents the magnitude of the squared deviation of a point from the black line. For example, the red rectangle has an area of $0.746\times 0.746 = 0.557$. The sum of all the areas of the rectangles is $47.70$. The sum of all the areas represents the sum of the squared deviations.
In this demonstration, you can move the black bar by clicking on it and dragging it up or down. In the second figure it has been moved up to $1.0$. The deviation of the red point from the black bar is now $0$ since they are both $1$. The sum of the deviations is now $10 (0 + 1 + 2 + 3 + 4)$ and the sum of squared deviations is $30 (0^2 + 1^2 + 2^2 + 3^2 + 4^2)$.
As you move the bar up and down, the value of the sum of absolute deviations and the sum of squared deviations changes.
See if you can find the placement of the black bar that produces the smallest value for the sum of the squared deviations.
To check and see if you found the smallest value, click the "OK" button at the bottom of the graph. It will move the bar to the location that produces the smallest sum of squared deviations.
For the initial data, the value that minimizes the sum of squared deviations is also the value that minimizes the sum of absolute deviations. This will not be true for most data.
You can also move the individual points. Click on one of the points and move it up or down and note the effect. Your goal for this demonstration is to discover a rule for determining what value will give you the smallest sum of squared deviations.
Illustrated Instructions
Below is a screen shot of the simulation's beginning screen. The "Total Dev." is the sum of the absolute deviations; the "Total Area" is the sum of the squared deviations.
Below is an example after the vertical line has been changed. The distances to the line have been recalculated.
You can change the data by clicking on a data point and dragging. An example with changed data is shown below. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.06%3A_Squared_Differences_Simulation.txt |
Learning Objectives
• State whether it is the mean or median that minimizes the mean absolute deviation
• State whether it is the mean or median that is the balance point on a balance scale
In the section "What is central tendency," we saw that the center of a distribution could be defined three ways:
1. the point on which a distribution would balance
2. the value whose average absolute deviation from all the other values is minimized
3. the value whose average squared difference from all the other values is minimized
From the simulation in this chapter, you discovered (we hope) that the mean is the point on which a distribution would balance, the median is the value that minimizes the sum of absolute deviations, and the mean is the value that minimizes the sum of the squared deviations.
Table \(1\) shows the absolute and squared deviations of the numbers \(2, 3, 4, 9\) and \(16\) from their median of \(4\) and their mean of \(6.8\). You can see that the sum of absolute deviations from the median (\(20\)) is smaller than the sum of absolute deviations from the mean (\(22.8\)). On the other hand, the sum of squared deviations from the median (\(174\)) is larger than the sum of squared deviations from the mean (\(134.8\)).
Table \(1\): Absolute and squared deviations from the median of 4 and the mean of 6.8
Value Absolute Deviation from Median Absolute Deviation from Mean Squared Deviation from Median Squared Deviation from Mean
2 2 4.8 4 23.04
3 1 3.8 1 14.44
4 0 2.8 0 7.84
9 5 2.2 25 4.84
16 12 9.2 144 84.64
Total 20 22.8 174 134.8
Figure \(1\) shows that the distribution balances at the mean of \(6.8\) and not at the median of \(4\). The relative advantages and disadvantages of the mean and median are discussed in the section "Comparing Measures" later in this chapter.
When a distribution is symmetric, then the mean and the median are the same. Consider the following distribution: \(1, 3, 4, 5, 6, 7, 9\). The mean and median are both \(5\). The mean, median, and mode are identical in the bell-shaped normal distribution.
3.08: Mean and Median Demo
Learning Objectives
• To study how the mean and median change with different distributions
Instructions
This demonstration shows how the relative size of the mean and the median depends on the skew of the distribution. The demonstration begins by showing a histogram of a symmetric distribution (no skew). The mean and median are both \(5.0\). The mean is shown on the histogram as a small blue line; the median is shown as a small purple line. The standard deviation is \(1.81\). A red line extends one sd in each direction from the mean. The standard deviation is calculated assuming the data portrayed in the graph represent the entire population. You can change the values of the data set by "painting" the histogram with the mouse. Change the distribution in various ways and note how the skew affects whether the mean is bigger than the median or vice versa.
Illustrated Instructions
The simulation starts out with a symmetrical distribution. As can be seen seen in the screenshot below the mean and median are equal and there is \(0\) skew.
The distribution can be changed "painting" it with the mouse. Below is an example of a negatively skewed distribution. Note that the mean and median are no longer equal to each other. Trying painting several different types of distributions to see how mean and median values are affected relative to each other. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.07%3A_Median_and_Mean.txt |
Learning Objectives
• Compute the trimean
• Compute the geometric mean directly
• Compute the geometric mean using logs
• Use the geometric to compute annual portfolio returns
• Compute a trimmed mean
Although the mean, median, and mode are by far the most commonly used measures of central tendency, they are by no means the only measures. This section defines three additional measures of central tendency: the trimean, the geometric mean, and the trimmed mean. These measures will be discussed again in the section "Comparing Measures of Central Tendency."
Trimean
The trimean is a weighted average of the $25^{th}$ percentile, the $50^{th}$ percentile, and the $75^{th}$ percentile. Letting $P25$ be the $25^{th}$ percentile, $P50$ be the $50^{th}$and $P75$ be the $75^{th}$ percentile, the formula for the trimean is:
$Trimean = \dfrac{P25 + 2P50 + P75}{4}$
As you can see from the formula, the median is weighted twice as much as the $25^{th}$ and $75^{th}$ percentiles. Table $1$ shows the number of touchdown (TD) passes thrown by each of the $31$ teams in the National Football League in the $2000$ season. The relevant percentiles are shown in Table $2$.
Table $1$: Number of touchdown passes
$\begin{matrix} 37 & 33 & 33 & 32 & 29 & 28 & 28 & 23\ 22 & 22 & 22 & 21 & 21 & 21 & 20 & 20\ 19 & 19 & 18 & 18 & 18 & 18 & 16 & 15\ 14 & 14 & 14 & 12 & 12 & 9 & 6 & \end{matrix}$
Table $2$: Percentiles
Percentile Value
25 15
50 20
75 23
The trimean is therefore
$\dfrac{15 + 2 \times 20 + 23}{4} = \dfrac{78}{4} = 19.5.$
Geometric Mean
The geometric mean is computed by multiplying all the numbers together and then taking the $n^{th}$ root of the product. For example, for the numbers $1, 10$ and $100$, the product of all the numbers is:
$1 \times 10 \times 100 = 1,000.$
Since there are three numbers, we take the cubed root of the product ($1,000$) which is equal to $10$. The formula for the geometric mean is therefore
$\large\left(\Pi \, X \right)^{1/N}$
where the symbol $Π$ means to multiply. Therefore, the equation says to multiply all the values of $X$ and then raise the result to the $1/N$th power. Raising a value to the $\dfrac{1}{N}^{th}$ power is, of course, the same as taking the $N^{th}$ root of the value. In this case, $1000^{1/3}$ is the cube root of $1,000$.
The geometric mean has a close relationship with logarithms. Table $3$ shows the logs (base $10$) of these three numbers. The arithmetic mean of the three logs is $1$. The anti-log of this arithmetic mean of $1$ is the geometric mean. The anti-log of $1$ is $10^1 = 10$. Note that the geometric mean only makes sense if all the numbers are positive.
Table $3$: Logarithms
X $\log _{10}(X)$
1 0
10 1
100 2
The geometric mean is an appropriate measure to use for averaging rates. For example, consider a stock portfolio that began with a value of $\1,000$ and had annual returns of ${13\%, 22\%, 12\%, -5\%, and -13\%}$. Table $4$ shows the value after each of the five years.
Table $4$: Portfolio Returns
Year Return Value
1 13% 1,130
2 22% 1,379
3 12% 1,544
4 -5% 1,467
5 -13% 1,276
The question is how to compute average annual rate of return. The answer is to compute the geometric mean of the returns. Instead of using the percents, each return is represented as a multiplier indicating how much higher the value is after the year. This multiplier is $1.13$ for a $13\%$ return and $0.95$ for a $5\%$ loss. The multipliers for this example are ${1.13, 1.22, 1.12, 0.95,\: and\; 0.87}$. The geometric mean of these multipliers is $1.05$. Therefore, the average annual rate of return is $5\%$. Table $5$ shows how a portfolio gaining $5\%$ a year would end up with the same value ($\1,276$) as shown in Table $4$.
Table $5$: Portfolio Returns
Year Return Value
1 5% 1,050
2 5% 1,103
3 5% 1,158
4 5% 1,216
5 5% 1,276
Trimmed Mean
To compute a trimmed mean, you remove some of the higher and lower scores and compute the mean of the remaining scores. A mean trimmed $10\%$ is a mean computed with $10\%$ of the scores trimmed off: $5\%$ from the bottom and $5\%$ from the top. A mean trimmed $50\%$ is computed by trimming the upper $25\%$ of the scores and the lower $25\%$ of the scores and computing the mean of the remaining scores. The trimmed mean is similar to the median which, in essence, trims the upper $49\%$ and the lower $49\%$ of the scores. Therefore the trimmed mean is a hybrid of the mean and the median. To compute the mean trimmed $20\%$ for the touchdown pass data shown in Table $1$, you remove the lower $10\%$ of the scores (${6, 9,\: and\; 12}$) as well as the upper $10\%$ of the scores (${33, 33,\: and\; 37}$) and compute the mean of the remaining $25$ scores. This mean is $20.16$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.09%3A_Additional_Measures.txt |
Learning Objectives
• State how the measures differ in symmetric distributions
• State which measure(s) should be used to describe the center of a skewed distribution
How do the various measures of central tendency compare with each other? For symmetric distributions, the mean, median, trimean, and trimmed mean are equal, as is the mode except in bimodal distributions. Differences among the measures occur with skewed distributions. Figure \(1\) shows the distribution of \(642\) scores on an introductory psychology test. Notice this distribution has a slight positive skew.
Measures of central tendency are shown in Table \(1\). Notice they do not differ greatly, with the exception that the mode is considerably lower than the other measures. When distributions have a positive skew, the mean is typically higher than the median, although it may not be in bimodal distributions. For these data, the mean of \(91.58\) is higher than the median of \(90\). Typically the trimean and trimmed mean will fall between the median and the mean, although in this case, the trimmed mean is slightly lower than the median. The geometric mean is lower than all measures except the mode.
Table \(1\): Measures of central tendency for the test scores
Measure Value
Mode 84.00
Median 90.00
Geometric Mean 89.70
Trimean 90.25
Mean Trimmed 50% 89.81
Mean 91.58
The distribution of baseball salaries (in \(1994\)) shown in Figure \(2\) has a much more pronounced skew than the distribution in Figure \(2\).
Table \(2\) shows the measures of central tendency for these data. The large skew results in very different values for these measures. No single measure of central tendency is sufficient for data such as these. If you were asked the very general question: "So, what do baseball players make?" and answered with the mean of \(\$1,183,000\), you would not have told the whole story since only about one third of baseball players make that much. If you answered with the mode of \(\$250,000\) or the median of \(\$500,000\), you would not be giving any indication that some players make many millions of dollars. Fortunately, there is no need to summarize a distribution with a single number. When the various measures differ, our opinion is that you should report the mean, median, and either the trimean or the mean trimmed \(50\%\). Sometimes it is worth reporting the mode as well. In the media, the median is usually reported to summarize the center of skewed distributions. You will hear about median salaries and median prices of houses sold, etc. This is better than reporting only the mean, but it would be informative to hear more statistics.
Table \(2\): Measures of central tendency for baseball salaries (in thousands of dollars)
Measure Value
Mode 250
Median 500
Geometric Mean 555
Trimean 792
Mean Trimmed 50% 619
Mean 1,183
3.11: Variability
Learning Objectives
• To study how much the numbers in a distribution differ from each other
Variability refers to how much the numbers in a distribution differ from each other. The most common measures are presented in "Measures of Variability." The "variability demo" allows you to change the standard deviation of a distribution and view a graph of the changed distribution.
One of the more counter-intuitive facts in introductory statistics is that the formula for variance when computed in a population is biased when applied in a sample. The "Estimating Variance Simulation" shows concretely why this is the case. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.10%3A_Comparing_Measures.txt |
Learning Objectives
• Compute the range
• Compute the variance in the population
• Compute the standard deviation from the variance
What is Variability?
Variability refers to how "spread out" a group of scores is. To see what we mean by spread out, consider graphs in Figure $1$. These graphs represent the scores on two quizzes. The mean score for each quiz is $7.0$. Despite the equality of means, you can see that the distributions are quite different. Specifically, the scores on $\text{Quiz 1}$ are more densely packed and those on $\text{Quiz 2}$ are more spread out. The differences among students were much greater on $\text{Quiz 2}$ than on $\text{Quiz 1}$.
$\text{Quiz 1}$
$\text{Quiz 1}$
Figure $1$: Bar charts of two quizzes
The terms variability, spread, and dispersion are synonyms, and refer to how spread out a distribution is. Just as in the section on central tendency where we discussed measures of the center of a distribution of scores, in this chapter we will discuss measures of the variability of a distribution. There are four frequently used measures of variability: the range, interquartile range, variance, and standard deviation. In the next few paragraphs, we will look at each of these four measures of variability in more detail.
Range
The range is the simplest measure of variability to calculate, and one you have probably encountered many times in your life. The range is simply the highest score minus the lowest score. Let’s take a few examples. What is the range of the following group of numbers: ${10, 2, 5, 6, 7, 3, 4}$? Well, the highest number is $10$, and the lowest number is $2$, so $10 - 2 = 8$. The range is $8$. Let’s take another example. Here’s a dataset with $10$ numbers: ${99, 45, 23, 67, 45, 91, 82, 78, 62, 51}$. What is the range? The highest number is $99$ and the lowest number is $23$, so $99 - 23$ equals $76$; the range is $76$. Now consider the two quizzes shown in Figure $1$. On $\text{Quiz 1}$, the lowest score is $5$ and the highest score is $9$. Therefore, the range is $4$. The range on $\text{Quiz 2}$ was larger: the lowest score was $4$ and the highest score was $10$. Therefore the range is $6$.
Interquartile Range
The interquartile range (IQR) is the range of the middle $50\%$ of the scores in a distribution. It is computed as follows:
$IQR = 75^{th} percentile - 25^{th} percentile$
For $\text{Quiz 1}$, the $75^{th}$ percentile is $8$ and the $25^{th}$ percentile is $6$. The interquartile range is therefore $2$. For $\text{Quiz 2}$, which has greater spread, the $75^{th}$ percentile is $9$, the $25^{th}$ percentile is $5$, and the interquartile range is $4$. Recall that in the discussion of box plots, the $75^{th}$ percentile was called the upper hinge and the $25^{th}$ percentile was called the lower hinge. Using this terminology, the interquartile range is referred to as the $H$-spread.
A related measure of variability is called the semi-interquartile range. The semi-interquartile range is defined simply as the interquartile range divided by $2$. If a distribution is symmetric, the median plus or minus the semi-interquartile range contains half the scores in the distribution.
Variance
Variability can also be defined in terms of how close the scores in the distribution are to the middle of the distribution. Using the mean as the measure of the middle of the distribution, the variance is defined as the average squared difference of the scores from the mean. The data from $\text{Quiz 1}$ are shown in Table $1$. The mean score is $7.0$. Therefore, the column "Deviation from Mean" contains the score minus $7$. The column "Squared Deviation" is simply the previous column squared.
Table $1$: Calculation of Variance for $\text{Quiz 1}$ scores
Scores Deviation from Mean Squared Deviation
9 2 4
9 2 4
9 2 4
8 1 1
8 1 1
8 1 1
8 1 1
7 0 0
7 0 0
7 0 0
7 0 0
7 0 0
6 -1 1
6 -1 1
6 -1 1
6 -1 1
6 -1 1
6 -1 1
5 -2 4
5 -2 4
Means
7 0 1.5
One thing that is important to notice is that the mean deviation from the mean is $0$. This will always be the case. The mean of the squared deviations is $1.5$. Therefore, the variance is $1.5$. Analogous calculations with $\text{Quiz 2}$ show that its variance is $6.7$. The formula for the variance is:
$s^2=\frac{\sum (X-\mu )^2}{N}$
where $\sigma ^2$ is the variance, $\mu$ is the mean, and $N$ is the number of numbers. For $\text{Quiz 1}$, $\mu = 7$ and $N = 20$.
If the variance in a sample is used to estimate the variance in a population, then the previous formula underestimates the variance and the following formula should be used:
$s^2=\frac{\sum (X-M)^2}{N-1}$
where $s^2$ is the estimate of the variance and $M$ is the sample mean. Note that $M$ is the mean of a sample taken from a population with a mean of $\mu$. Since, in practice, the variance is usually computed in a sample, this formula is most often used. The simulation "estimating variance" illustrates the bias in the formula with $N$ in the denominator.
Let's take a concrete example. Assume the scores ${1, 2, 4,\: and\; 5 }$ were sampled from a larger population. To estimate the variance in the population you would compute $s^2$ as follows:
$M = \dfrac {1 + 2 + 4 + 5}{4} = \dfrac {12}{4} = 3$
\begin{align*} s^2 &= \dfrac{[(1-3)^2 + (2-3)^2 + (4-3)^2 + (5-3)^2]}{(4-1)}\ &= \dfrac{(4 + 1 + 1 + 4)}{3}\ &= \dfrac{10}{3}\ &= 3.333 \end{align*}
There are alternate formulas that can be easier to use if you are doing your calculations with a hand calculator. You should note that these formulas are subject to rounding error if your values are very large and/or you have an extremely large number of observations.
$\sigma ^2=\cfrac{\sum X^2-\cfrac{(\sum X)^2}{N}}{N}$
and
$s^2=\cfrac{\sum X^2-\cfrac{(\sum X)^2}{N}}{N-1}$
For this example,
$\sum X^2=1^2+2^2+4^2+5^2=46$
$\dfrac {(\sum X)^2}{N}=\dfrac {(1+2+4+5)^2}{4}=\dfrac {144}{4}=36$
$\sigma ^2=\dfrac {(46-36)}{4}=2.5$
$s^2=\dfrac {(46-36)}{3}=3.333\; \; \text{as with the other formula}$
Standard Deviation
The standard deviation is simply the square root of the variance. This makes the standard deviations of the two quiz distributions $1.225$ and $2.588$. The standard deviation is an especially useful measure of variability when the distribution is normal or approximately normal (see Chapter on Normal Distributions) because the proportion of the distribution within a given number of standard deviations from the mean can be calculated. For example, $68\%$ of the distribution is within one standard deviation of the mean and approximately $95\%$ of the distribution is within two standard deviations of the mean. Therefore, if you had a normal distribution with a mean of $50$ and a standard deviation of $10$, then $68\%$ of the distribution would be between $50 - 10 = 40$ and $50 +10 =60$. Similarly, about $95\%$ of the distribution would be between $50 - 2 \times 10 = 30$ and $50 + 2 \times 10 = 70$. The symbol for the population standard deviation is $\sigma$; the symbol for an estimate computed in a sample is $s$. Figure $2$ shows two normal distributions. The red distribution has a mean of $40$ and a standard deviation of $5$; the blue distribution has a mean of $60$ and a standard deviation of $10$. For the red distribution, $68\%$ of the distribution is between $35$ and $45$; for the blue distribution, $68\%$ is between $50$ and $70$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.12%3A_Measures_of_Variability.txt |
Learning Objectives
• Identify differences in the means and standard deviations of distributions
Instructions
The demonstration shows a graph of two normal distributions. Both distributions have means of \(50\). The red distribution has a standard deviation of \(10\); the blue distribution has a standard deviation of \(5\). You can see that the red distribution is more spread out than the blue distribution. Note that about two thirds of the area of the distributions is within one standard deviation of the mean. For the red distribution, this is between \(40\) and \(60\); for the blue distribution, this is between \(45\) and \(55\). About \(95\%\) of a normal distribution is within two standard deviations from the mean. For the red distribution, this is between \(30\) and \(70\); for the blue it is between \(40\) and \(60\).
You can change the means and standard deviations of the distributions and see the results visually. For some values, the distributions will be off the graph. For example, if you give a distribution a mean of \(200\), it will not be shown.
Illustrated Instructions
The demonstration starts with \(2\) normal distributions with equal means and different standard deviations (see screenshot below).
The means and standard deviations for both distributions can be changed and these changes will be reflected in the graph. The screenshot below shows the distributions with different means and standard deviations.
3.14: Estimating Variance Simulation
Learning Objectives
• Learn which measure of central tendency will balance a distribution
Instructions
This simulation samples from the population of \(50\) numbers shown here. You can see that there are \(10\) instances of the values \(1, 2, 3, 4\), and \(5\). The mean of the population is therefore \(3\). The variance is the average squared deviation from the mean of \(3\). You can compute that this is exactly \(2\).
When you click on the button "Draw \(4\) numbers" four scores are sampled (with replacement) from the population. The four numbers are shown in red, as is the mean of the four numbers. The variance is then computed in two ways. The upper formula computes the variance by computing the mean of the squared deviations or the four sampled numbers from the sample mean. The lower formula computes the mean of the squared deviations or the four sampled numbers from the population mean of \(3.00\) (on rare occasions, the sample and population means will be equal). The computed variances are placed in the fields to the right of the formulas. The mean of the values in a field is shown at the bottom of the field. When there is only one value in the field, the mean will, of course, equal that value.
If you click the "Draw \(4\) numbers" button again, another four numbers will be sampled. The mean and variance will also be computed as before. The fields to the right of the formulas will hold both variances and the bottom of the field will show the mean of the variances.
The population variance is exactly \(2\). Use this fact to assess the relative value of the two formulas for variance. See which one, on average, approaches \(2\) and which one gives lower estimates. Explore whether either formula is always more accurate, or whether sometimes one is more accurate and at other times, the other formula is. If the variance based on the sample mean had been computed by dividing by \(N-1 = 3\) instead of \(4\), then the variance would be \(\tfrac{4}{3}\) times bigger. Does multiplying the variance by \(\tfrac{4}{3}\) lead to better estimates?
Illustrated Instructions
As can be seen in the screenshot below, the variance estimation simulation begins by displaying a population of \(50\) numbers ranging from \(1 - 5\).
Each time the "Draw \(4\) numbers" button is clicked four numbers are sampled from the population and the mean, the variance of the sample from the sample mean as well as the variance of the sample from the population mean are calculated. The variances are stored in fields next to their respective formula. The screenshot below shows the simulation after the "Draw \(4\) numbers" button has been clicked four times.
Use the simulation to explore whether either formula is on average more accurate than the other. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.13%3A_Variability_Demo.txt |
Learning Objectives
• Compute skew using two different formulas
• Compute kurtosis
We saw in the section on distributions in Chapter 1 that shapes of distributions can differ in skew and/or kurtosis. This section presents numerical indexes of these two measures of shape.
Skew
Figure $1$ shows a distribution with a very large positive skew. Recall that distributions with positive skew have tails that extend to the right.
Distributions with positive skew normally have larger means than medians. The mean and median of the baseball salaries shown in Figure $1$ are $\1,183,417$ and $\500,000$ respectively. Thus, for this highly-skewed distribution, the mean is more than twice as high as the median. The relationship between skew and the relative size of the mean and median led the statistician Pearson to propose the following simple and convenient numerical index of skew:
$\frac{3(Mean-Median)}{\sigma }$
The standard deviation of the baseball salaries is $1,390,922$. Therefore, Pearson's measure of skew for this distribution is $\dfrac {3(1,183,417 - 500,000)}{1,390,922} = 1.47$.
Just as there are several measures of central tendency, there is more than one measure of skew. Although Pearson's measure is a good one, the following measure is more commonly used. It is sometimes referred to as the third moment about the mean.
$\sum \dfrac {(X-\mu )^3}{\sigma ^3}$
Kurtosis
The following measure of kurtosis is similar to the definition of skew. The value "$3$" is subtracted to define "no kurtosis" as the kurtosis of a normal distribution. Otherwise, a normal distribution would have a kurtosis of $3$.
$\sum \dfrac {(X-\mu )^4}{\sigma ^4}-3$
3.16: Comparing Distributions Demo
Learning Objectives
• Learn how one might compare data from two distributions
Instructions
This demonstration collects response time data from you and computes various descriptive statistics. It also displays histograms and box plots. After you click the button to begin, a rectangle will appear to the right of the button. Move the cursor to the rectangle and click. The time it takes you to click on the rectangle is recorded. Half of the time the rectangle is small and half of the time it is big. After two practice trials and \(40\) regular trials, the descriptive statistics and graphs will be displayed.
Illustrated Instructions
The simulation is started by clicking the button in the lower left of the screen. Clicking this button begins a series of trials where a another button appears on the right and you will be asked to click this button as quickly as possible.
Half of the buttons will large and half will be small. Once you complete \(40\) trials, the simulation will show you descriptive statistics about the reaction time data for each button size (see screenshot below).
3.17: Effects of Linear Transformations
Learning Objectives
• Compute the mean of a transformed variable
• Compute the variance of a transformed variable
This section covers the effects of linear transformations on measures of central tendency and variability. Let's start with an example we saw before in the section that defined linear transformation: temperatures of cities. Table $1$ shows the temperatures of $5$ cities.
Table $1$: Temperatures in $5$ cities on $11/16/2002$
City Degrees Fahrenheit Degrees Centigrade
Houston
Chicago
Minneapolis
Miami
Phoenix
54
37
31
78
70
12.22
2.78
-0.56
25.56
21.11
Mean
Median
54.000
54.000
12.220
12.220
Variance 330.00 101.852
SD 18.166 10.092
Recall that to transform the degrees Fahrenheit to degrees Centigrade, we use the formula
$C = 0.556F - 17.778$
which means we multiply each temperature Fahrenheit by $0.556$ and then subtract $17.778$. As you might have expected, you multiply the mean temperature in Fahrenheit by $0.556$ and then subtract $17.778$ to get the mean in Centigrade. That is, $(0.556)(54) - 17.778 = 12.22$. The same is true for the median. Note that this relationship holds even if the mean and median are not identical as they are in Table $1$.
The formula for the standard deviation is just as simple: the standard deviation in degrees Centigrade is equal to the standard deviation in degrees Fahrenheit times $0.556$. Since the variance is the standard deviation squared, the variance in degrees Centigrade is equal to $0.5562^2$ times the variance in degrees Fahrenheit.
To sum up, if a variable $X$ has a mean of $\mu$, a standard deviation of $\sigma$, and a variance of $\sigma ^2$, then a new variable $Y$ created using the linear transformation
$Y = bX + A$
will have a mean of $b\mu + A$, a standard deviation of $b\sigma$, and a variance of $b^2\sigma ^2$.
It should be noted that the term "linear transformation" is defined differently in the field of linear algebra. For details, follow this link. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.15%3A_Shapes_of_Distributions.txt |
Learning Objectives
• Compute the variance of the sum of two uncorrelated variables
• Compute the variance of the difference between two uncorrelated variables
As you will see in later sections, there are many occasions in which it is important to know the variance of the sum of two variables. Consider the following situation:
1. you have two populations,
2. you sample one number from each population, and
3. you add the two numbers together.
The question is, "What is the variance of this sum?" For example, suppose the two populations are the populations of $8$-year-old males and $8$-year-old females in Houston, Texas, and that the variable of interest is memory span. You repeat the following steps thousands of times:
1. sample one male and one female
2. measure the memory span of each
3. sum the two memory spans.
After you have done this thousands of times, you compute the variance of the sum. It turns out that the variance of this sum can be computed according to the following formula:
$\sigma_{sum}^2 = \sigma_{M}^2 + \sigma_{F}^2$
where the first term is the variance of the sum, the second term is the variance of the males and the third term is the variance of the females. Therefore, if the variances on the memory span test for the males and females were $0.9$ and $0.8$ respectively, then the variance of the sum would be $1.7$.
The formula for the variance of the difference between the two variables (memory span in this example) is shown below. Notice that the expression for the difference is the same as the formula for the sum.
$\sigma_{difference}^2 = \sigma_{M}^2 + \sigma_{F}^2$
More generally, the variance sum law can be written as follows:
$\sigma_{Z \pm Y}^2 = \sigma_{X}^2 + \sigma_{Y}^2$
which is read: The variance of $X$ plus or minus $Y$ is equal to the variance of $X$ plus the variance of $Y$.
These formulas for the sum and difference of variables given above only apply when the variables are independent.
In this example, we have thousands of randomly-paired scores. Since the scores are paired randomly, there is no relationship between the memory span of one member of the pair and the memory span of the other. Therefore the two scores are independent. Contrast this situation with one in which thousands of people are sampled and two measures (such as verbal and quantitative SAT) are taken from each. In this case, there would be a relationship between the two variables since higher scores on the verbal SAT are associated with higher scores on the quantitative SAT (although there are many examples of people who score high on one test and low on the other). Thus the two variables are not independent and the variance of the total SAT score would not be the sum of the variances of the verbal SAT and the quantitative SAT. The general form of the variance sum law is presented in Section 4.7 in the chapter on correlation.
3.19: Statistical Literacy
Learning Objectives
• How to select between Mean and Median
The Mean or the Median?
Example \(1\)
The playbill for the Alley Theatre in Houston wants to appeal to advertisers. They reported the mean household income and the median age of theatergoers. What might have guided their choice of the mean or median?
Solution
It is likely that they wanted to emphasize that theatergoers had high income but de-emphasize how old they are. The distributions of income and age of theatergoers probably have positive skew. Therefore the mean is probably higher than the median, which results in higher income and lower age than if the median household income and mean age had been presented. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.18%3A_Variance_Sum_Law_I_-_Uncorrelated_Variables.txt |
General questions
Q1
Make up a dataset of $12$ numbers with a positive skew. Use a statistical program to compute the skew. Is the mean larger than the median as it usually is for distributions with a positive skew? What is the value for skew? (relevant section & relevant section )
Q2
Repeat Q1 only this time make the dataset have a negative skew. (relevant section & relevant section)
Q3
Make up three data sets with $5$ numbers each that have: (relevant section & relevant section)
1. the same mean but different standard deviations.
2. the same mean but different medians.
3. the same median but different means.
Q4
Find the mean and median for the following three variables: (relevant section)
$\begin{matrix} A & B & C\ 8 & 4 & 6\ 5 & 4 & 2\ 7 & 6 & 3\ 1 & 3 & 4\ 3 & 4 & 1 \end{matrix}$
Q5
A sample of $30$ distance scores measured in yards has a mean of $7$, a variance of $16$, and a standard deviation of $4$.
1. You want to convert all your distances from yards to feet, so you multiply each score in the sample by $3$. What are the new mean, variance, and standard deviation?
2. You then decide that you only want to look at the distance past a certain point. Thus, after multiplying the original scores by $3$, you decide to subtract $4$ feet from each of the scores. Now what are the new mean, variance, and standard deviation? (relevant section)
Q6
You recorded the time in seconds it took for $8$ participants to solve a puzzle. These times appear below. However, when the data was entered into the statistical program, the score that was supposed to be $22.1$ was entered as $21.2$. You had calculated the following measures of central tendency: the mean, the median, and the mean trimmed $25\%$. Which of these measures of central tendency will change when you correct the recording error? (relevant section & relevant section)
$\begin{matrix} 15.2\ 18.8\ 19.3\ 19.7\ 20.2\ 21.8\ 22.1\ 29.4 \end{matrix}$
Q7
For the test scores in question Q6, which measures of variability (range, standard deviation, variance) would be changed if the $22.1$ data point had been erroneously recorded as $21.2$? (relevant section)
Q8
You know the minimum, the maximum, and the $25^{th}$, $50^{th}$, and $75^{th}$ percentiles of a distribution. Which of the following measures of central tendency or variability can you determine? (relevant section, relevant section & relevant section)
mean, median, mode, trimean, geometric mean,
range, interquartile range, variance, standard deviation
Q9
1. Find the value ($v$) for which $\sum (X-v)^2$ is minimized.
2. Find the value ($v$) for which $\sum \left | X-v\right |$ is minimized.
Q10
Your younger brother comes home one day after taking a science test. He says that someone at school told him that "$60\%$ of the students in the class scored above the median test grade." What is wrong with this statement? What if he said "$60\%$ of the students scored below the mean?" (relevant section)
Q11
An experiment compared the ability of three groups of participants to remember briefly-presented chess positions. The data are shown below. The numbers represent the number of pieces correctly remembered from three chess positions. Compare the performance of each group. Consider spread as well as central tendency. (relevant section, relevant section & relevant section)
Non-players
Beginners
Tournament players
22.1
32.5
40.1
22.3
37.1
45.6
26.2
39.1
51.2
29.6
40.5
56.4
31.7
45.5
58.1
33.5
51.3
71.1
38.9
52.6
74.9
39.7
55.7
75.9
43.2
55.9
80.3
43.2
57.7
85.3
Q12
True/False: A bimodal distribution has two modes and two medians. (relevant section)
Q13
True/False: The best way to describe a skewed distribution is to report the mean. (relevant section)
Q14
True/False: When plotted on the same graph, a distribution with a mean of $50$ and a standard deviation of $10$ will look more spread out than will a distribution with a mean of $60$ and a standard deviation of $5$. (relevant section)
Q15
Compare the mean, median, trimean in terms of their sensitivity to extreme scores (relevant section).
Q16
If the mean time to respond to a stimulus is much higher than the median time to respond, what can you say about the shape of the distribution of response times? (relevant section)
Q17
A set of numbers is transformed by taking the log base $10$ of each number. The mean of the transformed data is $1.65$. What is the geometric mean of the untransformed data? (relevant section)
Q18
Which measure of central tendency is most often used for returns on investment?
Q19
The histogram is in balance on the fulcrum. What are the mean, median, and mode of the distribution (approximate where necessary)?
Questions from Case Studies
The following questions are from the Angry Moods (AM) case study.
Q20
(AM#4) Does Anger-Out have a positive skew, a negative skew, or no skew? (relevant section)
Q21
(AM#8) What is the range of the Anger-In scores? What is the interquartile range? (relevant section)
Q22
(AM#12) What is the overall mean Control-Out score? What is the mean Control-Out score for the athletes? What is the mean Control-Out score for the non-athletes? (relevant section)
Q23
(AM#15) What is the variance of the Control-In scores for the athletes? What is the variance of the Control-In scores for the non-athletes? (relevant section)
The following question is from the Flatulence (F) case study.
Q24
(F#2) Based on a histogram of the variable "perday", do you think the mean or median of this variable is larger? Calculate the mean and median to see if you are right. (relevant section & relevant section)
The following questions are from the Stroop (S) case study.
Q25
(S#1) Compute the mean for "words". (relevant section)
Q26
(S#2) Compute the mean and standard deviation for "colors". (relevant section & relevant section)
The following questions are from the Physicians' Reactions (PR) case study.
Q27
(PR#2) What is the mean expected time spent for the average-weight patients? What is the mean expected time spent for the overweight patients? (relevant section)
Q28
(PR#3) What is the difference in means between the groups? By approximately how many standard deviations do the means differ? (relevant section & relevant section)
The following question is from the Smiles and Leniency (SL) case study.
Q29
(SL#2) Find the mean, median, standard deviation, and interquartile range for the leniency scores of each of the four groups. (relevant section & relevant section)
The following questions are from the ADHD Treatment (AT) case study.
Q30
(AT#4) What is the mean number of correct responses of the participants after taking the placebo ($0$ mg/kg)? (relevant section)
Q31
(AT#7) What are the standard deviation and the interquartile range of the $d0$ condition? (relevant section)
Selected Answers
S4
Variable A: Mean = $4.8$, Median = $5$
S5
1. Mean = $21$, Var = $144$, SD = $12$
S9
1. $5.2$
S22
Non-athletes: $23.2$
S23
Athletes: $20.5$
S26
Mean = $20.2$
S27
Ave. weight: $31.4$
S29
False smile group:
Mean = $5.37$
Median = $5.50$
SD = $1.83$
IQR = $3.0$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/03%3A_Summarizing_Distributions/3.E%3A_Summarizing_Distributions_%28Exercises%29.txt |
Probability is an important and complex field of study. Fortunately, only a few basic issues in probability theory are essential for understanding statistics at the level covered in this book. These basic issues are covered in this chapter. The introductory section discusses the definitions of probability. This is not as simple as it may seem. The section on basic concepts covers how to compute probabilities in a variety of simple situations. The Gambler's Fallacy Simulation provides an opportunity to explore this fallacy by simulation. The Birthday Demonstration illustrates the probability of finding two or more people with the same birthday. The Binomial Demonstration shows the binomial distribution for different parameters. The section on base rates discusses an important but often-ignored factor in determining probabilities. It also presents Bayes' Theorem. The Bayes' Theorem Demonstration shows how a tree diagram and Bayes' Theorem result in the same answer. Finally, the Monty Hall Demonstration lets you play a game with a very counterintuitive result.
• 4.1: Introduction to Bivariate Data
In this chapter we consider bivariate data, which for now consists of two quantitative variables for each individual. Our first interest is in summarizing such data in a way that is analogous to summarizing univariate (single variable) data.
• 4.2: Values of the Pearson Correlation
The Pearson product-moment correlation coefficient is a measure of the strength of the linear relationship between two variables. It is referred to as Pearson's correlation or simply as the correlation coefficient. If the relationship between the variables is not linear, then the correlation coefficient does not adequately represent the strength of the relationship between the variables.
• 4.3: Guessing Correlations
This demonstration allows you to learn about Pearson's correlation by viewing scatter plots with different values of Pearson's r. In each case, you will have an opportunity to guess the correlation. With a little practice, you should get pretty good at it.
• 4.4: Properties of r
A basic property of Pearson's r is that its possible range is from -1 to 1. A correlation of -1 means a perfect negative linear relationship, a correlation of 0 means no linear relationship, and a correlation of 1 means a perfect positive linear relationship.
• 4.5: Computing r
There are several formulas that can be used to compute Pearson's correlation. Some formulas make more conceptual sense whereas others are easier to actually compute. We are going to begin with a formula that makes more conceptual sense.
• 4.6: Restriction of Range Demo
This demonstration illustrates the effect of restricting the range of scores on the the correlaton between variables.
• 4.7: Variance Sum Law II - Correlated Variables
When variables are correlated, the variance of the sum or difference includes a correlation factor.
• 4.8: Statistical Literacy
• 4.E: Describing Bivariate Data (Exercises)
04: Describing Bivariate Data
Learning Objectives
• Define "bivariate data"
• Define "scatter plot"
• Distinguish between a linear and a nonlinear relationship
• Identify positive and negative associations from a scatter plot
Measures of central tendency, variability, and spread summarize a single variable by providing important information about its distribution. Often, more than one variable is collected on each individual. For example, in large health studies of populations it is common to obtain variables such as age, sex, height, weight, blood pressure, and total cholesterol on each individual. Economic studies may be interested in, among other things, personal income and years of education. As a third example, most university admissions committees ask for an applicant's high school grade point average and standardized admission test scores (e.g., SAT). In this chapter we consider bivariate data, which for now consists of two quantitative variables for each individual. Our first interest is in summarizing such data in a way that is analogous to summarizing univariate (single variable) data.
By way of illustration, let's consider something with which we are all familiar: age. Let’s begin by asking if people tend to marry other people of about the same age. Our experience tells us "yes," but how good is the correspondence? One way to address the question is to look at pairs of ages for a sample of married couples. Table \(1\) below shows the ages of \(10\) married couples. Going across the columns we see that, yes, husbands and wives tend to be of about the same age, with men having a tendency to be slightly older than their wives. This is no big surprise, but at least the data bear out our experiences, which is not always the case.
Table \(1\): Sample of spousal ages of \(10\) White American Couples
Husband 36 72 37 36 51 50 47 50 37 41
Wife 35 67 33 35 50 46 47 42 36 41
The pairs of ages in Table \(1\) are from a dataset consisting of \(282\) pairs of spousal ages, too many to make sense of from a table. What we need is a way to summarize the \(282\) pairs of ages. We know that each variable can be summarized by a histogram (see Figure \(1\)) and by a mean and standard deviation (See Table \(2\)).
Table \(2\): Means and standard deviations of spousal ages
Mean Standard
Deviation
Husbands 49 11
Wives 47 11
Each distribution is fairly skewed with a long right tail. From Table \(1\) we see that not all husbands are older than their wives and it is important to see that this fact is lost when we separate the variables. That is, even though we provide summary statistics on each variable, the pairing within couple is lost by separating the variables. We cannot say, for example, based on the means alone what percentage of couples has younger husbands than wives. We have to count across pairs to find this out. Only by maintaining the pairing can meaningful answers be found about couples per se. Another example of information not available from the separate descriptions of husbands and wives' ages is the mean age of husbands with wives of a certain age. For instance, what is the average age of husbands with \(45\)-year-old wives? Finally, we do not know the relationship between the husband's age and the wife's age.
We can learn much more by displaying the bivariate data in a graphical form that maintains the pairing. Figure \(2\) shows a scatter plot of the paired ages. The \(x\)-axis represents the age of the husband and the \(y\)-axis the age of the wife.
There are two important characteristics of the data revealed by Figure \(2\). First, it is clear that there is a strong relationship between the husband's age and the wife's age: the older the husband, the older the wife. When one variable (\(Y\)) increases with the second variable (\(X\)), we say that \(X\) and \(Y\) have a positive association. Conversely, when \(Y\) decreases as \(X\) increases, we say that they have a negative association.
Second, the points cluster along a straight line. When this occurs, the relationship is called a linear relationship.
Figure \(3\) shows a scatter plot of Arm Strength and Grip Strength from \(149\) individuals working in physically demanding jobs including electricians, construction and maintenance workers, and auto mechanics. Not surprisingly, the stronger someone's grip, the stronger their arm tends to be. There is therefore a positive association between these variables. Although the points cluster along a line, they are not clustered quite as closely as they are for the scatter plot of spousal age.
Not all scatter plots show linear relationships. Figure \(4\) shows the results of an experiment conducted by Galileo on projectile motion. In the experiment, Galileo rolled balls down an incline and measured how far they traveled as a function of the release height. It is clear from Figure \(4\) that the relationship between "Release Height" and "Distance Traveled" is not described well by a straight line: If you drew a line connecting the lowest point and the highest point, all of the remaining points would be above the line. The data are better fit by a parabola.
D. Dickey and T. Arnold's description of the study including a movie
Scatter plots that show linear relationships between variables can differ in several ways including the slope of the line about which they cluster and how tightly the points cluster about the line. A statistical measure of the strength of the relationship between two quantitative variables that takes these factors into account is the subject of the section "Values of Pearson's Correlation."
Contributor
• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.
• Rudy Guerra and David M. Lane | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/04%3A_Describing_Bivariate_Data/4.01%3A_Introduction_to_Bivariate_Data.txt |
Learning Objectives
• Describe what Pearson's correlation measures
• Give the symbols for Pearson's correlation in the sample and in the population
• State the possible range for Pearson's correlation
• Identify a perfect linear relationship
The Pearson product-moment correlation coefficient is a measure of the strength of the linear relationship between two variables. It is referred to as Pearson's correlation or simply as the correlation coefficient. If the relationship between the variables is not linear, then the correlation coefficient does not adequately represent the strength of the relationship between the variables.
The symbol for Pearson's correlation is "\(ρ\)" when it is measured in the population and "\(r\)" when it is measured in a sample. Because we will be dealing almost exclusively with samples, we will use \(r\) to represent Pearson's correlation unless otherwise noted.
Pearson's \(r\) can range from \(-1\) to \(1\). An \(r\) of \(-1\) indicates a perfect negative linear relationship between variables, an \(r\) of \(0\) indicates no linear relationship between variables, and an \(r\) of \(1\) indicates a perfect positive linear relationship between variables. Figure \(1\) shows a scatter plot for which \(r = 1\).
With real data, you would not expect to get values of \(r\) of exactly \(-1, 0,\: or\; 1\). The data for spousal ages shown in Figure \(4\) and described in the introductory section has an \(r\) of \(0.97\).
The relationship between grip strength and arm strength depicted in Figure \(5\) (also described in the introductory section) is \(0.63\).
4.03: Guessing Correlations
Learning Objectives
• Learn how to estimate Pearson's correlation coefficient by viewing scatter plots
Instructions
This demonstration allows you to learn about Pearson's correlation by viewing scatter plots with different values of Pearson's \(r\). In each case, you will have an opportunity to guess the correlation. With a little practice, you should get pretty good at it.
Illustrated Instructions
The demonstration begins by displaying a scatterplot (shown below) and you guess what the correlation of the data in the scatterplot is by selecting one of five options below the graph.
As can be seen in screenshots below, once you select an answer you will be prompted whether the answer is correct or incorrect. Click the "New Data" to get a new dataset.
4.04: Properties of r
Learning Objectives
• State the range of values for Pearson's correlation
• State the values that represent perfect linear relationships
• State the relationship between the correlation of \(Y\) with \(X\) and the correlation of \(X\) with \(Y\)
• State the effect of linear transformations on Pearson's correlation
A basic property of Pearson's \(r\) is that its possible range is from \(-1\) to \(1\). A correlation of \(-1\) means a perfect negative linear relationship, a correlation of \(0\) means no linear relationship, and a correlation of \(1\) means a perfect positive linear relationship.
Pearson's correlation is symmetric in the sense that the correlation of \(X\) with \(Y\) is the same as the correlation of \(Y\) with \(X\). For example, the correlation of Weight with Height is the same as the correlation of Height with Weight.
A critical property of Pearson's \(r\) is that it is unaffected by linear transformations. This means that multiplying a variable by a constant and/or adding a constant does not change the correlation of that variable with other variables. For instance, the correlation of Weight and Height does not depend on whether Height is measured in inches, feet, or even miles. Similarly, adding five points to every student's test score would not change the correlation of the test score with other variables such as GPA. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/04%3A_Describing_Bivariate_Data/4.02%3A_Values_of_the_Pearson_Correlation.txt |
Learning Objectives
• Define $X$ and $x$
• State why $\sum xy=0$ when there is no relationship
• Calculate $r$
There are several formulas that can be used to compute Pearson's correlation. Some formulas make more conceptual sense whereas others are easier to actually compute. We are going to begin with a formula that makes more conceptual sense.
We are going to compute the correlation between the variables $X$ and $Y$ shown in Table $1$. We begin by computing the mean for $X$ and subtracting this mean from all values of $X$. The new variable is called "$x$". The variable "$y$" is computed similarly. The variables $x$ and $y$ are said to be deviation scores because each score is a deviation from the mean. Notice that the means of $x$ and $y$ are both $0$. Next we create a new column by multiplying $x$ and $y$.
Before proceeding with the calculations, let's consider why the sum of the $xy$ column reveals the relationship between $X$ and $Y$. If there were no relationship between $X$ and $Y$, then positive values of $x$ would be just as likely to be paired with negative values of $y$ as with positive values. This would make negative values of $xy$ as likely as positive values and the sum would be small. On the other hand, consider Table 1 in which high values of $X$ are associated with high values of $Y$ and low values of $X$ are associated with low values of $Y$. You can see that positive values of $x$ are associated with positive values of $y$ and negative values of $x$ are associated with negative values of $y$. In all cases, the product of $x$ and $y$ is positive, resulting in a high total for the $xy$ column. Finally, if there were a negative relationship then positive values of $x$ would be associated with negative values of $y$ and negative values of $x$ would be associated with positive values of $y$. This would lead to negative values for $xy$.
Table $1$: Calculation of $r$
X Y x y xy x2 y2
1 4 -3 -5 15 9 25
3 6 -1 -3 3 1 9
5 10 1 1 1 1 1
5 12 1 3 3 1 9
6 13 2 4 8 4 16
Total 20 45 0 0 30 16 60
Mean 4 9 0 0 6
Pearson's $r$ is designed so that the correlation between height and weight is the same whether height is measured in inches or in feet. To achieve this property, Pearson's correlation is computed by dividing the sum of the $xy$ column ($\sum xy$) by the square root of the product of the sum of the $x^2$ column ($\sum x^2$) and the sum of the $y^2$ column ($\sum y^2$). The resulting formula is:
$r=\dfrac{\sum xy}{\sqrt{\sum x^2\sum y^2}}$
and therefore
$r=\dfrac{30}{\sqrt{(16)(60)}}=\dfrac{30}{\sqrt{960}}=\dfrac{30}{30.984}=0.968$
An alternative computational formula that avoids the step of computing deviation scores is:
$r=\dfrac{\sum XY-\dfrac{\sum X\sum Y}{N}}{\sqrt{\left ( \sum X^2-\dfrac{\left ( \sum X \right )^2}{N} \right )}\sqrt{\left ( \sum Y^2-\dfrac{\left ( \sum Y \right )^2}{N} \right )}}$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/04%3A_Describing_Bivariate_Data/4.05%3A_Computing_r.txt |
Learning Objectives
• State the effect of choosing the upper and lower extremes on the correlation
Instructions
This demonstration illustrates the effect of restricting the range of scores on the the correlaton between variables. When the demonstration begins, the dataset "Strength" is displayed. The \(X\)-axis contains scores on a measure of grip strength; the \(Y\)-axis contains scores on a measure of arm strength. Notice that there is a strong relationship. Also note that the two scatterplots are identical and both show that for the entire dataset of \(147\) subjects, the correlaton is \(0.63\). The two vertical bars on the upper graph can be used to select a subset of scores to be plotted on the lower graph. If you click on the left-hand bar and drag it to the right, scores to the left of the bar will not be included on the lower graph. The correlaton shown for the lower graph shows only the correlation for the data in the subset. Similarly, if you drag the right-hand bar to the left, scores to the right of the bar will be excluded. If you click the "Data outside bars" button, the lower graph will, as you might expect, only include scores to the left of the left-hand bar and to the right of the right-hand bar.
Illustrated Instructions
The restriction of range simulation displays data from \(1\) of \(4\) datasets. You can change datasets by selecting one from the drop down menu at the top of the simulation. The top scatterplot displays all the data from the selected dataset.
Notice the two vertical bars on either side of this graph. You can "drag" these bars and restrict the range of data from the dataset that is displayed in the bottom scatterplot. You can include data on the inside or outside of the bars by selecting one of the options right above the graph.
For some browsers you will not see the bars move as you move them. They will move when you release the mouse button.
4.07: Variance Sum Law II - Correlated Variables
Learning Objectives
• State the variance sum law when $X$ and $Y$ are not assumed to be independent
• Compute the variance of the sum of two variables if the variance of each and their correlation is known
• Compute the variance of the difference between two variables if the variance of each and their correlation is known
Recall that when the variables $X$ and $Y$ are independent, the variance of the sum or difference between $X$ and $Y$ can be written as follows:
$\sigma_{X \pm Y}^2 = \sigma_{X }^2 + \sigma_{Y}^2 \label{eq1}$
which is read: "The variance of $X$ plus or minus $Y$ is equal to the variance of $X$ plus the variance of $Y$."
When $X$ and $Y$ are correlated, the following formula should be used:
$\sigma_{X \pm Y}^2 = \sigma_{X }^2 + \sigma_{Y}^2 \pm 2 \rho \sigma_X \sigma_Y \label{eq2}$
where $\rho$ is the correlation between $X$ and $Y$ in the population.
Example $1$
If the variance of verbal SAT were $10,000$, the variance of quantitative SAT were $11,000$ and the correlation between these two tests were $0.50$, what is the variance of total SAT (verbal + quantitative) and the difference (verbal - quantitative)?
Solution
Since the two variables are correlated, we use Equation \ref{eq2} instead of Equation \ref{eq1} for uncorrelated (independent) variables. Hence, the variance of the sum is
$\sigma^2_{verbal + quant} = 10,000 + 11,000 + 2\times 0.5\times \sqrt{10,000} \times \sqrt{11,000}$
which is equal to $31,488$. The variance of the difference is also determined by Equation \ref{eq2}:
$\sigma^2_{verbal - quant} = 10,000 + 11,000 - 2\times 0.5\times \sqrt{10,000} \times \sqrt{11,000}$
which is equal to $10,512$.
If the variances and the correlation are computed in a sample, then the following notation is used to express the variance sum law:
$s_{X \pm Y}^2 = s_{X }^2 + s_{Y}^2 \pm 2 r\, s_X \, s_Y$
4.08: Statistical Literacy
Learning Objectives
• Relationship between age and sleep
Age and Sleep
The graph below showing the relationship between age and sleep is based on a graph that appears on this web page.
Example \(1\)
Based on the graph in Figure \(1\) Why might Pearson's correlation not be a good way to describe the relationship?
Solution
Pearson's correlation measures the strength of the linear relationship between two variables. The relationship here is not linear. As age increases, hours slept decreases rapidly at first but then levels off. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/04%3A_Describing_Bivariate_Data/4.06%3A_Restriction_of_Range_Demo.txt |
General Questions
Q1
Describe the relationship between variables \(A\) and \(C\). Think of things these variables could represent in real life. (relevant section)
Q2
Make up a data set with \(10\) numbers that has a positive correlation. (relevant section & relevant section)
Q3
Make up a data set with \(10\) numbers that has a negative correlation. (relevant section & relevant section)
Q4
If the correlation between weight (in pounds) and height (in feet) is \(0.58\), find: (relevant section)
1. the correlation between weight (in pounds) and height (in yards)
2. the correlation between weight (in kilograms) and height (in meters)
Q5
Would you expect the correlation between High School GPA and College GPA to be higher when taken from your entire high school class or when taken from only the top \(20\) students? Why? (relevant section)
Q6
For a certain class, the relationship between the amount of time spent studying and the test grade earned was examined. It was determined that as the amount of time they studied increased, so did their grades. Is this a positive or negative association? (relevant section)
Q7
For this same class, the relationship between the amount of time spent studying and the amount of time spent socializing per week was also examined. It was determined that the more hours they spent studying, the fewer hours they spent socializing. Is this a positive or negative association? (relevant section)
Q8
For the following data: (relevant section)
A B
2 8
5 5
6 2
8 4
9 1
1. Find the deviation scores for Variable \(A\) that correspond to the raw scores of \(2\) and \(8\).
2. Find the deviation scores for Variable \(B\) that correspond to the raw scores of \(5\) and \(4\).
3. Just from looking at these scores, do you think these variable \(A\) and \(B\) are positively or negatively correlated? Why?
4. Now calculate the correlation. Were you right?
Q9
Students took two parts of a test, each worth \(50\) points. Part \(A\) has a variance of \(25\), and Part B has a variance of \(36\). The correlation between the test scores is \(0.8\).
1. If the teacher adds the grades of the two parts together to form a final test grade, what would the variance of the final test grades be?
2. What would the variance of Part \(A\) - Part \(B\) be? (relevant section)
Q10
True/False: The correlation in real life between height and weight is \(r=1\). (relevant section)
Q11
True/False: It is possible for variables to have \(r=0\) but still have a strong association. (relevant section & relevant section)
Q12
True/False: Two variables with a correlation of \(0.3\) have a stronger linear relationship than two variables with a correlation of \(-0.7\). (relevant section)
Q13
True/False: After polling a certain group of people, researchers found a \(0.5\) correlation between the number of car accidents per year and the driver's age. This means that older people get in more accidents. (relevant section)
Q14
True/False: The correlation between \(R \) and \(T\) is the same as the correlation between \(T\) and \(R\). (relevant section)
Q15
True/False: To examine bivariate data graphically, the best choice is two side by side histograms. (relevant section)
Q16
True/False: A correlation of \(r=1.2\) is not possible. (relevant section)
Questions from Case Studies
The following questions are from the Angry Moods (AM) case study.
Q17
(AM#21) What is the correlation between the Control-In and Control-Out scores? (relevant section)
Q18
(AM#22) Would you expect the correlation between the Anger-Out and Control-Out scores to be positive or negative? Compute this correlation. (relevant section & relevant section)
The following question is from the Flatulence (F) case study.
Q19
(F#4) Is there a relationship between the number of male siblings and embarrassment in front of romantic interests? Create a scatterplot and compute r. (relevant section & relevant section)
The following questions are from the Stroop (S) case study.
Q20
(S#6) Create a scatterplot showing "colors" on the \(Y\)-axis and "words" on the \(X\)-axis. (relevant section)
Q21
(S#7) Compute the correlation between "colors" and "words." (relevant section)
Q22
Sort the data by color-naming time. Choose only the \(20\) fastest color-namers and create a scatterplot. (relevant section)
1. What is the new correlation? (relevant section)
2. What is the technical term for the finding that this correlation is smaller than the correlation for the full dataset? (relevant section)
The following question is from the Animal Research (AR) case study.
Q23
(AR#11) What is the overall correlation between the belief that animal research is wrong and belief that animal research is necessary? (relevant section)
The following question is from the ADHD Treatment (AT) case study.
Q24
(AT#8) What is the correlation between the participants' correct number of responses after taking the placebo and their correct number of responses after taking \(0.60\) mg/kg of MPH? (relevant section)
Select Answers
1. \(0.58\)
1. \(-4, 2\)
1. \(109\)
\(r = 0.72\)
S19
\(r = -0.30\)
(larger numbers for the "romint" variable mean less embarrassment)
\(r = 0.70\) | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/04%3A_Describing_Bivariate_Data/4.E%3A_Describing_Bivariate_Data_%28Exercises%29.txt |
A dataset with two variables contains what is called bivariate data. This chapter discusses ways to describe the relationship between two variables. For example, you may wish to describe the relationship between the heights and weights of people to determine the extent to which taller people weigh more. The introductory section gives more examples of bivariate relationships and presents the most common way of portraying these relationships graphically. The next five sections discuss Pearson's correlation, the most common index of the relationship between two variables. The final section, "Variance Sum Law II," makes use of Pearson's correlation to generalize this law to bivariate data.
• 5.1: The Concept of "Probability"
Inferential statistics is built on the foundation of probability theory, and has been remarkably successful in guiding opinion about the conclusions to be drawn from data. Yet (paradoxically) the very idea of probability has been plagued by controversy from the beginning of the subject to the present day. In this section we provide a glimpse of the debate about the interpretation of the probability concept.
• 5.2: Basic Concepts of Probability
Probability is an important and complex field of study. Fortunately, only a few basic issues in probability theory are essential for understanding statistics at the level covered in this book. These basic issues are covered in this chapter.
• 5.3: Conditional Probability Demonstration
The simulation demonstrates how different conditional probabilities would be calculated from the provided data.
• 5.4: Gambler's Fallacy
The gambler's fallacy involves beliefs about sequences of independent events. By definition, if two events are independent, the occurrence of one event does not affect the occurrence of the second. The gambler's fallacy is mistakenly believing that certain events are dependent.
• 5.5: Permutations and Combinations
This section covers basic formulas for determining the number of various possible types of outcomes. The topics covered are: (1) counting the number of possible orders, (2) counting using the multiplication rule, (3) counting the number of permutations, and (4) counting the number of combinations.
• 5.6: Birthday Demo
Most people's intuition about the probability of two people in a group sharing a birthday is way off. This simulation allows you to approach this problem concretely.
• 5.7: Binomial Distribution
In the present section, we consider probability distributions for which there are just two possible outcomes with fixed probabilities summing to one. These distributions are called binomial distributions.
• 5.8: Binomial Demonstration
This demonstration allows you to explore the binomial distribution.
• 5.9: Poisson Distribution
The Poisson distribution can be used to calculate the probabilities of various numbers of "successes" based on the mean number of successes.
• 5.10: Multinomial Distribution
The multinomial distribution can be used to compute the probabilities in situations in which there are more than two possible outcomes.
• 5.11: Hypergeometric Distribution
The hypergeometric distribution is used to calculate probabilities when sampling without replacement.
• 5.12: Base Rates
Compute the probability of a condition from hits, false alarms, and base rates using a tree diagram. Compute the probability of a condition from hits, false alarms, and base rates using Bayes' Theorem
• 5.13: Bayes Demo
This demonstration lets you examine the effects of base rate, true positive rate, and false positive rate on the probability that a person diagnosed with disease X actually has the disease. The base rate is the proportion of people who have the disease. The true positive rate is the probability that a person with the disease will test positive. The false positive rate is the probability that someone who does not have the disease will test positive.
• 5.14: Monty Hall Problem
In the Monty Hall game, a contestant is shown three doors. Two of the doors have goats behind them and one has a car. The contestant chooses a door. Before opening the chosen door, Monty Hall opens a door that has a goat behind it. The contestant can then switch to the other unopened door, or stay with the original choice.
• 5.15: Statistical Literacy
• 5.E: Probability (Exercises)
05: Probability
Learning Objectives
• Define symmetrical outcomes
• Distinguish between frequentist and subjective approaches
• Determine whether the frequentist or subjective approach is better suited for a given situation
Inferential statistics is built on the foundation of probability theory, and has been remarkably successful in guiding opinion about the conclusions to be drawn from data. Yet (paradoxically) the very idea of probability has been plagued by controversy from the beginning of the subject to the present day. In this section we provide a glimpse of the debate about the interpretation of the probability concept.
One conception of probability is drawn from the idea of symmetrical outcomes. For example, the two possible outcomes of tossing a fair coin seem not to be distinguishable in any way that affects which side will land up or down. Therefore the probability of heads is taken to be \(1/2\), as is the probability of tails. In general, if there are \(N\) symmetrical outcomes, the probability of any given one of them occurring is taken to be \(1/N\). Thus, if a six-sided die is rolled, the probability of any one of the six sides coming up is \(1/6\).
Probabilities can also be thought of in terms of relative frequencies. If we tossed a coin millions of times, we would expect the proportion of tosses that came up heads to be pretty close to \(1/2\). As the number of tosses increases, the proportion of heads approaches \(1/2\). Therefore, we can say that the probability of a head is \(1/2\).
If it has rained in Seattle on \(62\%\) of the last \(100,000\) days, then the probability of it raining tomorrow might be taken to be \(0.62\). This is a natural idea but nonetheless unreasonable if we have further information relevant to whether it will rain tomorrow. For example, if tomorrow is August 1, a day of the year on which it seldom rains in Seattle, we should only consider the percentage of the time it rained on August 1. But even this is not enough since the probability of rain on the next August 1 depends on the humidity. (The chances are higher in the presence of high humidity.) So, we should consult only the prior occurrences of August 1 that had the same humidity as the next occurrence of August 1. Of course, wind direction also affects probability ... You can see that our sample of prior cases will soon be reduced to the empty set. Anyway, past meteorological history is misleading if the climate is changing.
For some purposes, probability is best thought of as subjective. Questions such as "What is the probability that Ms. Garcia will defeat Mr. Smith in an upcoming congressional election?" do not conveniently fit into either the symmetry or frequency approaches to probability. Rather, assigning probability \(0.7\) (say) to this event seems to reflect the speaker's personal opinion --- perhaps his willingness to bet according to certain odds. Such an approach to probability, however, seems to lose the objective content of the idea of chance; probability becomes mere opinion.
Two people might attach different probabilities to the election outcome, yet there would be no criterion for calling one "right" and the other "wrong." We cannot call one of the two people right simply because she assigned higher probability to the outcome that actually transpires. After all, you would be right to attribute probability \(1/6\) to throwing a six with a fair die, and your friend who attributes \(2/3\) to this event would be wrong. And you are still right (and your friend is still wrong) even if the die ends up showing a six! The lack of objective criteria for adjudicating claims about probabilities in the subjective perspective is an unattractive feature of it for many scholars.
Like most work in the field, the present text adopts the frequentist approach to probability in most cases. Moreover, almost all the probabilities we shall encounter will be nondogmatic, that is, neither zero nor one. An event with probability \(0\) has no chance of occurring; an event of probability \(1\) is certain to occur. It is hard to think of any examples of interest to statistics in which the probability is either \(0\) or \(1\). (Even the probability that the Sun will come up tomorrow is less than \(1\).)
The following example illustrates our attitude about probabilities. Suppose you wish to know what the weather will be like next Saturday because you are planning a picnic. You turn on your radio, and the weather person says, “There is a 10% chance of rain.” You decide to have the picnic outdoors and, lo and behold, it rains. You are furious with the weather person. But was she wrong? No, she did not say it would not rain, only that rain was unlikely. She would have been flatly wrong only if she said that the probability is \(0\) and it subsequently rained. However, if you kept track of her weather predictions over a long period of time and found that it rained on \(50\%\) of the days that the weather person said the probability was \(0.10\), you could say her probability assessments are wrong.
So when is it accurate to say that the probability of rain is \(0.10\)? According to our frequency interpretation, it means that it will rain \(10\%\) of the days on which rain is forecast with this probability. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/05%3A_Probability/5.01%3A_The_Concept_of_Probability.txt |
Learning Objectives
• Compute probability in a situation where there are equally-likely outcomes
• Apply concepts to cards and dice
• Compute the probability of two independent events both occurring
• Compute the probability of either of two independent events occurring
• Do problems that involve conditional probabilities
• Compute the probability that in a room of $N$ people, at least two share a birthday
• Describe the gambler's fallacy
Probability of a Single event
If you roll a six-sided die, there are six possible outcomes, and each of these outcomes is equally likely. A six is as likely to come up as a three, and likewise for the other four sides of the die. What, then, is the probability that a one will come up? Since there are six possible outcomes, the probability is $1/6$. What is the probability that either a one or a six will come up? The two outcomes about which we are concerned (a one or a six coming up) are called favorable outcomes. Given that all outcomes are equally likely, we can compute the probability of a one or a six using the formula:
$\text{probability}=\frac{\text{Number of favorable outcomes}}{\text{Number of possible equally-likely outcomes}}$
In this case there are two favorable outcomes and six possible outcomes. So the probability of throwing either a one or six is $1/3$. Don't be misled by our use of the term "favorable," by the way. You should understand it in the sense of "favorable to the event in question happening." That event might not be favorable to your well-being. You might be betting on a three, for example.
The above formula applies to many games of chance. For example, what is the probability that a card drawn at random from a deck of playing cards will be an ace? Since the deck has four aces, there are four favorable outcomes; since the deck has $52$ cards, there are $52$ possible outcomes. The probability is therefore $4/52 = 1/13$. What about the probability that the card will be a club? Since there are $13$ clubs, the probability is $13/52 = 1/4$.
Let's say you have a bag with $20$ cherries: $14$ sweet and $6$ sour. If you pick a cherry at random, what is the probability that it will be sweet? There are $20$ possible cherries that could be picked, so the number of possible outcomes is $20$. Of these $20$ possible outcomes, $14$ are favorable (sweet), so the probability that the cherry will be sweet is $14/20 = 7/10$. There is one potential complication to this example, however. It must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other. This wouldn't be true if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.
Here is a more complex example.
Example $1$
You throw $2$ dice. What is the probability that the sum of the two dice will be $6$? To solve this problem, list all the possible outcomes. There are $36$ of them since each die can come up one of six ways. The $36$ possibilities are shown below.
Table $1$: $36$ possibilities from $2$ dice throw
Die 1 Die 2 Total Die 1 Die 2 Total Die 1 Die 2 Total
1 1 2 3 1 4 5 1 6
1 2 3 3 2 5 5 2 7
1 3 4 3 3 6 5 3 8
1 4 5 3 4 7 5 4 9
1 5 6 3 5 8 5 5 10
1 6 7 3 6 9 5 6 11
2 1 3 4 1 5 6 1 7
2 2 4 4 2 6 6 2 8
2 3 5 4 3 7 6 3 9
2 4 6 4 4 8 6 4 10
2 5 7 4 5 9 6 5 11
2 6 8 4 6 10 6 6 12
You can see that $5$ of the $36$ possibilities total $6$. Therefore, the probability is $5/36$.
If you know the probability of an event occurring, it is easy to compute the probability that the event does not occur. If $P(A)$ is the probability of Event $A$, then $1-P(A)$ is the probability that the event does not occur. For the last example, the probability that the total is $6$ is $5/36$. Therefore, the probability that the total is not $6$ is $1 - 5/36 = 31/36$.
Probability of Two (or more) independent events
Events $A$ and $B$ are independent events if the probability of Event $B$ occurring is the same whether or not Event $A$ occurs. Let's take a simple example. A fair coin is tossed two times. The probability that a head comes up on the second toss is $1/2$ regardless of whether or not a head came up on the first toss. The two events are
1. first toss is a head and
2. second toss is a head.
So these events are independent.
Consider the two events:
1. "It will rain tomorrow in Houston" and
2. "It will rain tomorrow in Galveston" (a city near Houston)
These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.
Probability of A and B
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events. More formally, if events $A$ and $B$ are independent, then the probability of both $A$ and $B$ occurring is:
$P(A\; \text{and}\; B)=P(A)\times P(B)$
where $P(A\; \text{and}\; B)$ is the probability of events $A$ and $B$ both occurring, $P(A)$ is the probability of event $A$ occurring, and $P(B)$ is the probability of event $B$ occurring.
Example $2$
If you flip a coin twice, what is the probability that it will come up heads both times?
Solution
Event $A$ is that the coin comes up heads on the first flip and Event $B$ is that the coin comes up heads on the second flip. Since both $P(A)$ and $P(B)$ equal $1/2$, the probability that both events occur is
$\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$
Example $3$
If you flip a coin and roll a six-sided die, what is the probability that the coin comes up heads and the die comes up $1$?
Solution
Since the two events are independent, the probability is simply the probability of a head (which is $1/2$) times the probability of the die coming up $1$ (which is $1/6$). Therefore, the probability of both events occurring is
$\frac{1}{2}\times \frac{1}{6}=\frac{1}{12}$
Example $4$
You draw a card from a deck of cards, put it back, and then draw another card. What is the probability that the first card is a heart and the second card is black?
Solution
Since there are $52$ cards in a deck and $13$ of them are hearts, the probability that the first card is a heart is $13/52 = 1/4$. Since there are $26$ black cards in the deck, the probability that the second card is black is $26/52 = 1/2$. The probability of both events occurring is therefore
$\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}$
See the section on conditional probabilities on this page to see how to compute $P(A\; \text{and}\; B)$when $A$ and $B$ are not independent.
Probability of A or B
If Events $A$ and $B$ are independent, the probability that either Event $A$ or Event $B$ occurs is:
$P(A\; \text{or}\; B)=P(A)+P(B)-P(A\; \text{and}\; B)$
In this discussion, when we say "$A$ or $B$ occurs" we include three possibilities:
1. $A$ occurs and $B$ does not occur
2. $B$ occurs and $A$ does not occur
3. Both $A$ and $B$ occur
This use of the word "or" is technically called inclusive or because it includes the case in which both $A$ and $B$ occur. If we included only the first two cases, then we would be using an exclusive or.
(Optional) We can derive the law for $P(A\; \mathbf{or}\; B)$ from our law about $P(A\; \mathbf{and}\; B)$. The event "$\textbf{A-or-B}$" can happen in any of the following ways:
1. $\textbf{A-and-B}$ happens
2. $\textbf{A-and-not-B}$ happens
3. $\textbf{not-A-and-B}$ happens
The simple event $A$ can happen if either $\textbf{A-and-B}$ happens or $\textbf{A-and-not-B}$ happens. Similarly, the simple event $B$ happens if either $\textbf{A-and-B}$ happens or $\textbf{not-A-and-B}$ happens. $P(A) + P(B)$ is therefore $P(A-and-B) + P(A-and-not-B) + P(A-and-B) + P(not-A-and-B)$, whereas $P(A-or-B)$ is $P(A-and-B) + P(A-and-not-B) + P(not-A-and-B)$. We can make these two sums equal by subtracting one occurrence of $P(A-and-B)$ from the first. Hence, $P(A-or-B) = P(A) + P(B) - P(A-and-B)$.
Now for some examples.
Example $5$
If you flip a coin two times, what is the probability that you will get a head on the first flip or a head on the second flip (or both)?
Solution
Letting Event $A$ be a head on the first flip and Event $B$ be a head on the second flip, then $P(A) = 1/2$, $P(B) = 1/2$, and $P(A\; \text{and}\; B) = 1/4$. Therefore,
$P(A\; \text{or}\; B)=\frac{1}{2}+\frac{1}{2}-\frac{1}{4}=\frac{3}{4}$
Example $6$
If you throw a six-sided die and then flip a coin, what is the probability that you will get either a $6$ on the die or a head on the coin flip (or both)?
Solution
Using the formula,
\begin{align*} P(6\; \text{or head}) &= P(6)+P(\text{head})-P(6\; \text{and head})\ &= \frac{1}{6}+\frac{1}{2}-\left ( \frac{1}{6} \right )\left ( \frac{1}{2} \right )\ &= \frac{7}{12} \end{align*}
An alternate approach to computing this value is to start by computing the probability of not getting either a $6$ or a head. Then subtract this value from $1$ to compute the probability of getting a $6$ or a head. Although this is a complicated method, it has the advantage of being applicable to problems with more than two events. Here is the calculation in the present case. The probability of not getting either a $6$ or a head can be recast as the probability of
$(\text{not getting a 6})\; AND\; (\text{not getting a head})$
This follows because if you did not get a $6$ and you did not get a head, then you did not get a $6$ or a head. The probability of not getting a six is $1 - 1/6 = 5/6$. The probability of not getting a head is $1 - 1/2 = 1/2$. The probability of not getting a six and not getting a head is $5/6 \times 1/2 = 5/12$. This is therefore the probability of not getting a $6$ or a head. The probability of getting a six or a head is therefore (once again) $1 - 5/12 = 7/12$.
If you throw a die three times, what is the probability that one or more of your throws will come up with a $1$? That is, what is the probability of getting a $1$ on the first throw OR a $1$ on the second throw OR a $1$ on the third throw? The easiest way to approach this problem is to compute the probability of
• NOT getting a $1$ on the first throw
• AND not getting a $1$ on the second throw
• AND not getting a $1$ on the third throw
The answer will be $1$ minus this probability. The probability of not getting a $1$ on any of the three throws is $5/6 \times 5/6 \times 5/6 = 125/216$. Therefore, the probability of getting a $1$ on at least one of the throws is $1 - 125/216 = 91/216$.
Conditional Probabilities
Often it is required to compute the probability of an event given that another event has occurred. For example, what is the probability that two cards drawn at random from a deck of playing cards will both be aces? It might seem that you could use the formula for the probability of two independent events and simply multiply $4/52 \times 4/52 = 1/169$. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck.
Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case, the "condition" is that the first card is an ace. Symbolically, we write this as:
$P(\text{ace on second draw}\; |\; \text{an ace on the first draw})$
The vertical bar "|" is read as "given," so the above expression is short for: "The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw." What is this probability? Since after an ace is drawn on the first draw, there are $3$ aces out of $51$ total cards left. This means that the probability that one of these aces will be drawn is $3/51 = 1/17$.
If Events $A$ and $B$ are not independent, then $P(A\; \text{and} B) = P(A) \times P(B|A)$
Applying this to the problem of two aces, the probability of drawing two aces from a deck is $4/52 \times 3/51 = 1/221$.
Example $7$
If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card?
Solution
There are two ways you can satisfy this condition:
1. You can get the Ace of Diamonds first and then a black card
2. You can get a black card first and then the Ace of Diamonds
Let's calculate Case $A$.
The probability that the first card is the Ace of Diamonds is $1/52$. The probability that the second card is black given that the first card is the Ace of Diamonds is $26/51$ because $26$ of the remaining $51$ cards are black. The probability is therefore $1/52 \times 26/51 = 1/102$.
Now for Case $B$:
The probability that the first card is black is $26/52 = 1/2$. The probability that the second card is the Ace of Diamonds given that the first card is black is $1/51$. The probability of Case $B$ is therefore $1/2 \times 1/51 = 1/102$, the same as the probability of Case $A$. Recall that the probability of $A$ or $B$ is $P(A)+P(B)-P(A\; \text{and}\; B)$. In this problem, $P(A\; \text{and}\; B) = 0$ since a card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case $A$ or Case $B$ is $1/102 + 1/102 = 2/102 = 1/51$. So, $1/51$ is the probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck.
Birthday Problem
If there are $25$ people in a room, what is the probability that at least two of them share the same birthday. If your first thought is that it is $25/365 = 0.068$, you will be surprised to learn it is much higher than that. This problem requires the application of the sections on $P(A\; \text{and}\; B)$ and conditional probability.
This problem is best approached by asking what is the probability that no two people have the same birthday. Once we know this probability, we can simply subtract it from $1$ to find the probability that two people share a birthday.
If we choose two people at random, what is the probability that they do not share a birthday? Of the $365$ days on which the second person could have a birthday, $364$ of them are different from the first person's birthday. Therefore the probability is $364/365$. Let's define $P2$ as the probability that the second person drawn does not share a birthday with the person drawn previously. $P2$ is therefore $364/365$. Now define $P3$ as the probability that the third person drawn does not share a birthday with anyone drawn previously given that there are no previous birthday matches. $P3$ is therefore a conditional probability. If there are no previous birthday matches, then two of the $365$ days have been "used up," leaving $363$ non-matching days. Therefore $P3 = 363/365$. In like manner, $P4 = 362/365$, $P5 = 361/365$, and so on up to $P25 = 341/365$.
In order for there to be no matches, the second person must not match any previous person and the third person must not match any previous person, and the fourth person must not match any previous person, etc. Since $P(A\; \text{and}\; B) = P(A)P(B)$, all we have to do is multiply $P2, P3, P4 ...P25$ together. The result is $0.431$. Therefore the probability of at least one match is $0.569$.
Gambler's Fallacy
A fair coin is flipped five times and comes up heads each time. What is the probability that it will come up heads on the sixth flip? The correct answer is, of course, $1/2$. But many people believe that a tail is more likely to occur after throwing five heads. Their faulty reasoning may go something like this: "In the long run, the number of heads and tails will be the same, so the tails have some catching up to do." The flaws in this logic are exposed in the simulation in this chapter. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/05%3A_Probability/5.02%3A_Basic_Concepts_of_Probability.txt |
Learning Objectives
• Compute conditional probabilities
Instructions
The simulation shows a set of \(30\) objects varying in color (red, blue, and purple) and shape (\(X\) and \(O\)). One of the objects is to be chosen at random. The various possible conditional probabilities are shown below the objects. Click on one to see how the conditional probability is computed. For example, if you click on P(X|Red) which is read "the probability of \(X\) given Red" then a box is put around each of the red objects. Of these, those that are \(X's\) are shaded. The probability of \(X\) given it is red is the number of red \(X's\) (shaded boxes) divided by the total red items (the number of boxes).
Click "Another Example" for new distribution of objects.
Illustrated Instructions
This demonstration starts by presenting you with a dataset of red, blue and purple \(Xs\) and \(Os\) (see screenshot below).
Selecting one of the conditional probabilities changes the display to show you how this probability would be calculated from the provided data.
The screenshot below shows the conditional probability for getting a selection being Blue given that it is a \(O\). The demonstration shows Puts a square around all \(19\) \(Os\) in the data and a shaded square around the \(5\) blue \(Os\).
Click "Another Example" for new distribution of objects.
5.04: Gambler's Fallacy
Learning Objectives
• Understand the link between the number of trials and expected outcome of probability distributions
Instructions
The gambler's fallacy involves beliefs about sequences of independent events. By definition, if two events are independent, the occurrence of one event does not affect the occurrence of the second. For example, if a fair coin is flipped twice, the occurrence of a head on the first flip does not affect the outcome of the second flip.
What if a coin is flipped five times and comes up heads each time. Is a tail "due" and therefore more likely than not to occur on the next flip. Since the events are independent, the answer is "no."
The gambler's fallacy is mistakenly believing that the answer is "yes." Believers in the fallacy argue correctly that in the long run, the proportion of heads will be \(0.50\). Or, stated more precisely, as the number of flips approaches infinity, the proportion of heads approaches \(0.50\). Doesn't this imply that there must be some natural adjustment occurring, compensating for a string of heads with the later occurrence of more tails?
This simulation allows you to explore this question yourself. You can simulate the flipping of a single coin by clicking the "flip once" button. The number of flips (\(n\)), the number of heads, the number of tails, the difference between the number of heads and the number of tails, and the proportion of heads are all recorded and displayed.
To speed things up, you can click the "Flip \(100\) times" or the "Flip \(1000\) times" buttons. The "reset" button clears all the data. The lower portion of the simulation allows you to simulate \(25,000\) flips at a time. After you click the "Flip \(25,000\) times and draw graphs" button, the graph on the left will plot the difference between the number of heads and tails as a function of the number of coin flips. The graph on the right will plot the proportion of heads as a function of the number of coin flips.
Illustrated Instructions
The gambler's fallacy demonstration allows you to flip a fair coin in a variety of increments. Each time you click one of these buttons the total number of coin flips is increased by the increment on the respective button.
The screenshot below shows what happens when you click the "Flip \(25,000\) times...". The demonstration draws to graphs the first of which shows the Heads minus Tails for all trials and the second shows the proportion of heads for all trials. Clicking this button repeatedly increments the number of trials and adjusts the graphs accordingly. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/05%3A_Probability/5.03%3A_Conditional_Probability_Demonstration.txt |
Learning Objectives
• Calculate the probability of two independent events occurring
• Define permutations and combinations
• List all permutations and combinations
• Apply formulas for permutations and combinations
This section covers basic formulas for determining the number of various possible types of outcomes. The topics covered are:
• counting the number of possible orders
• counting using the multiplication rule
• counting the number of permutations
• counting the number of combinations
Possible Orders
Suppose you had a plate with three pieces of candy on it: one green, one yellow, and one red. You are going to pick up these three pieces one at a time. The question is: In how many different orders can you pick up the pieces? Table $1$ lists all the possible orders.
There are two orders in which red is first: red, yellow, green and red, green, yellow. Similarly, there are two orders in which yellow is first and two orders in which green is first. This makes six possible orders in which the pieces can be picked up.
Table $1$: Six Possible Orders.
Number First Second Third
1 red yellow green
2 red green yellow
3 yellow red green
4 yellow green red
5 green red yellow
6 green yellow red
The formula for the number of orders is shown below.
$\text{Number of orders} = n!$
where $n$ is the number of pieces to be picked up. The symbol "!" stands for factorial. Some examples are:
\begin{align} 3! &= 3 \times 2 \times 1 = 6 \ 4! &= 4 \times 3 \times 2 \times 1 = 24 \ 5! &= 5 \times 4 \times 3 \times 2 \times 1 = 120 \end{align}
This means that if there were $5$ pieces of candy to be picked up, they could be picked up in any of $5! = 120$ orders.
Multiplication Rule
Imagine a small restaurant whose menu has $3$ soups, $6$ entrées, and $4$ desserts. How many possible meals are there? The answer is calculated by multiplying the numbers to get $3 \times 6 \times 4 = 72$. You can think of it as first there is a choice among $3$ soups. Then, for each of these choices there is a choice among $6$ entrées resulting in $3 \times 6 = 18$ possibilities. Then, for each of these $18$ possibilities there are $4$ possible desserts yielding $18 \times 4 = 72$ total possibilities.
Permutations
Suppose that there were four pieces of candy (red, yellow, green, and brown) and you were only going to pick up exactly two pieces. How many ways are there of picking up two pieces? Table $2$ lists all the possibilities. The first choice can be any of the four colors. For each of these $4$ first choices there are $3$ second choices. Therefore there are $4 \times 3 = 12$ possibilities.
Table $2$: Twelve Possible Orders
Number First Second
1 red yellow
2 red green
3 red brown
4 yellow red
5 yellow green
6 yellow brown
7 green red
8 green yellow
9 green brown
10 brown red
11 brown yellow
12 brown green
More formally, this question is asking for the number of permutations of four things taken two at a time. The general formula is:
$_nP_r = \dfrac{n!}{(n-r)!}$
where $_nP_r$ is the number of permutations of $n$ things taken $r$ at a time. In other words, it is the number of ways $r$ things can be selected from a group of $n$ things. In this case,
$_4P_2 = \dfrac{4!}{(4-2)!} = \dfrac{4 \times 3 \times 3 \times 2 \times 1}{2 \times 1} = 12$
It is important to note that order counts in permutations. That is, choosing red and then yellow is counted separately from choosing yellow and then red. Therefore permutations refer to the number of ways of choosing rather than the number of possible outcomes. When order of choice is not considered, the formula for combinations is used.
Combinations
Now suppose that you were not concerned with the way the pieces of candy were chosen but only in the final choices. In other words, how many different combinations of two pieces could you end up with? In counting combinations, choosing red and then yellow is the same as choosing yellow and then red because in both cases you end up with one red piece and one yellow piece. Unlike permutations, order does not count. Table $3$ is based on Table $2$ but is modified so that repeated combinations are given an "$x$" instead of a number. For example, "yellow then red" has an "$x$" because the combination of red and yellow was already included as choice number $1$. As you can see, there are six combinations of the three colors.
Table $1$: Six Combinations.
Number First Second
1 red yellow
2 red green
3 red brown
x yellow red
4 yellow green
5 yellow brown
x green red
x green yellow
6 green brown
x brown red
x brown yellow
x brown green
The formula for the number of combinations is shown below where $_nC_r$ is the number of combinations for $n$ things taken $r$ at a time.
$_nC_r = \dfrac{n!}{(n-r)!r!}$
For our example,
$_4C_2 = \dfrac{4!}{(4-2)!2!} = \dfrac{4 \times 3 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = 6$
which is consistent with Table $3$.
As an example application, suppose there were six kinds of toppings that one could order for a pizza. How many combinations of exactly $3$ toppings could be ordered? Here $n = 6$ since there are $6$ toppings and $r = 3$ since we are taking $3$ at a time. The formula is then:
$_6C_3 = \dfrac{6!}{(6-3)!3!} = \dfrac{6\times 5 \times 4 \times 3 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(3 \times 2 \times 1)} = 30$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/05%3A_Probability/5.05%3A_Permutations_and_Combinations.txt |
Learning Objectives
• Become familiar with the properties of the birthday problem
Instructions
Most people's intuition about the probability of two people in a group sharing a birthday is way off. This simulation allows you to approach this problem concretely.
When you click the "Draw Birthday" button, a birthday is randomly drawn from the dates listed. The date drawn is listed on the left in red and shown in the list of dates in red. As you push the button, more dates will be shown. If a date is a duplicate, it will be shown in blue on the left. The "Show Duplicates" button shows allow duplicate dates drawn.
Illustrated Instructions
The screenshot below shows opening state of the birthday problem demonstration. Each time you click the "Draw Birthday" button a random date is drawn from the calendar and displayed under the button. Any duplicates are highlighted in blue and you can click the "Show Duplication Info" button to see summary data of all duplicates.
Below is an example of \(30\) drawn dates with one duplicate "\(10/18\)". The summary that appears when the "Show Duplication Info" button is clicked is also shown.
5.07: Binomial Distribution
Learning Objectives
• Define binomial outcomes
• Compute the probability of getting $X$ successes in $N$ trials
• Compute cumulative binomial probabilities
• Find the mean and standard deviation of a binomial distribution
When you flip a coin, there are two possible outcomes: heads and tails. Each outcome has a fixed probability, the same from trial to trial. In the case of coins, heads and tails each have the same probability of $1/2$. More generally, there are situations in which the coin is biased, so that heads and tails have different probabilities. In the present section, we consider probability distributions for which there are just two possible outcomes with fixed probabilities summing to one. These distributions are called binomial distributions.
A Simple Example
The four possible outcomes that could occur if you flipped a coin twice are listed below in Table $1$. Note that the four outcomes are equally likely: each has probability $1/4$. To see this, note that the tosses of the coin are independent (neither affects the other). Hence, the probability of a head on $\text{Flip 1}$ and a head on $\text{Flip 2}$ is the product of $P(H)$ and $P(H)$, which is $1/2 \times 1/2 = 1/4$. The same calculation applies to the probability of a head on $\text{Flip 1}$ and a tail on $\text{Flip 2}$. Each is $1/2 \times 1/2 = 1/4$.
Table $1$: Four Possible Outcomes
Outcome First Flip Second Flip
1 Heads Heads
2 Heads Tails
3 Tails Heads
4 Tails Tails
The four possible outcomes can be classified in terms of the number of heads that come up. The number could be two (Outcome $1$, one (Outcomes $2$ and $3$) or $0$ (Outcome $4$). The probabilities of these possibilities are shown in Table $2$ and in Figure $1$.
Since two of the outcomes represent the case in which just one head appears in the two tosses, the probability of this event is equal to $1/4 + 1/4 = 1/2$. Table $2$ summarizes the situation.
Table $2$: Probabilities of Getting $0$, $1$, or $2$ Heads
Number of Heads Probability
0 1/4
1 1/2
2 1/4
Figure $1$ is a discrete probability distribution: It shows the probability for each of the values on the $X$-axis. Defining a head as a "success," Figure $1$ shows the probability of $0$, $1$, and $2$ successes for two trials (flips) for an event that has a probability of $0.5$ of being a success on each trial. This makes Figure $1$ an example of a binomial distribution.
The Formula for Binomial Probabilities
The binomial distribution consists of the probabilities of each of the possible numbers of successes on $N$ trials for independent events that each have a probability of $\pi$ (the Greek letter pi) of occurring. For the coin flip example, $N = 2$ and $\pi =0.5$. The formula for the binomial distribution is shown below:
$P(x)=\dfrac{N!}{x!(N-x)!}\pi ^x(1-\pi )^{N-x}$
where $P(x)$ is the probability of x successes out of $N$ trials, $N$ is the number of trials, and $\pi$ is the probability of success on a given trial. Applying this to the coin flip example,
\begin{align} P(0)&=\dfrac{2!}{0!(2-0)!}(0.5)^0(1-0.5)^{2-0} \[5pt] &=\dfrac{2}{2}(1)(0.25) \[5pt] &=0.25 \end{align}
\begin{align}P(0)&=\dfrac{2!}{1!(2-1)!}(0.5)^1(1-0.5)^{2-1} \[5pt] &=\dfrac{2}{1}(0.5)(0.5) \[5pt] &=0.50\end{align}
\begin{align}P(0)&=\dfrac{2!}{2!(2-2)!}(0.5)^2(1-0.5)^{2-2} \[5pt] &=\dfrac{2}{2}(0.25)(1) \[5pt] &=0.25\end{align}
If you flip a coin twice, what is the probability of getting one or more heads? Since the probability of getting exactly one head is $0.50$ and the probability of getting exactly two heads is $0.25$, the probability of getting one or more heads is $0.50 + 0.25 = 0.75$.
Now suppose that the coin is biased. The probability of heads is only $0.4$. What is the probability of getting heads at least once in two tosses? Substituting into the general formula above, you should obtain the answer $0.64$.
Cumulative Probabilities
We toss a coin $12$ times. What is the probability that we get from $0$ to $3$ heads? The answer is found by computing the probability of exactly $0$ heads, exactly $1$ head, exactly $2$ heads, and exactly $3$ heads. The probability of getting from $0$ to $3$ heads is then the sum of these probabilities. The probabilities are: $0.0002$, $0.0029$, $0.0161$, and $0.0537$. The sum of the probabilities is $0.073$. The calculation of cumulative binomial probabilities can be quite tedious. Therefore we have provided a binomial calculator to make it easy to calculate these probabilities.
Mean and Standard Deviation of Binomial Distributions
Consider a coin-tossing experiment in which you tossed a coin $12$ times and recorded the number of heads. If you performed this experiment over and over again, what would the mean number of heads be? On average, you would expect half the coin tosses to come up heads. Therefore the mean number of heads would be $6$. In general, the mean of a binomial distribution with parameters $N$ (the number of trials) and $\pi$ (the probability of success on each trial) is:
$\mu =N\pi$
where $\mu$ is the mean of the binomial distribution. The variance of the binomial distribution is:
$\sigma ^2=N\pi (1-\pi )$
where $\sigma ^2$ is the variance of the binomial distribution.
Let's return to the coin-tossing experiment. The coin was tossed $12$ times, so $N = 12$. A coin has a probability of $0.5$ of coming up heads. Therefore, $\pi =0.5$. The mean and variance can therefore be computed as follows:
$\mu =N\pi =(12)(0.5)=6$
$\sigma ^2=N\pi (1-\pi )=(12)(0.5)(1.0-0.5)=3.0$
Naturally, the standard deviation ($\sigma$) is the square root of the variance ($\sigma ^2$).
$\sigma =\sqrt{N\pi (1-\pi )}$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/05%3A_Probability/5.06%3A_Birthday_Demo.txt |
Learning Objectives
• State how \(N\) affects the shape of a binomial distribution
Instructions
This demonstration allows you to explore the binomial distribution. Here you can specify the number of trials (\(N\)) and the proportion of successes (\(p\)). Note that the proportion of successes is sometimes referred to with the Greek letter \(p\); here it will be referred to as \(p\).
The starting values for \(N\) and \(p\) are \(10\) and \(0.5\). The graph therefore shows the probability of from \(0\) to \(10\) successes out of \(10\) trials. The probability of \(0\) successes is so small that it just barely shows up on the graph. The most likely number of successes is \(5\), which has a probability of about \(0.25\). Naturally, the sum of the probabilities is \(1.0\).
The demonstration also shows the mean and the standard deviation of the binomial distribution.
Illustrated Instructions
The screenshot below shows the binomial demonstration with its default data. You can adjust the number of trials (\(N\)) as well as the proportion of successes (\(p\)).
The example below shows a distribution of \(20\) trials with a probability of success of \(0.7\). Notice that the mean and standard deviation for the distribution are also shown.
5.09: Poisson Distribution
Learning Objectives
• To study the use of Poisson distribution
The Poisson distribution can be used to calculate the probabilities of various numbers of "successes" based on the mean number of successes. In order to apply the Poisson distribution, the various events must be independent. Keep in mind that the term "success" does not really mean success in the traditional positive sense. It just means that the outcome in question occurs.
Suppose you knew that the mean number of calls to a fire station on a weekday is $8$. What is the probability that on a given weekday there would be $11$ calls? This problem can be solved using the following formula based on the Poisson distribution:
$\mathit{p}=\frac{e^{-\mu }\mu ^x}{x!}$
where
• $e$ is the base of natural logarithms ($2.7183$)
• $\mu$ is the mean number of "successes"
• $x$ is the number of "successes" in question
For this example,
$\mathit{p}=\frac{e^{-8}8^{11}}{11!}=0.072$
since the mean is $8$ and the question pertains to $11$ fires.
The mean of the Poisson distribution is $\mu$. The variance is also equal to $\mu$. Thus, for this example, both the mean and the variance are equal to $8$.
5.10: Multinomial Distribution
Learning Objectives
• Define multinomial outcomes
• Compute probabilities using the multinomial distribution
The binomial distribution allows one to compute the probability of obtaining a given number of binary outcomes. For example, it can be used to compute the probability of getting $6$ heads out of $10$ coin flips. The flip of a coin is a binary outcome because it has only two possible outcomes: heads and tails. The multinomial distribution can be used to compute the probabilities in situations in which there are more than two possible outcomes.
For example, suppose that two chess players had played numerous games and it was determined that the probability that Player $A$ would win is $0.40$, the probability that Player $B$ would win is $0.35$, and the probability that the game would end in a draw is $0.25$. The multinomial distribution can be used to answer questions such as: "If these two chess players played $12$ games, what is the probability that Player $A$ would win $7$ games, Player $B$ would win $2$ games, and the remaining $3$ games would be drawn?" The following formula gives the probability of obtaining a specific set of outcomes when there are three possible outcomes for each event:
$p =\dfrac{n!}{n_1!n_2!n_3!} p_1^{n_1}p_2^{n_2}p_3^{n_3}$
where
• $p$ is the probability,
• $n$ is the total number of events
• $n_1$ is the number of times Outcome $1$ occurs,
• $n_2$ is the number of times Outcome $2$ occurs,
• $n_3$ is the number of times Outcome $3$ occurs,
• $p_1$ is the probability of Outcome $1$
• $p_2$ is the probability of Outcome $2$, and
• $p_3$ is t he probability of Outcome $3$.
Example $1$: Chess
For the chess example discussed above:
• $n$ = $12$ ($12$ games are played),
• $n_1$ = $7$ (number won by Player $A$),
• $n_2$ = $2$ (number won by Player $B$),
• $n_3$ = $3$ (the number drawn),
• $p_1$ = $0.40$ (probability Player $A$ wins)
• $p_2$ = $0.35$ (probability Player $B$ wins)
• $p_3$ = $0.25$ (probability of a draw)
$P= \dfrac{12!}{7!2!3!}\times 40^7\times 35^2 \times 25^3 = 0.0248$
The formula for $k$ outcomes is
$p =\dfrac{n!}{n_1!n_2!n_3!} p_1^{n_1}p_2^{n_2} \ldots p_k^{n_k}$
Note that the binomial distribution is a special case of the multinomial distribution when $k = 2$.
5.11: Hypergeometric Distribution
Learning Objectives
• To study the use of hypergeometric distribution
The hypergeometric distribution is used to calculate probabilities when sampling without replacement. For example, suppose you first randomly sample one card from a deck of $52$. Then, without putting the card back in the deck you sample a second and then (again without replacing cards) a third. Given this sampling procedure, what is the probability that exactly two of the sampled cards will be aces ($4$ of the $52$ cards in the deck are aces). You can calculate this probability using the following formula based on the hypergeometric distribution:
$p =\dfrac{ (_{k}C_{x})(_{(n-k)}C_{(n-x)}) }{ _{n}C_{n}}$
where
• $k$ is the number of "successes" in the population
• $x$ is the number of "successes" in the sample
• $N$ is the size of the population
• $n$ is the number sampled
• $p$ is the probability of obtaining exactly $x$ successes
• $_kC_x$ is the number of combinations of $k$ things taken $x$ at a time
In this example, $k = 4$ because there are four aces in the deck, $x = 2$ because the problem asks about the probability of getting two aces, $N = 52$ because there are $52$ cards in a deck, and $n = 3$ because $3$ cards were sampled. Therefore,
\begin{align} p &=\dfrac{(_4C_2) (_{(52-4)}C_{(3-2})}{_{52}C_3} \[5pt] &= \dfrac{\dfrac{4!}{2!2!}\dfrac{48!}{47!1!}}{\dfrac{52!}{49!3!}} = 0.013 \end{align}
The mean and standard deviation of the hypergeometric distribution are:
$mean = \dfrac{n\,k}{N}$
$\sigma_{hypergeometric} = \sqrt{\dfrac{n\,k(N-k)(N-m)}{N^2(N-1)}}$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/05%3A_Probability/5.08%3A_Binomial_Demonstration.txt |
Learning Objectives
• Compute the probability of a condition from hits, false alarms, and base rates using a tree diagram
• Compute the probability of a condition from hits, false alarms, and base rates using Bayes' Theorem
Suppose that at your regular physical exam you test positive for Disease $X$. Although Disease $X$ has only mild symptoms, you are concerned and ask your doctor about the accuracy of the test. It turns out that the test is $95\%$ accurate. It would appear that the probability that you have Disease $X$ is therefore $0.95$. However, the situation is not that simple.
For one thing, more information about the accuracy of the test is needed because there are two kinds of errors the test can make: misses and false positives. If you actually have Disease $X$ and the test failed to detect it, that would be a miss. If you did not have Disease $X$ and the test indicated you did, that would be a false positive. The miss and false positive rates are not necessarily the same. For example, suppose that the test accurately indicates the disease in $99\%$ of the people who have it and accurately indicates no disease in $91\%$ of the people who do not have it. In other words, the test has a miss rate of $0.01$ and a false positive rate of $0.09$. This might lead you to revise your judgment and conclude that your chance of having the disease is $0.91$. This would not be correct since the probability depends on the proportion of people having the disease. This proportion is called the base rate.
Assume that Disease $X$ is a rare disease, and only $2\%$ of people in your situation have it. How does that affect the probability that you have it? Or, more generally, what is the probability that someone who tests positive actually has the disease? Let's consider what would happen if one million people were tested. Out of these one million people, $2\%$ or $20,000$ people would have the disease. Of these $20,000$ with the disease, the test would accurately detect it in $99\%$ of them. This means that $19,800$ cases would be accurately identified. Now let's consider the $98\%$ of the one million people ($980,000$) who do not have the disease. Since the false positive rate is $0.09$, $9\%$ of these $980,000$ people will test positive for the disease. This is a total of $88,200$ people incorrectly diagnosed.
To sum up, $19,800$ people who tested positive would actually have the disease and $88,200$ people who tested positive would not have the disease. This means that of all those who tested positive, only
$\dfrac{19,800}{19,800 + 88,200} = 0.1833$
of them would actually have the disease. So the probability that you have the disease is not $0.95$, or $0.91$, but only $0.1833$.
These results are summarized in Table $1$. The numbers of people diagnosed with the disease are shown in red. Of the one million people tested, the test was correct for $891,800$ of those without the disease and for $19,800$ with the disease; the test was correct $91\%$ of the time. However, if you look only at the people testing positive (shown in red), only $19,800 (0.1833)$ of the $88,200 + 19,800 = 108,000$ testing positive actually have the disease.
Table $1$: Diagnosing Disease $X$
True Condition
No Disease
980,000
Disease
20,000
Test Result
Test Result
Positive
88,200
Negative
891,800
Positive
19,800
Negative
200
Bayes' Theorem
This same result can be obtained using Bayes' theorem. Bayes' theorem considers both the prior probability of an event and the diagnostic value of a test to determine the posterior probability of the event. For the current example, the event is that you have Disease $X$. Let's call this Event $D$. Since only $2\%$ of people in your situation have Disease $X$, the prior probability of Event $D$ is $0.02$. Or, more formally, $P(D) = 0.02$. If $P(D')$ represents the probability that Event $D$ is false, then $P(D') = 1 - P(D) = 0.98$.
To define the diagnostic value of the test, we need to define another event: that you test positive for Disease $X$. Let's call this Event $T$. The diagnostic value of the test depends on the probability you will test positive given that you actually have the disease, written as $P(T|D)$, and the probability you test positive given that you do not have the disease, written as $P(T|D')$. Bayes' theorem shown below allows you to calculate $P(D|T)$, the probability that you have the disease given that you test positive for it.
$P(D|T)=\frac{P(T|D)P(D)}{P(T|D)P(D)+P(T|D')P(D')}$
The various terms are:
$P(T|D) = 0.99$
$P(T|D') = 0.09$
$P(D) = 0.02$
$P(D') = 0.98$
Therefore,
$P(D|T)=\frac{(0.99)(0.02)}{(0.99)(0.02)+(0.09)(0.98)}=0.1833$
which is the same value computed previously.
5.13: Bayes Demo
Learning Objectives
• Calculate probabilities based on a tree diagram
• Calculate probabilities based on Bayes' theorem
Instructions
This demonstration lets you examine the effects of base rate, true positive rate, and false positive rate on the probability that a person diagnosed with disease \(X\) actually has the disease. The base rate is the proportion of people who have the disease. The true positive rate is the probability that a person with the disease will test positive. The false positive rate is the probability that someone who does not have the disease will test positive. The demonstration is based on \(10,000\) people being tested. A tree diagram showing the results and calculations based on Bayes' theorem are shown. They should always agree.
You can change the initial values and then press the "Calculate" button.
Illustrated Instructions
The Bayes' Theorem demonstration starts by displaying the results for the default base rate, true positive rate and the false positive rate as shown in the screenshot below. You can change any of these three numbers and click the "Calculate" button to get the results based on the changes you make.
5.14: Monty Hall Problem
Learning Objectives
• Monty Hall problem demonstration
Instructions
This demonstration lets you play the Monty Hall game. In the Monty Hall game, a contestant is shown three doors. Two of the doors have goats behind them and one has a car. The contestant chooses a door. Before opening the chosen door, Monty Hall opens a door that has a goat behind it. The contestant can then switch to the other unopened door, or stay with the original choice. In this demonstration, you begin by clicking on a door. A different door showing a goat will then appear. You then either click your original door again or switch to a different door. You will see whether you win a car or a goat. Your cumulative results will be shown.
5.15: Statistical Literacy
Learning Objectives
• Learning to use the Base Rate information to compute probability
School shooting: The warning signs
This webpage gives the FBI list of warning signs for school shooters.
Example \(1\): What do you think?
Do you think it is likely that someone showing a majority of these signs would actually shoot people in school?
Solution
Fortunately the vast majority of students do not become shooters. It is necessary to take this base rate information into account in order to compute the probability that any given student will be a shooter. The warning signs are unlikely to be sufficiently predictive to warrant the conclusion that a student will become a shooter. If an action is taken on the basis of these warning signs, it is likely that the student involved would never have become a shooter even without the action. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/05%3A_Probability/5.12%3A_Base_Rates.txt |
You may want to use the Binomial Calculator for some of these exercises.
General Questions
Q1
1. What is the probability of rolling a pair of dice and obtaining a total score of $9$ or more?
2. What is the probability of rolling a pair of dice and obtaining a total score of $7$? (relevant section)
Q2
A box contains four black pieces of cloth, two striped pieces, and six dotted pieces. A piece is selected randomly and then placed back in the box. A second piece is selected randomly. What is the probability that:
1. both pieces are dotted?
2. the first piece is black and the second piece is dotted?
3. one piece is black and one piece is striped?
(relevant section)
Q3
A card is drawn at random from a deck.
1. What is the probability that it is an ace or a king?
2. What is the probability that it is either a red card or a black card? (relevant section)
Q4
The probability that you will win a game is $0.45$.
1. If you play the game $80$ times, what is the most likely number of wins?
2. What are the mean and standard deviation of a binomial distribution with $\pi =0.45$ and $N = 80$? (relevant section)
Q5
A fair coin is flipped $9$ times. What is the probability of getting exactly $6$ heads? (relevant section)
Q6
When Susan and Jessica play a card game, Susan wins $60\%$ of the time. If they play $9$ games, what is the probability that Jessica will have won more games than Susan? (relevant section)
Q7
You flip a coin three times.
1. What is the probability of getting heads on only one of your flips?
2. What is the probability of getting heads on at least one flip? (relevant section & relevant section)
Q8
A test correctly identifies a disease in $95\%$ of people who have it. It correctly identifies no disease in $94\%$ of people who do not have it. In the population, $3\%$ of the people have the disease. What is the probability that you have the disease if you tested positive? (relevant section)
Q9
A jar contains $10$ blue marbles, $5$ red marbles, $4$ green marbles, and $1$ yellow marble. Two marbles are chosen (without replacement).
1. What is the probability that one will be green and the other red?
2. What is the probability that one will be blue and the other yellow? (relevant section)
Q10
You roll a fair die five times, and you get a $6$ each time. What is the probability that you get a $6$ on the next roll? (relevant section)
Q11
You win a game if you roll a die and get a $2$ or a $5$. You play this game $60$ times.
1. What is the probability that you win between $5$ and $10$ times (inclusive)?
2. What is the probability that you will win the game at least $15$ times?
3. What is the probability that you will win the game at least $40$ times?
4. What is the most likely number of wins.
5. What is the probability of obtaining the number of wins in d?
(relevant section)
Q12
In a baseball game, Tommy gets a hit $30\%$ of the time when facing this pitcher. Joey gets a hit $25\%$ of the time. They are both coming up to bat this inning.
1. What is the probability that Joey or Tommy (but not both) will get a hit?
2. What is the probability that neither player gets a hit?
3. What is the probability that they both get a hit? (relevant section)
Q13
An unfair coin has a probability of coming up heads of $0.65$. The coin is flipped $50$ times. What is the probability it will come up heads $25$ or fewer times? (Give answer to at least $3$ decimal places). (relevant section)
Q14
You draw two cards from a deck, what is the probability that
1. both of them are face cards (king, queen, or jack)?
2. What is the probability that you draw two cards from a deck and both of them are hearts? (relevant section)
Q15
True/False: You are more likely to get a pattern of $HTHHHTHTTH$ than $HHHHHHHHTT$ when you flip a coin $10$ times. (relevant section)
Q16
True/False: Suppose that at your regular physical exam you test positive for a relatively rare disease. You will need to start taking medicine if you have the disease, so you ask your doctor about the accuracy of the test. It turns out that the test is $98\%$ accurate. The probability that you have Disease $X$ is therefore $0.98$ and the probability that you do not have it is $0.02$. (relevant section)
Questions from Case Studies
The following questions are from the Diet and Health (DH) case study.
Q17
(DH#1)
1. What percentage of people on the AHA diet had some sort of illness or death?
2. What is the probability that if you randomly selected a person on the AHA diet, he or she would have some sort of illness or death? (relevant section)
3. If $3$ people on the AHA diet are chosen at random, what is the probability that they will all be healthy? (relevant section)
Q18
(DH#2)
1. What percentage of people on the Mediterranean diet had some sort of illness or death?
2. What is the probability that if you randomly selected a person on the Mediterranean diet, he or she would have some sort of illness or death? (relevant section)
3. What is the probability that if you randomly selected a person on the Mediterranean diet, he or she would have cancer? (relevant section)
4. If you randomly select five people from the Mediterranean diet, what is the probability that they would all be healthy? (relevant section)
The following questions are from (reproduced with permission)
Visit the site
Q19
Five faces of a fair die are painted black, and one face is painted white. The die is rolled six times. Which of the following results is more likely?
a. Black side up on five of the rolls; white side up on the other roll
b. Black side up on all six rolls
c. a and b are equally likely
Q20
One of the items on the student survey for an introductory statistics course was "Rate your intelligence on a scale of $1$ to $10$." The distribution of this variable for the $100$ women in the class is presented below. What is the probability of randomly selecting a woman from the class who has an intelligence rating that is LESS than seven ($7$)?
Intelligence Rating Count
5 12
6 24
7 38
8 23
9 2
10 1
a. $(12 + 24)/100 = 0.36$
b. $(12 + 24 + 38)/100 = 0.74$
c. $38/100 = 0.38$
d. $(23 + 2 + 1)/100 = 0.26$
e. None of the above.
Q21
You roll $2$ fair six-sided dice. Which of the following outcomes is most likely to occur on the next roll?
1. Getting double $3$.
2. Getting a 3 and a $4$.
3. They are equally likely. Explain your choice.
Q22
If Tahnee flips a coin $10$ times, and records the results (Heads or Tails), which outcome below is more likely to occur, $A$ or $B$? Explain your choice.
Throw Number 1 2 3 4 5 6 7 8 9 10
A H T T H T H H T T T
B H T H T H T H T H T
Q23
A bowl has $100$ wrapped hard candies in it. $20$ are yellow, $50$ are red, and $30$ are blue. They are well mixed up in the bowl. Jenny pulls out a handful of $10$ candies, counts the number of reds, and tells her teacher. The teacher writes the number of red candies on a list. Then, Jenny puts the candies back into the bowl, and mixes them all up again. Four of Jenny's classmates, Jack, Julie, Jason, and Jerry do the same thing. They each pick ten candies, count the reds, and the teacher writes down the number of reds. Then they put the candies back and mix them up again each time. The teacher's list for the number of reds is most likely to be (please select one):
a. ${8,9,7,10,9}$
b. ${3,7,5,8,5}$
c. ${5,5,5,5,5}$
d. ${2,4,3,4,3}$
e. ${3,0,9,2,8}$
Q24
An insurance company writes policies for a large number of newly-licensed drivers each year. Suppose $40\%$ of these are low-risk drivers, $40\%$ are moderate risk, and $20\%$ are high risk. The company has no way to know which group any individual driver falls in when it writes the policies. None of the low-risk drivers will have an at-fault accident in the next year, but $10\%$ of the moderate-risk and $20\%$ of the high-risk drivers will have such an accident. If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk?
Q25
You are to participate in an exam for which you had no chance to study, and for that reason cannot do anything but guess for each question (all questions being of the multiple choice type, so the chance of guessing the correct answer for each question is $1/d$, $d$ being the number of options per question; so in case of a $4$-choice question, your chance is $0.25$). Your instructor offers you the opportunity to choose amongst the following exam formats:
1. $6$ questions of the $4$-choice type; you pass when $5$ or more answers are correct
2. $5$ questions of the $5$-choice type; you pass when $4$ or more answers are correct
3. $4$ questions of the $10$-choice type; you pass when $3$ or more answers are correct.
Rank the three exam formats according to their attractiveness. It should be clear that the format with the highest probability to pass is the most attractive format. Which would you choose and why?
Q26
Consider the question of whether the home team wins more than half of its games in the National Basketball Association. Suppose that you study a simple random sample of $80$ professional basketball games and find that $52$ of them are won by the home team.
1. Assuming that there is no home court advantage and that the home team therefore wins $50\%$ of its games in the long run, determine the probability that the home team would win $65\%$ or more of its games in a simple random sample of $80$ games.
2. Does the sample information (that $52$ of a random sample of $80$ games are won by the home team) provide strong evidence that the home team wins more than half of its games in the long run? Explain.
Q27
A refrigerator contains $6$ apples, $5$ oranges, $10$ bananas, $3$ pears, $7$ peaches, $11$ plums, and $2$ mangos.
1. Imagine you stick your hand in this refrigerator and pull out a piece of fruit at random. What is the probability that you will pull out a pear?
2. Imagine now that you put your hand in the refrigerator and pull out a piece of fruit. You decide you do not want to eat that fruit so you put it back into the refrigerator and pull out another piece of fruit. What is the probability that the first piece of fruit you pull out is a banana and the second piece you pull out is an apple?
3. What is the probability that you stick your hand in the refrigerator one time and pull out a mango or an orange?
Select Answers
S1
1. $5/18 = 0.278$
S2
1. $1/6 = 0.167$
S3
1. $2/13 = 0.154$
S4
1. mean = $36$; SD = $4.24$
S5
$0.164$
S7
1. $7/8 = 0.875$
S9
1. $2/19 = 0.105$
S11
1. $0.937$
S12
1. $0.075$
S14
1. $1/17 = 0.0588$
S17
1. $0.493$
S18
1. $0.10$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/05%3A_Probability/5.E%3A_Probability_%28Exercises%29.txt |
A research design is the set of methods and procedures used in collecting and analyzing measures of the variables specified in the research problem research. The design of a study defines the study type (descriptive, correlational, semi-experimental, experimental, review, meta-analytic) and sub-type (e.g., descriptive-longitudinal case study), research problem, hypotheses, independent and dependent variables, experimental design, and, if applicable, data collection methods and a statistical analysis plan. Research design is the framework that has been created to find answers to research questions.
• 6.1: Scientific Method
To be a proper scientific investigation the data must be collected systematically. However, scientific investigation does not necessarily require experimentation in the sense of manipulating variables and observing the results. Observational studies in the fields of astronomy, developmental psychology, and ethology are common and provide valuable scientific information.
• 6.2: Measurement
The collection of data involves measurement. Measurement of some characteristics such as height and weight are relatively straightforward. The measurement of psychological attributes such as self esteem can be complex. A good measurement scale should be both reliable and valid. These concepts will be discussed in turn.
• 6.3: Data Collection
Most statistical analyses require that your data be in numerical rather than verbal form (you can’t punch letters into your calculator). Therefore, data collected in verbal form must be coded so that it is represented by numbers.
• 6.4: Sampling Bias
Descriptions of various types of sampling such as simple random sampling and stratified random sampling are covered in another section. This section discusses various types of sampling biases including self-selection bias and survivorship bias. Examples of other sampling biases that are not easily categorized will also be given. It is important to keep in mind that sampling bias refers to the method of sampling, not the sample itself.
• 6.5: Experimental Designs
There are many ways an experiment can be designed. For example, subjects can all be tested under each of the treatment conditions or a different group of subjects can be used for each treatment. An experiment might have just one independent variable or it might have several. This section describes basic experimental designs and their advantages and disadvantages.
• 6.6: Causation
The concept of causation is a complex one in the philosophy of science. Since a full coverage of this topic is well beyond the scope of this text, we focus on two specific topics: (1) the establishment of causation in experiments and (2) the establishment of causation in non-experimental designs.
• 6.7: Statistical Literacy
A low level of HDL have long been known to be a risk factor for heart disease. Taking niacin has been shown to increase HDL levels and has been recommended for patients with low levels of HDL. The assumption of this recommendation is that niacin causes HDL to increase thus causing a lower risk for heart disease. What experimental design involving niacin would test whether the relationship between HDL and heart disease is causal?
• 6.E: Research Design (Exercises)
Flowchart of four phases (enrollment, intervention allocation, follow-up, and data analysis) of a parallel randomized trial of two groups. Image use with permission (CC BYT-SA 3.0; PrevMedFellow).
Contributors and Attributions
• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.
• Wikipedia
06: Research Design
Learning Objectives
• Brief discussion of the most important principles of the scientific method
This section contains a brief discussion of the most important principles of the scientific method. A thorough treatment of the philosophy of science is beyond the scope of this work.
One of the hallmarks of the scientific method is that it depends on empirical data. To be a proper scientific investigation the data must be collected systematically. However, scientific investigation does not necessarily require experimentation in the sense of manipulating variables and observing the results. Observational studies in the fields of astronomy, developmental psychology, and ethology are common and provide valuable scientific information.
Theories and explanations are very important in science. Theories in science can never be proved since one can never be 100% certain that a new empirical finding inconsistent with the theory will never be found.
Scientific theories must be potentially disconfirmable. If a theory can accommodate all possible results then it is not a scientific theory. Therefore, a scientific theory should lead to testable hypotheses. If a hypothesis is disconfirmed, then the theory from which the hypothesis was deduced is incorrect. For example, the secondary reinforcement theory of attachment states that an infant becomes attached to its parent by means of a pairing of the parent with a primary reinforcer (food). It is through this "secondary reinforcement" that the child-parent bond forms. The secondary reinforcement theory has been disconfirmed by numerous experiments. Perhaps the most notable is one in which infant monkeys were fed by a surrogate wire mother while a surrogate cloth mother was available. The infant monkeys formed no attachment to the wire monkeys and frequently clung to the cloth surrogate mothers.
History of Attachment Theory
If a hypothesis derived from a theory is confirmed then the theory has survived a test and it becomes more useful and better thought of by the researchers in the field. A theory is not confirmed when correct hypotheses are derived from it.
A key difference between scientific explanations and faith-based explanations is simply that faith-based explanations are based on faith and do not need to be testable. This does not mean that an explanation that cannot be tested is incorrect in some cosmic sense. It just means that it is not a scientific explanation.
The method of investigation in which a hypothesis is developed from a theory and then confirmed or disconfirmed involves deductive reasoning. However, deductive reasoning does not explain where the theory came from in the first place. In general, a theory is developed by a scientist who is aware of many empirical findings on a topic of interest. Then, through a generally poorly understood process called "induction" the scientist develops a way to explain all or most of the findings within a relatively simple framework or theory.
An important attribute of a good scientific theory is that it is parsimonious. That is, that it is simple in the sense that it uses relatively few constructs to explain many empirical findings. A theory that it so complex that it has as many assumptions as it has predictions is not very valuable.
Although strictly speaking, disconfirming an hypothesis deduced from a theory disconfirms the theory, it rarely leads to the abandonment of the theory. Instead, the theory will probably be modified to accommodate the inconsistent finding. If the theory has to be modified over and over to accommodate new findings, the theory generally becomes less and less parsimonious. This can lead to discontent with the theory and the search for a new theory. If a new theory is developed that can explain the same facts in a more parsimonious way, then the new theory will eventually supercede the old theory. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/06%3A_Research_Design/6.01%3A_Scientific_Method.txt |
Learning Objectives
• Describe reliability in terms of true scores and error
• Define the standard error of measurement and state why it is valuable
• Distinguish between reliability and validity
• State the how reliability determines the upper limit to validity
The collection of data involves measurement. Measurement of some characteristics such as height and weight are relatively straightforward. The measurement of psychological attributes such as self esteem can be complex. A good measurement scale should be both reliable and valid. These concepts will be discussed in turn.
Reliability
The notion of reliability revolves around whether you would get at least approximately the same result if you measure something twice with the same measurement instrument. A common way to define reliability is the correlation between parallel forms of a test. Letting "test" represent a parallel form of the test, the symbol $r_{test,test}$is used to denote the reliability of the test.
True Scores and Error
Assume you wish to measure a person's mean response time to the onset of a stimulus. For simplicity, assume that there is no learning over tests which, of course, is not really true. The person is given $1,000$ trials on the task and you obtain the response time on each trial.
The mean response time over the $1,000$ trials can be thought of as the person's "true" score, or at least a very good approximation of it. Theoretically, the true score is the mean that would be approached as the number of trials increases indefinitely.
An individual response time can be thought of as being composed of two parts: the true score and the error of measurement. Thus if the person's true score were $345$ and their response on one of the trials were $358$, then the error of measurement would be $13$. Similarly, if the response time were $340$, the error of measurement would be $-5$.
Now consider the more realistic example of a class of students taking a $100$-point true/false exam. Let's assume that each student knows the answer to some of the questions and has no idea about the other questions. For the sake of simplicity, we are assuming there is no partial knowledge of any of the answers and for a given question a student either knows the answer or guesses. Finally, assume the test is scored such that a student receives one point for a correct answer and loses a point for an incorrect answer. In this example, a student's true score is the number of questions they know the answer to and their error score is their score on the questions they guessed on. For example, assume a student knew $90$ of the answers and guessed correctly on $7$ of the remaining $10$ (and therefore incorrectly on $3$). Their true score would be $90$ since that is the number of answers they knew. Their error score would be $7 - 3 = 4$ and therefore their actual test score would be $90 + 4$.
Every test score can be thought of as the sum of two independent components, the true score and the error score. This can be written as:
$y_{test}=y_{true}+y_{error}$
The following expression follows directly from the Variance Sum Law:
$\sigma _{Test}^{2}=\sigma _{True}^{2}+\sigma _{Error}^{2}$
Reliability in Terms of True Scores and Error
It can be shown that the reliability of a test, $r_{test,test}$, is the ratio of true-score variance to test-score variance. This can be written as:
$r_{test,test}=\frac{\sigma _{True}^{2}}{\sigma _{Test}^{2}}=\frac{\sigma _{True}^{2}}{\sigma _{True}^{2}+\sigma _{Error}^{2}}$
PDF of derivation
It is important to understand the implications of the role the variance of true scores plays in the definition of reliability: If a test were given in two populations for which the variance of the true scores differed, the reliability of the test would be higher in the population with the higher true-score variance. Therefore, reliability is not a property of a test per se but the reliability of a test in a given population.
Assessing Error of Measurement
The reliability of a test does not show directly how close the test scores are to the true scores. That is, it does not reveal how much a person's test score would vary across parallel forms of test. By definition, the mean over a large number of parallel tests would be the true score. The standard deviation of a person's test scores would indicate how much the test scores vary from the true score. This standard deviation is called the standard error of measurement. In practice, it is not practical to give a test over and over to the same person and/or assume that there are no practice effects. Instead, the following formula is used to estimate the standard error of measurement.
$S_{measurement}=S_{test}\sqrt{1-r_{test,test}}$
where $S_{measurement}$is the standard error of measurement, $S_{test}$ is the standard deviation of the test scores, and $r_{test,test}$ is the reliability of the test. Taking the extremes, if the reliability is $0$ then the standard error of measurement is equal to the standard deviation of the test; if the reliability is perfect ($1.0$) then the standard error of measurement is $0$.
Increasing Reliability
It is important to make measures as reliable as is practically possible. Suppose an investigator is studying the relationship between spatial ability and a set of other variables. The higher the reliability of the test of spatial ability, the higher the correlations will be. Similarly, if an experimenter seeks to determine whether a particular exercise regiment decreases blood pressure, the higher the reliability of the measure of blood pressure, the more sensitive the experiment. More precisely, the higher the reliability the higher the power of the experiment. Power is covered in detail here. Finally, if a test is being used to select students for college admission or employees for jobs, the higher the reliability of the test the stronger will be the relationship to the criterion.
Two basic ways of increasing reliability are
1. to improve the quality of the items and
2. to increase the number of items.
Items that are either too easy so that almost everyone gets them correct or too difficult so that almost no one gets them correct are not good items: they provide very little information. In most contexts, items which about half the people get correct are the best (other things being equal).
Items that do not correlate with other items can usually be improved. Sometimes the item is confusing or ambiguous.
Increasing the number of items increases reliability in the manner shown by the following formula:
$r_{new,new}=\frac{kr_{test,test}}{1+(k-1)r_{test,test}}$
where $k$ is the factor by which the test length is increased, $r_{new,new}$ is the reliability of the new longer test, and $r_{test,test}$ is the current reliability. For example, if a test with $50$ items has a reliability of $0.70$ then the reliability of a test that is $1.5$ times longer ($75$ items) would be calculated as follows:
$r_{new,new}=\frac{(1.5)(0.70)}{1+(1.5-1)(0.70)}$
which equals $0.78$. Thus increasing the number of items from $50$ to $75$ would increase the reliability from $0.70$ to $0.78$.
It is important to note that this formula assumes the new items have the same characteristics as the old items. Obviously adding poor items would not increase the reliability as expected and might even decrease the reliability.
More Information on Reliability from William Trochim's Knowledge Source
Validity
The validity of a test refers to whether the test measures what it is supposed to measure. The three most common types of validity are face validity, empirical validity, and construct validity. We consider these types of validity below.
• Face Validity: A test's face validity refers to whether the test appears to measure what it is supposed to measure. That is, does the test "on its face" appear to measure what it is supposed to be measuring. An Asian history test consisting of a series of questions about Asian history would have high face validity. If the test included primarily questions about American history then it would have little or no face validity as a test of Asian history.
• Predictive Validity: Predictive validity (sometimes called empirical validity) refers to a test's ability to predict the relevant behavior. For example, the main way in which SAT tests are validated is by their ability to predict college grades. Thus, to the extent these tests are successful at predicting college grades they are said to possess predictive validity.
• Construct Validity: Construct validity is more difficult to define. In general, a test has construct validity if its pattern of correlations with other measures is in line with the construct it is purporting to measure. Construct validity can be established by showing a test has both convergent and divergent validity. A test has convergent validity if it correlates with other tests that are also measures of the construct in question. Divergent validity is established by showing the test does not correlate highly with tests of other constructs. Of course, some constructs may overlap so the establishment of convergent and divergent validity can be complex.
To take an example, suppose one wished to establish the construct validity of a new test of spatial ability. Convergent and divergent validity could be established by showing the test correlates relatively highly with other measures of spatial ability but less highly with tests of verbal ability or social intelligence.
Reliability and Predictive Validity
The reliability of a test limits the size of the correlation between the test and other measures. In general, the correlation of a test with another measure will be lower than the test's reliability. After all, how could a test correlate with something else as high as it correlates with a parallel form of itself? Theoretically it is possible for a test to correlate as high as the square root of the reliability with another measure. For example, if a test has a reliability of $0.81$ then it could correlate as high as $0.90$ with another measure. This could happen if the other measure were a perfectly reliable test of the same construct as the test in question. In practice, this is very unlikely.
A correlation above the upper limit set by reliabilities can act as a red flag. For example, Vul, Harris, Winkielman, and Paschler ($2009$) found that in many studies the correlations between various fMRI activation patterns and personality measures were higher than their reliabilities would allow. A careful examination of these studies revealed serious flaws in the way the data were analyzed.
Vul, E., Harris, C., Winkielman, P., & Paschler, H. (2009) Puzzlingly High Correlations in fMRI Studies of Emotion, Personality, and Social Cognition. Perspectives on Psychological Science, 4, 274-290. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/06%3A_Research_Design/6.02%3A_Measurement.txt |
Learning Objectives
• Describe how a variable such as height should be recorded
• Choose a good response scale for a questionnaire
Most statistical analyses require that your data be in numerical rather than verbal form (you can’t punch letters into your calculator). Therefore, data collected in verbal form must be coded so that it is represented by numbers. To illustrate, consider the data in Table $1$.
Table $1$: Example Data
Student Name Hair Color Gender Major Height Computer Experience
Norma Brown Female Psychology 5’4” Lots
Amber Blonde Female Social Science 5’7” Very little
Paul Blonde Male History 6’1” Moderate
Christopher Black Male Biology 5’10” Lots
Sonya Brown Female Psychology 5’4” Little
Can you conduct statistical analyses on the above data or must you re-code it in some way? For example, how would you go about computing the average height of the $5$ students. You cannot enter students’ heights in their current form into a statistical program -- the computer would probably give you an error message because it does not understand notation such as $5’4”$. One solution is to change all the numbers to inches. So, $5’4”$ becomes $(5 \times 12 ) + 4 = 64$, and $6’1”$ becomes $(6 \times 12 ) + 1 = 73$, and so forth. In this way, you are converting height in feet and inches to simply height in inches. From there, it is very easy to ask a statistical program to calculate the mean height in inches for the $5$ students.
You may ask, “Why not simply ask subjects to write their height in inches in the first place?” Well, the number one rule of data collection is to ask for information in such a way as it will be most accurately reported. Most people know their height in feet and inches and cannot quickly and accurately convert it into inches “on the fly.” So, in order to preserve data accuracy, it is best for researchers to make the necessary conversions.
Let’s take another example. Suppose you wanted to calculate the mean amount of computer experience for the five students shown in Table $1$. One way would be to convert the verbal descriptions to numbers as shown in Table $2$. Thus, "Very Little" would be converted to "$1$" and "Little" would be converted to "$2$."
Table $2$: Conversion of verbal descriptions to numbers.
1 2 3 4 5
Very Little Little Moderate Lots Very Lots
Example $1$: How much information should I record?
Say you are volunteering at a track meet at your college, and your job is to record each runner’s time as they pass the finish line for each race. Their times are shown in large red numbers on a digital clock with eight digits to the right of the decimal point, and you are told to record the entire number in your tablet. Thinking eight decimal places is a bit excessive, you only record runners’ times to one decimal place. The track meet begins, and runner number one finishes with a time of $22.93219780$ seconds. You dutifully record her time in your tablet, but only to one decimal place, that is $22.9$. Race number two finishes and you record $32.7$ for the winning runner. The fastest time in Race number three is $25.6$. Race number four winning time is $22.9$, Race number five is…. But wait! You suddenly realize your mistake; you now have a tie between runner one and runner four for the title of Fastest Overall Runner! You should have recorded more information from the digital clock -- that information is now lost, and you cannot go back in time and record running times to more decimal places.
The point is that you should think very carefully about the scales and specificity of information needed in your research before you begin collecting data. If you believe you might need additional information later but are not sure, measure it; you can always decide to not use some of the data, or “collapse” your data down to lower scales if you wish, but you cannot expand your data set to include more information after the fact. In this example, you probably would not need to record eight digits to the right of the decimal point. But recording only one decimal digit is clearly too few.
Example $2$
Pretend for a moment that you are teaching five children in middle school (yikes!), and you are trying to convince them that they must study more in order to earn better grades. To prove your point, you decide to collect actual data from their recent math exams, and, toward this end, you develop a questionnaire to measure their study time and subsequent grades. You might develop a questionnaire which looks like the following:
1. Please write your name: ____________________________
2. Please indicate how much you studied for this math exam:
a lot……………moderate……….…….little
3. Please circle the grade you received on the math exam: $A\; B\; C\; D\; F$
Given the above questionnaire, your obtained data might look like the following:
Name Amount Studied Grade
John Little C
Sally Moderate B
Alexander Lots A
Linda Moderate A
Thomas Little B
Eyeballing the data, it seems as if the children who studied more received better grades, but it’s difficult to tell. “Little,” “lots,” and “$B$,” are imprecise, qualitative terms. You could get more precise information by asking specifically how many hours they studied and their exact score on the exam. The data then might look as follows:
Name Hours studied % Correct
John 5 71
Sally 9 83
Alexander 13 97
Linda 12 91
Thomas 7 85
Of course, this assumes the students would know how many hours they studied. Rather than trust the students' memories, you might ask them to keep a log of their study time as they study.
Contributors and Attributions
• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.
• Heidi Zeimer | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/06%3A_Research_Design/6.03%3A_Data_Collection.txt |
Learning Objectives
• Recognize sampling bias
• Distinguish among self-selection bias, undercoverage bias, and survivorship bias
Descriptions of various types of sampling such as simple random sampling and stratified random sampling are covered in another section. This section discusses various types of sampling biases including self-selection bias and survivorship bias. Examples of other sampling biases that are not easily categorized will also be given.
It is important to keep in mind that sampling bias refers to the method of sampling, not the sample itself. There is no guarantee that random sampling will result in a sample representative of the population just as not every sample obtained using a biased sampling method will be greatly non-representative of the population.
Self-Selection Bias
Imagine that a university newspaper ran an ad asking for students to volunteer for a study in which intimate details of their sex lives would be discussed. Clearly the sample of students who would volunteer for such a study would not be representative of the students at the university. Similarly, an online survey about computer use is likely to attract people more interested in technology than is typical. In both of these examples, people who "self-select" themselves for the experiment are likely to differ in important ways from the population the experimenter wishes to draw conclusions about. Many of the admittedly "non-scientific" polls taken on television or web sites suffer greatly from self-selection bias.
A self-selection bias can result when the non-random component occurs after the potential subject has enlisted in the experiment. Considering again the hypothetical experiment in which subjects are to be asked intimate details of their sex lives, assume that the subjects did not know what the experiment was going to be about until they showed up.Many of the subjects would then likely leave the experiment resulting in a biased sample.
Undercoverage Bias
A common type of sampling bias is to sample too few observations from a segment of the population. A commonly-cited example of undercoverage is the poll taken by the Literary Digest in \(1936\) that indicated that Landon would win an election against Roosevelt by a large margin when, in fact, it was Roosevelt who won by a large margin. A common explanation is that poorer people were undercovered because they were less likely to have telephones and that this group was more likely to support Roosevelt.
A detailed analysis by Squire (\(1988\)) showed that it was not just an undercoverage bias that resulted in the faulty prediction of the election results. He concluded that, in addition to the undercoverage described above, there was a nonresponse bias (a form of self-selection bias) such that those favoring Landon were more likely to return their survey than were those favoring Roosevelt.
Survivorship Bias
Survivorship bias occurs when the observations recorded at the end of the investigation are a non-random set of those present at the beginning of the investigation. The gains in stock funds is an area in which survivorship bias often plays a role. The basic problem is that poorly-performing funds are often either eliminated or merged into other funds. Suppose one considers a sample of stock funds that exist in the present and then calculates the mean \(10\)-year appreciation of those funds. Can these results be validly generalized to other stock funds of the same type? The problem is that the poorly-performing stock funds that are not still in existence (did not survive for \(10\) years) are not included and therefore there is a bias toward selecting better-performing funds. There is good evidence that this survivorship bias is substantial (Malkiel, \(1995\)).
In World War II, the statistician Abraham Wald analyzed the distribution of hits from anti-aircraft fire on aircraft returning from missions. The idea was that this information would be useful for deciding where to place extra armor. A naive approach would be to put armor at locations that were frequently hit to reduce the damage there. However, this would ignore the survivorship bias occurring because only a subset of aircraft return. Wald's approach was the opposite: if there were few hits in a certain location on returning planes, then hits in that location were likely to bring a plane down. Therefore, he recommended that locations without hits on the returning planes should be given extra armor. A detailed and mathematical description of Wald's work can be found in Mangel and Samaniego (\(1984\).) | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/06%3A_Research_Design/6.04%3A_Sampling_Bias.txt |
Learning Objectives
• Distinguish between between-subject and within-subject designs
• State the advantages of within-subject designs
• Define "multi-factor design" and "factorial design"
• Identify the levels of a variable in an experimental design
• Describe when counterbalancing is used
There are many ways an experiment can be designed. For example, subjects can all be tested under each of the treatment conditions or a different group of subjects can be used for each treatment. An experiment might have just one independent variable or it might have several. This section describes basic experimental designs and their advantages and disadvantages.
Between-Subjects Designs
In a between-subjects design, the various experimental treatments are given to different groups of subjects. For example, in the "Teacher Ratings" case study, subjects were randomly divided into two groups. Subjects were all told they were going to see a video of an instructor's lecture after which they would rate the quality of the lecture. The groups differed in that the subjects in one group were told that prior teaching evaluations indicated that the instructor was charismatic whereas subjects in the other group were told that the evaluations indicated the instructor was punitive. In this experiment, the independent variable is "Condition" and has two levels (charismatic teacher and punitive teacher). It is a between-subjects variable because different subjects were used for the two levels of the independent variable: subjects were in either the "charismatic teacher" or the "punitive teacher" condition. Thus the comparison of the charismatic-teacher condition with the punitive-teacher condition is a comparison between the subjects in one condition with the subjects in the other condition.
The two conditions were treated exactly the same except for the instructions they received. Therefore, it would appear that any difference between conditions should be attributed to the treatments themselves. However, this ignores the possibility of chance differences between the groups. That is, by chance, the raters in one condition might have, on average, been more lenient than the raters in the other condition. Randomly assigning subjects to treatments ensures that all differences between conditions are chance differences; it does not ensure there will be no differences. The key question, then, is how to distinguish real differences from chance differences. The field of inferential statistics answers just this question. The inferential statistics applicable to testing the difference between the means of the two conditions can be found here. Analyzing the data from this experiment reveals that the ratings in the charismatic-teacher condition were higher than those in the punitive-teacher condition. Using inferential statistics, it can be calculated that the probability of finding a difference as large or larger than the one obtained if the treatment had no effect is only \(0.018\). Therefore it seems likely that the treatment had an effect and it is not the case that all differences were chance differences.
Independent variables often have several levels. For example, in the "Smiles and Leniency" case study the independent variable is "type of smile" and there are four levels of this independent variable:
1. false smile
2. felt smile
3. miserable smile
4. a neutral control
Keep in mind that although there are four levels, there is only one independent variable. Designs with more than one independent variable are considered next.
Multi-Factor Between-Subject Designs
In the "Bias Against Associates of the Obese" experiment, the qualifications of potential job applicants were judged. Each applicant was accompanied by an associate. The experiment had two independent variables: the weight of the associate (obese or average) and the applicant's relationship to the associate (girl friend or acquaintance). This design can be described as an Associate's Weight (\(2\)) x Associate's Relationship (\(2\)) factorial design. The numbers in parentheses represent the number of levels of the independent variable. The design was a factorial design because all four combinations of associate's weight and associate's relationship were included. The dependent variable was a rating of the applicant's qualifications (on a \(9\)-point scale).
If two separate experiments had been conducted, one to test the effect of Associate's Weight and one to test the effect of Associate's Relationship then there would be no way to assess whether the effect of Associate's Weight depended on the Associate's Relationship. One might imagine that the Associate's Weight would have a larger effect if the associate were a girl friend rather than merely an acquaintance. A factorial design allows this question to be addressed. When the effect of one variable does differ depending on the level of the other variable then it is said that there is an interaction between the variables.
Factorial designs can have three or more independent variables. In order to be a between-subjects design there must be a separate group of subjects for each combination of the levels of the independent variables.
Within-Subjects Designs
A within-subjects design differs from a between-subjects design in that the same subjects perform at all levels of the independent variable. For example consider the "ADHD Treatment" case study. In this experiment, subjects diagnosed as having attention deficit disorder were each tested on a delay of gratification task after receiving methylphenidate (MPH). All subjects were tested four times, once after receiving one of the four doses. Since each subject was tested under each of the four levels of the independent variable "dose," the design is a within-subjects design and dose is a within-subjects variable. Within-subjects designs are sometimes called repeated-measures designs.
Counterbalancing
In a within-subject design it is important not to confound the order in which a task is performed with the experimental treatment. For example, consider the problem that would have occurred if, in the ADHD study, every subject had received the doses in the same order starting with the lowest and continuing to the highest. It is not unlikely that experience with the delay of gratification task would have an effect. If practice on this task leads to better performance, then it would appear that higher doses caused the better performance when, in fact, it was the practice that caused the better performance.
One way to address this problem is to counterbalance the order of presentations. In other words, subjects would be given the doses in different orders in such a way that each dose was given in each sequential position an equal number of times. An example of counterbalancing is shown in Table \(1\).
Table \(1\): Counterbalanced order for four subjects
Subject 0 mg/kg 0.15 mg/kg 0.30 mg/kg 0.60 mg/kg
1 First Second Third Fourth
2 Second Third Fourth First
3 Third Fourth First Second
4 Fourth First Second Third
It should be kept in mind that counterbalancing is not a satisfactory solution if there are complex dependencies between which treatment precedes which and the dependent variable. In these cases, it is usually better to use a between-subjects design than a within-subjects design.
Advantage of Within-Subjects Designs
An advantage of within-subjects designs is that individual differences in subjects' overall levels of performance are controlled. This is important because subjects invariably will differ greatly from one another. In an experiment on problem solving, some subjects will be better than others regardless of the condition they are in. Similarly, in a study of blood pressure some subjects will have higher blood pressure than others regardless of the condition. Within-subjects designs control these individual differences by comparing the scores of a subject in one condition to the scores of the same subject in other conditions. In this sense each subject serves as his or her own control. This typically gives within-subjects designs considerably more power than between-subjects designs. That is, this makes within-subjects designs more able to detect an effect of the independent variable than are between-subjects designs.
Within-subjects designs are often called "repeated-measures" designs since repeated measurements are taken for each subject. Similarly, a within-subject variable can be called a repeated-measures factor.
Complex Designs
Designs can contain combinations of between-subject and within-subject variables. For example, the "Weapons and Aggression" case study has one between-subject variable (gender) and two within-subject variables (the type of priming word and the type of word to be responded to). | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/06%3A_Research_Design/6.05%3A_Experimental_Designs.txt |
Learning Objectives
• Explain how experimentation allows causal inferences
• Explain the role of unmeasured variables
• Explain the "third-variable" problem
• Explain how causation can be inferred in non-experimental designs
The concept of causation is a complex one in the philosophy of science. Since a full coverage of this topic is well beyond the scope of this text, we focus on two specific topics:
1. the establishment of causation in experiments
2. the establishment of causation in non-experimental designs
Stanford's Encyclopedia of Philosophy: Causation Topics
Establishing Causation in Experiments
Consider a simple experiment in which subjects are sampled randomly from a population and then assigned randomly to either the experimental group or the control group. Assume the condition means on the dependent variable differed. Does this mean the treatment caused the difference?
To make this discussion more concrete, assume that the experimental group received a drug for insomnia, the control group received a placebo, and the dependent variable was the number of minutes the subject slept that night. An obvious obstacle to inferring causality is that there are many unmeasured variables that affect how many hours someone sleeps. Among them are how much stress the person is under, physiological and genetic factors, how much caffeine they consumed, how much sleep they got the night before, etc. Perhaps differences between the groups on these factors are responsible for the difference in the number of minutes slept.
At first blush it might seem that the random assignment eliminates differences in unmeasured variables. However, this is not the case. Random assignment ensures that differences on unmeasured variables are chance differences. It does not ensure that there are no differences. Perhaps, by chance, many subjects in the control group were under high stress and this stress made it more difficult to fall asleep. The fact that the greater stress in the control group was due to chance does not mean it could not be responsible for the difference between the control and the experimental groups. In other words, the observed difference in "minutes slept" could have been due to a chance difference between the control group and the experimental group rather than due to the drug's effect.
This problem seems intractable since, by definition, it is impossible to measure an "unmeasured variable" just as it is impossible to measure and control all variables that affect the dependent variable. However, although it is impossible to assess the effect of any single unmeasured variable, it is possible to assess the combined effects of all unmeasured variables. Since everyone in a given condition is treated the same in the experiment, differences in their scores on the dependent variable must be due to the unmeasured variables. Therefore, a measure of the differences among the subjects within a condition is a measure of the sum total of the effects of the unmeasured variables. The most common measure of differences is the variance. By using the within-condition variance to assess the effects of unmeasured variables, statistical methods determine the probability that these unmeasured variables could produce a difference between conditions as large or larger than the difference obtained in the experiment. If that probability is low, then it is inferred (that's why they call it inferential statistics) that the treatment had an effect and that the differences are not entirely due to chance. Of course, there is always some nonzero probability that the difference occurred by chance so total certainty is not a possibility.
Causation in Non-Experimental Designs
It is almost a cliché that correlation does not mean causation. The main fallacy in inferring causation from correlation is called the "third variable problem" and means that a third variable is responsible for the correlation between two other variables. An excellent example used by Li (\(1975\)) to illustrate this point is the positive correlation in Taiwan in the \(1970's\) between the use of contraception and the number of electric appliances in one's house. Of course, using contraception does not induce you to buy electrical appliances or vice versa. Instead, the third variable of education level affects both.
Does the possibility of a third-variable problem make it impossible to draw causal inferences without doing an experiment? One approach is to simply assume that you do not have a third-variable problem. This approach, although common, is not very satisfactory. However, be aware that the assumption of no third-variable problem may be hidden behind a complex causal model that contains sophisticated and elegant mathematics.
A better though, admittedly more difficult approach, is to find converging evidence. This was the approach taken to conclude that smoking causes cancer. The analysis included converging evidence from retrospective studies, prospective studies, lab studies with animals, and theoretical understandings of cancer causes.
A second problem is determining the direction of causality. A correlation between two variables does not indicate which variable is causing which. For example, Reinhart and Rogoff (\(2010\)) found a strong correlation between public debt and GDP growth. Although some have argued that public debt slows growth, most evidence supports the alternative that slow growth increases public debt.
Excellent Video on Causality Featuring Evidence that Smoking Causes Cancer(See Chapter 11)
1. Li, C. (1975) Path analysis: A primer. Boxwood Press, Pacific Grove. CA .
2. Reinhart, C. M. and Rogoff, K. S. (2010). Growth in a Time of Debt. Working Paper 15639, National Bureau of Economic Research, http://www.nber.org/papers/w15639
6.07: Statistical Literacy
Learning Objectives
• Design a statistical study involving niacin, HDL and heart disease
Low HDL and Niacin
A low level of High-density lipoproteins (HDL) have long been known to be a risk factor for heart disease. Taking niacin has been shown to increase HDL levels and has been recommended for patients with low levels of HDL. The assumption of this recommendation is that niacin causes HDL to increase thus causing a lower risk for heart disease.
Example \(1\): What do you think?
What experimental design involving niacin would test whether the relationship between HDL and heart disease is causal?
Solution
You could randomly assign patients with low levels of HDL to a condition in which they received niacin or to one in which they did not. A finding that niacin increased HDL without decreasing heart disease would cast doubt on the causal relationship. This is exactly what was found in a study conducted by the NIH. See the description of the results here.
6.E: Research Design (Exercises)
General Questions
Q1
To be a scientific theory, the theory must be potentially ______________.
Q2
What is the difference between a faith-based explanation and a scientific explanation?
Q3
What does it mean for a theory to be parsimonious?
Q4
Define reliability in terms of parallel forms.
Q5
Define true score.
Q6
What is the reliability if the true score variance is \(80\) and the test score variance is \(100\)?
Q7
What statistic relates to how close a score on one test will be to a score on a parallel form?
Q8
What is the effect of test length on the reliability of a test?
Q9
Distinguish between predictive validity and construct validity.
Q10
What is the theoretical maximum correlation of a test with a criterion if the test has a reliability of \(0.81\)?
Q11
An experiment solicits subjects to participate in a highly stressful experiment. What type of sampling bias is likely to occur?
Q12
Give an example of survivorship bias not presented in this text.
Q13
Distinguish "between-subject" variables from "within-subjects" variables.
Q14
Of the variables "gender" and "trials," which is likely to be a between-subjects variable and which a within-subjects variable?
Q15
Define interaction.
Q16
What is counterbalancing used for?
Q17
How does randomization deal with the problem of pre-existing differences between groups?
Q18
Give an example of the "third variable problem" other than those in this text. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/06%3A_Research_Design/6.06%3A_Causation.txt |
Most of the statistical analyses presented in this book are based on the bell-shaped or normal distribution. The introductory section defines what it means for a distribution to be normal and presents some important properties of normal distributions. The interesting history of the discovery of the normal distribution is described in the second section. Methods for calculating probabilities based on the normal distribution are described in Areas of Normal Distributions. The Varieties of Normal Distribution Demo allows you to enter values for the mean and standard deviation of a normal distribution and see a graph of the resulting distribution. A frequently used normal distribution is called the Standard Normal distribution and is described in the section with that name. The binomial distribution can be approximated by a normal distribution. The section Normal Approximation to the Binomial shows this approximation. The Normal Approximation Demo allows you to explore the accuracy of this approximation.
• 7.1: Introduction to Normal Distributions
The normal distribution is the most important and most widely used distribution in statistics. It is sometimes called the "bell curve," although the tonal qualities of such a bell would be less than pleasing. It is also called the "Gaussian curve" after the mathematician Karl Friedrich Gauss. As you will see in the section on the history of the normal distribution, although Gauss played an important role in its history, Abraham de Moivre first discovered the normal distribution.
• 7.2: History of the Normal Distribution
In the section on the history of the normal distribution, we saw that the normal distribution can be used to approximate the binomial distribution. This section shows how to compute these approximations.
• 7.3: Areas Under Normal Distributions
Areas under portions of a normal distribution can be computed by using calculus. Since this is a non-mathematical treatment of statistics, we will rely on computer programs and tables to determine these areas.
• 7.4: Varieties Demonstration
This demonstration allows you to change the mean and standard deviation of two normal distributions and observe the effects on the shapes of the distributions.
• 7.5: Standard Normal Distribution
Normal distributions do not necessarily have the same means and standard deviations. A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution.
• 7.6: Normal Approximation to the Binomial
In the section on the history of the normal distribution, we saw that the normal distribution can be used to approximate the binomial distribution. This section shows how to compute these approximations.
• 7.7: Normal Approximation Demonstration
The normal distribution can be used to approximate the binomial distribution. This demonstration allows you to explore the accuracy of the approximation under a variety of conditions.
• 7.8: Statistical Literacy
Risk analyses often are based on the assumption of normal distributions. Critics have said that extreme events in reality are more frequent than would be expected assuming normality. The assumption has even been called a "Great Intellectual Fraud."
• 7.E: Normal Distribution (Exercises)
07: Normal Distribution
Learning Objectives
• Describe the shape of normal distributions
• State $7$ features of normal distributions
The normal distribution is the most important and most widely used distribution in statistics. It is sometimes called the "bell curve," although the tonal qualities of such a bell would be less than pleasing. It is also called the "Gaussian curve" after the mathematician Karl Friedrich Gauss. As you will see in the section on the history of the normal distribution, although Gauss played an important role in its history, Abraham de Moivre first discovered the normal distribution.
Strictly speaking, it is not correct to talk about "the normal distribution" since there are many normal distributions. Normal distributions can differ in their means and in their standard deviations. Figure $1$ shows three normal distributions. The green (left-most) distribution has a mean of $-3$ and a standard deviation of $0.5$, the distribution in red (the middle distribution) has a mean of $0$ and a standard deviation of $1$, and the distribution in black (right-most) has a mean of $2$ and a standard deviation of $3$. These as well as all other normal distributions are symmetric with relatively more values at the center of the distribution and relatively few in the tails.
The density of the normal distribution (the height for a given value on the $x$ axis) is shown below. The parameters $\mu$ and $\sigma$ are the mean and standard deviation, respectively, and define the normal distribution. The symbol $e$ is the base of the natural logarithm and $\pi$ is the constant pi.
$\frac{1}{\sqrt{2\pi \sigma ^2}}\boldsymbol{e}^{\tfrac{-(x-\mu )^2}{2\sigma ^2}}$
Since this is a non-mathematical treatment of statistics, do not worry if this expression confuses you. We will not be referring back to it in later sections.
Seven features of normal distributions are listed below. These features are illustrated in more detail in the remaining sections of this chapter.
1. Normal distributions are symmetric around their mean
2. The mean, median, and mode of a normal distribution are equal.
3. The area under the normal curve is equal to $1.0$.
4. Normal distributions are denser in the center and less dense in the tails.
5. Normal distributions are defined by two parameters, the mean ($\mu$) and the standard deviation ($\sigma$).
6. $68\%$ of the area of a normal distribution is within one standard deviation of the mean.
7. Approximately $95\%$ of the area of a normal distribution is within two standard deviations of the mean. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/07%3A_Normal_Distribution/7.01%3A_Introduction_to_Normal_Distributions.txt |
Learning Objectives
• Learn the interesting history of a normal curve
In the chapter on probability, we saw that the binomial distribution could be used to solve problems such as "If a fair coin is flipped $100$ times, what is the probability of getting $60$ or more heads?" The probability of exactly $x$ heads out of $N$ flips is computed using the formula:
$P(x)=\frac{N!}{x!(N-x)!}\pi ^x(1-\pi )^{N-x}$
where $x$ is the number of heads ($60$), $N$ is the number of flips ($100$), and $\pi$ is the probability of a head ($0.5$). Therefore, to solve this problem, you compute the probability of $60$ heads, then the probability of $61$ heads, $62$ heads, etc., and add up all these probabilities. Imagine how long it must have taken to compute binomial probabilities before the advent of calculators and computers.
Abraham de Moivre, an $18^{th}$ century statistician and consultant to gamblers, was often called upon to make these lengthy computations. de Moivre noted that when the number of events (coin flips) increased, the shape of the binomial distribution approached a very smooth curve. Binomial distributions for $2$, $4$, and $12$ flips are shown in Figure $1$.
de Moivre reasoned that if he could find a mathematical expression for this curve, he would be able to solve problems such as finding the probability of $60$ or more heads out of $100$ coin flips much more easily. This is exactly what he did, and the curve he discovered is now called the "normal curve."
The smooth curve in Figure $2$ is the normal distribution. Note how well it approximates the binomial probabilities represented by the heights of the blue lines.
The importance of the normal curve stems primarily from the fact that the distributions of many natural phenomena are at least approximately normally distributed. One of the first applications of the normal distribution was to the analysis of errors of measurement made in astronomical observations, errors that occurred because of imperfect instruments and imperfect observers. Galileo in the $17^{th}$ century noted that these errors were symmetric and that small errors occurred more frequently than large errors. This led to several hypothesized distributions of errors, but it was not until the early $19^{th}$ century that it was discovered that these errors followed a normal distribution. Independently, the mathematicians Adrain in $1808$ and Gauss in $1809$ developed the formula for the normal distribution and showed that errors were fit well by this distribution.
This same distribution had been discovered by Laplace in $1778$ when he derived the extremely important central limit theorem, the topic of a later section of this chapter. Laplace showed that even if a distribution is not normally distributed, the means of repeated samples from the distribution would be very nearly normally distributed, and that the larger the sample size, the closer the distribution of means would be to a normal distribution.
Most statistical procedures for testing differences between means assume normal distributions. Because the distribution of means is very close to normal, these tests work well even if the original distribution is only roughly normal.
Quételet was the first to apply the normal distribution to human characteristics. He noted that characteristics such as height, weight, and strength were normally distributed. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/07%3A_Normal_Distribution/7.02%3A_History_of_the_Normal_Distribution.txt |
Learning Objectives
• State the proportion of a normal distribution within \(1\) and within \(2\) standard deviations of the mean
• Use the calculator "Calculate Area for a given \(X\)"
• Use the calculator "Calculate \(X\) for a given Area"
Areas under portions of a normal distribution can be computed by using calculus. Since this is a non-mathematical treatment of statistics, we will rely on computer programs and tables to determine these areas. Figure \(1\) shows a normal distribution with a mean of \(50\) and a standard deviation of \(10\). The shaded area between \(40\) and \(60\) contains \(68\%\) of the distribution.
\(68\%\) of the area is within one standard deviation (\(10\)) of the mean (\(50\))
Figure \(2\) shows a normal distribution with a mean of \(100\) and a standard deviation of \(20\). As in Figure \(1\), \(68\%\) of the distribution is within one standard deviation of the mean.
\(68\%\) of the area is within one standard deviation (\(20\)) of the mean (\(100\)).
The normal distributions shown in Figures \(1\) and \(2\) are specific examples of the general rule that \(68\%\) of the area of any normal distribution is within one standard deviation of the mean.
Figure \(3\) shows a normal distribution with a mean of \(75\) and a standard deviation of \(10\). The shaded area contains \(95\%\) of the area and extends from \(55.4\) to \(94.6\). For all normal distributions, \(95\%\) of the area is within \(1.96\) standard deviations of the mean. For quick approximations, it is sometimes useful to round off and use \(2\) rather than \(1.96\) as the number of standard deviations you need to extend from the mean so as to include \(95\%\) of the area.
The normal calculator can be used to calculate areas under the normal distribution. For example, you can use it to find the proportion of a normal distribution with a mean of \(90\) and a standard deviation of \(12\) that is above \(110\). Set the mean to \(90\) and the standard deviation to \(12\). Then enter "\(110\)" in the box to the right of the radio button "Above." At the bottom of the display you will see that the shaded area is \(0.0478\). See if you can use the calculator to find that the area between \(115\) and \(120\) is \(0.0124\).
Say you wanted to find the score corresponding to the \(75^{th}\) percentile of a normal distribution with a mean of \(90\) and a standard deviation of \(12\). Using the inverse normal calculator, you enter the parameters as shown in Figure \(5\) and find that the area below \(98.09\) is \(0.75\).
Normal and Inverse Normal Calculator
7.04: Varieties Demonstration
Learning Objectives
• Be able to describe differences in distributions based on their shape and statistics
Instructions
This demonstration allows you to change the mean and standard deviation of two normal distributions and observe the effects on the shapes of the distributions.When the demonstration begins, the red distribution has a mean of \(50\) and a standard deviaton of \(10\). The blue distribution has a mean of \(50\) and a standard deviation of \(5\). Notice how much more spread out the red distribution is than the blue distribution.
Choose different values for the two distributions and investigate their effects.
Illustrated Instructions
The demonstration starts by displaying two curves both with a mean of \(50\) and with standard deviations of \(10\) and \(5\) respectively. You can change the values for the mean and standard deviation for both curves.
The screenshot below shows the distributions with different means and standard deviations. Note that the curves are on an axis that ranges from \(0 - 100\), keep this in mind when changing the values of the mean and standard deviations as some values will not be displayed correctly. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/07%3A_Normal_Distribution/7.03%3A_Areas_Under_Normal_Distributions.txt |
Learning Objectives
• State the mean and standard deviation of the standard normal distribution
• Use a $Z$ table
• Use the normal calculator
• Transform raw data to $Z$ scores
As discussed in the introductory section, normal distributions do not necessarily have the same means and standard deviations. A normal distribution with a mean of $0$ and a standard deviation of $1$ is called a standard normal distribution.
Areas of the normal distribution are often represented by tables of the standard normal distribution. A portion of a table of the standard normal distribution is shown in Table $1$.
Table $1$: A portion of a table of the standard normal distribution
Z Area below
-2.5 0.0062
-2.49 0.0064
-2.48 0.0066
-2.47 0.0068
-2.46 0.0069
-2.45 0.0071
-2.44 0.0073
-2.43 0.0075
-2.42 0.0078
-2.41 0.008
-2.4 0.0082
-2.39 0.0084
-2.38 0.0087
-2.37 0.0089
-2.36 0.0091
-2.35 0.0094
-2.34 0.0096
-2.33 0.0099
-2.32 0.0102
The first column titled "$Z$" contains values of the standard normal distribution; the second column contains the area below $Z$. Since the distribution has a mean of $0$ and a standard deviation of $1$, the $Z$ column is equal to the number of standard deviations below (or above) the mean. For example, a $Z$ of $-2.5$ represents a value $2.5$ standard deviations below the mean. The area below $Z$ is $0.0062$.
The same information can be obtained using the following Java applet. Figure $1$ shows how it can be used to compute the area below a value of $-2.5$ on the standard normal distribution. Note that the mean is set to $0$ and the standard deviation is set to $1$.
Calculate Areas
A value from any normal distribution can be transformed into its corresponding value on a standard normal distribution using the following formula:
$Z=\frac{X-\mu }{\sigma }$
where $Z$ is the value on the standard normal distribution, $X$ is the value on the original distribution, $\mu$ is the mean of the original distribution, and $\sigma$ is the standard deviation of the original distribution.
Example $1$
As a simple application, what portion of a normal distribution with a mean of $50$ and a standard deviation of $10$ is below $26$?
Solution
Applying the formula, we obtain
$Z = \frac{26 - 50}{10} = -2.4$
From Table $1$, we can see that $0.0082$ of the distribution is below $-2.4$. There is no need to transform to $Z$ if you use the applet as shown in Figure $2$.
If all the values in a distribution are transformed to $Z$ scores, then the distribution will have a mean of $0$ and a standard deviation of $1$. This process of transforming a distribution to one with a mean of $0$ and a standard deviation of $1$ is called standardizing the distribution.
7.06: Normal Approximation to the Binomial
Learning Objectives
• State the relationship between the normal distribution and the binomial distribution
• Use the normal distribution to approximate the binomial distribution
• State when the approximation is adequate
In the section on the history of the normal distribution, we saw that the normal distribution can be used to approximate the binomial distribution. This section shows how to compute these approximations.
Let's begin with an example. Assume you have a fair coin and wish to know the probability that you would get $8$ heads out of $10$ flips. The binomial distribution has a mean of $\mu =N\pi =(10)(0.5)=5$ and a variance of $\sigma ^2=N\pi (1-\pi )=(10)(0.5)(0.5)=2.5$. The standard deviation is therefore $1.5811$. A total of $8$ heads is $(8 - 5)/1.5811 = 1.897$ standard deviations above the mean of the distribution. The question then is, "What is the probability of getting a value exactly $1.897$ standard deviations above the mean?" You may be surprised to learn that the answer is $0$: The probability of any one specific point is $0$. The problem is that the binomial distribution is a discrete probability distribution, whereas the normal distribution is a continuous distribution.
The solution is to round off and consider any value from $7.5$ to $8.5$ to represent an outcome of $8$ heads. Using this approach, we figure out the area under a normal curve from $7.5$ to $8.5$. The area in green in Figure $1$ is an approximation of the probability of obtaining $8$ heads.
The solution is therefore to compute this area. First we compute the area below $8.5$ and then subtract the area below $7.5$.
The results of using the normal area calculator to find the area below $8.5$ are shown in Figure $2$. The results for $7.5$ are shown in Figure $3$.
The difference between the areas is $0.044$, which is the approximation of the binomial probability. For these parameters, the approximation is very accurate. The demonstration in the next section allows you to explore its accuracy with different parameters.
If you did not have the normal area calculator, you could find the solution using a table of the standard normal distribution (a $Z$ table) as follows:
1. Find a $Z$ score for $8.5$ using the formula $Z = (8.5 - 5)/1.5811 = 2.21$.
2. Find the area below a $Z$ of $2.21 = 0.987$.
3. Find a $Z$ score for $7.5$ using the formula $Z = (7.5 - 5)/1.5811 = 1.58$.
4. Find the area below a $Z$ of $1.58 = 0.943$.
5. Subtract the value in step $4$ from the value in step $2$ to get $0.044$.
The same logic applies when calculating the probability of a range of outcomes. For example, to calculate the probability of $8$ to $10$ flips, calculate the area from $7.5$ to $10.5$.
The accuracy of the approximation depends on the values of $N$ and $\pi$. A rule of thumb is that the approximation is good if both $N\pi$ and $N(1-\pi )$ are both greater than $10$.
7.07: Normal Approximation Demonstration
Learning Objectives
• State the relationship between sample size and the accuracy of normal approximation of the binomial distribution
Instructions
The normal distribution can be used to approximate the binomial distribution. This demonstration allows you to explore the accuracy of the approximation under a variety of conditions.
Illustrated Instructions
The demonstration displays the probability of success over a specific number of trials based on the entered total number of trials (\(N\)) and the probability of success on a given trial. You can change \(N\) as well as \(p\) and then select the proportion for which you would like the probability to be calculated.
By clicking the appropriate radio button, you can choose to calculate the probability above a specified value or between two values. Try out various values of \(N\) and \(p\) to calculate various approximations to the binomial to explore the accuracy of the approximation.
The calculation based on the normal approximation to the binomial is shown in green below and is equal to \(0.1714\). The actual binomial probability of \(0.1719\) is shown in red.
The screenshot below displays results for the probability of greater than \(10\) successful trials with \(15\) total trials and a \(0.5\) probability of success. The probability based on the normal approximation is displayed at the bottom of the screen in green next to the binomial probability displayed in red.
7.08: Statistical Literacy
Learning Objectives
• Evaluating "Tail Risk"
Risk analyses often are based on the assumption of normal distributions. Critics have said that extreme events in reality are more frequent than would be expected assuming normality. The assumption has even been called a "Great Intellectual Fraud."
A recent article discussing how to protect investments against extreme events defined "tail risk" as "A tail risk, or extreme shock to financial markets, is technically defined as an investment that moves more than three standard deviations from the mean of a normal distribution of investment returns."
Example \(1\): what do you think?
Tail risk can be evaluated by assuming a normal distribution and computing the probability of such an event. Is that how "tail risk" should be evaluated?
Solution
Events more than three standard deviations from the mean are very rare for normal distributions. However, they are not as rare for other distributions such as highly-skewed distributions. If the normal distribution is used to assess the probability of tail events defined this way, then the "tail risk" will be underestimated. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/07%3A_Normal_Distribution/7.05%3A_Standard_Normal_Distribution.txt |
You may want to use the "Calculate Area for a given X" and the "Calculate X for a given Area" calculators for some of these exercises.
General Questions
Q1
If scores are normally distributed with a mean of $35$ and a standard deviation of $10$, what percent of the scores is: (relevant section)
1. greater than $34$
2. smaller than $42$
3. between $28$ and $34$
Q2
1. What are the mean and standard deviation of the standard normal distribution?
2. What would be the mean and standard deviation of a distribution created by multiplying the standard normal distribution by $8$ and then adding $75$? (relevant section & here)
Q3
The normal distribution is defined by two parameters. What are they? (relevant section)
Q4
1. What proportion of a normal distribution is within one standard deviation of the mean?
2. What proportion is more than $2.0$ standard deviations from the mean?
3. What proportion is between $1.25$ and $2.1$ standard deviations above the mean? (relevant section)
Q5
A test is normally distributed with a mean of $70$ and a standard deviation of $8$.
1. What score would be needed to be in the $85^{th}$ percentile?
2. What score would be needed to be in the $22^{nd}$ percentile? (relevant section)
Q6
Assume a normal distribution with a mean of $70$ and a standard deviation of $12$. What limits would include the middle $65\%$ of the cases? (relevant section)
Q7
A normal distribution has a mean of $20$ and a standard deviation of $4$. Find the $Z$ scores for the following numbers: (relevant section)
1. $28$
2. $18$
3. $10$
4. $23$
Q8
Assume the speed of vehicles along a stretch of $I-10$ has an approximately normal distribution with a mean of $71$ mph and a standard deviation of $8$ mph.
1. The current speed limit is $65$ mph. What is the proportion of vehicles less than or equal to the speed limit?
2. What proportion of the vehicles would be going less than $50$ mph?
3. A new speed limit will be initiated such that approximately $10\%$ of vehicles will be over the speed limit. What is the new speed limit based on this criterion?
4. In what way do you think the actual distribution of speeds differs from a normal distribution? (relevant section)
Q9
A variable is normally distributed with a mean of $120$ and a standard deviation of $5$. One score is randomly sampled. What is the probability it is above $127$? (relevant section)
Q10
You want to use the normal distribution to approximate the binomial distribution. Explain what you need to do to find the probability of obtaining exactly $7$ heads out of $12$ flips. (relevant section)
Q11
A group of students at a school takes a history test. The distribution is normal with a mean of $25$, and a standard deviation of $4$.
1. Everyone who scores in the top $30\%$ of the distribution gets a certificate. What is the lowest score someone can get and still earn a certificate?
2. The top $5\%$ of the scores get to compete in a statewide history contest. What is the lowest score someone can get and still go onto compete with the rest of the state? (relevant section)
Q12
Use the normal distribution to approximate the binomial distribution and find the probability of getting $15$ to $18$ heads out of $25$ flips. Compare this to what you get when you calculate the probability using the binomial distribution. Write your answers out to four decimal places. (relevant section & relevant section)
Q13
True/false: For any normal distribution, the mean, median, and mode will have the same value. (relevant section)
Q14
True/false: In a normal distribution, $11.5\%$ of scores are greater than $Z = 1.2$. (relevant section)
Q15
True/false: The percentile rank for the mean is $50\%$ for any normal distribution. (relevant section)
Q16
True/false: The larger the $\pi$, the better the normal distribution approximates the binomial distribution. (relevant section & relevant section)
Q17
True/false: A $Z$-score represents the number of standard deviations above or below the mean. (relevant section)
Q18
True/false: Abraham de Moivre, a consultant to gamblers, discovered the normal distribution when trying to approximate the binomial distribution to make his computations easier. (relevant section)
Q19
True/false: The standard deviation of the blue distribution shown below is about $10$. (relevant section)
Q20
True/false: In the figure below, the red distribution has a larger standard deviation than the blue distribution. (relevant section)
Q21
True/false: The red distribution has more area underneath the curve than the blue distribution does. (relevant section)
Questions from Case Studies
The following question uses data from the Angry Moods (AM) case study.
Q22
For this problem, use the Anger Expression (AE) scores.
1. Compute the mean and standard deviation.
2. Then, compute what the $25^{th}$, $50^{th}$ and $75^{th}$ percentiles would be if the distribution were normal.
3. Compare the estimates to the actual $25^{th}$, $50^{th}$ and $75^{th}$ percentiles. (relevant section)
The following question uses data from the Physicians Reaction (PR) case study.
Q23
For this problem, use the time spent with the overweight patients.
1. Compute the mean and standard deviation of this distribution.
2. What is the probability that if you chose an overweight participant at random, the doctor would have spent $31$ minutes or longer with this person?
3. Now assume this distribution is normal (and has the same mean and standard deviation). Now what is the probability that if you chose an overweight participant at random, the doctor would have spent $31$ minutes or longer with this person? (relevant section)
The following questions are from ARTIST (reproduced with permission)
Visit the site
Q24
A set of test scores are normally distributed. Their mean is $100$ and standard deviation is $20$. These scores are converted to standard normal $z$ scores. What would be the mean and median of this distribution?
1. $0$
2. $1$
3. $50$
4. $100$
Q25
Suppose that weights of bags of potato chips coming from a factory follow a normal distribution with mean $12.8$ ounces and standard deviation $0.6$ ounces. If the manufacturer wants to keep the mean at $12.8$ ounces but adjust the standard deviation so that only $1\%$ of the bags weigh less than $12$ ounces, how small does he/she need to make that standard deviation?
Q26
A student received a standardized ($z$) score on a test that was $-0. 57$. What does this score tell about how this student scored in relation to the rest of the class? Sketch a graph of the normal curve and shade in the appropriate area.
Q27
Suppose you take $50$ measurements on the speed of cars on Interstate $5$, and that these measurements follow roughly a Normal distribution. Do you expect the standard deviation of these $50$ measurements to be about $1$ mph, $5$ mph, $10$ mph, or $20$ mph? Explain.
Q28
Suppose that combined verbal and math SAT scores follow a normal distribution with mean $896$ and standard deviation $174$. Suppose further that Peter finds out that he scored in the top $3\%$ of SAT scores. Determine how high Peter's score must have been.
Q29
Heights of adult women in the United States are normally distributed with a population mean of $\mu =63.5$ inches and a population standard deviation of $\sigma =2.5$. A medical researcher is planning to select a large random sample of adult women to participate in a future study. What is the standard value, or $z$-value, for an adult woman who has a height of $68.5$ inches?
Q30
An automobile manufacturer introduces a new model that averages $27$ miles per gallon in the city. A person who plans to purchase one of these new cars wrote the manufacturer for the details of the tests, and found out that the standard deviation is $3$ miles per gallon. Assume that in-city mileage is approximately normally distributed.
1. What is the probability that the person will purchase a car that averages less than $20$ miles per gallon for in-city driving?
2. What is the probability that the person will purchase a car that averages between $25$ and $29$ miles per gallon for in-city driving?
Select Answers
S1
1. $75.8\%$
S2
1. Mean = $75$
S4
1. $0.088$
S5
1. $78.3$
S7
1. $2.0$
S8
1. $0.227$
S11
1. $27.1$
S12
$0.2037$ (normal approximation)
S22
$25^{th}$ percentile:
1. $28.27$
2. $26.75$
S23
1. $0.053$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/07%3A_Normal_Distribution/7.E%3A_Normal_Distribution_%28Exercises%29.txt |
• 8.1: Q-Q Plots
The quantile-quantile or q-q plot is an exploratory graphical device used to check the validity of a distributional assumption for a data set. In general, the basic idea is to compute the theoretically expected value for each data point based on the distribution in question. If the data indeed follow the assumed distribution, then the points on the q-q plot will fall approximately on a straight line.
• 8.2: Contour Plots
Contour plots portray data for three variables in two dimensions. The plot contains a number of contour lines. Each contour line is shown in an X-Y plot and has a constant value on a third variable.
• 8.3: 3D Plots
Just as two-dimensional scatter plots show the data in two dimensions, 3D plots show data in three dimensions.
• 8.4: Statistical Literacy
Reading a Weather Map
• 8.E: Advanced Graphs (Exercises)
08: Advanced Graphs
Learning Objectives
• State what $q-q$ plots are used for.
• Describe the shape of a $q-q$ plot when the distributional assumption is met.
• Be able to create a normal $q-q$ plot.
The quantile-quantile or $q-q$ plot is an exploratory graphical device used to check the validity of a distributional assumption for a data set. In general, the basic idea is to compute the theoretically expected value for each data point based on the distribution in question. If the data indeed follow the assumed distribution, then the points on the $q-q$ plot will fall approximately on a straight line.
Before delving into the details of $q-q$ plots, we first describe two related graphical methods for assessing distributional assumptions: the histogram and the cumulative distribution function (CDF). As will be seen, $q-q$ plots are more general than these alternatives.
Assessing Distributional Assumptions
As an example, consider data measured from a physical device such as the spinner depicted in Figure $1$. The red arrow is spun around the center, and when the arrow stops spinning, the number between $0$ and $1$ is recorded. Can we determine if the spinner is fair?
If the spinner is fair, then these numbers should follow a uniform distribution. To investigate whether the spinner is fair, spin the arrow $n$ times, and record the measurements by ${\mu _1, \mu _2, ..., \mu _n}$. In this example, we collect $n = 100$ samples. The histogram provides a useful visualization of these data. In Figure $2$, we display three different histograms on a probability scale. The histogram should be flat for a uniform sample, but the visual perception varies depending on whether the histogram has $10$, $5$, or $3$ bins. The last histogram looks flat, but the other two histograms are not obviously flat. It is not clear which histogram we should base our conclusion on.
Alternatively, we might use the cumulative distribution function (CDF), which is denoted by $F(\mu )$. The CDF gives the probability that the spinner gives a value less than or equal to $\mu$, that is, the probability that the red arrow lands in the interval $[0, \mu ]$. By simple arithmetic, $F(\mu ) = \mu$, which is the diagonal straight line $y = x$. The CDF based upon the sample data is called the empirical CDF (ECDF), is denoted by $\widehat{F}_n(u)$, and is defined to be the fraction of the data less than or equal to $\mu$; that is,
$\widehat{F}_n(u)=\frac{\#u_i\leq u}{n}$
In general, the ECDF takes on a ragged staircase appearance.
For the spinner sample analyzed in Figure $2$, we computed the ECDF and CDF, which are displayed in Figure $3$. In the left frame, the ECDF appears close to the line $y = x$, shown in the middle frame. In the right frame, we overlay these two curves and verify that they are indeed quite close to each other. Observe that we do not need to specify the number of bins as with the histogram.
q-q plot for uniform data
The $q-q$ plot for uniform data is very similar to the empirical CDF graphic, except with the axes reversed. The $q-q$ plot provides a visual comparison of the sample quantiles to the corresponding theoretical quantiles. In general, if the points in a $q-q$ plot depart from a straight line, then the assumed distribution is called into question.
Here we define the qth quantile of a batch of n numbers as a number $ξ_q$ such that a fraction q x n of the sample is less than $ξ_q$, while a fraction $(1 - q) \times n$ of the sample is greater than $ξ_q$. The best known quantile is the median, $ξ_{0.5}$, which is located in the middle of the sample.
Consider a small sample of $5$ numbers from the spinner:
$\mu _1 = 0.41,\; \mu _2 =0.24,\; \mu _3 =0.59,\; \mu _4 =0.03,\; \mu _5 =0.67$
Based upon our description of the spinner, we expect a uniform distribution to model these data. If the sample data were “perfect,” then on average there would be an observation in the middle of each of the $5$ intervals: $0$ to $0.2$, $0.2$ to $0.4$, $0.4$ to $0.6$, and so on. Table $1$ shows the $5$ data points (sorted in ascending order) and the theoretically expected value of each based on the assumption that the distribution is uniform (the middle of the interval).
Table $1$: Computing the Expected Quantile Values
Data (μ) Rank (i) Middle of the ith Interval
0.03 1 0.1
0.24 2 0.3
0.41 3 0.5
0.59 4 0.7
0.67 5 0.9
The theoretical and empirical CDFs are shown in Figure $4$ and the $q-q$ plot is shown in the left frame of Figure $5$.
In general, we consider the full set of sample quantiles to be the sorted data values
$μ_{(1)} < μ_{(2)} < μ_{(3)} < \ldots < μ_{(n-1)} < μ_{(n)} ,$
where the parentheses in the subscript indicate the data have been ordered. Roughly speaking, we expect the first ordered value to be in the middle of the interval $(0, 1/n)$, the second to be in the middle of the interval $(1/n, 2/n)$, and the last to be in the middle of the interval $(\tfrac{n - 1}{n}, 1)$. Thus, we take as the theoretical quantile the value
$\xi _q=q\approx \frac{i-0.5}{n}$
where $q$ corresponds to the $i^{th}$ ordered sample value. We subtract the quantity $0.5$ so that we are exactly in the middle of the interval $(\tfrac{i-1}{n}, \tfrac{i}{n})$. These ideas are depicted in the right frame of Figure $4$ for our small sample of size $n = 5$.
We are now prepared to define the $q-q$ plot precisely. First, we compute the n expected values of the data, which we pair with the n data points sorted in ascending order. For the uniform density, the $q-q$ plot is composed of the $n$ ordered pairs
$(\tfrac{i-0.5}{n}, u_i),\; for\; i=1,2,\cdots ,n$
This definition is slightly different from the ECDF, which includes the points $(u_i, \tfrac{i}{n})$. In the left frame of Figure $5$, we display the $q-q$ plot of the $5$ points in Table $1$. In the right two frames of Figure $5$, we display the $q-q$ plot of the same batch of numbers used in Figure $2$. In the final frame, we add the diagonal line $y = x$ as a point of reference.
The sample size should be taken into account when judging how close the $q-q$ plot is to the straight line. We show two other uniform samples of size $n = 10$ and $n = 1000$ in Figure $6$. Observe that the $q-q$ plot when $n = 1000$ is almost identical to the line $y = x$, while such is not the case when the sample size is only $n = 10$.
In Figure $7$, we show the $q-q$ plots of two random samples that are not uniform. In both examples, the sample quantiles match the theoretical quantiles only at the median and at the extremes. Both samples seem to be symmetric around the median. But the data in the left frame are closer to the median than would be expected if the data were uniform. The data in the right frame are further from the median than would be expected if the data were uniform.
In fact, the data were generated in the $R$ language from beta distributions with parameters $a = b = 3$ on the left and $a = b =0.4$ on the right. In Figure $8$ we display histograms of these two data sets, which serve to clarify the true shapes of the densities. These are clearly non-uniform.
q-q plot for normal data
The definition of the $q-q$ plot may be extended to any continuous density. The $q-q$ plot will be close to a straight line if the assumed density is correct. Because the cumulative distribution function of the uniform density was a straight line, the $q-q$ plot was very easy to construct. For data that are not uniform, the theoretical quantiles must be computed in a different manner.
Let ${z_1, z_2, ..., z_n}$ denote a random sample from a normal distribution with mean $\mu =0$ and standard deviation $\sigma =1$. Let the ordered values be denoted by
$z_{(1)} < z_{(2)} < z_{(3)} < \ldots < z_{(n-1)} <z_{(n)}$
These n ordered values will play the role of the sample quantiles.
Let us consider a sample of $5$ values from a distribution to see how they compare with what would be expected for a normal distribution. The $5$ values in ascending order are shown in the first column of Table $2$.
Table $2$: Computing the expected quantile values for normal data.
Data (z) Rank (i) Middle of the ith Interval Normal(z)
-1.96 1 0.1 -1.28
-0.78 2 0.3 -0.52
0.31 3 0.5 0.00
1.15 4 0.7 0.52
1.62 5 0.9 1.28
Just as in the case of the uniform distribution, we have $5$ intervals. However, with a normal distribution the theoretical quantile is not the middle of the interval but rather the inverse of the normal distribution for the middle of the interval. Taking the first interval as an example, we want to know the $z$ value such that $0.1$ of the area in the normal distribution is below $z$. This can be computed using the Inverse Normal Calculator as shown in Figure $9$. Simply set the “Shaded Area” field to the middle of the interval ($0.1$) and click on the “Below” button. The result is $-1.28$. Therefore, $10\%$ of the distribution is below a $z$ value of $-1.28$.
The $q-q$ plot for the data in Table $2$ is shown in the left frame of Figure $11$.
In general, what should we take as the corresponding theoretical quantiles? Let the cumulative distribution function of the normal density be denoted by $\Phi (z)$. In the previous example, $\Phi (-1.28)=0.10$ and $\Phi (0.00)=0.50$. Using the quantile notation, if $\xi _q$ is the $q^{th}$ quantile of a normal distribution, then
$Φ(ξ_q)= q$
That is, the probability a normal sample is less than $\xi _q$ is in fact just $q$.
Consider the first ordered value, $z_1$. What might we expect the value of $\Phi (z_1)$ to be? Intuitively, we expect this probability to take on a value in the interval $(0, 1/n)$. Likewise, we expect $\Phi (z_2)$ to take on a value in the interval $(1/n, 2/n)$. Continuing, we expect $\Phi (z_n)$ to fall in the interval $((n - 1)/n, 1)$. Thus, the theoretical quantile we desire is defined by the inverse (not reciprocal) of the normal CDF. In particular, the theoretical quantile corresponding to the empirical quantile $z_i$ should be
$ξ_q \approx \dfrac{i-0.5}{n}$
for $i = 1, 2, \ldots, n$.
The empirical CDF and theoretical quantile construction for the small sample given in Table $2$ are displayed in Figure $10$. For the larger sample of size $100$, the first few expected quantiles are $-2.576$, $-2.170$, and $-1.960$.
In the left frame of Figure $11$, we display the $q-q$ plot of the small normal sample given in Table $2$. The remaining frames in Figure $11$ display the $q-q$ plots of normal random samples of size $n = 100$ and $n = 1000$. As the sample size increases, the points in the $q-q$ plots lie closer to the line $y = x$.
As before, a normal $q-q$ plot can indicate departures from normality. The two most common examples are skewed data and data with heavy tails (large kurtosis). In Figure $12$, we show normal $q-q$ plots for a chi-squared (skewed) data set and a Student’s-$t$ (kurtotic) data set, both of size $n = 1000$. The data were first standardized. The red line is again $y = x$. Notice, in particular, that the data from the $t$ distribution follow the normal curve fairly closely until the last dozen or so points on each extreme.
q-q plots for normal data with general mean and scale
Our previous discussion of $q-q$ plots for normal data all assumed that our data were standardized. One approach to constructing $q-q$ plots is to first standardize the data and then proceed as described previously. An alternative is to construct the plot directly from raw data.
In this section, we present a general approach for data that are not standardized. Why did we standardize the data in Figure $12$? The $q-q$ plot is comprised of the $n$ points
$\left ( \Phi ^{-1}\left ( \frac{i-5}{n} \right ),z_i \right )\; for\; i=1,2,...,n$
If the original data {$z_i$} are normal, but have an arbitrary mean $\mu$ and standard deviation $\sigma$, then the line $y = x$ will not match the expected theoretical quantiles. Clearly, the linear transformation
$μ + σ ξ_q$
would provide the $q^{th}$ theoretical quantile on the transformed scale. In practice, with a new data set $\{x_1,x_2, \ldots, x_n\}$,
the normal $q-q$ plot would consist of the n points
$\left ( \Phi ^{-1}\left ( \frac{i-5}{n} \right ),x_i \right )\; for\; i=1,2,...,n$
Instead of plotting the line $y = x$ as a reference line, the line
$y = M + s · x$
should be composed, where $M$ and $s$ are the sample moments (mean and standard deviation) corresponding to the theoretical moments $\mu$ and $\sigma$. Alternatively, if the data are standardized, then the line $y = x$ would be appropriate, since now the sample mean would be $0$ and the sample standard deviation would be $1$.
Example $1$: SAT Case Study
The SAT case study followed the academic achievements of $105$ college students majoring in computer science. The first variable is their verbal SAT score and the second is their grade point average (GPA) at the university level. Before we compute inferential statistics using these variables, we should check if their distributions are normal. In Figure $13$, we display the $q-q$ plots of the verbal SAT and university GPA variables.
The verbal SAT seems to follow a normal distribution reasonably well, except in the extreme tails. However, the university GPA variable is highly non-normal. Compare the GPA $q-q$ plot to the simulation in the right frame of Figure $7$. These figures are very similar, except for the region where $x\approx -1$. To follow these ideas, we computed histograms of the variables and their scatter diagram in Figure $14$. These figures tell quite a different story. The university GPA is bimodal, with about $20\%$ of the students falling into a separate cluster with a grade of $C$. The scatter diagram is quite unusual. While the students in this cluster all have below average verbal SAT scores, there are as many students with low SAT scores whose GPAs were quite respectable. We might speculate as to the cause(s): different distractions, different study habits, but it would only be speculation. But observe that the raw correlation between verbal SAT and GPA is a rather high $0.65$, but when we exclude the cluster, the correlation for the remaining $86$ students falls a little to $0.59$.
Discussion
Parametric modeling usually involves making assumptions about the shape of data, or the shape of residuals from a regression fit. Verifying such assumptions can take many forms, but an exploration of the shape using histograms and $q-q$ plots is very effective. The $q-q$ plot does not have any design parameters such as the number of bins for a histogram.
In an advanced treatment, the $q-q$ plot can be used to formally test the null hypothesis that the data are normal. This is done by computing the correlation coefficient of the $n$ points in the $q-q$ plot. Depending upon $n$, the null hypothesis is rejected if the correlation coefficient is less than a threshold. The threshold is already quite close to $0.95$ for modest sample sizes.
We have seen that the $q-q$ plot for uniform data is very closely related to the empirical cumulative distribution function. For general density functions, the so-called probability integral transform takes a random variable $X$ and maps it to the interval ($0, 1$) through the CDF of $X$ itself, that is,
$Y = F_X(X)$
which has been shown to be a uniform density. This explains why the $q-q$ plot on standardized data is always close to the line $y = x$ when the model is correct.
Finally, scientists have used special graph paper for years to make relationships linear (straight lines). The most common example used to be semi-log paper, on which points following the formula $y=ae^{bx}$ appear linear. This follows of course since $log(y) = log(a) + bx$, which is the equation for a straight line. The $q-q$ plots may be thought of as being “probability graph paper” that makes a plot of the ordered data values into a straight line. Every density has its own special probability graph paper. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/08%3A_Advanced_Graphs/8.01%3A_Q-Q_Plots.txt |
Learning Objectives
• Describe a contour plot
• Interpret a contour plot
Contour plots portray data for three variables in two dimensions. The plot contains a number of contour lines. Each contour line is shown in an \(X-Y\) plot and has a constant value on a third variable. Consider the Figure \(1\) that contains data on the fat, non-sugar carbohydrates, and calories present in a variety of breakfast cereals. Each line shows the carbohydrate and fat levels for cereals with the same number of calories. Note that the number of calories is not determined exactly by the fat and non-sugar carbohydrates since cereals also differ in sugar and protein.
An alternative way to draw the plot is shown in Figure \(2\). The areas with the same number of calories are shaded.
Contour plots are used in many disciplines. For example, in cartography, contour lines can indicate areas of equal elevation, whereas in meteorological maps, contour lines can show equal temperature or equal barometric pressure.
The following links explain the drawing and interpretation of countour maps.
Application in Geology
Application in Meteorology
8.03: 3D Plots
Learning Objectives
• Describe a \(3D\) Plot
• Give an example of the value of a \(3D\) plot
Just as two-dimensional scatter plots show the data in two dimensions, \(3D\) plots show data in three dimensions. Figure \(1\) shows a \(3D\) scatter plot of the fat, non-sugar carbohydrates, and calories from a variety of cereal types.
Many statistical packages allow you to rotate the axes interactively to view the data from a different vantage point. Figure \(2\) is an example.
A fourth dimension can be represented as long as it is represented as a nominal variable. Figure \(3\) represents the different manufacturers by using different colors.
Interactively rotating \(3D\) plots can sometimes reveal aspects of the data not otherwise apparent. Figure \(4\) shows data from a pseudo random number generator. Figure \(4\) does not show anything systematic and the random number generator appears to generate data with properties similar to those of true random numbers.
Figure \(5\) shows a different perspective on these data. Clearly they were not generated by a random process.
Figures \(4\) and \(5\) are reproduced with permission from R snippets by Bogumil Kaminski.
8.04: Statistical Literacy
Learning Objectives
• Reading a Weather Map
This web page portrays altitudes in the United States.
Example \(1\): What do you think?
What part of the state of Texas (North, South, East, or West) contains the highest elevation?
Solution
West Texas.
8.E: Advanced Graphs (Exercises)
General Questions
Q1
What are \(Q-Q\) plots useful for?
Q2
For the following data, plot the theoretically expected \(z\) score as a function of the actual \(z\) score (a \(Q-Q\) plot).
0 1.2
0 1.2
0 1.2
0 1.2
0 1.3
0 1.3
0.1 1.3
0.1 1.3
0.1 1.3
0.1 1.4
0.1 1.4
0.2 1.5
0.2 1.6
0.3 1.7
0.3 1.7
0.4 1.7
0.5 1.8
0.6 1.8
0.6 1.9
0.6 1.9
0.6 2.0
0.6 2.0
0.6 2.0
0.6 2.1
0.6 2.1
0.6 2.1
0.7 2.1
0.7 2.1
0.8 2.1
0.8 2.1
0.8 2.3
0.8 2.5
0.8 2.7
0.9 3.0
1.0 4.2
1.0 5.0
1.1 5.7
1.1 12.4
1.2 15.2
1.2
Q3
For the data in problem 2, describe how the data differ from a normal distribution.
Questions from Case studies
Q4
For the "SAT and College GPA" case study data, create a contour plot looking at University GPA as a function of Math SAT and High School GPA. Naturally, you should use a computer to do this.
Q5
For the "SAT and College GPA" case study data, create a \(3D\) plot using the variables University GPA, Math SAT, and High School GPA. Naturally, you should use a computer to do this. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/08%3A_Advanced_Graphs/8.02%3A_Contour_Plots.txt |
The concept of a sampling distribution is perhaps the most basic concept in inferential statistics. It is also a difficult concept because a sampling distribution is a theoretical distribution rather than an empirical distribution. The introductory section defines the concept and gives an example for both a discrete and a continuous distribution. It also discusses how sampling distributions are used in inferential statistics. The Basic Demo is an interactive demonstration of sampling distributions. It is designed to make the abstract concept of sampling distributions more concrete. The Sample Size Demo allows you to investigate the effect of sample size on the sampling distribution of the mean. The Central Limit Theorem (CLT) Demo is an interactive illustration of a very important and counter-intuitive characteristic of the sampling distribution of the mean. The remaining sections of the chapter concern the sampling distributions of important statistics: the Sampling Distribution of the Mean, the Sampling Distribution of the Difference Between Means, the Sampling Distribution of r, and the Sampling Distribution of a Proportion.
09: Sampling Distributions
Learning Objectives
• Define inferential statistics
• Graph a probability distribution for the mean of a discrete variable
• Describe a sampling distribution in terms of "all possible outcomes"
• Describe a sampling distribution in terms of repeated sampling
• Describe the role of sampling distributions in inferential statistics
• Define the standard error of the mean
Suppose you randomly sampled \(10\) people from the population of women in Houston, Texas, between the ages of \(21\) and \(35\) years and computed the mean height of your sample. You would not expect your sample mean to be equal to the mean of all women in Houston. It might be somewhat lower or it might be somewhat higher, but it would not equal the population mean exactly. Similarly, if you took a second sample of \(10\) people from the same population, you would not expect the mean of this second sample to equal the mean of the first sample.
Recall that inferential statistics concern generalizing from a sample to a population. A critical part of inferential statistics involves determining how far sample statistics are likely to vary from each other and from the population parameter. (In this example, the sample statistics are the sample means and the population parameter is the population mean.) As the later portions of this chapter show, these determinations are based on sampling distributions.
Discrete Distributions
We will illustrate the concept of sampling distributions with a simple example. Figure \(1\) shows three pool balls, each with a number on it. Two of the balls are selected randomly (with replacement) and the average of their numbers is computed.
All possible outcomes are shown below in Table \(1\).
Table \(1\): All possible outcomes when two balls are sampled with replacement
Outcome Ball 1 Ball 2 Mean
1 1 1 1.0
2 1 2 1.5
3 1 3 2.0
4 2 1 1.5
5 2 2 2.0
6 2 3 2.5
7 3 1 2.0
8 3 2 2.5
9 3 3 3.0
Notice that all the means are either \(1.0\), \(1.5\), \(2.0\), \(2.5\), or \(3.0\). The frequencies of these means are shown in Table \(2\). The relative frequencies are equal to the frequencies divided by nine because there are nine possible outcomes.
Table \(2\): Frequencies of means for \(N = 2\)
Mean Frequency Relative Frequency
1.0 1 0.111
1.5 2 0.222
2.0 3 0.333
2.5 2 0.222
3.0 1 0.111
Figure \(2\) shows a relative frequency distribution of the means based on Table \(2\). This distribution is also a probability distribution since the \(Y\)-axis is the probability of obtaining a given mean from a sample of two balls in addition to being the relative frequency.
The distribution shown in Figure \(2\) is called the sampling distribution of the mean. Specifically, it is the sampling distribution of the mean for a sample size of \(2\) (\(N = 2\)). For this simple example, the distribution of pool balls and the sampling distribution are both discrete distributions. The pool balls have only the values \(1\), \(2\), and \(3\), and a sample mean can have one of only five values shown in Table \(2\).
There is an alternative way of conceptualizing a sampling distribution that will be useful for more complex distributions. Imagine that two balls are sampled (with replacement) and the mean of the two balls is computed and recorded. Then this process is repeated for a second sample, a third sample, and eventually thousands of samples. After thousands of samples are taken and the mean computed for each, a relative frequency distribution is drawn. The more samples, the closer the relative frequency distribution will come to the sampling distribution shown in Figure \(2\). As the number of samples approaches infinity, the relative frequency distribution will approach the sampling distribution. This means that you can conceive of a sampling distribution as being a relative frequency distribution based on a very large number of samples. To be strictly correct, the relative frequency distribution approaches the sampling distribution as the number of samples approaches infinity.
Table \(3\): All possible outcomes when two balls are sampled with replacement
Outcome Ball 1 Ball 2 Range
1 1 1 0
2 1 2 1
3 1 3 2
4 2 1 1
5 2 2 0
6 2 3 1
7 3 1 2
8 3 2 1
9 3 3 0
It is important to keep in mind that every statistic, not just the mean, has a sampling distribution. For example, Table \(3\) shows all possible outcomes for the range of two numbers (larger number minus the smaller number). Table \(4\) shows the frequencies for each of the possible ranges and Figure \(3\) shows the sampling distribution of the range.
Table \(4\): Frequencies of ranges for \(N = 2\)
Range Frequency Relative Frequency
0 3 0.333
1 4 0.444
2 2 0.222
It is also important to keep in mind that there is a sampling distribution for various sample sizes. For simplicity, we have been using \(N = 2\). The sampling distribution of the range for \(N = 3\) is shown in Figure \(4\).
Continuous Distributions
In the previous section, the population consisted of three pool balls. Now we will consider sampling distributions when the population distribution is continuous. What if we had a thousand pool balls with numbers ranging from \(0.001\) to \(1.000\) in equal steps? (Although this distribution is not really continuous, it is close enough to be considered continuous for practical purposes.) As before, we are interested in the distribution of means we would get if we sampled two balls and computed the mean of these two balls. In the previous example, we started by computing the mean for each of the nine possible outcomes. This would get a bit tedious for this example since there are \(1,000,000\) possible outcomes (\(1,000\) for the first ball x \(1,000\) for the second). Therefore, it is more convenient to use our second conceptualization of sampling distributions which conceives of sampling distributions in terms of relative frequency distributions. Specifically, the relative frequency distribution that would occur if samples of two balls were repeatedly taken and the mean of each sample computed.
When we have a truly continuous distribution, it is not only impractical but actually impossible to enumerate all possible outcomes. Moreover, in continuous distributions, the probability of obtaining any single value is zero. Therefore, as discussed in the section "Introduction to Distributions," these values are called probability densities rather than probabilities.
Sampling Distributions and Inferential Statistics
As we stated in the beginning of this chapter, sampling distributions are important for inferential statistics. In the examples given so far, a population was specified and the sampling distribution of the mean and the range were determined. In practice, the process proceeds the other way: you collect sample data and from these data you estimate parameters of the sampling distribution. This knowledge of the sampling distribution can be very useful. For example, knowing the degree to which means from different samples would differ from each other and from the population mean would give you a sense of how close your particular sample mean is likely to be to the population mean. Fortunately, this information is directly available from a sampling distribution. The most common measure of how much sample means differ from each other is the standard deviation of the sampling distribution of the mean. This standard deviation is called the standard error of the mean. If all the sample means were very close to the population mean, then the standard error of the mean would be small. On the other hand, if the sample means varied considerably, then the standard error of the mean would be large.
To be specific, assume your sample mean were \(125\) and you estimated that the standard error of the mean were \(5\) (using a method shown in a later section). If you had a normal distribution, then it would be likely that your sample mean would be within \(10\) units of the population mean since most of a normal distribution is within two standard deviations of the mean.
Keep in mind that all statistics have sampling distributions, not just the mean. In later sections we will be discussing the sampling distribution of the variance, the sampling distribution of the difference between means, and the sampling distribution of Pearson's correlation, among others. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/09%3A_Sampling_Distributions/9.01%3A_Introduction_to_Sampling_Distributions.txt |
Learning Objectives
• Develop a basic understanding of the properties of a sampling distribution based on the properties of the population
Instructions
This simulation illustrates the concept of a sampling distribution. Depicted on the top graph is the population from which we are going to sample. There are \(33\) different values in the population: the integers from \(0\) to \(32\) (inclusive). You can think of the population as consisting of having an extremely large number of balls with \(0's\), an extremely large number with \(1's\), etc. on them. The height of the distribution shows the relative number of balls of each number. There is an equal number of balls for each number, so the distribution is a rectangle.
If you push the "animated sampling" button, five balls are selected and and are plotted on the second graph. The mean of this sample of five is then computed and plotted on the third graph. If you push the "animated sampling" button again, another sample of five will be taken, and again plotted on the second graph. The mean will be computed and plotted on the third graph. This third graph is labeled "Distribution of Sample Means, \(N = 5\)" because each value plotted is a sample mean based on a sample of five. At this point, you should have two means plotted in this graph.
The mean is depicted graphically on the distributions themselves by a blue vertical bar below the \(X\)-axis. A red line starts from this mean value and extends one standard deviation in length in both directions. The values of both the mean and the standard deviation are given to the left of the graph. Notice that the numeric form of a property matches its graphical form.
The sampling distribution of a statistic is the relative frequency distribution of that statistic that is approached as the number of samples (not the sample size!) approaches infinity. To approximate a sampling distribution, click the "\(5,000\) samples" button several times. The bottom graph is then a relative frequency distribution of the thousands of means. It is not truly a sampling distribution because it is based on a finite number of samples. Nonetheless, it is a very good approximation.
The simulation has been explained in terms of the sampling distribution of the mean for \(N = 5\). All statistics, not just the mean, have sampling distributions. Moreover, there is a different sampling distribution for each value of \(N\). For the sake of simplicity, this simulation only uses \(N = 5\). Finally, the default is to sample from a distribution for which each value has an equal chance of occurring. Other shapes of the distribution are possible. In this simulation, you can make the population normally distributed as well.
In this simulation, you can specify a sample statistic (the default is mean) and then sample a sufficiently large number of samples until the sampling distribution stabilizes. Make sure you understand the difference between the sample size (which here is \(5\)) and the number of samples included in a distribution.
You should also compare the value of a statistic in the population and the mean of the sampling distribution of that statistic. For some statistics, the mean of the sampling distribution will be very close to the corresponding population parameter; for at least one, there will be a large difference. Also note how the overall shape of sampling distribution differs from that of the population.
Finally, as shown in the video demo, you can change the parent population to a normal distribution from a uniform distribution.
Video Instructions
Video Demo
The video above demonstrates the use of the Sampling distribution Demonstration. The first graph represents the distribution of the population from which the sample will be drawn. In the video this distribution is changed to normal. Each time the "Animated Sample" button is clicked a random sample of five elements is drawn from the population. You can draw multiple samples of \(5\) by clicking on the buttons directly below "Animated Sample". The mean of each of these sample is displayed in the third graph on at the bottom. The graph can also be set to display other descriptive statistics besides the mean.
If you look to the left of this third graph you can see the mean and standard deviation of the sampling distribution. Try drawing \(50,000\) samples from both types of population distributions and compare the sampling distribution statistics to their equivalent population statistics to see if you can discover any trends. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/09%3A_Sampling_Distributions/9.02%3A_Basic_Demo.txt |
Learning Objectives
• Compare the sampling distributions of different sample sizes
Instructions
This simulation demonstrates the effect of sample size on the sampling distribution.
Depicted on the top graph is the population distribution. By default it is a uniform distribution (all values are equally likely). The sampling distributions for two different sample sizes are shown in the lower two graphs. The starting values are \(2\) and \(10\). By default, the statistic to be computed is the mean, although you can also specify to compute the median. For both the population distribution and the sampling distribution, the mean and the standard deviation are depicted graphically on the frequency distribution itself. The blue-colored vertical bar below the \(X\)-axis indicates the mean value. The red line starts from this mean value and extends one standard deviation in length in both directions. The values of both the mean and the standard deviation are also given to the left of the graph. Notice that the numeric form of a property matches its graphical form in color. In this simulation, you specify two sample sizes (the defaults are set at \(N = 2\) and \(N = 10\)), and then sample a sufficiently large number of samples until the sampling distributions stabilize. Compare the mean and standard deviaiton of the two sampling distributions. Repeat the process a couple times and watch the results. Do you observe a general rule regarding the effect of sample size on the mean and the standard deviation of the sampling distribution? You may also test the effect of sample size with a normal population or with a different sample statistic (the median). When you have discovered the rule, go back and answer the questions again.
Illustrated Instructions
Video Demonstration
The video below changes the population distribution from uniform to normal and then draws \(30,000\) samples with \(N = 2\) and \(30,000\) samples with \(N = 10\) by clicking the "\(10,000\) Samples" \(3\) times. Notice the differences in the means and standard deviations of the two sample distributions. How do these compare to the population?
The vertical bar to the right of each sampling distribution can be dragged along the \(x\)-axis and once the mouse is released the area of the curve to the left of the line is displayed above the chart.
For some browsers you will not see the bars move as you move them. They will move when you release the mouse button.
9.04: Central Limit Theorem Demonstration
Learning Objectives
• Develop a basic understanding of the properties of a sampling distribution based on the properties of the population
Instructions
This simulation demonstrates the effect of sample size on the shape of the sampling distribution of the mean.
Depicted on the top graph is the population which is sometimes referred to as the parent distribution. Two sampling distributions of the mean, associated with their respective sample size will be created on the second and third graphs.
For both the population distribution and the sampling distributions, their mean and the standard deviation are depicted graphically on the frequency distribution itself. The blue-colored vertical bar below the \(X\)-axis indicates where the mean value falls. The red line starts from this mean value and extends one standard deviation in length in both directions. The values of both the mean and the standard deviation are also given to the left of the graph. Notice that the numeric form of a property matches its graphical form in color. In addition, the skew and the kurtosis of each distribution are also provided to the left. These two variables are determined by the shape of distribution. The skew and kurtosis for a normal distribution are both \(0\).
In this simulation, you need to first specify a population (the default is uniform distribution). Take note of the skew and kurtosis of the population. Then pick two different sample sizes (the defaults are \(N=2\) and \(N=10\)), and sample a sufficiently large number of samples until the sampling distributions change relatively little with additional samples (about \(50,000\) samples.) Observe the overall shape of the two sampling distributions, and further compare their means, standard deviations, skew and kurtosis. Change the sample sizes and repeat the process a few times. Do you observe a general rule regarding the effect of sample size on the shape of the sampling distribution?
You may also test the effect of sample size with populations of other shape (uniform, skewed or custom ones).
Illustrated Instructions
Central Limit Theorem Video Demo
The video below changes the population distribution to skewed and draws \(100,000\) samples with \(N = 2\) and \(N = 10\) with the "\(10,000\) Samples" button. Note the statistics and shape of the two sample distributions how do these compare to each other and to the population? | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/09%3A_Sampling_Distributions/9.03%3A_Sample_Size_Demo.txt |
Learning Objectives
• State the mean and variance of the sampling distribution of the mean
• Compute the standard error of the mean
• State the central limit theorem
The sampling distribution of the mean was defined in the section introducing sampling distributions. This section reviews some important properties of the sampling distribution of the mean introduced in the demonstrations in this chapter.
Sampling Mean
The mean of the sampling distribution of the mean is the mean of the population from which the scores were sampled. Therefore, if a population has a mean $\mu$, then the mean of the sampling distribution of the mean is also $\mu$. The symbol $\mu _M$ is used to refer to the mean of the sampling distribution of the mean. Therefore, the formula for the mean of the sampling distribution of the mean can be written as:
$\mu _M = \mu$
Sampling Variance
The variance of the sampling distribution of the mean is computed as follows:
$\sigma_M^2 = \dfrac{\sigma^2}{N}$
That is, the variance of the sampling distribution of the mean is the population variance divided by $N$, the sample size (the number of scores used to compute a mean). Thus, the larger the sample size, the smaller the variance of the sampling distribution of the mean.
Note
(optional) This expression can be derived very easily from the variance sum law. Let's begin by computing the variance of the sampling distribution of the sum of three numbers sampled from a population with variance $\sigma ^2$. The variance of the sum would be:
$\sigma ^2 + \sigma ^2 + \sigma ^2$
For $N$ numbers, the variance would be $N\sigma ^2$. Since the mean is $1/N$ times the sum, the variance of the sampling distribution of the mean would be $1/N^2$ times the variance of the sum, which equals $\sigma ^2/N$.
The standard error of the mean is the standard deviation of the sampling distribution of the mean. It is therefore the square root of the variance of the sampling distribution of the mean and can be written as:
$\sigma_M = \dfrac{\sigma}{\sqrt{N}}$
The standard error is represented by a $\sigma$ because it is a standard deviation. The subscript ($M$) indicates that the standard error in question is the standard error of the mean.
Central Limit Theorem
Definition
The central limit theorem states that:
Given a population with a finite mean $\mu$ and a finite non-zero variance $\sigma ^2$, the sampling distribution of the mean approaches a normal distribution with a mean of $\mu$ and a variance of $\sigma ^2/N$ as $N$, the sample size, increases.
The expressions for the mean and variance of the sampling distribution of the mean are not new or remarkable. What is remarkable is that regardless of the shape of the parent population, the sampling distribution of the mean approaches a normal distribution as $N$ increases. If you have used the "Central Limit Theorem Demo," you have already seen this for yourself. As a reminder, Figure $1$ shows the results of the simulation for $N = 2$ and $N = 10$. The parent population was a uniform distribution. You can see that the distribution for $N = 2$ is far from a normal distribution. Nonetheless, it does show that the scores are denser in the middle than in the tails. For $N = 10$ the distribution is quite close to a normal distribution. Notice that the means of the two distributions are the same, but that the spread of the distribution for $N = 10$ is smaller.
Figure $2$ shows how closely the sampling distribution of the mean approximates a normal distribution even when the parent population is very non-normal. If you look closely you can see that the sampling distributions do have a slight positive skew. The larger the sample size, the closer the sampling distribution of the mean would be to a normal distribution. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/09%3A_Sampling_Distributions/9.05%3A_Sampling_Distribution_of_the_Mean.txt |
Learning Objectives
• State the mean and variance of the sampling distribution of the difference between means
• Compute the standard error of the difference between means
• Compute the probability of a difference between means being above a specified value
Statistical analyses are very often concerned with the difference between means. A typical example is an experiment designed to compare the mean of a control group with the mean of an experimental group. Inferential statistics used in the analysis of this type of experiment depend on the sampling distribution of the difference between means.
The sampling distribution of the difference between means can be thought of as the distribution that would result if we repeated the following three steps over and over again:
1. sample $n_1$ scores from $\text {Population 1}$ and $n_2$ scores from $\text {Population 2}$
2. compute the means of the two samples ($M_1$and $M_2$)
3. compute the difference between means, $M_1-M_2$
The distribution of the differences between means is the sampling distribution of the difference between means.
As you might expect, the mean of the sampling distribution of the difference between means is:
$\mu _{M_1-M_2}=\mu _1-\mu _2$
which says that the mean of the distribution of differences between sample means is equal to the difference between population means. For example, say that the mean test score of all $12$-year-olds in a population is $34$ and the mean of $10$-year-olds is $25$. If numerous samples were taken from each age group and the mean difference computed each time, the mean of these numerous differences between sample means would be $34-25=9$.
From the variance sum law, we know that:
$\sigma _{M_1-M_2}^{2}=\sigma _{M_1}^{2}+\sigma _{M_2}^{2}$
which says that the variance of the sampling distribution of the difference between means is equal to the variance of the sampling distribution of the mean for $\text {Population 1}$ plus the variance of the sampling distribution of the mean for $\text {Population 2}$. Recall the formula for the variance of the sampling distribution of the mean:
$\sigma _{M}^{2}=\frac{\sigma ^2}{N}$
Since we have two populations and two samples sizes, we need to distinguish between the two variances and sample sizes. We do this by using the subscripts $1$ and $2$. Using this convention, we can write the formula for the variance of the sampling distribution of the difference between means as:
$\sigma _{M_1-M_2}^{2}=\frac{\sigma _1^2}{n_1}+\frac{\sigma _2^2}{n_2}$
Since the standard error of a sampling distribution is the standard deviation of the sampling distribution, the standard error of the difference between means is:
$\sigma _{M_1-M_2}=\sqrt{\frac{\sigma _1^2}{n_1}+\frac{\sigma _2^2}{n_2}}$
Just to review the notation, the symbol on the left contains a sigma ($\sigma$), which means it is a standard deviation. The subscripts $M_1-M_2$ indicate that it is the standard deviation of the sampling distribution of $M_1-M_2$.
Now let's look at an application of this formula.
Example $1$
Assume there are two species of green beings on Mars. The mean height of $\text{Species 1}$ is $32$ while the mean height of $\text{Species 2}$ is $22$. The variances of the two species are $60$ and $70$, respectively and the heights of both species are normally distributed. You randomly sample $10$ members of $\text{Species 1}$ and $14$ members of $\text{Species 2}$. What is the probability that the mean of the $10$ members of $\text{Species 1}$ will exceed the mean of the $14$ members of $\text{Species 2}$ by $5$ or more? Without doing any calculations, you probably know that the probability is pretty high since the difference in population means is $10$. But what exactly is the probability?
Solution
First, let's determine the sampling distribution of the difference between means. Using the formulas above, the mean is
$\mu _{M_1-M_2}=32-22=10$
The standard error is:
$\sigma _{M_1-M_2}=\sqrt{\frac{60}{10}+\frac{70}{14}}=3.317$
The sampling distribution is shown in Figure $1$. Notice that it is normally distributed with a mean of $10$ and a standard deviation of $3.317$. The area above $5$ is shaded blue.
The last step is to determine the area that is shaded blue. Using either a $Z$ table or the normal calculator, the area can be determined to be $0.934$. Thus the probability that the mean of the sample from $\text{Species 1}$ will exceed the mean of the sample from $\text{Species 2}$ by $5$ or more is $0.934$.
As shown below, the formula for the standard error of the difference between means is much simpler if the sample sizes and the population variances are equal. When the variances and samples sizes are the same, there is no need to use the subscripts $1$ and $2$ to differentiate these terms.
$\sigma _{M_1-M_2}=\sqrt{\frac{\sigma _1^2}{n_1}+\frac{\sigma _2^2}{n_2}}=\sqrt{\frac{\sigma ^2}{n}+\frac{\sigma ^2}{n}}=\sqrt{\frac{2\sigma ^2}{n}}$
This simplified version of the formula can be used for the following problem.
Example $2$
The mean height of $15$-year-old boys (in cm) is $175$ and the variance is $64$. For girls, the mean is $165$ and the variance is $64$. If eight boys and eight girls were sampled, what is the probability that the mean height of the sample of girls would be higher than the mean height of the sample of boys? In other words, what is the probability that the mean height of girls minus the mean height of boys is greater than $0$?
Solution
As before, the problem can be solved in terms of the sampling distribution of the difference between means (girls - boys). The mean of the distribution is 165 - 175 = -10. The standard deviation of the distribution is:
$\sigma _{M_1-M_2}=\sqrt{\frac{2\sigma ^2}{n}}=\sqrt{\frac{(2)(64)}{8}}=4$
A graph of the distribution is shown in Figure $2$. It is clear that it is unlikely that the mean height for girls would be higher than the mean height for boys since in the population boys are quite a bit taller. Nonetheless it is not inconceivable that the girls' mean could be higher than the boys' mean.
A difference between means of 0 or higher is a difference of $10/4 = 2.5$ standard deviations above the mean of $-10$. The probability of a score $2.5$ or more standard deviations above the mean is $0.0062$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/09%3A_Sampling_Distributions/9.06%3A_Difference_Between_Means.txt |
Learning Objectives
• Transform $r$ to $z'$
• Calculate the probability of obtaining an $r$ above a specified value
Assume that the correlation between quantitative and verbal SAT scores in a given population is $0.60$. In other words, $\rho =0.60$. If $12$ students were sampled randomly, the sample correlation, $r$, would not be exactly equal to $0.60$. Naturally different samples of $12$ students would yield different values of $r$. The distribution of values of $r$ after repeated samples of $12$ students is the sampling distribution of $r$.
The shape of the sampling distribution of $r$ for the above example is shown in Figure $1$. You can see that the sampling distribution is not symmetric: it is negatively skewed. The reason for the skew is that $r$ cannot take on values greater than $1.0$ and therefore the distribution cannot extend as far in the positive direction as it can in the negative direction. The greater the value of $\rho$, the more pronounced the skew.
Figure $2$ shows the sampling distribution for $\rho =0.90$. This distribution has a very short positive tail and a long negative tail.
Referring back to the SAT example, suppose you wanted to know the probability that in a sample of $12$ students, the sample value of $r$ would be $0.75$ or higher. You might think that all you would need to know to compute this probability is the mean and standard error of the sampling distribution of $r$. However, since the sampling distribution is not normal, you would still not be able to solve the problem. Fortunately, the statistician Fisher developed a way to transform $r$ to a variable that is normally distributed with a known standard error. The variable is called $z'$ and the formula for the transformation is given below.
$z' = 0.5\: \ln \frac{1+r}{1-r}$
The details of the formula are not important here since normally you will use either a table, Pearson r to Fisher z' or calculator to do the transformation. What is important is that $z'$ is normally distributed and has a standard error of
$\frac{1}{\sqrt{N-3}}$
where $N$ is the number of pairs of scores.
Example $1$
Let's return to the question of determining the probability of getting a sample correlation of $0.75$ or above in a sample of $12$ from a population with a correlation of $0.60$.
Solution
The first step is to convert both $0.60$ and $0.75$ to their $z'$ values, which are $0.693$ and $0.973$, respectively.
The standard error of $z'$ for $N = 12$ is $0.333$.
Therefore the question is reduced to the following: given a normal distribution with a mean of $0.693$ and a standard deviation of $0.333$, what is the probability of obtaining a value of $0.973$ or higher?
The answer can be found directly from the applet "Calculate Area for a given X" to be $0.20$.
Alternatively, you could use the formula:
$z = \frac{X - \mu }{\sigma } = \frac{0.973 - 0.693}{0.333} = 0.841$
and use a table to find that the area above $0.841$ is $0.20$.
9.08: Sampling Distribution of p
Learning Objectives
• State the relationship between the sampling distribution of p and the normal distribution
Assume that in an election race between $\text{Candidate A}$ and $\text{Candidate B}$, $0.60$ of the voters prefer $\text{Candidate A}$. If a random sample of $10$ voters were polled, it is unlikely that exactly $60\%$ of them ($6$) would prefer $\text{Candidate A}$. By chance the proportion in the sample preferring $\text{Candidate A}$ could easily be a little lower than $0.60$ or a little higher than $0.60$. The sampling distribution of $p$ is the distribution that would result if you repeatedly sampled $10$ voters and determined the proportion ($p$) that favored $\text{Candidate A}$.
The sampling distribution of $p$ is a special case of the sampling distribution of the mean. Table $1$ shows a hypothetical random sample of $10$ voters. Those who prefer $\text{Candidate A}$ are given scores of $1$ and those who prefer $\text{Candidate B}$ are given scores of $0$. Note that seven of the voters prefer $\text{Candidate A}$ so the sample proportion ($p$) is
$p = \frac{7}{10} = 0.70$
As you can see, $p$ is the mean of the $10$ preference scores.
Table $1$: Sample of voters
Voter Preference
1 1
2 0
3 1
4 1
5 1
6 0
7 1
8 0
9 1
10 1
The distribution of $p$ is closely related to the binomial distribution. The binomial distribution is the distribution of the total number of successes (favoring $\text{Candidate A}$, for example) whereas the distribution of $p$ is the distribution of the mean number of successes. The mean, of course, is the total divided by the sample size, $N$. Therefore, the sampling distribution of $p$ and the binomial distribution differ in that $p$ is the mean of the scores ($0.70$) and the binomial distribution is dealing with the total number of successes ($7$).
The binomial distribution has a mean of
$\mu =N\pi$
Dividing by $N$ to adjust for the fact that the sampling distribution of $p$ is dealing with means instead of totals, we find that the mean of the sampling distribution of $p$ is:
$\mu _p=\pi$
The standard deviation of the binomial distribution is:
$\sqrt{N\pi(1-\pi )}$
Dividing by $N$ because $p$ is a mean not a total, we find the standard error of $p$:
$\sigma _p=\frac{\sqrt{N\pi(1-\pi )}}{N}=\sqrt{\frac{\pi(1-\pi )}{N}}$
Returning to the voter example, $\pi =0.60$ and $N = 10$. (Don't confuse $\pi =0.60$, the population proportion and $p = 0.70$, the sample proportion.) Therefore, the mean of the sampling distribution of $p$ is $0.60$. The standard error is
$\sigma _p=\sqrt{\frac{0.60(1-0.60)}{10}}=0.155$
The sampling distribution of $p$ is a discrete rather than a continuous distribution. For example, with an $N$ of $10$, it is possible to have a $p$ of $0.50$ or a $p$ of $0.60$ but not a $p$ of $0.55$.
The sampling distribution of $p$ is approximately normally distributed if $N$ is fairly large and $\pi$ is not close to $0$ or $1$. A rule of thumb is that the approximation is good if both $N\pi$ and $N(1-\pi )$ are greater than $10$. The sampling distribution for the voter example is shown in Figure $1$. Note that even though $N(1-\pi )$ is only $4$, the approximation is quite good.
9.09: Statistical Literacy
Learning Objectives
• Accuracy of Employment Figures
The monthly jobs report always gets a lot of attention. Presidential candidates refer to the report when it favors their position. Referring to the August \(2012\) report in which only \(96,000\) jobs were created, Republican presidential challenger Mitt Romney stated "the weak jobs report is devastating news for American workers and American families ... a harsh indictment of the president's handling of the economy." When the September \(2012\) report was released showing \(114,000\) jobs were created (and the previous report was revised upwards), some supporters of Romney claimed the data were tampered with for political reasons. The most famous statement, "Unbelievable jobs numbers...these Chicago guys will do anything..can't debate so change numbers," was made by former Chairman and CEO of General Electric.
Example \(1\): what do you think?
The standard error of the monthly estimate is \(55,000\). Given that, what do you think of the difference between the two job reports?
Solution
The difference between the two reports is very small given that the standard error is \(55,000\). It is not sensible to take any single jobs report too seriously. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/09%3A_Sampling_Distributions/9.07%3A_Sampling_Distribution_of_Pearson%27s_r.txt |
You may want to use the "r to z' calculator" and the "Calculate Area for a given X" applet for some of these exercises.
General Questions
Q1
A population has a mean of $50$ and a standard deviation of $6$.
1. What are the mean and standard deviation of the sampling distribution of the mean for $N = 16$?
2. What are the mean and standard deviation of the sampling distribution of the mean for $N = 20$? (relevant section)
Q2
Given a test that is normally distributed with a mean of $100$ and a standard deviation of $12$, find:
1. the probability that a single score drawn at random will be greater than $110$ ( relevant section)
2. the probability that a sample of $25$ scores will have a mean greater than $105$ (relevant section)
3. the probability that a sample of $64$ scores will have a mean greater than $105$ (relevant section)
4. the probability that the mean of a sample of $16$ scores will be either less than $95$ or greater than $105$ (relevant section)
Q3
What term refers to the standard deviation of the sampling distribution? (relevant section)
Q4
1. If the standard error of the mean is $10$ for $N = 12$, what is the standard error of the mean for $N = 22$?
2. If the standard error of the mean is $50$ for $N = 25$, what is it for $N = 64$? (relevant section)
Q5
A questionnaire is developed to assess women's and men's attitudes toward using animals in research. One question asks whether animal research is wrong and is answered on a $7$-point scale. Assume that in the population, the mean for women is $5$, the mean for men is $4$, and the standard deviation for both groups is $1.5$. Assume the scores are normally distributed. If $12$ women and $12$ men are selected randomly, what is the probability that the mean of the women will be more than $1.5$ points higher than the mean of the men? (relevant section)
Q6
If the correlation between reading achievement and math achievement in the population of fifth graders were $0.60$, what would be the probability that in a sample of $28$ students, the sample correlation coefficient would be greater than $0.65$? (relevant section)
Q7
If numerous samples of $N = 15$ are taken from a uniform distribution and a relative frequency distribution of the means is drawn, what would be the shape of the frequency distribution? (relevant section & relevant section)
Q8
A normal distribution has a mean of $20$ and a standard deviation of $10$. Two scores are sampled randomly from the distribution and the second score is subtracted from the first. What is the probability that the difference score will be greater than $5$? Hint: Read the Variance Sum Law section of Chapter 3. (relevant section & relevant section)
Q9
What is the shape of the sampling distribution of $r$? In what way does the shape depend on the size of the population correlation? (relevant section)
Q10
If you sample one number from a standard normal distribution, what is the probability it will be $0.5$? (relevant section & relevant section)
Q11
A variable is normally distributed with a mean of $120$ and a standard deviation of $5$. Four scores are randomly sampled. What is the probability that the mean of the four scores is above $127$? (relevant section)
Q12
The correlation between self esteem and extraversion is $0.30$. A sample of $84$ is taken.
1. What is the probability that the correlation will be less than $0.10$?
2. What is the probability that the correlation will be greater than $0.25$? (relevant section)
Q13
The mean GPA for students in $\text{School A}$ is $3.0$; the mean GPA for students in $\text{School B}$ is $2.8$. The standard deviation in both schools is $0.25$. The GPAs of both schools are normally distributed. If $9$ students are randomly sampled from each school, what is the probability that:
1. the sample mean for $\text{School A}$ will exceed that of $\text{School A}$ by $0.5$ or more? (relevant section)
2. the sample mean for $\text{School B}$ will be greater than the sample mean for $\text{School A}$? (relevant section)
Q14
In a city, $70\%$ of the people prefer $\text{Candidate A}$. Suppose $30$ people from this city were sampled.
1. What is the mean of the sampling distribution of $p$?
2. What is the standard error of $p$?
3. What is the probability that $80\%$ or more of this sample will prefer $\text{Candidate A}$?
4. What is the probability that $45\%$ or more of this sample will prefer some other candidate? (relevant section)
Q15
When solving problems where you need the sampling distribution of $r$, what is the reason for converting from $r$ to $z'$? (relevant section)
Q16
In the population, the mean SAT score is $1000$. Would you be more likely (or equally likely) to get a sample mean of $1200$ if you randomly sampled $10$ students or if you randomly sampled $30$ students? Explain. (relevant section & relevant section)
Q17
True/false: The standard error of the mean is smaller when $N = 20$ than when $N = 10$. (relevant section)
Q18
True/false: The sampling distribution of $r =0.8$ becomes normal as $N$ increases. (relevant section)
Q19
True/false: You choose $20$ students from the population and calculate the mean of their test scores. You repeat this process $100$ times and plot the distribution of the means. In this case, the sample size is $100$. (relevant section & relevant section)
Q20
True/false: In your school, $40\%$ of students watch TV at night. You randomly ask $5$ students every day if they watch TV at night. Every day, you would find that $2$ of the $5$ do watch TV at night. (relevant section & relevant section)
Q21
True/false: The median has a sampling distribution. (relevant section)
Q22
True/false: Refer to the figure below. The population distribution is shown in black, and its corresponding sampling distribution of the mean for $N = 10$ is labeled "$A$" (relevant section & relevant section)
Questions from Case Studies
The following questions use data from the Angry Moods (AM) case study.
Q23
1. How many men were sampled?
2. How many women were sampled?
Q24
What is the mean difference between men and women on the Anger-Out scores?
Q25
Suppose in the population, the Anger-Out score for men is two points higher than it is for women. The population variances for men and women are both $20$. Assume the Anger-Out scores for both genders are normally distributed. Given this information about the population parameters:
1. What is the mean of the sampling distribution of the difference between means? (relevant section)
2. What is the standard error of the difference between means? (relevant section)
3. What is the probability that you would have gotten this mean difference (see #24) or less in your sample? (relevant section)
The following questions use data from the Animal Research (AR) case study.
Q26
How many people were sampled to give their opinions on animal research?
Q27
(AR#11) What is the correlation in this sample between the belief that animal research is wrong and belief that animal research is necessary? (Ch. 4.E)
Q28
Suppose the correlation between the belief that animal research is wrong and the belief that animal research is necessary is $-0.68$ in the population.
1. Convert $-0.68$ to $z'$. (relevant section)
2. Find the standard error of this sampling distribution. (relevant section)
3. In a new sample, what is the probability that you would get the correlation found in the original sample (see #27) or a lower correlation (closer to $-1$)? (relevant section)
Selected Answers
S1
1. Mean = $50$, SD = $1.5$
S2
1. $0.019$
S4
1. $7.39$
S11
$0.0026$
S12
1. $0.690$
S13
1. $0.0055$
S14
1. $0.116$
S23
1. $30$
S25
1. $2$
S28
1. $0.603$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/09%3A_Sampling_Distributions/9.E%3A_Sampling_Distributions_%28Exercises%29.txt |
One of the major applications of statistics is estimating population parameters from sample statistics. For example, a poll may seek to estimate the proportion of adult residents of a city that support a proposition to build a new sports stadium. Out of a random sample of 200 people, 106 say they support the proposition. Thus in the sample, 0.53 of the people supported the proposition. This value of 0.53 is called a point estimate of the population proportion. It is called a point estimate because the estimate consists of a single value or point
• 10.1: Introduction to Estimation
One of the major applications of statistics is estimating population parameters from sample statistics. For example, a poll may seek to estimate the proportion of adult residents of a city that support a proposition to build a new sports stadium. Out of a random sample of 200 people, 106 say they support the proposition. Thus in the sample, 0.53 of the people supported the proposition. This value of 0.53 is called a point estimate of the population proportion. This is called a point estimate.
• 10.2: Degrees of Freedom
Some estimates are based on more information than others. For example, an estimate of the variance based on a sample size of 100 is based on more information than an estimate of the variance based on a sample size of 5. The degrees of freedom (df) of an estimate is the number of independent pieces of information on which the estimate is based.
• 10.3: Characteristics of Estimators
This section discusses two important characteristics of statistics used as point estimates of parameters: bias and sampling variability. Bias refers to whether an estimator tends to either over or underestimate the parameter. Sampling variability refers to how much the estimate varies from sample to sample.
• 10.4: Bias and Variability Simulation
This simulation lets you explore various aspects of sampling distributions. When it begins, a histogram of a normal distribution is displayed at the topic of the screen.
• 10.5: Confidence Intervals
• 10.6: Confidence Intervals Intro
Confidence intervals provide more information than point estimates. Confidence intervals for means are intervals constructed using a procedure that will contain the population mean a specified proportion of the time, typically either 95% or 99% of the time. These intervals are referred to as 95% and 99% confidence intervals respectively.
• 10.7: Confidence Interval for Mean
When you compute a confidence interval on the mean, you compute the mean of a sample in order to estimate the mean of the population. Clearly, if you already knew the population mean, there would be no need for a confidence interval. However, to explain how confidence intervals are constructed, we are going to work backwards and begin by assuming characteristics of the population. Then we will show how sample data can be used to construct a confidence interval.
• 10.8: t Distribution
The tt distribution is very similar to the normal distribution when the estimate of variance is based on many degrees of freedom, but has relatively more scores in its tails when there are fewer degrees of freedom. The tt distribution is therefore leptokurtic. The t distribution approaches the normal distribution as the degrees of freedom increase.
• 10.9: Confidence Interval Simulation
• 10.10: Difference between Means
It is much more common for a researcher to be interested in the difference between means than in the specific values of the means themselves.
• 10.11: Correlation
• 10.12: Proportion
• 10.13: Statistical Literacy
• 10.E: Estimation (Exercises)
10: Estimation
Learning Objectives
• Define statistic
• Define parameter
• Define point estimate
• Define interval estimate
• Define margin of error
One of the major applications of statistics is estimating population parameters from sample statistics. For example, a poll may seek to estimate the proportion of adult residents of a city that support a proposition to build a new sports stadium. Out of a random sample of $200$ people, $106$ say they support the proposition. Thus in the sample, $0.53$ of the people supported the proposition. This value of $0.53$ is called a point estimate of the population proportion. It is called a point estimate because the estimate consists of a single value or point.
Point estimates are usually supplemented by interval estimates called confidence intervals. Confidence intervals are intervals constructed using a method that contains the population parameter a specified proportion of the time. For example, if the pollster used a method that contains the parameter $95\%$ of the time it is used, he or she would arrive at the following $95\%$ confidence interval: $0.46 < \pi < 0.60$. The pollster would then conclude that somewhere between $0.46$ and $0.60$ of the population supports the proposal. The media usually reports this type of result by saying that $53\%$ favor the proposition with a margin of error of $7\%$.
In an experiment on memory for chess positions, the mean recall for tournament players was $63.8$ and the mean for non-players was $33.1$. Therefore a point estimate of the difference between population means is $30.7$. The $95\%$ confidence interval on the difference between means extends from $19.05$ to $42.35$. You will see how to compute this kind of interval in another section. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/10%3A_Estimation/10.01%3A_Introduction_to_Estimation.txt |
Learning Objectives
• Define degrees of freedom
• Estimate the variance from a sample of $1$ if the population mean is known
• State why deviations from the sample mean are not independent
• State the general formula for degrees of freedom in terms of the number of values and the number of estimated parameters
• Calculate $s^2$
Some estimates are based on more information than others. For example, an estimate of the variance based on a sample size of $100$ is based on more information than an estimate of the variance based on a sample size of $5$. The degrees of freedom ($df$) of an estimate is the number of independent pieces of information on which the estimate is based.
As an example, let's say that we know that the mean height of Martians is $6$ and wish to estimate the variance of their heights. We randomly sample one Martian and find that its height is $8$. Recall that the variance is defined as the mean squared deviation of the values from their population mean. We can compute the squared deviation of our value of $8$ from the population mean of $6$ to find a single squared deviation from the mean. This single squared deviation from the mean, $(8-6)^2 = 4$, is an estimate of the mean squared deviation for all Martians. Therefore, based on this sample of one, we would estimate that the population variance is $4$. This estimate is based on a single piece of information and therefore has $1\; df$. If we sampled another Martian and obtained a height of $5$, then we could compute a second estimate of the variance, $(5-6)^2 = 1$. We could then average our two estimates ($4$ and $1$) to obtain an estimate of $2.5$. Since this estimate is based on two independent pieces of information, it has two degrees of freedom. The two estimates are independent because they are based on two independently and randomly selected Martians. The estimates would not be independent if after sampling one Martian, we decided to choose its brother as our second Martian.
As you are probably thinking, it is pretty rare that we know the population mean when we are estimating the variance. Instead, we have to first estimate the population mean ($\mu$) with the sample mean ($M$). The process of estimating the mean affects our degrees of freedom as shown below.
Returning to our problem of estimating the variance in Martian heights, let's assume we do not know the population mean and therefore we have to estimate it from the sample. We have sampled two Martians and found that their heights are $8$ and $5$. Therefore $M$, our estimate of the population mean, is
$M = \frac{(8+5)}{2} = 6.5$
We can now compute two estimates of variance:
• Estimate $1 = (8-6.5)^2 = 2.25$
• Estimate $2 = (5-6.5)^2 = 2.25$
Now for the key question: Are these two estimates independent? The answer is no because each height contributed to the calculation of $M$. Since the first Martian's height of $8$ influenced $M$, it also influenced Estimate $2$. If the first height had been, for example, $10$, then $M$ would have been $7.5$ and Estimate $2$ would have been $(5-7.5)^2 = 6.25$ instead of $2.25$. The important point is that the two estimates are not independent and therefore we do not have two degrees of freedom. Another way to think about the non-independence is to consider that if you knew the mean and one of the scores, you would know the other score. For example, if one score is $5$ and the mean is $6.5$, you can compute that the total of the two scores is $13$ and therefore that the other score must be $13-5 = 8$.
In general, the degrees of freedom for an estimate is equal to the number of values minus the number of parameters estimated en route to the estimate in question. In the Martians example, there are two values ($8$ and $5$) and we had to estimate one parameter ($\mu$) on the way to estimating the parameter of interest ($\sigma ^2$). Therefore, the estimate of variance has $2 - 1 = 1$ degree of freedom. If we had sampled $12$ Martians, then our estimate of variance would have had $11$ degrees of freedom. Therefore, the degrees of freedom of an estimate of variance is equal to $N - 1$, where $N$ is the number of observations.
Recall from the section on variability that the formula for estimating the variance in a sample is:
$s^2 =\dfrac{\sum (X-M)^2}{N-1}$
The denominator of this formula is the degrees of freedom. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/10%3A_Estimation/10.02%3A_Degrees_of_Freedom.txt |
Learning Objectives
1. Define bias
2. Define sampling variability
3. Define expected value
4. Define relative efficiency
This section discusses two important characteristics of statistics used as point estimates of parameters: bias and sampling variability. Bias refers to whether an estimator tends to either over or underestimate the parameter. Sampling variability refers to how much the estimate varies from sample to sample.
Have you ever noticed that some bathroom scales give you very different weights each time you weigh yourself? With this in mind, let's compare two scales. Scale $1$ is a very high-tech digital scale and gives essentially the same weight each time you weigh yourself; it varies by at most $0.02$ pounds from weighing to weighing. Although this scale has the potential to be very accurate, it is calibrated incorrectly and, on average, overstates your weight by one pound. Scale $2$ is a cheap scale and gives very different results from weighing to weighing. However, it is just as likely to underestimate as overestimate your weight. Sometimes it vastly overestimates it and sometimes it vastly underestimates it. However, the average of a large number of measurements would be your actual weight. Scale $1$ is biased since, on average, its measurements are one pound higher than your actual weight. Scale $2$, by contrast, gives unbiased estimates of your weight. However, Scale $2$ is highly variable and its measurements are often very far from your true weight. Scale $1$, in spite of being biased, is fairly accurate. Its measurements are never more than $1.02$ pounds from your actual weight.
We now turn to more formal definitions of variability and precision. However, the basic ideas are the same as in the bathroom scale example.
Bias
A statistic is biased if the long-term average value of the statistic is not the parameter it is estimating. More formally, a statistic is biased if the mean of the sampling distribution of the statistic is not equal to the parameter. The mean of the sampling distribution of a statistic is sometimes referred to as the expected value of the statistic.
As we saw in the section on the sampling distribution of the mean, the mean of the sampling distribution of the (sample) mean is the population mean ($μ$). Therefore the sample mean is an unbiased estimate of $μ$. Any given sample mean may underestimate or overestimate $\mu$, but there is no systematic tendency for sample means to either under or overestimate $μ$.
In the section on variability, we saw that the formula for the variance in a population is
$\sigma^2 = \dfrac{\displaystyle \sum (X-\mu)^2}{N}$
whereas the formula to estimate the variance from a sample is
$s^2 = \dfrac{\displaystyle \sum (X-M)^2}{N-1}$
Notice that the denominators of the formulas are different: $N$ for the population and $N-1$ for the sample. We saw in the "Estimating Variance Simulation" that if $N$ is used in the formula for $s^2$, then the estimates tend to be too low and therefore biased. The formula with $N-1$ in the denominator gives an unbiased estimate of the population variance. Note that $N-1$ is the degrees of freedom.
Sampling Variability
The sampling variability of a statistic refers to how much the statistic varies from sample to sample and is usually measured by its standard error ; the smaller the standard error, the less the sampling variability. For example, the standard error of the mean is a measure of the sampling variability of the mean. Recall that the formula for the standard error of the mean is
$\sigma_M = \dfrac{\sigma}{\sqrt{N}}$
The larger the sample size ($N$), the smaller the standard error of the mean and therefore the lower the sampling variability.
Statistics differ in their sampling variability even with the same sample size. For example, for normal distributions, the standard error of the median is larger than the standard error of the mean. The smaller the standard error of a statistic, the more efficient the statistic. The relative efficiency of two statistics is typically defined as the ratio of their standard errors. However, it is sometimes defined as the ratio of their squared standard errors. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/10%3A_Estimation/10.03%3A_Characteristics_of_Estimators.txt |
Learning Objectives
• To explore various aspects of sampling distributions
This simulation lets you explore various aspects of sampling distributions. When it begins, a histogram of a normal distribution is displayed at the topic of the screen.
The distribution portrayed at the top of the screen is the population from which samples are taken. The mean of the distribution is indicated by a small blue line and the median is indicated by a small purple line. Since the mean and median are the same, the two lines overlap. The red line extends from the mean one standard deviation in each direction. Note the correspondence between the colors used on the histogram and the statistics displayed to the left of the histogram.
The second histogram displays the sample data. This histogram is initially blank. The third and fourth histograms show the distribution of statistics computed from the sample data. The number of samples (replications) that the third and fourth histograms are based on is indicated by the label "Reps=."
Basic operations
The simulation is set to initially sample five numbers from the population, compute the mean of the five numbers, and plot the mean. Click the "Animated sample" button and you will see the five numbers appear in the histogram. The mean of the five numbers will be computed and the mean will be plotted in the third histogram. Do this several times to see the distribution of means begin to be formed. Once you see how this works, you can speed things up by taking \(5\), \(1,000\), or \(10,000\) samples at a time.
Choosing a statistic
The following statistics can be computed from the samples by choosing form the pop-up menu:
1. Mean: Mean
2. SD: Standard deviation of the sample (\(N\) is used in the denominator)
3. Variance: Variance of the sample (\(N\) is used in the denominator)
4. Variance (\(U\)): Unbiased estimate of variance (\(N-1\) is used in denominator)
5. MAD: Mean absolute value of the deviation from the mean
6. Range: Range
Selecting a sample size
The size of each sample can be set to \(2, 5, 10, 16, 20\; or\; 25\) from the pop-up menu. Be sure not to confuse sample size with number of samples.
Comparison to a normal distribution
By clicking the "Fit normal" button you can see a normal distribution superimposed over the simulated sampling distribution.
Changing the population distribution
You can change the population by clicking on the top histogram with the mouse and dragging
10.05: Confidence Intervals
Learning Objectives
• To learn how to compute confidence intervals for various parameters
The next few sections show how to compute confidence intervals for a variety of parameters listed below.
• Confidence Interval for the Mean
• t distribution
• Confidence Interval Simulation
• Confidence Interval for the Difference Between Means
• Confidence Interval for Pearson's Correlation
• Confidence Interval for a Proportion
The "Confidence Interval Simulation" allows you to explore various properties of confidence intervals.
10.06: Confidence Intervals Intro
Learning Objectives
• Define confidence interval
• State why a confidence interval is not the probability the interval contains the parameter
Say you were interested in the mean weight of $10$-year-old girls living in the United States. Since it would have been impractical to weigh all the $10$-year-old girls in the United States, you took a sample of $16$ and found that the mean weight was $90$ pounds. This sample mean of $90$ is a point estimate of the population mean. A point estimate by itself is of limited usefulness because it does not reveal the uncertainty associated with the estimate; you do not have a good sense of how far this sample mean may be from the population mean. For example, can you be confident that the population mean is within $5$ pounds of $90$? You simply do not know.
Confidence intervals provide more information than point estimates. Confidence intervals for means are intervals constructed using a procedure (presented in the next section) that will contain the population mean a specified proportion of the time, typically either $95\%$ or $99\%$ of the time. These intervals are referred to as $95\%$ and $99\%$ confidence intervals respectively. An example of a $95\%$ confidence interval is shown below:
$72.85 < \mu < 107.15$
There is good reason to believe that the population mean lies between these two bounds of $72.85$ and $107.15$ since $95\%$ of the time confidence intervals contain the true mean.
If repeated samples were taken and the $95\%$ confidence interval computed for each sample, $95\%$ of the intervals would contain the population mean. Naturally, $5\%$ of the intervals would not contain the population mean.
It is natural to interpret a $95\%$ confidence interval as an interval with a $0.95$ probability of containing the population mean. However, the proper interpretation is not that simple. One problem is that the computation of a confidence interval does not take into account any other information you might have about the value of the population mean. For example, if numerous prior studies had all found sample means above $110$, it would not make sense to conclude that there is a $0.95$ probability that the population mean is between $72.85$ and $107.15$. What about situations in which there is no prior information about the value of the population mean? Even here the interpretation is complex. The problem is that there can be more than one procedure that produces intervals that contain the population parameter $95\%$ of the time. Which procedure produces the "true" $95\%$ confidence interval? Although the various methods are equal from a purely mathematical point of view, the standard method of computing confidence intervals has two desirable properties: each interval is symmetric about the point estimate and each interval is contiguous. Recall from the introductory section in the chapter on probability that, for some purposes, probability is best thought of as subjective. It is reasonable, although not required by the laws of probability, that one adopt a subjective probability of $0.95$ that a $95\%$ confidence interval, as typically computed, contains the parameter in question.
Confidence intervals can be computed for various parameters, not just the mean. For example, later in this chapter you will see how to compute a confidence interval for $\rho$, the population value of Pearson's $r$, based on sample data. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/10%3A_Estimation/10.04%3A_Bias_and_Variability_Simulation.txt |
Learning Objectives
• Use the inverse normal distribution calculator to find the value of $z$ to use for a confidence interval
• Compute a confidence interval on the mean when $\sigma$ is known
• Determine whether to use a $t$ distribution or a normal distribution
• Compute a confidence interval on the mean when $\sigma$ is estimated
When you compute a confidence interval on the mean, you compute the mean of a sample in order to estimate the mean of the population. Clearly, if you already knew the population mean, there would be no need for a confidence interval. However, to explain how confidence intervals are constructed, we are going to work backwards and begin by assuming characteristics of the population. Then we will show how sample data can be used to construct a confidence interval.
Assume that the weights of $10$-year-old children are normally distributed with a mean of $90$ and a standard deviation of $36$. What is the sampling distribution of the mean for a sample size of $9$? Recall from the section on the sampling distribution of the mean that the mean of the sampling distribution is $\mu$ and the standard error of the mean is
$\sigma _M=\frac{\sigma }{\sqrt{N}}$
For the present example, the sampling distribution of the mean has a mean of $90$ and a standard deviation of $36/3 = 12$. Note that the standard deviation of a sampling distribution is its standard error. Figure $1$ shows this distribution. The shaded area represents the middle $95\%$ of the distribution and stretches from $66.48$ to $113.52$. These limits were computed by adding and subtracting $1.96$ standard deviations to/from the mean of $90$ as follows:
$90 - (1.96)(12) = 66.48$
$90 + (1.96)(12) = 113.52$
The value of $1.96$ is based on the fact that $95\%$ of the area of a normal distribution is within $1.96$ standard deviations of the mean; $12$ is the standard error of the mean.
Figure $1$ shows that $95\%$ of the means are no more than $23.52$ units ($1.96$ standard deviations) from the mean of $90$. Now consider the probability that a sample mean computed in a random sample is within $23.52$ units of the population mean of $90$. Since $95\%$ of the distribution is within $23.52$ of $90$, the probability that the mean from any given sample will be within $23.52$ of $90$ is $0.95$. This means that if we repeatedly compute the mean ($M$) from a sample, and create an interval ranging from $M - 23.52$ to $M + 23.52$, this interval will contain the population mean $95\%$ of the time. In general, you compute the $95\%$ confidence interval for the mean with the following formula:
$\text{Lower limit} = M - Z_{0.95}\sigma _M$
$\text{Upper limit} = M + Z_{0.95}\sigma _M$
where $Z_{0.95}$ is the number of standard deviations extending from the mean of a normal distribution required to contain $0.95$ of the area and $\sigma _M$ is the standard error of the mean.
If you look closely at this formula for a confidence interval, you will notice that you need to know the standard deviation ($\sigma$) in order to estimate the mean. This may sound unrealistic, and it is. However, computing a confidence interval when $\sigma$ is known is easier than when $\sigma$ has to be estimated, and serves a pedagogical purpose. Later in this section we will show how to compute a confidence interval for the mean when $\sigma$ has to be estimated.
Suppose the following five numbers were sampled from a normal distribution with a standard deviation of $2.5: 2, 3, 5, 6,\; and\; 9$. To compute the $95\%$ confidence interval, start by computing the mean and standard error:
$M = \frac{2 + 3 + 5 + 6 + 9}{5} = 5$
$\sigma _M=\frac{2.5}{\sqrt{5}}=1.118$
$Z_{0.95}$ can be found using the normal distribution calculator and specifying that the shaded area is $0.95$ and indicating that you want the area to be between the cutoff points. As shown in Figure $2$, the value is $1.96$. If you had wanted to compute the $99\%$ confidence interval, you would have set the shaded area to $0.99$ and the result would have been $2.58$.
The confidence interval can then be computed as follows:
$\text{Lower limit} = 5 - (1.96)(1.118)= 2.81$
$\text{Upper limit} = 5 + (1.96)(1.118)= 7.19$
You should use the $t$ distribution rather than the normal distribution when the variance is not known and has to be estimated from sample data. When the sample size is large, say $100$ or above, the t distribution is very similar to the standard normal distribution. However, with smaller sample sizes, the $t$ distribution is leptokurtic, which means it has relatively more scores in its tails than does the normal distribution. As a result, you have to extend farther from the mean to contain a given proportion of the area. Recall that with a normal distribution, $95\%$ of the distribution is within $1.96$ standard deviations of the mean. Using the $t$ distribution, if you have a sample size of only $5$, $95\%$ of the area is within $2.78$ standard deviations of the mean. Therefore, the standard error of the mean would be multiplied by $2.78$ rather than $1.96$.
The values of $t$ to be used in a confidence interval can be looked up in a table of the $t$ distribution. A small version of such a table is shown in Table $1$. The first column, $df$, stands for degrees of freedom, and for confidence intervals on the mean, $df$ is equal to $N - 1$, where $N$ is the sample size.
Table $1$: Abbreviated $t$ table
df 0.95 0.99
2 4.303 9.925
3 3.182 5.841
4 2.776 4.604
5 2.571 4.032
8 2.306 3.355
10 2.228 3.169
20 2.086 2.845
50 2.009 2.678
100 1.984 2.626
You can also use the "inverse t distribution" calculator to find the $t$ values to use in confidence intervals. You will learn more about the $t$ distribution in the next section.
Assume that the following five numbers are sampled from a normal distribution: $2, 3, 5, 6,\; and\; 9$ and that the standard deviation is not known. The first steps are to compute the sample mean and variance:
$M=5\; \text{and}\; S^2=7.5$
The next step is to estimate the standard error of the mean. If we knew the population variance, we could use the following formula:
$\sigma _M=\frac{\sigma }{\sqrt{N}}$
Instead we compute an estimate of the standard error ($s_M$):
$s _M=\frac{s}{\sqrt{N}}=1.225$
The next step is to find the value of $t$. As you can see from Table $1$, the value for the $95\%$ interval for $df = N - 1 = 4$ is $2.776$. The confidence interval is then computed just as it is when $\sigma _M$. The only differences are that $s_M$ and t rather than $\sigma _M$ and $Z$ are used.
$\text{Lower limit} = 5 - (2.776)(1.225) = 1.60$
$\text{Upper limit} = 5 + (2.776)(1.225) = 8.40$
More generally, the formula for the $95\%$ confidence interval on the mean is:
$\text{Lower limit} = M - (t_{CL})(s_M)$
$\text{Upper limit} = M + (t_{CL})(s_M)$
where $M$ is the sample mean, $t_{CL}$ is the $t$ for the confidence level desired ($0.95$ in the above example), and $s_M$ is the estimated standard error of the mean.
We will finish with an analysis of the Stroop Data. Specifically, we will compute a confidence interval on the mean difference score. Recall that $47$ subjects named the color of ink that words were written in. The names conflicted so that, for example, they would name the ink color of the word "blue" written in red ink. The correct response is to say "red" and ignore the fact that the word is "blue." In a second condition, subjects named the ink color of colored rectangles.
Table $2$: Response times in seconds for $10$ subjects
Naming Colored Rectangle Interference Difference
17 38 21
15 58 43
18 35 17
20 39 19
18 33 15
20 32 12
20 45 25
19 52 33
17 31 14
21 29 8
Table $2$ shows the time difference between the interference and color-naming conditions for $10$ of the $47$ subjects. The mean time difference for all $47$ subjects is $16.362$ seconds and the standard deviation is $7.470$ seconds. The standard error of the mean is $1.090$. A $t$ table shows the critical value of $t$ for $47 - 1 = 46$ degrees of freedom is $2.013$ (for a $95\%$ confidence interval). Therefore the confidence interval is computed as follows:
$\text{Lower limit} = 16.362 - (2.013)(1.090) = 14.17$
$\text{Upper limit} = 16.362 - (2.013)(1.090) = 18.56$
Therefore, the interference effect (difference) for the whole population is likely to be between $14.17$ and $18.56$ seconds. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/10%3A_Estimation/10.07%3A_Confidence_Interval_for_Mean.txt |
Learning Objectives
• State the difference between the shape of the $t$ distribution and the normal distribution
• State how the difference between the shape of the $t$ distribution and normal distribution is affected by the degrees of freedom
• Use a $t$ table to find the value of $t$ to use in a confidence interval
• Use the $t$ calculator to find the value of $t$ to use in a confidence interval
In the introduction to normal distributions it was shown that $95\%$ of the area of a normal distribution is within $1.96$ standard deviations of the mean. Therefore, if you randomly sampled a value from a normal distribution with a mean of $100$, the probability it would be within $1.96\sigma$ of $100$ is $0.95$. Similarly, if you sample $N$ values from the population, the probability that the sample mean ($M$) will be within $1.96\sigma _M$ of $100$ is $0.95$.
Now consider the case in which you have a normal distribution but you do not know the standard deviation. You sample $N$ values and compute the sample mean ($M$) and estimate the standard error of the mean ($\sigma _M$) with $s_M$. What is the probability that $M$ will be within $1.96 s_M$ of the population mean ($\mu$)?
This is a difficult problem because there are two ways in which $M$ could be more than $1.96 s_M$ from $\mu$:
1. $M$ could, by chance, be either very high or very low and
2. $s_M$ could, by chance, be very low.
Intuitively, it makes sense that the probability of being within $1.96$ standard errors of the mean should be smaller than in the case when the standard deviation is known (and cannot be underestimated). But exactly how much smaller? Fortunately, the way to work out this type of problem was solved in the early $20^{th}$ century by W. S. Gosset who determined the distribution of a mean divided by an estimate of its standard error. This distribution is called the Student's $t$ distribution or sometimes just the $t$ distribution. Gosset worked out the $t$ distribution and associated statistical tests while working for a brewery in Ireland. Because of a contractual agreement with the brewery, he published the article under the pseudonym "Student." That is why the $t$ test is called the "Student's $t$ test."
The $t$ distribution is very similar to the normal distribution when the estimate of variance is based on many degrees of freedom, but has relatively more scores in its tails when there are fewer degrees of freedom. Figure $1$ shows $t$ distributions with $2$, $4$, and $10$ degrees of freedom and the standard normal distribution. Notice that the normal distribution has relatively more scores in the center of the distribution and the $t$ distribution has relatively more in the tails. The $t$ distribution is therefore leptokurtic. The $t$ distribution approaches the normal distribution as the degrees of freedom increase.
Since the $t$ distribution is leptokurtic, the percentage of the distribution within $1.96$ standard deviations of the mean is less than the $95\%$ for the normal distribution. Table $1$ shows the number of standard deviations from the mean required to contain $95\%$ and $99\%$ of the area of the $t$ distribution for various degrees of freedom. These are the values of $t$ that you use in a confidence interval. The corresponding values for the normal distribution are $1.96$ and $2.58$ respectively. Notice that with few degrees of freedom, the values of $t$ are much higher than the corresponding values for a normal distribution and that the difference decreases as the degrees of freedom increase. The values in Table $1$ can be obtained from the "Find $t$ for a confidence interval" calculator.
Table $1$: Abbreviated $t$ table
df 0.95 0.99
2 4.303 9.925
3 3.182 5.841
4 2.776 4.604
5 2.571 4.032
8 2.306 3.355
10 2.228 3.169
20 2.086 2.845
50 2.009 2.678
100 1.984 2.626
Returning to the problem posed at the beginning of this section, suppose you sampled $9$ values from a normal population and estimated the standard error of the mean ($\sigma _M$) with $s_M$. What is the probability that $M$ would be within $1.96 s_M$ of $\mu$? Since the sample size is $9$, there are $N - 1 = 8 df$. From Table $1$ you can see that with $8 df$ the probability is $0.95$ that the mean will be within $2.306 s_M$ of $\mu$. The probability that it will be within $1.96 s_M$ of $\mu$ is therefore lower than $0.95$.
As shown in Figure $2$, the "$t$ distribution" calculator can be used to find that $0.086$ of the area of a $t$ distribution is more than $1.96$ standard deviations from the mean, so the probability that $M$ would be less than $1.96 s_M$ from $\mu$ is $1 - 0.086 = 0.914$.
As expected, this probability is less than $0.95$ that would have been obtained if $\sigma _M$ had been known instead of estimated.
10.09: Confidence Interval Simulation
Learning Objectives
• Develop a basic understanding of the properties of a sampling distribution based on the properties of the population
Instructions
This simulation illustrates confidence intervals. For each run of the simulation, \(100\) sample experiments are conducted and a confidence interval on the mean is computed for each experiment. In each experiment, scores are sampled from a population with a mean of \(50\) and a standard deviation of \(10\). Therefore the parameter being estimated is always \(50\). Both \(95\%\) and \(99\%\) confidence intervals are computed for each experiment. If the \(95\%\) confidence interval contains the population mean of \(50\), then the confidence interval is shown as an orange line. If the interval does not contain \(50\), it is shown as a red line. The \(99\%\) confidence intervals are shown by extending the \(95\%\) intervals. The extension is in blue if the \(99\%\) interval contains \(50\) and in white if it does not. You can choose to make the sample size for each experiment \(10\), \(15\), or \(20\). One hundred simulated experiments are conducted when you click the "Sample" button. The cumulative results are shown at the bottom of the display. You can reset the cumulative results by clicking the "Clear" button.
Illustrated Instructions
The demonstration generates confidence intervals for sample experiments taken from a population with a mean of \(50\) and a standard deviation of \(10\). You can choose from various sample sizes but as can be seen from the figure below the default size is \(10\). Each time you click the sample button confidence intervals for \(100\) experiments are generated and displayed on the graph on the left.
The figure below displays the results of \(300\) experiments with a sample size of \(10\). The \(95\%\) confidence intervals that contain the mean of \(50\) are shown in orange and the those that do not are shown in red. The \(99\%\) confidence intervals are shown in blue if they contain \(50\) and white if they do not. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/10%3A_Estimation/10.08%3A_t_Distribution.txt |
Learning Objectives
• State the assumptions for computing a confidence interval on the difference between means
• Compute a confidence interval on the difference between means
• Format data for computer analysis
It is much more common for a researcher to be interested in the difference between means than in the specific values of the means themselves. We take as an example the data from the "Animal Research" case study. In this experiment, students rated (on a $7$-point scale) whether they thought animal research is wrong. The sample sizes, means, and variances are shown separately for males and females in Table $1$.
Table $1$: Means and Variances in Animal Research study
Condition n Mean Variance
Females 17 5.353 2.743
Males 17 3.882 2.985
As you can see, the females rated animal research as more wrong than did the males. This sample difference between the female mean of $5.35$ and the male mean of $3.88$ is $1.47$. However, the gender difference in this particular sample is not very important. What is important is the difference in the population. The difference in sample means is used to estimate the difference in population means. The accuracy of the estimate is revealed by a confidence interval.
In order to construct a confidence interval, we are going to make three assumptions:
1. The two populations have the same variance. This assumption is called the assumption of homogeneity of variance.
2. The populations are normally distributed.
3. Each value is sampled independently from each other value.
The consequences of violating these assumptions are discussed in a later section. For now, suffice it to say that small-to-moderate violations of assumptions $1$ and $2$ do not make much difference.
A confidence interval on the difference between means is computed using the following formula:
$\text{Lower Limit} = M_1 - M_2 -(t_{CL})(S_{M_1-M_2})$
$\text{Upper Limit} = M_1 - M_2 +(t_{CL})(S_{M_1-M_2})$
where $M_1 - M_2$ is the difference between sample means, $t_{CL}$ is the t for the desired level of confidence, and $S_{M_1-M_2}$ is the estimated standard error of the difference between sample means. The meanings of these terms will be made clearer as the calculations are demonstrated.
We continue to use the data from the "Animal Research" case study and will compute a confidence interval on the difference between the mean score of the females and the mean score of the males. For this calculation, we will assume that the variances in each of the two populations are equal.
The first step is to compute the estimate of the standard error of the difference between means ($S_{M_1-M_2}$). Recall from the relevant section in the chapter on sampling distributions that the formula for the standard error of the difference in means in the population is:
$\sigma _{M_1-M_2}=\sqrt{\frac{\sigma _{1}^{2}}{n_1}+\frac{\sigma _{2}^{2}}{n_2}}=\sqrt{\frac{\sigma ^{2}}{n}+\frac{\sigma ^{2}}{n}}=\sqrt{\frac{2\sigma ^2}{n}}$
In order to estimate this quantity, we estimate $\sigma ^2$ and use that estimate in place of $\sigma ^2$. Since we are assuming the population variances are the same, we estimate this variance by averaging our two sample variances. Thus, our estimate of variance is computed using the following formula:
$MSE=\frac{s_{1}^{2}+s_{2}^{2}}{2}$
where $MSE$ is our estimate of $\sigma ^2$. In this example,
$MSE=\frac{2.743 + 2.985}{2}=2.864$
Note that $MSE$ stands for "mean square error" and is the mean squared deviation of each score from its group's mean.
Since $n$ (the number of scores in each condition) is $17$,
$S_{M_1-M_2}=\sqrt{\frac{2MSE}{n}}=\sqrt{\frac{(2)(2.864)}{17}}=0.5805$
The next step is to find the $t$ to use for the confidence interval ($t_{CL}$). To calculate $t_{CL}$, we need to know the degrees of freedom. The degrees of freedom is the number of independent estimates of variance on which $MSE$ is based. This is equal to $(n_1 - 1) + (n_2 - 1)$ where $n_1$ is the sample size of the first group and $n_2$ is the sample size of the second group. For this example, $n_1= n_2 = 17$. When $n_1= n_2$, it is conventional to use "$n$" to refer to the sample size of each group. Therefore, the degrees of freedom is $16 + 16 = 32$.
Calculator: Find t for confidence interval
From either the above calculator or a $t$ table, you can find that the $t$ for a $95\%$ confidence interval for $32 df$ is $2.037$.
We now have all the components needed to compute the confidence interval. First, we know the difference between means:
$M_1 - M_2 = 5.353 - 3.882 = 1.471$
We know the standard error of the difference between means is
$S_{M_1 - M2} = 0.5805$
and that the $t$ for the $95\%$ confidence interval with $32 df$ is
$t_{CL}=2.037$
Therefore, the $95\%$ confidence interval is
$\text{Lower Limit} = 1.471 - (2.037)(0.5805) = 0.29$
$\text{Upper Limit} = 1.471 + (2.037)(0.5805) = 2.65$
We can write the confidence interval as:
$0.29 \leq \mu _f - \mu _m \leq 2.65$
where $\mu _f$ is the population mean for females and $\mu _m$ is the population mean for males. This analysis provides evidence that the mean for females is higher than the mean for males, and that the difference between means in the population is likely to be between $0.29$ and $2.65$.
Formatting data for Computer Analysis
Most computer programs that compute $t$ tests require your data to be in a specific form. Consider the data in Table $2$.
Table $2$: Example Data
Group 1 Group 2
3 5
4 6
5 7
Here there are two groups, each with three observations. To format these data for a computer program, you normally have to use two variables: the first specifies the group the subject is in and the second is the score itself. For the data in Table $2$, the reformatted data look as follows:
Table $3$: Reformatted Data
G Y
1 3
1 4
1 5
2 5
2 6
2 7
To use Analysis Lab to do the calculations, you would copy the data and then
Click the "Enter/Edit User Data" button. (You may be warned that for security reasons you must use the keyboard shortcut for pasting data.)
1. Paste your data.
2. Click "Accept Data."
3. Set the Dependent Variable to $Y$.
4. Set the Grouping Variable to $G$.
5. Click the $t$-test confidence interval button.
The $95\%$ confidence interval on the difference between means extends from $-4.267$ to $0.267$.
Computations for Unequal Sample Sizes (optional)
The calculations are somewhat more complicated when the sample sizes are not equal. One consideration is that $MSE$, the estimate of variance, counts the sample with the larger sample size more than the sample with the smaller sample size. Computationally this is done by computing the sum of squares error ($SSE$) as follows:
$SSE=\sum (X-M_1)^2+\sum (X-M_2)^2$
where $M_1$ is the mean for group $1$ and $M_2$ is the mean for group $2$. Consider the following small example:
Table $4$: Example Data
Group 1 Group 2
3 2
4 4
5
$M_1 = 4\; \; and\; \; M_2 = 3$
$SSE = (3-4)^2 + (4-4)^2 + (5-4)^2 + (2-3)^2 + (4-3)^2 = 4$
Then, $MSE$ is computed by: $MSE=\frac{SSE}{df}$
where the degrees of freedom ($df$) is computed as before:
$df = (n_1 -1) + (n_2 -1) = (3-1) + (2-1) = 3$
$MSE=\frac{SSE}{df}=\frac{4}{3}=1.333$
The formula
$S_{M_1-M_2}=\sqrt{\frac{2MSE}{n}}$
is replaced by
$S_{M_1-M_2}=\sqrt{\frac{2MSE}{n_h}}$
where $n_h$ is the harmonic mean of the sample sizes and is computed as follows:
$n_h=\frac{2}{\tfrac{1}{n_1}+\tfrac{1}{n_2}}=\frac{2}{\tfrac{1}{3}+\tfrac{1}{2}}=2.4$
and
$S_{M_1-M_2}=\sqrt{\frac{(2)(1.333)}{2.4}}=1.054$
$t_{CL}$ for $3 df$ and the $0.05$ level equals $3.182$.
Therefore the $95\%$ confidence interval is
$\text{Lower Limit} = 1 - (3.182)(1.054)= -2.35$
$\text{Upper Limit} = 1 + (3.182)(1.054)= 4.35$
We can write the confidence interval as:
$-2.35 \leq \mu _1 - \mu _2 \leq 4.35$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/10%3A_Estimation/10.10%3A_Difference_between_Means.txt |
Learning Objectives
• State the standard error of $z'$
• Compute a confidence interval on $\rho$
The computation of a confidence interval on the population value of Pearson's correlation ($\rho$) is complicated by the fact that the sampling distribution of $r$ is not normally distributed. The solution lies with Fisher's $z'$ transformation described in the section on the sampling distribution of Pearson's r. The steps in computing a confidence interval for $\rho$ are:
1. Convert $r$ to $z'$.
2. Compute a confidence interval in terms of $z'$.
3. Convert the confidence interval back to $r$.
Let's take the data from the case study Animal Research as an example. In this study, students were asked to rate the degree to which they thought animal research is wrong and the degree to which they thought it is necessary. As you might have expected, there was a negative relationship between these two variables: the more that students thought animal research is wrong, the less they thought it is necessary. The correlation based on $34$ observations is $-0.654$. The problem is to compute a $95\%$ confidence interval on $\rho$ based on this $r$ of $-0.654$.
The conversion of $r$ to $z'$ can be done using a calculator. This calculator shows that the $z'$ associated with an $r$ of $-0.654$ is $-0.78$.
The sampling distribution of $z'$ is approximately normally distributed and has a standard error of
$\frac{1}{\sqrt{N-3}}$
For this example, $N = 34$ and therefore the standard error is $0.180$. The $Z$ for a $95\%$ confidence interval ($Z_{0.95}$) is $1.96$, as can be found using the normal distribution calculator (setting the shaded area to $0.95$ and clicking on the "Between" button). The confidence interval is therefore computed as:
$\text{Lower limit} = -0.775 - (1.96)(0.18) = -1.13$
$\text{Upper limit} = -0.775 + (1.96)(0.18) = -0.43$
The final step is to convert the endpoints of the interval back to $r$ using a calculator. The $r$ associated with a $z'$ of $-1.13$ is $-0.81$ and the $r$ associated with a $z'$ of $-0.43$ is $-0.40$. Therefore, the population correlation ($\rho$) is likely to be between $-0.81$ and $-0.40$. The $95\%$ confidence interval is:
$-0.81 \leq \rho \leq -0.40$
To calculate the $99\%$ confidence interval, you use the $Z$ for a $99\%$ confidence interval of $2.58$ as follows:
$\text{Lower limit} = -0.775 - (2.58)(0.18) = -1.24$
$\text{Upper limit} = -0.775 + (2.58)(0.18) = -0.32$
Converting back to $r$, the confidence interval is:
$-0.84 \leq \rho \leq -0.31$
Naturally, the $99\%$ confidence interval is wider than the $95\%$ confidence interval.
10.12: Proportion
Learning Objectives
• Estimate the population proportion from sample proportions
• Apply the correction for continuity
• Compute a confidence interval
A candidate in a two-person election commissions a poll to determine who is ahead. The pollster randomly chooses $500$ registered voters and determines that $260$ out of the $500$ favor the candidate. In other words, $0.52$ of the sample favors the candidate. Although this point estimate of the proportion is informative, it is important to also compute a confidence interval. The confidence interval is computed based on the mean and standard deviation of the sampling distribution of a proportion. The formulas for these two parameters are shown below:
$\mu _p=\pi$
$\sigma _p=\sqrt{\frac{\pi (1-\pi )}{N}}$
Since we do not know the population parameter $\pi$, we use the sample proportion $p$ as an estimate. The estimated standard error of $p$ is therefore
$s _p=\sqrt{\frac{p(1-p)}{N}}$
We start by taking our statistic ($p$) and creating an interval that ranges ($Z_{0.95}$)($s_p$) in both directions, where $Z_{0.95}$ is the number of standard deviations extending from the mean of a normal distribution required to contain $0.95$ of the area (see the section on the confidence interval for the mean). The value of $Z_{0.95}$ is computed with the normal calculator and is equal to $1.96$. We then make a slight adjustment to correct for the fact that the distribution is discrete rather than continuous.
Normal Distribution Calculator
$s_p$ is calculated as shown below:
$s _p=\sqrt{\frac{(0.52)(1-0.52)}{300}}=0.0223$
To correct for the fact that we are approximating a discrete distribution with a continuous distribution (the normal distribution), we subtract $0.5/N$ from the lower limit and add $0.5/N$ to the upper limit of the interval. Therefore the confidence interval is
$p\pm Z_{0.95}\sqrt{\frac{p(1-p)}{N}}\pm \frac{0.5}{N}$
$\text{Lower limit}: 0.52 - (1.96)(0.0223) - 0.001 = 0.475$
$\text{Upper limit}: 0.52 + (1.96)(0.0223) + 0.001 = 0.565$
$0.475 \leq \pi \leq 0.565$
Since the interval extends $0.045$ in both directions, the margin of error is $0.045$. In terms of percent, between $47.5\%$ and $56.5\%$ of the voters favor the candidate and the margin of error is $4.5\%$. Keep in mind that the margin of error of $4.5\%$ is the margin of error for the percent favoring the candidate and not the margin of error for the difference between the percent favoring the candidate and the percent favoring the opponent. The margin of error for the difference is $9\%$, twice the margin of error for the individual percent. Keep this in mind when you hear reports in the media; the media often get this wrong.
10.13: Statistical Literacy
Learning Objectives
• No "Large Conclusions" from "Tiny" Samples?
In July of \(2011\), Gene Munster of Piper Jaffray reported the results of a survey in a note to clients. This research was reported throughout the media. Perhaps the fullest description was presented on the CNNMoney website (A service of CNN, Fortune, and Money) in an article entitled “Survey: iPhone retention \(94\%\) vs. Android \(47\%\).” The data were collected by asking people in food courts and baseball stadiums what their current phone was and what phone they planned to buy next. The data were collected in the summer of \(2011\). Below is a portion of the data:
Table \(1\): Sample phone retention data
Phone Keep Change Proportion
iPhone 58 4 0.94
Android 17 19 0.47
The article contains the strong caution: “It's only a tiny sample, so large conclusions must not be drawn.” This caution appears to be a welcome change from the overstating of findings typically found in the media. But has this report understated the importance of the study? Perhaps it is valid to draw some "large conclusions."
Example \(1\): what do you think?
Is it possible to conclude the vast majority of iPhone owners in the population sampled plan to buy another iPhone or is the sample size too small to justify this conclusion?
Solution
The confidence interval on the proportion extends from \(0.87\) to \(1.0\) (some methods give the interval from \(0.85\) to \(0.97\)). Even the lower bound indicates the vast majority of iPhone owners plan to buy another iPhone. A strong conclusion can be made even with this sample size. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/10%3A_Estimation/10.11%3A_Correlation.txt |
You may want to use the Analysis Lab and various calculators for some of these exercises.
Calculators:
• Inverse t Distribution: Finds t for a confidence interval.
• t Distribution: Computes areas of the t distribution.
• Fisher's r to z': Computes transformations in both directions.
• Inverse Normal Distribution: Use for confidence intervals.
General Questions
Q1
When would the mean grade in a class on a final exam be considered a statistic? When would it be considered a parameter? (relevant section)
Q2
Define bias in terms of expected value. (relevant section)
Q3
Is it possible for a statistic to be unbiased yet very imprecise? How about being very accurate but biased? (relevant section)
Q4
Why is a \(99\%\) confidence interval wider than a \(95\%\) confidence interval? (relevant section & relevant section)
Q5
When you construct a \(95\%\) confidence interval, what are you \(95\%\) confident about? (relevant section)
Q6
What is the difference in the computation of a confidence interval between cases in which you know the population standard deviation and cases in which you have to estimate it? (relevant section & relevant section)
Q7
Assume a researcher found that the correlation between a test he or she developed and job performance was \(0.55\) in a study of \(28\) employees. If correlations under \(0.35\) are considered unacceptable, would you have any reservations about using this test to screen job applicants? (relevant section)
Q8
What is the effect of sample size on the width of a confidence interval? (relevant section & relevant section)
Q9
How does the \(t\) distribution compare with the normal distribution? How does this difference affect the size of confidence intervals constructed using \(z\) relative to those constructed using \(t\)? Does sample size make a difference? (relevant section)
Q10
The effectiveness of a blood-pressure drug is being investigated. How might an experimenter demonstrate that, on average, the reduction in systolic blood pressure is \(20\) or more? (relevant section & relevant section)
Q11
A population is known to be normally distributed with a standard deviation of \(2.8\).
1. Compute the \(95\%\) confidence interval on the mean based on the following sample of nine: \(8, 9, 10, 13, 14, 16, 17, 20, 21\).
2. Now compute the \(99\%\) confidence interval using the same data. (relevant section)
Q12
A person claims to be able to predict the outcome of flipping a coin. This person is correct \(16/25\) times. Compute the \(95\%\) confidence interval on the proportion of times this person can predict coin flips correctly. What conclusion can you draw about this test of his ability to predict the future? (relevant section)
Q13
What does it mean that the variance (computed by dividing by \(N\)) is a biased statistic? (relevant section)
Q14
A confidence interval for the population mean computed from an \(N\) of \(16\) ranges from \(12\) to \(28\). A new sample of \(36\) observations is going to be taken. You can't know in advance exactly what the confidence interval will be because it depends on the random sample. Even so, you should have some idea of what it will be. Give your best estimation. (relevant section)
Q15
You take a sample of \(22\) from a population of test scores, and the mean of your sample is \(60\).
1. You know the standard deviation of the population is \(10\). What is the \(99\%\) confidence interval on the population mean.
2. Now assume that you do not know the population standard deviation, but the standard deviation in your sample is \(10\). What is the \(99\%\) confidence interval on the mean now? (relevant section)
Q16
You read about a survey in a newspaper and find that \(70\%\) of the \(250\) people sampled prefer Candidate \(A\). You are surprised by this survey because you thought that more like \(50\%\) of the population preferred this candidate. Based on this sample, is \(50\%\) a possible population proportion? Compute the \(95\%\) confidence interval to be sure. (relevant section)
Q17
Heights for teenage boys and girls were calculated. The mean height for the sample of \(12\) boys was \(174\) cm and the variance was \(62\). For the sample of \(12\) girls, the mean was \(166\) cm and the variance was \(65\).
1. What is the \(95\%\) confidence interval on the difference between population means?
2. What is the \(99\%\) confidence interval on the difference between population means?
3. Do you think the mean difference in the population could be about \(5\)? Why or why not? (relevant section)
Q18
You were interested in how long the average psychology major at your college studies per night, so you asked \(10\) psychology majors to tell you the amount they study. They told you the following times: \(2, 1.5, 3, 2, 3.5, 1, 0.5, 3, 2, 4\).
1. Find the \(95\%\) confidence interval on the population mean.
2. Find the \(90\%\) confidence interval on the population mean. (relevant section)
Q19
True/false: As the sample size gets larger, the probability that the confidence interval will contain the population mean gets higher. (relevant section & relevant section)
Q20
True/false: You have a sample of \(9\) men and a sample of \(8\) women. The degrees of freedom for the \(t\) value in your confidence interval on the difference between means is \(16\). (relevant section & relevant section)
Q21
True/false: Greek letters are used for statistics as opposed to parameters. (relevant section)
Q22
True/false: In order to construct a confidence interval on the difference between means, you need to assume that the populations have the same variance and are both normally distributed. (relevant section)
Q23
True/false: The red distribution represents the \(t\) distribution and the blue distribution represents the normal distribution. (relevant section)
Questions from Case Studies
The following questions are from the Angry Moods (AM) case study.
Q24
(AM#6c) Is there a difference in how much males and females use aggressive behavior to improve an angry mood? For the "Anger-Out" scores, compute a \(99\%\) confidence interval on the difference between gender means. (relevant section)
Q25
(AM#10) Calculate the \(95\%\) confidence interval for the difference between the mean Anger-In score for the athletes and non-athletes. What can you conclude? (relevant section)
Q26
Find the \(95\%\) confidence interval on the population correlation between the Anger-Out and Control-Out scores. (relevant section)
The following questions are from the Flatulence (F) case study.
Q27
(F#8) Compare men and women on the variable "perday." Compute the \(95\%\) confidence interval on the difference between means. (relevant section)
Q28
(F#10) What is the \(95\%\) confidence interval of the mean time people wait before farting in front of a romantic partner. (relevant section)
The following questions use data from the Animal Research (AR) case study.
Q29
(AR#3) What percentage of the women studied in this sample strongly agreed (gave a rating of \(7\)) that using animals for research is wrong?
Q30
Use the proportion you computed in #29. Compute the \(95\%\) confidence interval on the population proportion of women who strongly agree that animal research is wrong. (relevant section)
Q31
Compute a \(95\%\) confidence interval on the difference between the gender means with respect to their beliefs that animal research is wrong. (relevant section)
The following question is from the ADHD Treatment (AT) case study.
Q32
(AT#8) What is the correlation between the participants' correct number of responses after taking the placebo and their correct number of responses after taking \(0.60\) mg/kg of MPH? Compute the \(95\%\) confidence interval on the population correlation. (relevant section)
The following question is from the Weapons and Aggression (WA) case study.
Q33
(WA#4) Recall that the hypothesis is that a person can name an aggressive word more quickly if it is preceded by a weapon word prime than if it is preceded by a neutral word prime. The first step in testing this hypothesis is to compute the difference between
1. the naming time of aggressive words when preceded by a neutral word prime and
2. the naming time of aggressive words when preceded by a weapon word prime separately for each of the \(32\) participants. That is, compute an \(- aw\) for each participant.
1. Would the hypothesis of this study be supported if the difference were positive or if it were negative?
2. What is the mean of this difference score? (relevant section)
3. What is the standard deviation of this difference score? (relevant section)
4. What is the 95% confidence interval of the mean difference score? (relevant section)
5. What does the confidence interval computed in (d) say about the hypothesis.
The following question is from the Diet and Health (WA) case study.
Q34
Compute a \(95\%\) confidence interval on the proportion of people who are healthy on the AHA diet.
Cancers Deaths Nonfatal illness Healthy Total
AHA
15 24 25 239 303 Mediterranean 7 14 8 273 302 Total 22 38 33 512 605
The following questions are from (reproduced with permission) Visit the site
Q35
Suppose that you take a random sample of \(10,000\) Americans and find that \(1,111\) are left-handed. You perform a test of significance to assess whether the sample data provide evidence that more than \(10\%\) of all Americans are left-handed, and you calculate a test statistic of \(3.70\) and a \(p\)-value of \(0.0001\). Furthermore, you calculate a \(99\%\) confidence interval for the proportion of left-handers in America to be \((0.103,0.119)\). Consider the following statements: The sample provides strong evidence that more than \(10\%\) of all Americans are left-handed. The sample provides evidence that the proportion of left-handers in America is much larger than \(10\%\). Which of these two statements is the more appropriate conclusion to draw? Explain your answer based on the results of the significance test and confidence interval.
Q36
A student wanted to study the ages of couples applying for marriage licenses in his county. He studied a sample of \(94\) marriage licenses and found that in \(67\) cases the husband was older than the wife. Do the sample data provide strong evidence that the husband is usually older than the wife among couples applying for marriage licenses in that county? Explain briefly and justify your answer.
Q37
Imagine that there are \(100\) different researchers each studying the sleeping habits of college freshmen. Each researcher takes a random sample of size \(50\) from the same population of freshmen. Each researcher is trying to estimate the mean hours of sleep that freshmen get at night, and each one constructs a \(95\%\) confidence interval for the mean. Approximately how many of these \(100\) confidence intervals will NOT capture the true mean?
1. None
2. \(1\) or \(2\)
3. \(3\) to \(7\)
4. about half
5. \(95\) to \(100\)
6. other
Selected Answers
S11
1. \((12.39, 16.05)\)
\((0.43, 0.85)\)
S15
1. \((53.96, 66.04)\)
S17
1. \((1.25, 14.75)\)
S18
1. \((1.45, 3.05)\)
S26
\((-0.713, -0.414)\)
S27
\((-0.98, 3.09)\)
\(41\%\)
1. \(7.16\) | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/10%3A_Estimation/10.E%3A_Estimation_%28Exercises%29.txt |
When interpreting an experimental finding, a natural question arises as to whether the finding could have occurred by chance. Hypothesis testing is a statistical procedure for testing whether chance is a plausible explanation of an experimental finding. Misconceptions about hypothesis testing are common among practitioners as well as students. To help prevent these misconceptions, this chapter goes into more detail about the logic of hypothesis testing than is typical for an introductory-level text.
• 11.1: Introduction to Hypothesis Testing
• 11.2: Significance Testing
It is conventional to conclude the null hypothesis is false if the probability value is less than 0.05. More conservative researchers conclude the null hypothesis is false only if the probability value is less than 0.01. When a researcher concludes that the null hypothesis is false, the researcher is said to have rejected the null hypothesis. The probability value below which the null hypothesis is rejected is called the α (alpha) level or simply α. It is also called the significance level.
• 11.3: Type I and II Errors
A Type I error occurs when a significance test results in the rejection of a true null hypothesis. A Type II error that can be made in significance testing is failing to reject a false null hypothesis. Unlike a Type I error, a Type II error is not really an error. When a statistical test is not significant, it means that the data do not provide strong evidence that the null hypothesis is false. Lack of significance does not support the conclusion that the null hypothesis is true.
• 11.4: One- and Two-Tailed Tests
A probability calculated in only one tail of the distribution is called a "one-tailed probability" and probability calculated in both tails of a distribution is called a "two-tailed probability." You should always decide whether you are going to use a one-tailed or a two-tailed probability before looking at the data.
• 11.5: Significant Results
When a probability value is below the α level, the effect is statistically significant and the null hypothes is is rejected. However, not all statistically significant effects should be treated the same way. For example, you should have less confidence that the null hypothesis is false if p = 0.049 than p = 0.003. Thus, rejecting the null hypothesis is not an all-or-none proposition.
• 11.6: Non-Significant Results
• 11.7: Steps in Hypothesis Testing
• 11.8: Significance Testing and Confidence Intervals
• 11.9: Misconceptions of Hypothesis Testing
• 11.10: Statistical Literacy
• 11.E: Logic of Hypothesis Testing (Exercises)
11: Logic of Hypothesis Testing
Learning Objectives
• Describe the logic by which it can be concluded that someone can distinguish between two things
• State whether random assignment ensures that all uncontrolled sources of variation will be equal
• Define precisely what the probability is that is computed to reach the conclusion that a difference is not due to chance
• Distinguish between the probability of an event and the probability of a state of the world
• Define "null hypothesis"
• Be able to determine the null hypothesis from a description of an experiment
• Define "alternative hypothesis"
The statistician R. Fisher explained the concept of hypothesis testing with a story of a lady tasting tea. Here we will present an example based on James Bond who insisted that martinis should be shaken rather than stirred. Let's consider a hypothetical experiment to determine whether Mr. Bond can tell the difference between a shaken and a stirred martini. Suppose we gave Mr. Bond a series of $16$ taste tests. In each test, we flipped a fair coin to determine whether to stir or shake the martini. Then we presented the martini to Mr. Bond and asked him to decide whether it was shaken or stirred. Let's say Mr. Bond was correct on $13$ of the $16$ taste tests. Does this prove that Mr. Bond has at least some ability to tell whether the martini was shaken or stirred?
This result does not prove that he does; it could be he was just lucky and guessed right $13$ out of $16$ times. But how plausible is the explanation that he was just lucky? To assess its plausibility, we determine the probability that someone who was just guessing would be correct $13/16$ times or more. This probability can be computed from the binomial distribution, and the binomial distribution calculator shows it to be $0.0106$. This is a pretty low probability, and therefore someone would have to be very lucky to be correct $13$ or more times out of $16$ if they were just guessing. So either Mr. Bond was very lucky, or he can tell whether the drink was shaken or stirred. The hypothesis that he was guessing is not proven false, but considerable doubt is cast on it. Therefore, there is strong evidence that Mr. Bond can tell whether a drink was shaken or stirred.
Binomial Calculator
Let's consider another example. The case study Physicians' Reactions sought to determine whether physicians spend less time with obese patients. Physicians were sampled randomly and each was shown a chart of a patient complaining of a migraine headache. They were then asked to estimate how long they would spend with the patient. The charts were identical except that for half the charts, the patient was obese and for the other half, the patient was of average weight. The chart a particular physician viewed was determined randomly. Thirty-three physicians viewed charts of average-weight patients and $38$ physicians viewed charts of obese patients.
The mean time physicians reported that they would spend with obese patients was $24.7$ minutes as compared to a mean of $31.4$ minutes for average-weight patients. How might this difference between means have occurred? One possibility is that physicians were influenced by the weight of the patients. On the other hand, perhaps by chance, the physicians who viewed charts of the obese patients tend to see patients for less time than the other physicians. Random assignmentof charts does not ensure that the groups will be equal in all respects other than the chart they viewed. In fact, it is certain the two groups differed in many ways by chance. The two groups could not have exactly the same mean age (if measured precisely enough such as in days). Perhaps a physician's age affects how long physicians see patients. There are innumerable differences between the groups that could affect how long they view patients. With this in mind, is it plausible that these chance differences are responsible for the difference in times?
To assess the plausibility of the hypothesis that the difference in mean times is due to chance, we compute the probability of getting a difference as large or larger than the observed difference ($31.4 - 24.7 = 6.7$ minutes) if the difference were, in fact, due solely to chance. Using methods presented in another section,this probability can be computed to be $0.0057$. Since this is such a low probability, we have confidence that the difference in times is due to the patient's weight and is not due to chance.
The Probability Value
It is very important to understand precisely what the probability values mean. In the James Bond example, the computed probability of $0.0106$ is the probability he would be correct on $13$ or more taste tests (out of $16$) if he were just guessing.
It is easy to mistake this probability of $0.0106$ as the probability he cannot tell the difference. This is not at all what it means.
The probability of $0.0106$ is the probability of a certain outcome ($13$ or more out of $16$) assuming a certain state of the world (James Bond was only guessing). It is not the probability that a state of the world is true. Although this might seem like a distinction without a difference, consider the following example. An animal trainer claims that a trained bird can determine whether or not numbers are evenly divisible by $7$. In an experiment assessing this claim, the bird is given a series of $16$ test trials. On each trial, a number is displayed on a screen and the bird pecks at one of two keys to indicate its choice. The numbers are chosen in such a way that the probability of any number being evenly divisible by $7$ is $0.50$. The bird is correct on $9/16$ choices. Using the binomial calculator, we can compute that the probability of being correct nine or more times out of $16$ if one is only guessing is $0.40$. Since a bird who is only guessing would do this well $40\%$ of the time, these data do not provide convincing evidence that the bird can tell the difference between the two types of numbers. As a scientist, you would be very skeptical that the bird had this ability. Would you conclude that there is a $0.40$ probability that the bird can tell the difference? Certainly not! You would think the probability is much lower than $0.0001$.
To reiterate, the probability value is the probability of an outcome ($9/16$ or better) and not the probability of a particular state of the world (the bird was only guessing). In statistics, it is conventional to refer to possible states of the world as hypotheses since they are hypothesized states of the world. Using this terminology, the probability value is the probability of an outcome given the hypothesis. It is not the probability of the hypothesis given the outcome.
This is not to say that we ignore the probability of the hypothesis. If the probability of the outcome given the hypothesis is sufficiently low, we have evidence that the hypothesis is false. However, we do not compute the probability that the hypothesis is false. In the James Bond example, the hypothesis is that he cannot tell the difference between shaken and stirred martinis. The probability value is low ($0.0106$), thus providing evidence that he can tell the difference. However, we have not computed the probability that he can tell the difference. A branch of statistics called Bayesian statistics provides methods for computing the probabilities of hypotheses. These computations require that one specify the probability of the hypothesis before the data are considered and, therefore, are difficult to apply in some contexts.
The Null Hypothesis
The hypothesis that an apparent effect is due to chance is called the null hypothesis. In the Physicians' Reactions example, the null hypothesis is that in the population of physicians, the mean time expected to be spent with obese patients is equal to the mean time expected to be spent with average-weight patients. This null hypothesis can be written as:
$\mu _{obese}=\mu _{average}$
or as
$\mu _{obese}-\mu _{average}=0$
The null hypothesis in a correlational study of the relationship between high school grades and college grades would typically be that the population correlation is $0$. This can be written as
$\rho =0$
where $\rho$ is the population correlation (not to be confused with $r$, the correlation in the sample).
Although the null hypothesis is usually that the value of a parameter is $0$, there are occasions in which the null hypothesis is a value other than $0$. For example, if one were testing whether a subject differed from chance in their ability to determine whether a flipped coin would come up heads or tails, the null hypothesis would be that $\pi =0.5$.
Keep in mind that the null hypothesis is typically the opposite of the researcher's hypothesis. In the Physicians' Reactions study, the researchers hypothesized that physicians would expect to spend less time with obese patients. The null hypothesis that the two types of patients are treated identically is put forward with the hope that it can be discredited and therefore rejected. If the null hypothesis were true, a difference as large or larger than the sample difference of $6.7$ minutes would be very unlikely to occur. Therefore, the researchers rejected the null hypothesis of no difference and concluded that in the population, physicians intend to spend less time with obese patients.
If the null hypothesis is rejected, then the alternative to the null hypothesis (called the alternative hypothesis) is accepted. The alternative hypothesis is simply the reverse of the null hypothesis. If the null hypothesis $\mu _{obese}=\mu _{average}$ is rejected, then there are two alternatives:
$\mu _{obese}<\mu _{average}$
or
$\mu _{obese}>\mu _{average}$
Naturally, the direction of the sample means determines which alternative is adopted. Some textbooks have incorrectly argued that rejecting the null hypothesis that two population means are equal does not justify a conclusion about which population mean is larger. Kaiser ($1960$) showed how it is justified to draw a conclusion about the direction of the difference.
1. Kaiser, H. F. (1960) Directional statistical decisions. Psychological Review, 67, 160-167. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/11%3A_Logic_of_Hypothesis_Testing/11.01%3A_Introduction_to_Hypothesis_Testing.txt |
Learning Objectives
• Describe how a probability value is used to cast doubt on the null hypothesis
• Define "statistically significant"
• Distinguish between statistical significance and practical significance
• Distinguish between two approaches to significance testing
A low probability value casts doubt on the null hypothesis. How low must the probability value be in order to conclude that the null hypothesis is false? Although there is clearly no right or wrong answer to this question, it is conventional to conclude the null hypothesis is false if the probability value is less than $0.05$. More conservative researchers conclude the null hypothesis is false only if the probability value is less than $0.01$. When a researcher concludes that the null hypothesis is false, the researcher is said to have rejected the null hypothesis. The probability value below which the null hypothesis is rejected is called the $\alpha$ (alpha) level or simply $\alpha$. It is also called the significance level.
When the null hypothesis is rejected, the effect is said to be statistically significant. For example, in the Physicians' Reactions case study, the probability value is $0.0057$. Therefore, the effect of obesity is statistically significant and the null hypothesis that obesity makes no difference is rejected. It is very important to keep in mind that statistical significance means only that the null hypothesis of exactly no effect is rejected; it does not mean that the effect is important, which is what "significant" usually means. When an effect is significant, you can have confidence the effect is not exactly zero. Finding that an effect is significant does not tell you about how large or important the effect is.
Do not confuse statistical significance with practical significance. A small effect can be highly significant if the sample size is large enough.
Why does the word "significant" in the phrase "statistically significant" mean something so different from other uses of the word? Interestingly, this is because the meaning of "significant" in everyday language has changed. It turns out that when the procedures for hypothesis testing were developed, something was "significant" if it signified something. Thus, finding that an effect is statistically significant signifies that the effect is real and not due to chance. Over the years, the meaning of "significant" changed, leading to the potential misinterpretation.
There are two approaches (at least) to conducting significance tests. In one (favored by R. Fisher), a significance test is conducted and the probability value reflects the strength of the evidence against the null hypothesis. If the probability is below $0.01$, the data provide strong evidence that the null hypothesis is false. If the probability value is below $0.05$ but larger than $0.01$, then the null hypothesis is typically rejected, but not with as much confidence as it would be if the probability value were below $0.01$. Probability values between $0.05$ and $0.10$ provide weak evidence against the null hypothesis and, by convention, are not considered low enough to justify rejecting it. Higher probabilities provide less evidence that the null hypothesis is false.
The alternative approach (favored by the statisticians Neyman and Pearson) is to specify an α level before analyzing the data. If the data analysis results in a probability value below the $\alpha$ level, then the null hypothesis is rejected; if it is not, then the null hypothesis is not rejected. According to this perspective, if a result is significant, then it does not matter how significant it is. Moreover, if it is not significant, then it does not matter how close to being significant it is. Therefore, if the $0.05$ level is being used, then probability values of $0.049$ and $0.001$ are treated identically. Similarly, probability values of $0.06$ and $0.34$ are treated identically.
The former approach (preferred by Fisher) is more suitable for scientific research and will be adopted here. The latter is more suitable for applications in which a yes/no decision must be made. For example, if a statistical analysis were undertaken to determine whether a machine in a manufacturing plant were malfunctioning, the statistical analysis would be used to determine whether or not the machine should be shut down for repair. The plant manager would be less interested in assessing the weight of the evidence than knowing what action should be taken. There is no need for an immediate decision in scientific research where a researcher may conclude that there is some evidence against the null hypothesis, but that more research is needed before a definitive conclusion can be drawn. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/11%3A_Logic_of_Hypothesis_Testing/11.02%3A_Significance_Testing.txt |
Learning Objectives
• Define Type I and Type II errors
• Interpret significant and non-significant differences
• Explain why the null hypothesis should not be accepted when the effect is not significant
In the Physicians' Reactions case study, the probability value associated with the significance test is $0.0057$. Therefore, the null hypothesis was rejected, and it was concluded that physicians intend to spend less time with obese patients. Despite the low probability value, it is possible that the null hypothesis of no true difference between obese and average-weight patients is true and that the large difference between sample means occurred by chance. If this is the case, then the conclusion that physicians intend to spend less time with obese patients is in error. This type of error is called a Type I error. More generally, a Type I error occurs when a significance test results in the rejection of a true null hypothesis.
By one common convention, if the probability value is below $0.05$, then the null hypothesis is rejected. Another convention, although slightly less common, is to reject the null hypothesis if the probability value is below $0.01$. The threshold for rejecting the null hypothesis is called the $\alpha$ (alpha) level or simply $\alpha$. It is also called the significance level. As discussed in the section on significance testing, it is better to interpret the probability value as an indication of the weight of evidence against the null hypothesis than as part of a decision rule for making a reject or do-not-reject decision. Therefore, keep in mind that rejecting the null hypothesis is not an all-or-nothing decision.
The Type I error rate is affected by the $\alpha$ level: the lower the $\alpha$ level, the lower the Type I error rate. It might seem that $\alpha$ is the probability of a Type I error. However, this is not correct. Instead, $\alpha$ is the probability of a Type I error given that the null hypothesis is true. If the null hypothesis is false, then it is impossible to make a Type I error.
The second type of error that can be made in significance testing is failing to reject a false null hypothesis. This kind of error is called a Type II error. Unlike a Type I error, a Type II error is not really an error. When a statistical test is not significant, it means that the data do not provide strong evidence that the null hypothesis is false. Lack of significance does not support the conclusion that the null hypothesis is true. Therefore, a researcher should not make the mistake of incorrectly concluding that the null hypothesis is true when a statistical test was not significant. Instead, the researcher should consider the test inconclusive. Contrast this with a Type I error in which the researcher erroneously concludes that the null hypothesis is false when, in fact, it is true.
A Type II error can only occur if the null hypothesis is false. If the null hypothesis is false, then the probability of a Type II error is called $\beta$ (beta). The probability of correctly rejecting a false null hypothesis equals $1-\beta$ and is called power. Power is covered in detail in another section. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/11%3A_Logic_of_Hypothesis_Testing/11.03%3A_Type_I_and_II_Errors.txt |
Learning Objectives
• Define Type I and Type II errors
• Interpret significant and non-significant differences
• Explain why the null hypothesis should not be accepted when the effect is not significant
In the James Bond case study, Mr. Bond was given $16$ trials on which he judged whether a martini had been shaken or stirred. He was correct on $13$ of the trials. From the binomial distribution, we know that the probability of being correct $13$ or more times out of $16$ if one is only guessing is $0.0106$. Figure $1$ shows a graph of the binomial distribution. The red bars show the values greater than or equal to $13$. As you can see in the figure, the probabilities are calculated for the upper tail of the distribution. A probability calculated in only one tail of the distribution is called a "one-tailed probability."
Binomial Calculator
A slightly different question can be asked of the data: "What is the probability of getting a result as extreme or more extreme than the one observed?" Since the chance expectation is $8/16$, a result of $3/16$ is equally as extreme as $13/16$. Thus, to calculate this probability, we would consider both tails of the distribution. Since the binomial distribution is symmetric when $\pi =0.5$, this probability is exactly double the probability of $0.0106$ computed previously. Therefore, $p = 0.0212$. A probability calculated in both tails of a distribution is called a "two-tailed probability" (see Figure $2$).
Should the one-tailed or the two-tailed probability be used to assess Mr. Bond's performance? That depends on the way the question is posed. If we are asking whether Mr. Bond can tell the difference between shaken or stirred martinis, then we would conclude he could if he performed either much better than chance or much worse than chance. If he performed much worse than chance, we would conclude that he can tell the difference, but he does not know which is which. Therefore, since we are going to reject the null hypothesis if Mr. Bond does either very well or very poorly, we will use a two-tailed probability.
On the other hand, if our question is whether Mr. Bond is better than chance at determining whether a martini is shaken or stirred, we would use a one-tailed probability. What would the one-tailed probability be if Mr. Bond were correct on only $3$ of the $16$ trials? Since the one-tailed probability is the probability of the right-hand tail, it would be the probability of getting $3$ or more correct out of $16$. This is a very high probability and the null hypothesis would not be rejected.
The null hypothesis for the two-tailed test is $\pi =0.5$. By contrast, the null hypothesis for the one-tailed test is $\pi \leq 0.5$. Accordingly, we reject the two-tailed hypothesis if the sample proportion deviates greatly from $0.5$ in either direction. The one-tailed hypothesis is rejected only if the sample proportion is much greater than $0.5$. The alternative hypothesis in the two-tailed test is $\pi \neq 0.5$. In the one-tailed test it is $\pi > 0.5$.
You should always decide whether you are going to use a one-tailed or a two-tailed probability before looking at the data. Statistical tests that compute one-tailed probabilities are called one-tailed tests; those that compute two-tailed probabilities are called two-tailed tests. Two-tailed tests are much more common than one-tailed tests in scientific research because an outcome signifying that something other than chance is operating is usually worth noting. One-tailed tests are appropriate when it is not important to distinguish between no effect and an effect in the unexpected direction. For example, consider an experiment designed to test the efficacy of a treatment for the common cold. The researcher would only be interested in whether the treatment was better than a placebo control. It would not be worth distinguishing between the case in which the treatment was worse than a placebo and the case in which it was the same because in both cases the drug would be worthless.
Some have argued that a one-tailed test is justified whenever the researcher predicts the direction of an effect. The problem with this argument is that if the effect comes out strongly in the non-predicted direction, the researcher is not justified in concluding that the effect is not zero. Since this is unrealistic, one-tailed tests are usually viewed skeptically if justified on this basis alone. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/11%3A_Logic_of_Hypothesis_Testing/11.04%3A_One-_and_Two-Tailed_Tests.txt |
Learning Objectives
• Discuss whether rejection of the null hypothesis should be an all-or-none proposition
• State the usefulness of a significance test when it is extremely likely that the null hypothesis of no difference is false even before doing the experiment
When a probability value is below the $\alpha$ level, the effect is statistically significant and the null hypothesis is rejected. However, not all statistically significant effects should be treated the same way. For example, you should have less confidence that the null hypothesis is false if $p = 0.049$ than $p = 0.003$. Thus, rejecting the null hypothesis is not an all-or-none proposition.
If the null hypothesis is rejected, then the alternative to the null hypothesis (called the alternative hypothesis) is accepted. Consider the one-tailed test in the James Bond case study: Mr. Bond was given $16$ trials on which he judged whether a martini had been shaken or stirred and the question is whether he is better than chance on this task. The null hypothesis for this one-tailed test is that $\pi \leq 0.5$, where $\pi$ is the probability of being correct on any given trial. If this null hypothesis is rejected, then the alternative hypothesis that $\pi >0.5$ is accepted. If $\pi$ is greater than $0.5$, then Mr. Bond is better than chance on this task.
Now consider the two-tailed test used in the Physicians' Reactions case study. The null hypothesis is:
$μ_{obese} = μ_{average}$
If this null hypothesis is rejected, then there are two alternatives:
$μ_{obese} < μ_{average}$
$μ_{obese} > μ_{average}$
Naturally, the direction of the sample means determines which alternative is adopted. If the sample mean for the obese patients is significantly lower than the sample mean for the average-weight patients, then one should conclude that the population mean for the obese patients is lower than the population mean for the average-weight patients.
There are many situations in which it is very unlikely two conditions will have exactly the same population means. For example, it is practically impossible that aspirin and acetaminophen provide exactly the same degree of pain relief. Therefore, even before an experiment comparing their effectiveness is conducted, the researcher knows that the null hypothesis of exactly no difference is false. However, the researcher does not know which drug offers more relief. If a test of the difference is significant, then the direction of the difference is established. This point is also made in the section on the relationship between confidence intervals and significance tests.
Note
Some textbooks have incorrectly stated that rejecting the null hypothesis that two population means are equal does not justify a conclusion about which population mean is larger. Instead, they say that all one can conclude is that the population means differ. The validity of concluding the direction of the effect is clear if you note that a two-tailed test at the $0.05$ level is equivalent to two separate one-tailed tests each at the $0.025$ level. The two null hypotheses are then
$μ_{obese} ≥ μ_{average}$
$μ_{obese} ≤ μ_{average}$
If the former of these is rejected, then the conclusion is that the population mean for obese patients is lower than that for average-weight patients. If the latter is rejected, then the conclusion is that the population mean for obese patients is higher than that for average-weight patients. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/11%3A_Logic_of_Hypothesis_Testing/11.05%3A_Significant_Results.txt |
Learning Objectives
• Explain why the null hypothesis should not be accepted
• Discuss the problems of affirming a negative conclusion
When a significance test results in a high probability value, it means that the data provide little or no evidence that the null hypothesis is false. However, the high probability value is not evidence that the null hypothesis is true. The problem is that it is impossible to distinguish a null effect from a very small effect.
Example $1$
For example, in the James Bond Case Study, suppose Mr. Bond is, in fact, just barely better than chance at judging whether a martini was shaken or stirred. Assume he has a $0.51$ probability of being correct on a given trial $\pi=0.51$. Let's say Experimenter Jones (who did not know $\pi=0.51$ tested Mr. Bond and found he was correct $49$ times out of $100$ tries. How would the significance test come out?
Solution
The experimenter’s significance test would be based on the assumption that Mr. Bond has a $0.50$ probability of being correct on each trial $\pi=0.50$. Given this assumption, the probability of his being correct $49$ or more times out of $100$ is $0.62$. This means that the probability value is $0.62$, a value very much higher than the conventional significance level of $0.05$. This result, therefore, does not give even a hint that the null hypothesis is false. However, we know (but Experimenter Jones does not) that $\pi=0.51$ and not $0.50$ and therefore that the null hypothesis is false. So, if Experimenter Jones had concluded that the null hypothesis was true based on the statistical analysis, he or she would have been mistaken.
Concluding that the null hypothesis is true is called accepting the null hypothesis. To do so is a serious error.
Binomial Calculator
Further argument for not accepting the null hypothesis
Do not accept the null hypothesis when you do not reject it.
So how should the non-significant result be interpreted? The experimenter should report that there is no credible evidence Mr. Bond can tell whether a martini was shaken or stirred, but that there is no proof that he cannot. It is generally impossible to prove a negative. What if I claimed to have been Socrates in an earlier life? Since I have no evidence for this claim, I would have great difficulty convincing anyone that it is true. However, no one would be able to prove definitively that I was not.
Often a non-significant finding increases one's confidence that the null hypothesis is false. Consider the following hypothetical example.
Example $2$
A researcher develops a treatment for anxiety that he or she believes is better than the traditional treatment. A study is conducted to test the relative effectiveness of the two treatments: $20$ subjects are randomly divided into two groups of 10. One group receives the new treatment and the other receives the traditional treatment. The mean anxiety level is lower for those receiving the new treatment than for those receiving the traditional treatment. However, the difference is not significant. The statistical analysis shows that a difference as large or larger than the one obtained in the experiment would occur $11\%$ of the time even if there were no true difference between the treatments. In other words, the probability value is $0.11$. A naive researcher would interpret this finding as evidence that the new treatment is no more effective than the traditional treatment. However, the sophisticated researcher, although disappointed that the effect was not significant, would be encouraged that the new treatment led to less anxiety than the traditional treatment. The data support the thesis that the new treatment is better than the traditional one even though the effect is not statistically significant. This researcher should have more confidence that the new treatment is better than he or she had before the experiment was conducted. However, the support is weak and the data are inconclusive. What should the researcher do?
Solution
A reasonable course of action would be to do the experiment again. Let's say the researcher repeated the experiment and again found the new treatment was better than the traditional treatment. However, once again the effect was not significant and this time the probability value was $0.07$. The naive researcher would think that two out of two experiments failed to find significance and therefore the new treatment is unlikely to be better than the traditional treatment. The sophisticated researcher would note that two out of two times the new treatment was better than the traditional treatment. Moreover, two experiments each providing weak support that the new treatment is better, when taken together, can provide strong support. Using a method for combining probabilities, it can be determined that combining the probability values of $0.11$ and $0.07$ results in a probability value of $0.045$. Therefore, these two non-significant findings taken together result in a significant finding.
Although there is never a statistical basis for concluding that an effect is exactly zero, a statistical analysis can demonstrate that an effect is most likely small. This is done by computing a confidence interval. If all effect sizes in the interval are small, then it can be concluded that the effect is small. For example, suppose an experiment tested the effectiveness of a treatment for insomnia. Assume that the mean time to fall asleep was $2$ minutes shorter for those receiving the treatment than for those in the control group and that this difference was not significant. If the $95\%$ confidence interval ranged from $-4$ to $8$ minutes, then the researcher would be justified in concluding that the benefit is eight minutes or less. However, the researcher would not be justified in concluding the null hypothesis is true, or even that it was supported. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/11%3A_Logic_of_Hypothesis_Testing/11.06%3A_Non-Significant_Results.txt |
Learning Objectives
• Be able to state the null hypothesis for both one-tailed and two-tailed tests
• Differentiate between a significance level and a probability level
• State the four steps involved in significance testing
1. The first step is to specify the null hypothesis. For a two-tailed test, the null hypothesis is typically that a parameter equals zero although there are exceptions. A typical null hypothesis is $\mu _1-\mu _2=0$ which is equivalent to $\mu _1=\mu _2$. For a one-tailed test, the null hypothesis is either that a parameter is greater than or equal to zero or that a parameter is less than or equal to zero. If the prediction is that $\mu _1$ is larger than $\mu _2$, then the null hypothesis (the reverse of the prediction) is $\mu _1-\mu _2\geq 0$. This is equivalent to $\mu _1\leq \mu _2$.
2. The second step is to specify the $\alpha$ level which is also known as the significance level. Typical values are $0.05$ and $0.01$.
3. The third step is to compute the probability value (also known as the $p$ value). This is the probability of obtaining a sample statistic as different or more different from the parameter specified in the null hypothesis given that the null hypothesis is true.
4. Finally, compare the probability value with the $\alpha$ level. If the probability value is lower then you reject the null hypothesis. Keep in mind that rejecting the null hypothesis is not an all-or-none decision. The lower the probability value, the more confidence you can have that the null hypothesis is false. However, if your probability value is higher than the conventional $\alpha$ level of $0.05$, most scientists will consider your findings inconclusive. Failure to reject the null hypothesis does not constitute support for the null hypothesis. It just means you do not have sufficiently strong data to reject it.
11.08: Significance Testing and Confidence Intervals
Learning Objectives
• Explain why a confidence interval makes clear that one should not accept the null hypothesis
There is a close relationship between confidence intervals and significance tests. Specifically, if a statistic is significantly different from \(0\) at the \(0.05\) level, then the \(95\%\) confidence interval will not contain \(0\). All values in the confidence interval are plausible values for the parameter, whereas values outside the interval are rejected as plausible values for the parameter. In the Physicians' Reactions case study, the \(95\%\) confidence interval for the difference between means extends from \(2.00\) to \(11.26\). Therefore, any value lower than \(2.00\) or higher than \(11.26\) is rejected as a plausible value for the population difference between means. Since zero is lower than \(2.00\), it is rejected as a plausible value and a test of the null hypothesis that there is no difference between means is significant. It turns out that the \(p\) value is \(0.0057\). There is a similar relationship between the \(99\%\) confidence interval and significance at the \(0.01\) level.
Whenever an effect is significant, all values in the confidence interval will be on the same side of zero (either all positive or all negative). Therefore, a significant finding allows the researcher to specify the direction of the effect. There are many situations in which it is very unlikely two conditions will have exactly the same population means. For example, it is practically impossible that aspirin and acetaminophen provide exactly the same degree of pain relief. Therefore, even before an experiment comparing their effectiveness is conducted, the researcher knows that the null hypothesis of exactly no difference is false. However, the researcher does not know which drug offers more relief. If a test of the difference is significant, then the direction of the difference is established because the values in the confidence interval are either all positive or all negative.
If the \(95\%\) confidence interval contains zero (more precisely, the parameter value specified in the null hypothesis), then the effect will not be significant at the \(0.05\) level. Looking at non-significant effects in terms of confidence intervals makes clear why the null hypothesis should not be accepted when it is not rejected: Every value in the confidence interval is a plausible value of the parameter. Since zero is in the interval, it cannot be rejected. However, there is an infinite number of other values in the interval (assuming continuous measurement), and none of them can be rejected either.
11.09: Misconceptions of Hypothesis Testing
Learning Objectives
• State why the probability value is not the probability the null hypothesis is false
• Explain why a low probability value does not necessarily mean there is a large effect
• Explain why a non-significant outcome does not mean the null hypothesis is probably true
Misconceptions about significance testing are common. This section lists three important ones.
1. Misconception: The probability value is the probability that the null hypothesis is false.
Proper interpretation: The probability value is the probability of a result as extreme or more extreme given that the null hypothesis is true. It is the probability of the data given the null hypothesis. It is not the probability that the null hypothesis is false.
2. Misconception: A low probability value indicates a large effect.
Proper interpretation: A low probability value indicates that the sample outcome (or one more extreme) would be very unlikely if the null hypothesis were true. A low probability value can occur with small effect sizes, particularly if the sample size is large.
3. Misconception: A non-significant outcome means that the null hypothesis is probably true.
Proper interpretation: A non-significant outcome means that the data do not conclusively demonstrate that the null hypothesis is false.
11.10: Statistical Literacy
Learning Objectives
• Evidence for the Higgs Boson
Research in March, 2012 reported here found evidence for the existence of the Higgs Boson particle. However, the evidence for the existence of the particle was not statistically significant.
Example \(1\):what do you think?
Did the researchers conclude that their investigation had been a failure or did they conclude they have evidence of the particle, just not strong enough evidence to draw a confident conclusion?
Solution
One of the investigators stated, "We see some tantalizing evidence but not significant enough to make a stronger statement." Therefore, they were encouraged by the result. In a subsequent study, the evidence was significant. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/11%3A_Logic_of_Hypothesis_Testing/11.07%3A_Steps_in_Hypothesis_Testing.txt |
You may want to use the Binomial Calculator for some of these exercises.
General Questions
Q1
An experiment is conducted to test the claim that James Bond can taste the difference between a Martini that is shaken and one that is stirred. What is the null hypothesis? (relevant section)
Q2
The following explanation is incorrect. What three words should be added to make it correct? (relevant section)
The probability value is the probability of obtaining a statistic as different from the parameter specified in the null hypothesis as the statistic obtained in the experiment. The probability value is computed assuming that the null hypothesis is true.
Q3
Why do experimenters test hypotheses they think are false? (relevant section)
Q4
State the null hypothesis for:
1. An experiment testing whether echinacea decreases the length of colds.
2. A correlational study on the relationship between brain size and intelligence.
3. An investigation of whether a self-proclaimed psychic can predict the outcome of a coin flip.
4. A study comparing a drug with a placebo on the amount of pain relief. (A one-tailed test was used.)
(relevant section & relevant section)
Q5
Assume the null hypothesis is that $\mu =50$ and that the graph shown below is the sampling distribution of the mean ($M$). Would a sample value of $M= 60$ be significant in a two-tailed test at the $0.05$ level? Roughly what value of $M$ would be needed to be significant? (relevant section & relevant section)
Q6
A researcher develops a new theory that predicts that vegetarians will have more of a particular vitamin in their blood than non-vegetarians. An experiment is conducted and vegetarians do have more of the vitamin, but the difference is not significant. The probability value is $0.13$. Should the experimenter's confidence in the theory increase, decrease, or stay the same? (relevant section)
Q7
A researcher hypothesizes that the lowering in cholesterol associated with weight loss is really due to exercise. To test this, the researcher carefully controls for exercise while comparing the cholesterol levels of a group of subjects who lose weight by dieting with a control group that does not diet. The difference between groups in cholesterol is not significant. Can the researcher claim that weight loss has no effect? (relevant section)
Q8
A significance test is performed and $p = 0.20$. Why can't the experimenter claim that the probability that the null hypothesis is true is $0.20$? (relevant section, relevant section & relevant section)
Q9
For a drug to be approved by the FDA, the drug must be shown to be safe and effective. If the drug is significantly more effective than a placebo, then the drug is deemed effective. What do you know about the effectiveness of a drug once it has been approved by the FDA (assuming that there has not been a Type I error)? (relevant section)
Q10
When is it valid to use a one-tailed test? What is the advantage of a one-tailed test? Give an example of a null hypothesis that would be tested by a one-tailed test. (relevant section)
Q11
Distinguish between probability value and significance level. (relevant section)
Q12
Suppose a study was conducted on the effectiveness of a class on "How to take tests." The SAT scores of an experimental group and a control group were compared. (There were $100$ subjects in each group.) The mean score of the experimental group was $503$ and the mean score of the control group was $499$. The difference between means was found to be significant, $p = 0.037$. What do you conclude about the effectiveness of the class? (relevant section & relevant section)
Q13
Is it more conservative to use an alpha level of $0.01$ or an alpha level of $0.05$? Would beta be higher for an alpha of $0.05$ or for an alpha of $0.01$? (relevant section)
Q14
Why is $H_o: M_1 = M_2$ not a proper null hypothesis? (relevant section)
Q15
An experimenter expects an effect to come out in a certain direction. Is this sufficient basis for using a one-tailed test? Why or why not? (relevant section)
Q16
How do the Type I and Type II error rates of one-tailed and two-tailed tests differ? (relevant section & relevant section)
Q17
A two-tailed probability is $0.03$. What is the one-tailed probability if the effect were in the specified direction? What would it be if the effect were in the other direction? (relevant section)
Q18
You choose an alpha level of $0.01$ and then analyze your data.
1. What is the probability that you will make a Type I error given that the null hypothesis is true?
2. What is the probability that you will make a Type I error given that the null hypothesis is false? (relevant section)
Q19
Why doesn't it make sense to test the hypothesis that the sample mean is $42$? (relevant section & relevant section)
Q20
True/false: It is easier to reject the null hypothesis if the researcher uses a smaller alpha ($\alpha$) level. (relevant section & relevant section)
Q21
True/false: You are more likely to make a Type I error when using a small sample than when using a large sample. (relevant section)
Q22
True/false: You accept the alternative hypothesis when you reject the null hypothesis. (relevant section)
Q23
True/false: You do not accept the null hypothesis when you fail to reject it. (relevant section)
Q24
True/false: A researcher risks making a Type I error any time the null hypothesis is rejected. (relevant section) | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/11%3A_Logic_of_Hypothesis_Testing/11.E%3A_Logic_of_Hypothesis_Testing_%28Exercises%29.txt |
Many, if not most experiments are designed to compare means. The experiment may involve only one sample mean that is to be compared to a specific value. Or the experiment could be testing differences among many different experimental conditions, and the experimenter could be interested in comparing each mean with each of the other means. This chapter covers methods of comparing means in many different experimental situations. The topics covered here in sections E, F, I, and J are typically covered in other texts in a chapter on Analysis of Variance. We prefer to cover them here since they bear no necessary relationship to analysis of variance. As discussed by Wilkinson (1999), it is not logical to consider the procedures in this chapter as tests to be performed subsequent to an analysis of variance. Nor is it logical to call them post-hoc tests as some computer programs do.
Thumbnail: Student's t-distribution with 2 degrees of freedom. (CC BY-SA 3.0; IkamusumeFan).
12: Tests of Means
Learning Objectives
• Compute the probability of a sample mean being at least as high as a specified value when $\sigma$ is known
• Compute a two-tailed probability
• Compute the probability of a sample mean being at least as high as a specified value when $\sigma$ is estimated
• State the assumptions required for item $3$ above
This section shows how to test the null hypothesis that the population mean is equal to some hypothesized value. For example, suppose an experimenter wanted to know if people are influenced by a subliminal message and performed the following experiment. Each of nine subjects is presented with a series of $100$ pairs of pictures. As a pair of pictures is presented, a subliminal message is presented suggesting the picture that the subject should choose. The question is whether the (population) mean number of times the suggested picture is chosen is equal to $50$. In other words, the null hypothesis is that the population mean ($\mu$) is $50$. The (hypothetical) data are shown in Table $1$. The data in Table $1$ have a sample mean ($M$) of $51$. Thus the sample mean differs from the hypothesized population mean by $1$.
Table $1$: Distribution of scores
Frequency
45
48
49
49
51
52
53
55
57
The significance test consists of computing the probability of a sample mean differing from $\mu$ by one (the difference between the hypothesized population mean and the sample mean) or more. The first step is to determine the sampling distribution of the mean. As shown in a previous section, the mean and standard deviation of the sampling distribution of the mean are
$μ_M = μ$
and
$\sigma_M = \dfrac{\sigma}{\sqrt{N}}$
respectively. It is clear that $\mu _M=50$. In order to compute the standard deviation of the sampling distribution of the mean, we have to know the population standard deviation ($\sigma$).
The current example was constructed to be one of the few instances in which the standard deviation is known. In practice, it is very unlikely that you would know $\sigma$ and therefore you would use $s$, the sample estimate of $\sigma$. However, it is instructive to see how the probability is computed if $\sigma$ is known before proceeding to see how it is calculated when $\sigma$ is estimated.
For the current example, if the null hypothesis is true, then based on the binomial distribution, one can compute that variance of the number correct is
$\sigma ^2 = N\pi (1-\pi ) = 100(0.5)(1-0.5) = 25$
Therefore, $\sigma =5$. For a $\sigma$ of $5$ and an $N$ of $9$, the standard deviation of the sampling distribution of the mean is $5/3 = 1.667$. Recall that the standard deviation of a sampling distribution is called the standard error.
To recap, we wish to know the probability of obtaining a sample mean of $51$ or more when the sampling distribution of the mean has a mean of $50$ and a standard deviation of $1.667$. To compute this probability, we will make the assumption that the sampling distribution of the mean is normally distributed. We can then use the normal distribution calculator as shown in Figure $1$.
Notice that the mean is set to $50$, the standard deviation to $1.667$, and the area above $51$ is requested and shown to be $0.274$.
Therefore, the probability of obtaining a sample mean of $51$ or larger is $0.274$. Since a mean of $51$ or higher is not unlikely under the assumption that the subliminal message has no effect, the effect is not significant and the null hypothesis is not rejected.
The test conducted above was a one-tailed test because it computed the probability of a sample mean being one or more points higher than the hypothesized mean of $50$ and the area computed was the area above $51$. To test the two-tailed hypothesis, you would compute the probability of a sample mean differing by one or more in either direction from the hypothesized mean of $50$. You would do so by computing the probability of a mean being less than or equal to $49$ or greater than or equal to $51$.
The results of the normal distribution calculator are shown in Figure $2$.
As you can see, the probability is $0.548$ which, as expected, is twice the probability of $0.274$ shown in Figure $1$.
Before normal calculators such as the one illustrated above were widely available, probability calculations were made based on the standard normal distribution. This was done by computing $Z$ based on the formula
$Z=\frac{M-\mu }{\sigma _M}$
where $Z$ is the value on the standard normal distribution, $M$ is the sample mean, $\mu$ is the hypothesized value of the mean, and $\sigma _M$ is the standard error of the mean. For this example, $Z = (51-50)/1.667 = 0.60$. Use the normal calculator, with a mean of $0$ and a standard deviation of $1$, as shown below.
Notice that the probability (the shaded area) is the same as previously calculated (for the one-tailed test).
As noted, in real-world data analysis it is very rare that you would know $\sigma$ and wish to estimate $\mu$. Typically $\sigma$ is not known and is estimated in a sample by $s$, and $\sigma _M$ is estimated by $s_M$.
Example $1$: adhd treatment
Consider the data in the "ADHD Treatment" case study. These data consist of the scores of $24$ children with ADHD on a delay of gratification (DOG) task. Each child was tested under four dosage levels. Table $2$ shows the data for the placebo ($0$ mg) and highest dosage level ($0.6$ mg) of methylphenidate. Of particular interest here is the column labeled "Diff" that shows the difference in performance between the $0.6$ mg ($D60$) and the $0$ mg ($D0$) conditions. These difference scores are positive for children who performed better in the $0.6$ mg condition than in the control condition and negative for those who scored better in the control condition. If methylphenidate has a positive effect, then the mean difference score in the population will be positive. The null hypothesis is that the mean difference score in the population is $0$.
Table $2$: DOG scores as a function of dosage
D0 D60 Diff
57 62 5
27 49 22
32 30 -2
31 34 3
34 38 4
38 36 -2
71 77 6
33 51 18
34 45 11
53 42 -11
36 43 7
42 57 15
26 36 10
52 58 6
36 35 -1
55 60 5
36 33 -3
42 49 7
36 33 -3
54 59 5
34 35 1
29 37 8
33 45 12
33 29 -4
Solution
To test this null hypothesis, we compute $t$ using a special case of the following formula:
$t=\frac{\text{statistic-hypothesized value}}{\text{estimated standard error of the statistic}}$
The special case of this formula applicable to testing a single mean is
$t=\frac{M-\mu }{s_M}$
where $t$ is the value we compute for the significance test, $M$ is the sample mean, $\mu$ is the hypothesized value of the population mean, and $s_M$ is the estimated standard error of the mean. Notice the similarity of this formula to the formula for Z.
In the previous example, we assumed that the scores were normally distributed. In this case, it is the population of difference scores that we assume to be normally distributed.
The mean ($M$) of the $N = 24$ difference scores is $4.958$, the hypothesized value of $\mu$ is $0$, and the standard deviation ($s$) is $7.538$. The estimate of the standard error of the mean is computed as:
$s_M=\frac{s}{\sqrt{N}}=\frac{7.5382}{\sqrt{24}}=1.54$
Therefore,
$t =\frac{4.96}{1.54} = 3.22$
The probability value for $t$ depends on the degrees of freedom. The number of degrees of freedom is equal to $N - 1 = 23$. As shown below, the t distribution calculator finds that the probability of a $t$ less than $-3.22$ or greater than $3.22$ is only $0.0038$. Therefore, if the drug had no effect, the probability of finding a difference between means as large or larger (in either direction) than the difference found is very low. Therefore the null hypothesis that the population mean difference score is zero can be rejected. The conclusion is that the population mean for the drug condition is higher than the population mean for the placebo condition.
Review of Assumptions
1. Each value is sampled independently from each other value.
2. The values are sampled from a normal distribution. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/12%3A_Tests_of_Means/12.01%3A_Testing_a_Single_Mean.txt |
Learning Objectives
• State how the degrees of freedom affect the difference between the \(t\) and normal distributions
Instructions
This demonstration allows you to compare the \(t\) distribution to the standard normal distribution. At the start, the standard normal distribution is compared to a \(t\) distribution with three degrees of freedom. You can change the degrees of freedom of the \(t\) distribution with the slider. The \(3\) and the \(50\) mark the ends of the slider. The "current" degrees of freedom are shown at the bottom where it says "\(t\) distribution with \(df=3\)." As you change the slider, the df are shown by this last line. The "zoom in" button allows you to change the scale of the graph to see the tails of the distribution in more detail.
Illustrated Instructions
Video Demo
In the video the slider is used to increase the degrees of freedom. Notice how the \(t\)-distribution change as the degrees of freedom increase. The video concludes with the tails of the tails of the graphs being zoomed into via the "zoom in" button and the degrees of freedom being decreased.
12.03: Difference between Two Means
Learning Objectives
• State the assumptions for testing the difference between two means
• Estimate the population variance assuming homogeneity of variance
• Compute the standard error of the difference between means
• Compute $t$ and $p$ for the difference between means
• Format data for computer analysis
It is much more common for a researcher to be interested in the difference between means than in the specific values of the means themselves. This section covers how to test for differences between means from two separate groups of subjects. A later section describes how to test for differences between the means of two conditions in designs where only one group of subjects is used and each subject is tested in each condition.
We take as an example the data from the "Animal Research" case study. In this experiment, students rated (on a $7$-point scale) whether they thought animal research is wrong. The sample sizes, means, and variances are shown separately for males and females in Table $1$.
Table $1$: Means and Variances in Animal Research study
Group n Mean Variance
Females 17 5.353 2.743
Males 17 3.882 2.985
As you can see, the females rated animal research as more wrong than did the males. This sample difference between the female mean of $5.35$ and the male mean of $3.88$ is $1.47$. However, the gender difference in this particular sample is not very important. What is important is whether there is a difference in the population means.
In order to test whether there is a difference between population means, we are going to make three assumptions:
1. The two populations have the same variance. This assumption is called the assumption of homogeneity of variance.
2. The populations are normally distributed.
3. Each value is sampled independently from each other value. This assumption requires that each subject provide only one value. If a subject provides two scores, then the scores are not independent. The analysis of data with two scores per subject is shown in the section on the correlated t test later in this chapter.
The consequences of violating the first two assumptions are investigated in the simulation in the next section. For now, suffice it to say that small-to-moderate violations of assumptions $1$ and $2$ do not make much difference. It is important not to violate assumption $3$.
We saw the following general formula for significance testing in the section on testing a single mean:
$t=\frac{\text{statistic-hypothesized value}}{\text{estimated standard error of the statistic}}$
In this case, our statistic is the difference between sample means and our hypothesized value is $0$. The hypothesized value is the null hypothesis that the difference between population means is $0$.
We continue to use the data from the "Animal Research" case study and will compute a significance test on the difference between the mean score of the females and the mean score of the males. For this calculation, we will make the three assumptions specified above.
The first step is to compute the statistic, which is simply the difference between means.
$M_1 - M_2 = 5.3529 - 3.8824 = 1.4705$
Since the hypothesized value is $0$, we do not need to subtract it from the statistic.
The next step is to compute the estimate of the standard error of the statistic. In this case, the statistic is the difference between means, so the estimated standard error of the statistic is ($S_{M_1 - M_2}$). Recall from the relevant section in the chapter on sampling distributions that the formula for the standard error of the difference between means is:
$\sigma _{M_1 - M_2}=\sqrt{\frac{\sigma _{1}^{2}}{n_1}+\frac{\sigma _{2}^{2}}{n_2}}=\sqrt{\frac{\sigma ^2}{n}+\frac{\sigma ^2}{n}}=\sqrt{\frac{2\sigma ^2}{n}}$
In order to estimate this quantity, we estimate $\sigma ^2$ and use that estimate in place of $\sigma ^2$. Since we are assuming the two population variances are the same, we estimate this variance by averaging our two sample variances. Thus, our estimate of variance is computed using the following formula:
$MSE=\frac{s_{1}^{2}+s_{2}^{2}}{2}$
where $MSE$ is our estimate of $\sigma ^2$. In this example,
$MSE = \frac{2.743 + 2.985}{2} = 2.864$
Since $n$ (the number of scores in each group) is $17$,
$S_{M_1-M_2}=\sqrt{\frac{2MSE}{n}}=\sqrt{\frac{(2)(2.864)}{17}}=0.5805$
The next step is to compute $t$ by plugging these values into the formula:
$t = \frac{1.4705}{0.5805} = 2.533$
Finally, we compute the probability of getting a $t$ as large or larger than $2.533$ or as small or smaller than $-2.533$. To do this, we need to know the degrees of freedom. The degrees of freedom is the number of independent estimates of variance on which $MSE$ is based. This is equal to $(n_1 - 1) + (n_2 - 1)$, where $n_1$ is the sample size of the first group and $n_2$ is the sample size of the second group. For this example, $n_1 = n_2 = 17$. When $n_1 = n_2$, it is conventional to use "$n$" to refer to the sample size of each group. Therefore, the degrees of freedom is $16 + 16 = 32$.
Once we have the degrees of freedom, we can use the t distribution calculator to find the probability. Figure $1$ shows that the probability value for a two-tailed test is $0.0164$. The two-tailed test is used when the null hypothesis can be rejected regardless of the direction of the effect. As shown in Figure $1$, it is the probability of a $t < -2.533$ or a $t > 2.533$.
The results of a one-tailed test are shown in Figure $2$. As you can see, the probability value of $0.0082$ is half the value for the two-tailed test.
Formatting Data for Computer Analysis
Most computer programs that compute $t$ tests require your data to be in a specific form. Consider the data in Table $2$.
Table $2$: Example Data
Group 1 Group 2
3 2
4 6
5 8
Here there are two groups, each with three observations. To format these data for a computer program, you normally have to use two variables: the first specifies the group the subject is in and the second is the score itself. The reformatted version of the data in Table $2$ is shown in Table $3$.
Table $3$: Reformatted Data
G Y
1 3
1 4
1 5
2 2
2 6
2 8
To use Analysis Lab to do the calculations, you would copy the data and then
1. Click the "Enter/Edit Data" button. (You may be warned that for security reasons you must use the keyboard shortcut for pasting data.)
2. Paste your data.
3. Click "Accept Data."
4. Set the Dependent Variable to $Y$.
5. Set the Grouping Variable to $G$.
6. Click the "$t$-test/confidence interval" button.
The $t$ value is $-0.718$, the $df = 4$, and $p = 0.512$.
Computations for Unequal Sample Sizes (optional)
The calculations are somewhat more complicated when the sample sizes are not equal. One consideration is that $MSE$, the estimate of variance, counts the group with the larger sample size more than the group with the smaller sample size. Computationally, this is done by computing the sum of squares error ($SSE$) as follows:
$SSE=\sum (X-M_1)^2+\sum (X-M_2)^2$
where $M_1$ is the mean for group $1$ and $M_2$ is the mean for group $2$. Consider the following small example:
Table $4$: Unequal $n$
Group 1 Group 2
3 2
4 4
5
$M_1 = 4 \; \text{and}\; M_2 = 3$
$SSE = (3-4)^2 + (4-4)^2 + (5-4)^2 + (2-3)^2 + (4-3)^2 = 4$
Then, $MSE$ is computed by:
$MSE = \frac{SSE}{df}$
where the degrees of freedom ($df$) is computed as before:
$df = (n_1 - 1) + (n_2 - 1) = (3 - 1) + (2 - 1) = 3$
$MSE = \frac{SSE}{df}=\frac{4}{3}=1.333$
The formula
$S_{M_1-M_2}=\sqrt{\frac{2MSE}{n}}$
is replaced by
$S_{M_1-M_2}=\sqrt{\frac{2MSE}{\mathfrak{n} _h}}$
where $\mathfrak{n} _h$ is the harmonic mean of the sample sizes and is computed as follows:
$\mathfrak{n} _h=\frac{2}{\tfrac{1}{\mathfrak{n_1}}+\tfrac{1}{\mathfrak{n_2}}}=\frac{2}{\tfrac{1}{3}+\tfrac{1}{2}}=2.4$
and
$S_{M_1-M_2}=\sqrt{\frac{(2)(1.333)}{2.4}}=1.054$
Therefore,
$t = \frac{4-3}{1.054} = 0.949$
and the two-tailed $p = 0.413$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/12%3A_Tests_of_Means/12.02%3A_t_Distribution_Demo.txt |
Learning Objectives
• State when heterogenety of variance can lead to a very high Type I error rate
• State the effect of skew of on the Type I error rate
Instructions
This demonstration allows you to explore the effects of violating the assumptions of normality and homogeneity of variance. When the simulation starts you see the distributions of two populations. By default, they are both normally distributed, have means of \(0\) and standard deviations of \(1\). The default sample size for the simulations is \(5\) per group. If you push the "simulate" button, \(2,000\) simulated experiments are conducted. You can adjust the number of simulations from \(2,000\) to \(10,000\). A \(t\)-test is computed for each experiment and the number of tests that were significant, not significant, and the Type I error rate (the proportion significant) are displayed.
Since the null hypothesis is true and all assumptions are met with these default values, the Type I error rate should be close to \(0.05\), especially if you ran a large number of simulations. It will not equal \(0.05\) because of random variation. However, the larger the number of simulations you run, the closer the Type I error rate should come to \(0.05\).
You can explore the effects of violating the assumptions of the test by making one or both of the distributions skewed and/or by making the standard deviations of the distributions different. You can also explore the effects of sample size and of the significance level used (\(0.05\) or \(0.01\)).
By exploring various distributions, sample sizes, and significance levels, you can get a feeling for how well the test works when its violations are violated. A test that is relatively unaffected by violations of its assumptions is said to be "robust."
Illustrated Instructions
Video Demo
The video below begins by running \(2000\) simulations with the two populations each with means of \(0\), standard deviations of \(2\) no skewness and sample sizes of \(5\). The video continues by varying different aspect of the distributions and running more simulations. Note the number of significant tests after each set of simulations.
12.05: Pairwise Comparisons
Learning Objectives
• Define pairwise comparison
• Describe the problem with doing $t$ tests among all pairs of means
• Calculate the Tukey HSD test
• Explain why the Tukey test should not necessarily be considered a follow-up test
Many experiments are designed to compare more than two conditions. We will take as an example the case study "Smiles and Leniency." In this study, the effect of different types of smiles on the leniency shown to a person was investigated. An obvious way to proceed would be to do a t test of the difference between each group mean and each of the other group means. This procedure would lead to the six comparisons shown in Table $1$.
Table $1$: Six Comparisons among Means
false vs felt
false vs miserable
false vs neutral
felt vs miserable
felt vs neutral
miserable vs neutral
The problem with this approach is that if you did this analysis, you would have six chances to make a Type I error. Therefore, if you were using the $0.05$ significance level, the probability that you would make a Type I error on at least one of these comparisons is greater than $0.05$. The more means that are compared, the more the Type I error rate is inflated. Figure $1$ shows the number of possible comparisons between pairs of means (pairwise comparisons) as a function of the number of means. If there are only two means, then only one comparison can be made. If there are $12$ means, then there are $66$ possible comparisons.
Figure $2$ shows the probability of a Type I error as a function of the number of means. As you can see, if you have an experiment with $12$ means, the probability is about $0.70$ that at least one of the $66$ comparisons among means would be significant even if all $12$ population means were the same.
The Tukey Honestly Significant Difference Test
The Type I error rate can be controlled using a test called the Tukey Honestly Significant Difference test or Tukey HSD for short. The Tukey HSD is based on a variation of the $t$ distribution that takes into account the number of means being compared. This distribution is called the studentized range distribution.
Let's return to the leniency study to see how to compute the Tukey HSD test. You will see that the computations are very similar to those of an independent-groups t test. The steps are outlined below:
1. Compute the means and variances of each group. They are shown below.
Table $2$: Means and variances of each group
Condition Mean Variance
False 5.37 3.34
Felt 4.91 2.83
Miserable 4.91 2.11
Neutral 4.12 2.32
1. Compute $MSE$, which is simply the mean of the variances. It is equal to $2.65$.
2. Compute $Q=\frac{M_i-M_j}{\sqrt{\tfrac{MSE}{n}}}$ for each pair of means, where $M_i$ is one mean, $M_j$ is the other mean, and $n$ is the number of scores in each group. For these data, there are $34$ observations per group. The value in the denominator is $0.279$.
3. Compute $p$ for each comparison using the Studentized Range Calculator. The degrees of freedom is equal to the total number of observations minus the number of means. For this experiment, $df = 136 - 4 = 132$.
The tests for these data are shown in Table $2$.
Table $2$: Six Pairwise Comparisons
Comparison Mi-Mj Q p
False - Felt 0.46 1.65 0.649
False - Miserable 0.46 1.65 0.649
False - Neutral 1.25 4.48 0.010
Felt - Miserable 0.00 0.00 1.000
Felt - Neutral 0.79 2.83 0.193
Miserable - Neutral 0.79 2.83 0.193
The only significant comparison is between the false smile and the neutral smile.
It is not unusual to obtain results that on the surface appear paradoxical. For example, these results appear to indicate that
1. the false smile is the same as the miserable smile,
2. the miserable smile is the same as the neutral control, and
3. the false smile is different from the neutral control.
This apparent contradiction is avoided if you are careful not to accept the null hypothesis when you fail to reject it. The finding that the false smile is not significantly different from the miserable smile does not mean that they are really the same. Rather it means that there is not convincing evidence that they are different. Similarly, the non-significant difference between the miserable smile and the control does not mean that they are the same. The proper conclusion is that the false smile is higher than the control and that the miserable smile is either
1. equal to the false smile,
2. equal to the control, or
3. somewhere in-between.
The assumptions of the Tukey test are essentially the same as for an independent-groups t test: normality, homogeneity of variance, and independent observations. The test is quite robust to violations of normality. Violating homogeneity of variance can be more problematical than in the two-sample case since the $MSE$ is based on data from all groups. The assumption of independence of observations is important and should not be violated.
Computer Analysis
For most computer programs, you should format your data the same way you do for an independent-groups t test. The only difference is that if you have, say, four groups, you would code each group as $1$, $2$, $3$, or $4$ rather than just $1$ or $2$.
Although full-featured statistics programs such as SAS, SPSS, R, and others can compute Tukey's test, smaller programs (including Analysis Lab) may not. However, these programs are generally able to compute a procedure known as Analysis of Variance (ANOVA). This procedure will be described in detail in a later chapter. Its relevance here is that an ANOVA computes the $MSE$ that is used in the calculation of Tukey's test. For example, the following shows the ANOVA summary table for the "Smiles and Leniency" data.
The column labeled MS stands for "Mean Square" and therefore the value $2.6489$ in the "Error" row and the MS column is the "Mean Square Error" or MSE. Recall that this is the same value computed here ($2.65$) when rounded off.
Tukey's Test Need Not be a Follow-Up to ANOVA
Some textbooks introduce the Tukey test only as a follow-up to an analysis of variance. There is no logical or statistical reason why you should not use the Tukey test even if you do not compute an ANOVA (or even know what one is). If you or your instructor do not wish to take our word for this, see the excellent article on this and other issues in statistical analysis by Leland Wilkinson and the APA Board of Scientific Affairs' Task Force on Statistical Inference, published in the American Psychologist, August 1999, Vol. 54, No. 8, 594–604.
Computations for Unequal Sample Sizes (optional)
The calculation of $MSE$ for unequal sample sizes is similar to its calculation in an independent-groups t test. Here are the steps:
1. Compute a Sum of Squares Error ($SSE$) using the following formula $SSE=\sum (X-M_1)^2+\sum (X-M_2)^2+\cdots +\sum (X-M_k)^2$ where $M_i$ is the mean of the $i^{th}$ group and $k$ is the number of groups.
2. Compute the degrees of freedom error ($dfe)$ by subtracting the number of groups ($k$) from the total number of observations ($N$). Therefore, $dfe = N - k$
3. Compute $MSE$ by dividing $SSE$ by $dfe$:$MSE = \frac{SSE}{dfe}$
4. For each comparison of means, use the harmonic mean of the $n's$ for the two means ($\mathfrak{n_h}$).
All other aspects of the calculations are the same as when you have equal sample sizes. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/12%3A_Tests_of_Means/12.04%3A_Robustness_Simulation.txt |
Learning Objectives
• Define linear combination
• Specify a linear combination in terms of coefficients
• Do a significance test for a specific comparison
There are many situations in which the comparisons among means are more complicated than simply comparing one mean with another. This section shows how to test these more complex comparisons. The methods in this section assume that the comparison among means was decided on before looking at the data. Therefore these comparisons are called planned comparisons. A different procedure is necessary for unplanned comparisons.
Let's begin with the made-up data from a hypothetical experiment shown in Table $1$. Twelve subjects were selected from a population of high-self-esteem subjects ($esteem = 1$) and an additional $12$ subjects were selected from a population of low-self-esteem subjects ($esteem = 2$). Subjects then performed on a task and (independent of how well they really did) half in each esteem category were told they succeeded ($outcome = 1$) and the other half were told they failed ($outcome = 2$). Therefore, there were six subjects in each of the four esteem/outcome combinations and $24$ subjects in all.
After the task, subjects were asked to rate (on a $10$-point scale) how much of their outcome (success or failure) they attributed to themselves as opposed to being due to the nature of the task.
Table $1$: Data from Hypothetical Experiment
outcome esteem attrib
1 1 7
1 1 8
1 1 7
1 1 8
1 1 9
1 1 5
1 2 6
1 2 5
1 2 7
1 2 4
1 2 5
1 2 6
2 1 4
2 1 6
2 1 5
2 1 4
2 1 7
2 1 3
2 2 9
2 2 8
2 2 9
2 2 8
2 2 7
2 2 6
The means of the four conditions are shown in Table $2$.
Table $2$: Mean ratings of self-attributions of success or failure
Outcome Esteem Mean
Success High Self-Esteem 7.333
Low Self-Esteem 5.500
Failure High Self-Esteem 4.833
Low Self-Esteem 7.833
There are several questions we can ask about the data. We begin by asking whether, on average, subjects who were told they succeeded differed significantly from subjects who were told they failed. The means for subjects in the success condition are $7.333$ for the high-self-esteem subjects and $5.500$ for the low-self-esteem subjects. Therefore, the mean for all subjects in the success condition is $\tfrac{7.3333 + 5.5000}{2} = 6.4167$. Similarly, the mean for all subjects in the failure condition is $\tfrac{4.8333 + 7.8333}{2} = 6.3333$. The question is: How do we do a significance test for this difference of $6.4167-6.3333 = 0.083$?
The first step is to express this difference in terms of a linear combination using a set of coefficients and the means. This may sound complex, but it is really pretty easy. We can compute the mean of the success conditions by multiplying each success mean by $0.5$ and then adding the result. In other words, we compute
$(0.5)(7.333) + (0.5)(5.500)= 3.67 + 2.75= 6.42$
Similarly, we can compute the mean of the failure conditions by multiplying each "failure" mean by $0.5$ and then adding the result:
$(0.5)(4.833) + (0.5)(7.833)= 2.417 + 3.917= 6.33$
The difference between the two means can be expressed as
$0.5 \times 7.333 + 0.5 \times 5.500 -(0.5 \times 4.833 + 0.5 \times 7.833) = 0.5 \times 7.333 + 0.5 \times 5.500 - 0.5 \times 4.833 - 0.5 \times 7.833$
We therefore can compute the difference between the "success" mean and the "failure" mean by multiplying each "success" mean by $0.5$, each failure mean by $-0.5$, and adding the results. In Table $3$, the coefficient column is the multiplier and the product column is the result of the multiplication. If we add up the four values in the product column, we get
$L = 3.667 + 2.750 - 2.417 - 3.917 = 0.083$
This is the same value we got when we computed the difference between means previously (within rounding error). We call the value "$L$" for "linear combination."
Table $3$: Coefficients for comparing low and high self-esteem
Outcome Esteem Mean Coeff Product
Success High Self-Esteem 7.333 0.5 3.667
Low Self-Esteem 5.500 0.5 2.750
Failure High Self-Esteem 4.833 -0.5 -2.417
Low Self-Esteem 7.833 -0.5 -3.917
Now, the question is whether our value of $L$ is significantly different from $0$. The general formula for $L$ is
$L=\sum c_iM_i$
where $c_i$ is the $i_{th}$ coefficient and $M_i$ is the $i_{th}$ mean. As shown above, $L = 0.083$. The formula for testing $L$ for significance is shown below:
$t=\cfrac{L}{\sqrt{\cfrac{\sum c_{i}^{2}MSE}{n}}}$
In this example,
$\sum c_i^2 = 0.5^2 + 0.5^2 + (-0.5)^2 + (-0.5)^2 = 1.00$
$MSE$ is the mean of the variances. The four variances are shown in Table $4$. Their mean is $1.625$. Therefore $MSE = 1.625$.
Table $4$: Variances of attributions of success or failure to oneself
Outcome Esteem Variance
Success High Self-Esteem 1.867
Low Self-Esteem 1.100
Failure High Self-Esteem 2.167
Low Self-Esteem 1.367
The value of n is the number of subjects in each group. Here $n = 6$.
Putting it all together,
$t= \cfrac{0.083}{\sqrt{\cfrac{(1)(1.625)}{6}}} = 0.16$
We need to know the degrees of freedom in order to compute the probability value. The degrees of freedom is
$df = N - k$
where $N$ is the total number of subjects ($24$) and $k$ is the number of groups ($4$). Therefore, $df = 20$. Using the Online Calculator, we find that the two-tailed probability value is $0.874$. Therefore, the difference between the "success" condition and the "failure" condition is not significant.
t distribution Calculator
A more interesting question about the results is whether the effect of outcome (success or failure) differs depending on the self-esteem of the subject. For example, success may make high-self-esteem subjects more likely to attribute the outcome to themselves, whereas success may make low-self-esteem subjects less likely to attribute the outcome to themselves.
To test this, we have to test a difference between differences. Specifically, is the difference between success and failure outcomes for the high-self-esteem subjects different from the difference between success and failure outcomes for the low-self-esteem subjects? The means shown in Table 5 show that this is the case. For the high-self-esteem subjects, the difference between the success and failure attribution scores is $7.333-4.833 = 2.500$. For low-self-esteem subjects, the difference is $5.500-7.833 = -2.333$. The difference between differences is $2.500 - (-2.333) = 4.833$.
The coefficients to test this difference between differences are shown in Table $5$.
Table $5$: Coefficients for testing difference between differences
Self-Esteem Outcome Mean Coeff Product
High Success 7.333 1 7.333
Failure 4.833 -1 -4.833
Low Success 5.500 -1 -5.500
Failure 7.833 1 7.83
If it is hard to see where these coefficients came from, consider that our difference between differences was computed this way:
\begin{align*} (7.33 - 4.83) - (5.50 - 7.83) &= 7.33 - 4.83 - 5.50 + 7.83\ &= (1)7.33 + (-1)4.83 + (-1)5.50 + (1)7.83 \end{align*}
The values in parentheses are the coefficients.
To continue the calculations,
$L=4.83$
$\sum c_i^2 = 1^2 + (-1)^2 + (-1)^2 + 1^2 = 4.00$
$t= \cfrac{4.83}{\sqrt{\cfrac{(4)(1.625)}{6}}} = 4.64$
The two-tailed $p$ value is $0.0002$. Therefore, the difference between differences is highly significant.
In a later chapter on Analysis of Variance, you will see that comparisons such as this are testing what is called an interaction. In general, there is an interaction when the effect of one variable differs as a function of the level of another variable. In this example, the effect of the outcome variable is different depending on the subject's self-esteem. For the high-self-esteem subjects, success led to more self-attribution than did failure; for the low-self-esteem subjects, success led to less self-attribution than did failure.
Multiple Comparisons
The more comparisons you make, the greater your chance of a Type I error. It is useful to distinguish between two error rates:
1. the per-comparison error rate and
2. the familywise error rate
The per-comparison error rate is the probability of a Type I error for a particular comparison. The familywise error rate is the probability of making one or more Type I errors in a family or set of comparisons. In the attribution experiment discussed above, we computed two comparisons. If we use the $0.05$ level for each comparison, then the per-comparison rate is simply $0.05$. The familywise rate can be complex. Fortunately, there is a simple approximation that is fairly accurate when the number of comparisons is small. Defining $\alpha$ as the per-comparison error rate and $c$ as the number of comparisons, the following inequality always holds true for the familywise error rate ($FW$):
$FW \leq c\alpha$
This inequality is called the Bonferroni inequality. In practice, $FW$ can be approximated by $c\alpha$. This is a conservative approximation since $FW$ can never be greater than $c\alpha$ and is generally less than $c\alpha$.
The Bonferroni inequality can be used to control the familywise error rate as follows: If you want the familywise error rate to be $\alpha$, you use $\alpha /c$ as the per-comparison error rate. This correction, called the Bonferroni correction, will generally result in a familywise error rate less than $\alpha$. Alternatively, you could multiply the probability value by $c$ and use the original $\alpha$ level.
Should the familywise error rate be controlled? Unfortunately, there is no clear-cut answer to this question. The disadvantage of controlling the familywise error rate is that it makes it more difficult to obtain a significant result for any given comparison: The more comparisons you do, the lower the per-comparison rate must be and therefore the harder it is to reach significance. That is, the power is lower when you control the familywise error rate. The advantage is that you have a lower chance of making a Type I error.
One consideration is the definition of a family of comparisons. Let's say you conducted a study in which you were interested in whether there was a difference between male and female babies in the age at which they started crawling. After you finished analyzing the data, a colleague of yours had a totally different research question: Do babies who are born in the winter differ from those born in the summer in the age they start crawling? Should the familywise rate be controlled or should it be allowed to be greater than $0.05$? Our view is that there is no reason you should be penalized (by lower power) just because your colleague used the same data to address a different research question. Therefore, the familywise error rate need not be controlled. Consider the two comparisons done on the attribution example at the beginning of this section: These comparisons are testing completely different hypotheses. Therefore, controlling the familywise rate is not necessary.
Now consider a study designed to investigate the relationship between various variables and the ability of subjects to predict the outcome of a coin flip. One comparison is between males and females; a second comparison is between those over $40$ and those under $40$; a third is between vegetarians and non-vegetarians; and a fourth is between firstborns and others. The question of whether these four comparisons are testing different hypotheses depends on your point of view. On the one hand, there is nothing about whether age makes a difference that is related to whether diet makes a difference. In that sense, the comparisons are addressing different hypotheses. On the other hand, the whole series of comparisons could be seen as addressing the general question of whether anything affects the ability to predict the outcome of a coin flip. If nothing does, then allowing the familywise rate to be high means that there is a high probability of reaching the wrong conclusion.
Orthogonal Comparisons
In the preceding sections, we talked about comparisons being independent. Independent comparisons are often called orthogonal comparisons. There is a simple test to determine whether two comparisons are orthogonal: If the sum of the products of the coefficients is 0, then the comparisons are orthogonal. Consider again the experiment on the attribution of success or failure. Table $6$ shows the coefficients previously presented in Table $3$ and in Table $5$. The column "$C1$" contains the coefficients from the comparison shown in Table $3$; the column "$C2$" contains the coefficients from the comparison shown in Table $5$. The column labeled "Product" is the product of these two columns. Note that the sum of the numbers in this column is $0$. Therefore, the two comparisons are orthogonal.
Table $6$: Coefficients for two orthogonal comparisons
Outcome Esteem C1 C2 Product
Success High Self-Esteem 0.5 1 0.5
Low Self-Esteem 0.5 -1 -0.5
Failure High Self-Esteem -0.5 -1 0.5
Low Self-Esteem -0.5 1 -0.5
Table $7$ shows two comparisons that are not orthogonal. The first compares the high-self-esteem subjects to low-self-esteem subjects; the second considers only those in the success group and compares high-self-esteem subjects to low-self-esteem subjects. The failure group is ignored by using $0's$ as coefficients. Clearly the comparison of these two groups of subjects for the whole sample is not independent of the comparison of them for the success group only. You can see that the sum of the products of the coefficients is $0.5$ and not $0$.
Table $1$: Coefficients for two non-orthogonal comparisons
Outcome Esteem C1 C2 Product
Success High Self-Esteem 0.5 0.5 0.25
Low Self-Esteem -0.5 -0.5 0.25
Failure High Self-Esteem 0.5 0.0 0.0
Low Self-Esteem -0.5 0.0 0.0 | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/12%3A_Tests_of_Means/12.06%3A_Specific_Comparisons.txt |
Learning Objectives
• Determine whether you have correlated pairs or independent groups
• Compute a t test for correlated pairs
Let's consider how to analyze the data from the "ADHD Treatment" case study. These data consist of the scores of $24$ children with ADHD on a delay of gratification (DOG) task. Each child was tested under four dosage levels. In this section, we will be concerned only with testing the difference between the mean of the placebo ($D0$) condition and the mean of the highest dosage condition ($D60$). The first question is why the difference between means should not be tested using the procedure described in the section Difference Between Two Means (Independent Groups). The answer lies in the fact that in this experiment we do not have independent groups. The scores in the $D0$ condition are from the same subjects as the scores in the $D60$ condition. There is only one group of subjects, each subject being tested in both the $D0$ and $D60$ conditions.
Figure $1$ shows a scatter plot of the $60$-mg scores ($D60$) as a function of the $0$-mg scores ($D0$). It is clear that children who get more correct in the $D0$ condition tend to get more correct in the $D60$ condition. The correlation between the two conditions is high: $r = 0.80$. Clearly these two variables are not independent.
Computations
You may recall that the method to test the difference between these means was presented in the section on "Testing a Single Mean." The computational procedure is to compute the difference between the $D60$ and the $D0$ conditions for each child and test whether the mean difference is significantly different from $0$. The difference scores are shown in Table $1$. As shown in the section on testing a single mean, the mean difference score is $4.96$ which is significantly different from $0$: $t = 3.22,\; df = 23,\; p = 0.0038$. This $t$ test has various names including "correlated $t$ test" and "related-pairs $t$ test".
In general, the correlated $t$ test is computed by first computing the difference between the two scores for each subject. Then, a test of a single mean is computed on the mean of these difference scores.
Table $1$: DOG scores as a function of dosage
D0 D60 D60-D0
57 62 5
27 49 22
32 30 -2
31 34 3
34 38 4
38 36 -2
71 77 6
33 51 18
34 45 11
53 42 -11
36 43 7
42 57 15
26 36 10
52 58 6
36 35 -1
55 60 5
36 33 -3
42 49 7
36 33 -3
54 59 5
34 35 1
29 37 8
33 45 12
33 29 -4
If you had mistakenly used the method for an independent-groups $t$ test with these data, you would have found that $t = 1.42$, $df = 46$, and $p = 0.15$. That is, the difference between means would not have been found to be statistically significant. This is a typical result: correlated $t$ tests almost always have greater power than independent-groups $t$ tests. This is because in correlated $t$ tests, each difference score is a comparison of performance in one condition with the performance of that same subject in another condition. This makes each subject "their own control" and keeps differences between subjects from entering into the analysis. The result is that the standard error of the difference between means is smaller in the correlated $t$ test and, since this term is in the denominator of the formula for $t$, results in a larger $t$.
Details about the Standard Error of the Difference between Means (Optional)
To see why the standard error of the difference between means is smaller in a correlated $t$ test, consider the variance of difference scores. As shown in the section on the Variance Sum Law, the variance of the sum or difference of the two variables $X$ and $Y$ is:
$S_{X\pm Y}^{2} = S_{X}^{2} + S_{Y}^{2} \pm 2rS_XS_Y$
Therefore, the variance of difference scores is the variance in the first condition ($X$) plus the variance in the second condition ($Y$) minus twice the product of
1. the correlation,
2. the standard deviation of $X$, and
3. the standard deviation of $Y$. For the current example, $r = 0.80$ and the variances and standard deviations are shown in Table $2$.
Table $2$: Variances and Standard Deviations
D0 D60 D60 - D0
Variance 128.02 151.78 56.82
Sd 11.31 12.32 7.54
The variance of the difference scores of $56.82$ can be computed as:
$128.02 + 151.78 - (2)(0.80)(11.31)(12.32)$
which is equal to $56.82$ except for rounding error. Notice that the higher the correlation, the lower the standard error of the mean. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/12%3A_Tests_of_Means/12.07%3A_Correlated_Pairs.txt |
Learning Objectives
• State the effect of increasing \(r\) on power
Instructions
This simulation demonstrates the \(t\) test for correlated observations. When you click the "sample" button, \(12\) subjects with scores in two conditions are sampled from a population. Each population is normally distributed with a standard deviation of \(4\). The mean for condition \(1\) is \(15\) and the mean for condition \(2\) is \(13\). The initial population correlation between conditions is \(0.5\), although you can change that. The data from the sample of \(12\) subjects is graphed with a line connecting the two data points for a subject. The difference scores (Condition \(A\) - Condition \(B\)) are computed and graphed as well and shown in a table. The bottom of simulation displays the mean difference score, the standard deviation of the difference scores, the standard error of the mean difference score, the value of \(t\), and the value of \(p\).
Illustrated Instructions
Video Demo
The video begins by running simulations with the population correlation set to \(0.5\). More simulation are run with the correlation set to \(0.1\). As you watch the video and run simulations for yourself see if you can determine a correspondence between an aspect of the graph and the standard deviation of the difference scores. Specifically, look at the degree to which the lines diverge from being parallel and the standard deviation. What effect does increasing the correlation have on the parallel lines and the standard deviation? What effect does increasing the correlation have?
12.09: Specific Comparisons (Correlated Observations)
Learning Objectives
• Compute t for a comparison for repeated-measures data
In the "Weapons and Aggression" case study, subjects were asked to read words presented on a computer screen as quickly as they could. Some of the words were aggressive words such as injure or shatter. Others were control words such as relocate or consider. These two types of words were preceded by words that were either the names of weapons, such as shotgun or grenade, or non-weapon words, such as rabbit or fish. For each subject, the mean reading time across words was computed for these four conditions. The four conditions are labeled as shown in Table $1$. Table $2$ shows the data from five subjects.
Table $1$: Description of Conditions
Variable Description
aw The time in milliseconds (msec) to name an aggressive word following a weapon word prime.
an The time in milliseconds (msec) to name an aggressive word following a non-weapon word prime.
cw The time in milliseconds (msec) to name a control word following a weapon word prime.
cn The time in milliseconds (msec) to name a control word following a non-weapon word prime.
Table $2$: Data from Five Subjects
Subject aw an cw cn
1 447 440 432 452
2 427 437 469 451
3 417 418 445 434
4 348 371 353 344
5 471 443 462 463
One question was whether reading times would be shorter when the preceding word was a weapon word ($aw$ and $cw$ conditions) than when it was a non-weapon word ($an$ and $cn$ conditions). In other words,
Is
$L_1 = (an + cn) - (aw + cw)$
greater than $0$?
This is tested for significance by computing $L_1$ for each subject and then testing whether the mean value of $L_1$ is significantly different from $0$. Table $3$ shows $L_1$ for the first five subjects. $L_1$ for Subject 1 was computed by:
$L1 = (440 + 452) - (447 + 432) = 892 - 879 = 13$
Table $3$: $L_1$ for Five Subjects
Subject aw an cw cn L1
1 447 440 432 452 13
2 427 437 469 451 -8
3 417 418 445 434 -10
4 348 371 353 344 14
5 471 443 462 463 -27
Once $L_1$ is computed for each subject, the significance test described in the section "Testing a Single Mean" can be used. First we compute the mean and the standard error of the mean for $L_1$. There were $32$ subjects in the experiment. Computing $L_1$ for the $32$ subjects, we find that the mean and standard error of the mean are $5.875$ and $4.2646$, respectively. We then compute
$t=\frac{M-\mu }{S_M}$
where $M$ is the sample mean, $\mu$ is the hypothesized value of the population mean ($0$ in this case), and $s_M$ is the estimated standard error of the mean. The calculations show that $t = 1.378$. Since there were $32$ subjects, the degrees of freedom is $32 - 1 = 31$. The t distribution calculator shows that the two-tailed probability is $0.178$.
A more interesting question is whether the priming effect (the difference between words preceded by a non-weapon word and words preceded by a weapon word) is different for aggressive words than it is for non-aggressive words. That is, do weapon words prime aggressive words more than they prime non-aggressive words? The priming of aggressive words is ($an - aw$). The priming of non-aggressive words is ($cn - cw$). The comparison is the difference:
$L_2 = (an - aw) - (cn - cw)$
Table $4$ shows $L_2$ for five of the $32$ subjects.
Table $4$: $L_2$ for Five Subjects
Subject aw an cw cn L2
1 447 440 432 452 -27
2 427 437 469 451 28
3 417 418 445 434 12
4 348 371 353 344 32
5 471 443 462 463 -29
The mean and standard error of the mean for all $32$ subjects are $8.4375$ and $3.9128$, respectively. Therefore, $t = 2.156$ and $p = 0.039$.
Multiple Comparisons
Issues associated with doing multiple comparisons are the same for related observations as they are for multiple comparisons among independent groups.
Orthogonal Comparisons
The most straightforward way to assess the degree of dependence between two comparisons is to correlate them directly. For the weapons and aggression data, the comparisons $L_1$ and $L_2$ are correlated $0.24$. Of course, this is a sample correlation and only estimates what the correlation would be if $L_1$ and $L_2$ were correlated in the population. Although mathematically possible, orthogonal comparisons with correlated observations are very rare. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/12%3A_Tests_of_Means/12.08%3A_Correlated_t_Simulation.txt |
Learning Objectives
• Compute the Bonferroni correction
• Calculate pairwise comparisons using the Bonferroni correction
In the section on all pairwise comparisons among independent groups, the Tukey HSD test was the recommended procedure. However, when you have one group with several scores from the same subjects, the Tukey test makes an assumption that is unlikely to hold: The variance of difference scores is the same for all pairwise differences between means.
The standard practice for pairwise comparisons with correlated observations is to compare each pair of means using the method outlined in the section "Difference Between Two Means (Correlated Pairs)" with the addition of the Bonferroni correction described in the section "Specific Comparisons." For example, suppose you were going to do all pairwise comparisons among four means and hold the familywise error rate at \(0.05\). Since there are six possible pairwise comparisons among four means, you would use \(0.05/6 = 0.0083\) for the per-comparison error rate.
As an example, consider the case study "Stroop Interference." There were three tasks, each performed by \(47\) subjects. In the "words" task, subjects read the names of \(60\) color words written in black ink; in the "color" task, subjects named the colors of \(60\) rectangles; in the "interference" task, subjects named the ink color of \(60\) conflicting color words. The times to read the stimuli were recorded. In order to compute all pairwise comparisons, the difference in times for each pair of conditions for each subject is calculated. Table \(1\) shows these scores for five of the \(47\) subjects.
Table \(1\): Pairwise Differences
W-C W-I C-I
-3 -24 -21
2 -41 -43
-1 -18 -17
-4 -23 -19
-2 -17 -15
Table \(2\) shows data for all \(47\) subjects.
Table \(2\): Pairwise Differences for \(47\) subjects
W-C W-I C-I
-3 -24 -21
2 -41 -43
-1 -18 -17
-4 -23 -19
-2 -17 -15
-3 -15 -12
-3 -28 -25
-3 -36 -33
-3 -17 -14
-2 -10 -8
-1 -11 -10
-1 -10 -9
-3 -26 -23
0 -4 -4
-4 -28 -24
-5 -19 -14
-5 -18 -13
-8 -23 -15
-7 -22 -15
-3 -28 -25
-9 -30 -21
-5 -21 -16
-7 -23 -16
-9 -21 -12
-7 -35 -28
-4 -27 -23
-4 -25 -21
-2 -16 -14
-3 -14 -11
-5 -14 -9
-3 -14 -11
-5 -15 -10
-12 -25 -13
-1 -9 -8
-2 -13 -11
-8 -17 -9
3 -13 -16
-6 -33 -27
-3 -12 -9
-7 -19 -12
-8 -19 -11
-6 -31 -25
-1 -19 -18
-5 -13 -8
-3 -18 -15
-9 -28 -19
-5 -22 -17
The means, standard deviations (\(Sd\)), and standard error of the mean (\(Sem\)), \(t\), and \(p\) for all \(47\) subjects are shown in Table \(3\). The \(t's\) are computed by dividing the means by the standard errors of the mean. Since there are \(47\) subjects, the degrees of freedom is \(46\). Notice how different the standard deviations are. For the Tukey test to be valid, all population values of the standard deviation would have to be the same.
Table \(3\): Pairwise Comparisons
Comparison Mean Sd Sem t p
W-C -4.15 2.99 0.44 -9.53 <0.001
W-I -20.51 7.84 1.14 -17.93 <0.001
C-I -16.36 7.47 1.09 -15.02 <0.001
Using the Bonferroni correction for three comparisons, the \(p\) value has to be below \(0.05/3 = 0.0167\) for an effect to be significant at the \(0.05\) level. For these data, all \(p\) values are far below that, and therefore all pairwise differences are significant.
12.11: Statistical Literacy
Learning Objectives
• Effectiveness of Surgery for Weight Loss
Research on the effectiveness of surgery for weight loss reported here found that "The surgery was associated with significantly greater weight loss [than the control group who dieted] through \(2\) years (\(61.3\) versus \(11.2\) pounds, \(P<0.001\))."
Example \(1\): what do you think
What test could have been used and how would it have been computed?
Solution
For each subject a difference score between their initial weight and final weight could be computed. A t test of whether the mean difference score differs significantly from \(0\) could then be computed. The mean difference score will equal the difference between the mean weight losses of the two groups (\(61.3 - 11.2 = 50.1\)). | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/12%3A_Tests_of_Means/12.10%3A_Pairwise_%28Correlated%29.txt |
General Questions
Q1
The scores of a random sample of \(8\) students on a physics test are as follows: \(60, 62, 67, 69, 70, 72, 75, 78\).
1. Test to see if the sample mean is significantly different from \(65\) at the \(0.05\) level. Report the \(t\) and \(p\) values.
2. The researcher realizes that she accidentally recorded the score that should have been \(76\) as \(67\). Are these corrected scores significantly different from \(65\) at the \(0.05\) level? (relevant section)
Q2
A (hypothetical) experiment is conducted on the effect of alcohol on perceptual motor ability. Ten subjects are each tested twice, once after having two drinks and once after having two glasses of water. The two tests were on two different days to give the alcohol a chance to wear off. Half of the subjects were given alcohol first and half were given water first. The scores of the \(10\) subjects are shown below. The first number for each subject is their performance in the "water" condition. Higher scores reflect better performance. Test to see if alcohol had a significant effect. Report the \(t\) and \(p\) values. (relevant section)
water
alcohol
16
13
15
13
11
10
20
18
19
17
14
11
13
10
15
15
14
11
16
16
Q3
The scores on a (hypothetical) vocabulary test of a group of \(20\) year olds and a group of \(60\) year olds are shown below.
20 yr olds
60 yr olds
27
26
26
29
21
29
24
29
15
27
18
16
17
20
12
27
13
1. Test the mean difference for significance using the \(0.05\) level. (relevant section).
2. List the assumptions made in computing your answer.(relevant section)
Q4
The sampling distribution of a statistic is normally distributed with an estimated standard error of \(12\), (\(df = 20)\).
1. What is the probability that you would have gotten a mean of \(107\) (or more extreme) if the population parameter were \(100\)? Is this probability significant at the \(0.05\) level (two-tailed)?
2. What is the probability that you would have gotten a mean of \(95\) or less (one-tailed)? Is this probability significant at the \(0.05\) level? You may want to use the t Distribution calculator for this problem. (relevant section)
Q5
How do you decide whether to use an independent groups \(t\) test or a correlated \(t\) test (test of dependent means)? (relevant section & relevant section)
Q6
An experiment compared the ability of three groups of subjects to remember briefly-presented chess positions. The data are shown below.
Non-players
Beginners
Tournament players
22.1
32.5
40.1
22.3
37.1
45.6
26.2
39.1
51.2
29.6
40.5
56.4
31.7
45.5
58.1
33.5
51.3
71.1
38.9
52.6
74.9
39.7
55.7
75.9
43.2
55.9
80.3
43.2
57.7
85.3
1. Using the Tukey HSD procedure, determine which groups are significantly different from each other at the \(0.05\) level. (relevant section)
2. Now compare each pair of groups using \(t\)-tests. Make sure to control for the familywise error rate (at \(0.05\)) by using the Bonferroni correction. Specify the alpha level you used.
Q7
Below are data showing the results of six subjects on a memory test. The three scores per subject are their scores on three trials (\(a\), \(b\), and \(c\)) of a memory task. Are the subjects getting better each trial? Test the linear effect of trial for the data.
a
b
c
4
6
7
3
7
8
2
8
5
1
4
7
4
6
9
2
4
2
1. Compute \(L\) for each subject using the contrast weights \(-1\), \(0\), and \(1\). That is, compute \((-1)(a) + (0)(b) + (1)(c)\) for each subject.
2. Compute a one-sample \(t\)-test on this column (with the \(L\) values for each subject) you created. (relevant section)
Q8
Participants threw darts at a target. In one condition, they used their preferred hand; in the other condition, they used their other hand. All subjects performed in both conditions (the order of conditions was counterbalanced). Their scores are shown below.
Preferred
Non-preferred
12
7
7
9
11
8
13
10
10
9
1. Which kind of \(t\)-test should be used?
2. Calculate the two-tailed \(t\) and \(p\) values using this \(t\) test.
3. Calculate the one-tailed \(t\) and \(p\) values using this \(t\) test.
Q9
Assume the data in the previous problem were collected using two different groups of subjects: One group used their preferred hand and the other group used their non-preferred hand. Analyze the data and compare the results to those for the previous problem (relevant section)
Q10
You have \(4\) means, and you want to compare each mean to every other mean.
1. How many tests total are you going to compute?
2. What would be the chance of making at least one Type I error if the Type I error for each test was \(0.05\) and the tests were independent? (relevant section & relevant section)
3. Are the tests independent and how does independence/non-independence affect the probability in (b).
Q11
In an experiment, participants were divided into \(4\) groups. There were \(20\) participants in each group, so the degrees of freedom (error) for this study was \(80 - 4 = 76\). Tukey's HSD test was performed on the data.
1. Calculate the \(p\) value for each pair based on the \(Q\) value given below. You will want to use the Studentized Range Calculator.
2. Which differences are significant at the \(0.05\) level? (relevant section)
Comparison of Groups
Q
A - B
3.4
A - C
3.8
A - D
4.3
B - C
1.7
B - D
3.9
C - D
3.7
Q12
If you have \(5\) groups in your study, why shouldn't you just compute a \(t\) test of each group mean with each other group mean? (relevant section)
Q13
You are conducting a study to see if students do better when they study all at once or in intervals. One group of \(12\) participants took a test after studying for one hour continuously. The other group of \(12\) participants took a test after studying for three twenty minute sessions. The first group had a mean score of \(75\) and a variance of \(120\). The second group had a mean score of \(86\) and a variance of \(100\).
1. What is the calculated \(t\) value? Are the mean test scores of these two groups significantly different at the \(0.05\) level?
2. What would the \(t\) value be if there were only \(6\) participants in each group? Would the scores be significant at the \(0.05\) level?
Q14
A new test was designed to have a mean of \(80\) and a standard deviation of \(10\). A random sample of \(20\) students at your school take the test, and the mean score turns out to be \(85\). Does this score differ significantly from \(80\)? To answer this problem, you may want to use the Normal Distribution Calculator.(relevant section)
Q15
You perform a one-sample \(t\) test and calculate a \(t\) statistic of \(3.0\). The mean of your sample was \(1.3\) and the standard deviation was \(2.6\). How many participants were used in this study? (relevant section)
Q16
True/false: The contrasts \((-3, 1 1 1)\) and \((0, 0 , -1, 1)\) are orthogonal. (relevant section)
Q17
True/false: If you are making \(4\) comparisons between means, then based on the Bonferroni correction, you should use an alpha level of \(0.01\) for each test. (relevant section)
Q18
True/false: Correlated \(t\) tests almost always have greater power than independent \(t\) tests. (relevant section)
Q19
True/false:The graph below represents a violation of the homogeneity of variance assumption. (relevant section)
Q20
True/false: When you are conducting a one-sample \(t\) test and you know the population standard deviation, you look up the critical \(t\) value in the table based on the degrees of freedom. (relevant section)
Questions from Case Studies
The following questions use data from the Angry Moods (AM) case study.
Q21
(AM#17) Do athletes or non-athletes calm down more when angry? Conduct a \(t\) test to see if the difference between groups in Control-In scores is statistically significant.
Q22
Do people in general have a higher Anger-Out or Anger-In score? Conduct a \(t\) test on the difference between means of these two scores. Are these two means independent or dependent? (relevant section)
The following questions use data from the Smiles and Leniency (SL) case study.
Q23
Compare each mean to the neutral mean. Be sure to control for the familywise error rate. (relevant section)
Q24
Does a "felt smile" lead to more leniency than other types of smiles?
1. Calculate \(L\) (the linear combination) using the following contrast weights \(false: -1,\; felt: 2,\; miserable: -1,\; neutral: 0\).
2. Perform a significance test on this value of \(L\). (relevant section)
The following questions are from the Animal Research (AR) case study.
Q25
(AR#8) Conduct an independent samples \(t\) test comparing males to females on the belief that animal research is necessary. (relevant section)
Q26
(AR#9) Based on the \(t\) test you conducted in the previous problem, are you able to reject the null hypothesis if \(alpha = 0.05\)? What about if \(alpha = 0.1\)? (relevant section)
Q27
(AR#10) Is there any evidence that the \(t\) test assumption of homogeneity of variance is violated in the \(t\) test you computed in #25? (relevant section)
The following questions use data from the ADHD Treatment (AT) case study.
Q28
Compare each dosage with the dosage below it (compare \(d0\) and \(d15\), \(d15\) and \(d30\), and \(d30\) and \(d60\)). Remember that the patients completed the task after every dosage.
1. If the familywise error rate is \(0.05\), what is the alpha level you will use for each comparison when doing the Bonferroni correction?
2. Which differences are significant at this level? (relevant section)
Q29
Does performance increase linearly with dosage?
1. Plot a line graph of this data.
2. Compute \(L\) for each patient. To do this, create a new variable where you multiply the following coefficients by their corresponding dosages and then sum up the total: \((-3)d0 + (-1)d15 + (1)d30 + (3)d60\) (see #8). What is the mean of \(L\)?
3. Perform a significance test on \(L\). Compute the \(95\%\) confidence interval for \(L\). (relevant section)
Select Answers
S1
1. \(t(7) = 1.91\)
1. \(0.035\)
S7
1. two-tailed \(p = 0.0088\)
S8
1. \(p = 0.1662\)
S11
1. \(A - B: p = 0.085\)
S13
1. \(t(22) = 2.57\)
\(t(76) = 3.04\)
S25
1. \(p = 0.0745\)
S29
1. \(p = 0.0006\) | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/12%3A_Tests_of_Means/12.E%3A_Tests_of_Means_%28Exercises%29.txt |
Power is defined as the probability of correctly rejecting a false null hypothesis. For example, it can be the probability that given there is a difference between the population means of the new method and the standard method, the sample means will be significantly different. The probability of failing to reject a false null hypothesis is often referred to as β. Therefore power can be defined as:
power=1−β
It is very important to consider power while designing an experiment. You should avoid spending a lot of time and/or money on an experiment that has little chance of finding a significant effect.
• 13.1: Introduction to Power
Power is defined as the probability of correctly rejecting a false null hypothesis. In terms of our example, it is the probability that given there is a difference between the population means of the new method and the standard method, the sample means will be significantly different. It is very important to consider power while designing an experiment. You should avoid spending a lot of time and/or money on an experiment that has little chance of finding a significant effect.
• 13.2: Example Calculations
• 13.3: Power Demo
• 13.4: Power Demo II
• 13.5: Factors Affecting Power
Several factors affect the power of a statistical test. Some of the factors are under the control of the experimenter, whereas others are not. The following example will be used to illustrate the various factors.
• 13.6: Statistical Literacy
• 13.E: Power (Exercises)
13: Power
Learning Objectives
• Identify situations in which it is important to estimate power
Suppose you work for a foundation whose mission is to support researchers in mathematics education and your role is to evaluate grant proposals and decide which ones to fund. You receive a proposal to evaluate a new method of teaching high-school algebra. The research plan is to compare the achievement of students taught by the new method with the achievement of students taught by the traditional method. The proposal contains good theoretical arguments why the new method should be superior and the proposed methodology is sound. In addition to these positive elements, there is one important question still to be answered: Does the experiment have a high probability of providing strong evidence that the new method is better than the standard method if, in fact, the new method is actually better? It is possible, for example, that the proposed sample size is so small that even a fairly large population difference would be difficult to detect. That is, if the sample size is small, then even a fairly large difference in sample means might not be significant. If the difference is not significant, then no strong conclusions can be drawn about the population means. It is not justified to conclude that the null hypothesis that the population means are equal is true just because the difference is not significant. Of course, it is not justified to conclude that this null hypothesis is false. Therefore, when an effect is not significant, the result is inconclusive. You may prefer that your foundation's money be used to fund a project that has a higher probability of being able to make a strong conclusion.
Power is defined as the probability of correctly rejecting a false null hypothesis. In terms of our example, it is the probability that given there is a difference between the population means of the new method and the standard method, the sample means will be significantly different. The probability of failing to reject a false null hypothesis is often referred to as $\beta$. Therefore power can be defined as:
$\text{power} = 1 - β$
It is very important to consider power while designing an experiment. You should avoid spending a lot of time and/or money on an experiment that has little chance of finding a significant effect. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/13%3A_Power/13.01%3A_Introduction_to_Power.txt |
skills to develop
• Compute power using the binomial distribution
• Compute power using the normal distribution
• Use a power calculator to compute power for the $t$ distribution
In the "Shaking and Stirring Martinis" case study, the question was whether Mr. Bond could tell the difference between martinis that were stirred and martinis that were shaken. For the sake of this example, assume he can tell the difference and is able to correctly state whether a martini had been shaken or stirred $0.75$ of the time. Now, suppose an experiment is being conducted to investigate whether Mr. Bond can tell the difference. Specifically, is Mr. Bond correct more than $0.50$ of the time? We know that he is (that's an assumption of the example). However, the experimenter does not know and asks Mr. Bond to judge $16$ martinis. The experimenter will do a significance test based on the binomial distribution. Specifically, if a one tailed test is significant at the $0.05$ level, then he or she will conclude that Mr. Bond can tell the difference. The probability value is computed assuming the null hypothesis is true ($\pi =0.50$). Therefore, the experimenter will determine how many times Mr. Bond is correct, and compute the probability of being correct that many or more times given that the null hypothesis is true. The question is: what is the probability the experimenter will correctly reject the null hypothesis that $\pi =0.50$? In other words, what is the power of this experiment?
The binomial distribution for $N = 16$ and $\pi =0.50$ is shown in Figure $1$. The probability of being correct on $11$ or more trials is $0.105$ and the probability of being correct on $12$ or more trials is $0.038$. Therefore, the probability of being correct on $12$ or more trials is less than $0.05$. This means that the null hypothesis will be rejected if Mr. Bond is correct on $12$ or more trials and will not be rejected otherwise.
Binomial Calculator
We know that Mr. Bond is correct $0.75$ of the time. (Obviously the experimenter does not know this or there would be no need for an experiment.) The binomial distribution with $N = 16$ and $\pi =0.75$ is shown in Figure $2$.
The probability of being correct on $12$ or more trials is $0.63$. Therefore, the power of the experiment is $0.63$.
To sum up, the probability of being correct on $12$ or more trials given that the null hypothesis is true is less than $0.05$. Therefore, if Mr. Bond is correct on $12$ or more trials, the null hypothesis will be rejected. Given Mr. Bond's true ability to be correct on $0.75$ of the trials, the probability he will be correct on $12$ or more trials is $0.63$. Therefore power is $0.63$.
In the section on Testing a Single Mean for significance, the first example was based on the assumption that the experimenter knew the population variance. Although this is rarely true in practice, the example is very useful for pedagogical purposes. For the same reason, the following example assumes the experimenter knows the population variance. Power calculators are available for situations in which the experimenter does not know the population variance.
Suppose a math achievement test were known to have a mean of $75$ and a standard deviation of $10$. A researcher is interested in whether a new method of teaching results in a higher mean. Assume that although the experimenter does not know it, the population mean for the new method is $80$. The researcher plans to sample $25$ subjects and do a one-tailed test of whether the sample mean is significantly higher than $75$. What is the probability that the researcher will correctly reject the false null hypothesis that the population mean for the new method is $75$ or lower? The following shows how this probability is computed.
The researcher assumes that the population standard deviation with the new method is the same as with the old method ($10$) and that the distribution is normal. Since the population standard deviation is assumed to be known, the researcher can use the normal distribution rather than the t distribution to compute the $p$ value. Recall that the standard error of the mean ($\sigma _M$) is
$\sigma _M=\frac{\sigma }{\sqrt{N}}$
which is equal to $10/5 = 2$ in this example. As can be seen in Figure $3$, if the null hypothesis that the population mean equals $75$ is true, then the probability of a sample mean being greater than or equal to $78.29$ is $0.05$. Therefore, the experimenter will reject the null hypothesis if the sample mean, $M$, is $78.29$ or larger.
The question, then, is what is the probability the experimenter gets a sample mean greater than $78.29$ given that the population mean is $80$? Figure $4$shows that this probability is $0.80$.
Therefore, the probability that the experimenter will reject the null hypothesis that the population mean of the new method is $75$ is $0.80$. In other words, $power = 0.80$.
Calculation of power is more complex for $t$ tests and for Analysis of Variance. The power calculator computes power for a $t$ test of independent groups. Calculators for other types of designs are linked to below:
Russ Lenth's Power Calculators | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/13%3A_Power/13.02%3A_Example_Calculations.txt |
Learning Objectives
• State the effect of one- versus two-tailed tests on power
• State the effect of $\alpha$ level on power
Instructions
This simulation illustrates the effect of
1. sample size
2. the difference between population mean and hypothesized mean
3. the standard deviation
4. the type of test (one-tailed or two)
5. significance level on the power of a two-sample $t$ test
You specify the difference between the population mean and hypothesized mean by either entering text in the box or by moving the slider. The population standard deviation can be entered in the box or specified by the pop-up menu. Finally, you specify the number of tails of the test. A power graph for the significance level $0.10$, $0.05$, and $0.01$ significance levels as a function of sample size is displayed.
1. Examine the power curves. The $X$ axis shows sample size, the $Y$ axis shows power. Note the effect of sample size and significance level on power.
2. The default difference between the population mean and hypothesized mean is $1.55$. Use the slider to change this value and note the effect on the power curves.
3. Change the population standard deviation ($sd$) and notice its effect on power.
4. Compare the power of one-tailed and two-tailed tests when the difference between the population mean and hypothesized mean is $2$. Determine which is higher.
5. Set the difference between the population mean and hypothesized mean to zero. Since the null hypothesis is true, the $Y$ axis no shows the Type I error rate, not power. Note the effect of sample size on the Type I error rate.
6. Determine the effect of changing the standard deviation on the Type I error rate.
Illustrated Instructions
Video Demo
The video demonstration begins by changing the population mean and hypothesized mean difference to $2$ and then the population standard deviation to $3.5$. Notice how the power curves change with each adjustment. The video by switching between one-tailed and two tailed tests.
13.04: Power Demo II
Learning Objectives
• State the effect of effect size on power
• State the effect of one- versus two-tailed tests on power
• State the effect of the standard deviation on power
• State the effect of $\alpha$ level on power
Instructions
This demonstration shows various aspects of power. The graph displays power for a one-sample $Z$-test of the null hypothesis that the population mean is $50$. The red distribution is the sampling distribution of the mean assuming the null hypothesis is true. The blue distribution is the sampling distribution of the mean based on the "true" mean. The default value of the true mean is $70$. The cutoff for significance is based on the red distribution. For a one-tailed test at the $0.05$ level, the cutoff point is determined so that $5\%$ of the area is to the right of the cutoff. For the default values, the cutoff is $58.22$. Therefore, any $\text{sample mean} = 58.22$ would be significant.
The shaded area of the blue distribution shows power. It is the probability that the sample mean will be $58.22$ if the true mean is $70$. For the default values it is $0.991$.
You specify the true mean by either entering text in the box. The population standard deviation can be specified by the pop-up menu. Finally, you can specify the number of tails of the test and the significance level. Power and the cut-off points are displayed.
1. Notice the effect on power of changing the true mean to $60$. Look at the distributions to find the area that represents power.
2. Change the sample size and notice its effect on the cutoff point and on power.
3. Change the population standard deviation ($sd$) and notice its effect on the cutoff point and on power.
4. Compare the power of one-tailed and two-tailed tests. What is the power for each when the true mean is $55$. What about $45$? (The one-tailed tests are based on the expectation that the $\text{population mean is} > 50$.)
5. Set the value of the true mean to $50$. Does power make sense in this situation?
6. Determine the effect of the significance level on the red and blue distributions. On what (in addition to power) does it have its effect?
Illustrated Instructions
Video Demo
The video demonstration starts by changing the true mean to $55$ and then the standard to $28$. Notice how the probability of rejecting the null hypothesis varies with these changes.
The video concludes by changing the test to two-tailed and decreasing the significance level to $0.01$. Again notice the changes in the probability of rejection. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/13%3A_Power/13.03%3A_Power_Demo.txt |
Learning Objectives
• State what the effect of each of the factors is
Several factors affect the power of a statistical test. Some of the factors are under the control of the experimenter, whereas others are not. The following example will be used to illustrate the various factors.
Suppose a math achievement test were known to be normally distributed with a mean of $75$ and a standard deviation of $\sigma$. A researcher is interested in whether a new method of teaching results in a higher mean. Assume that although the experimenter does not know it, the population mean $\mu$ for the new method is larger than $75$. The researcher plans to sample $N$ subjects and do a one-tailed test of whether the sample mean is significantly higher than $75$. In this section, we consider factors that affect the probability that the researcher will correctly reject the false null hypothesis that the population mean is $75$ or lower. In other words, factors that affect power.
Sample Size
Figure $1$ shows that the larger the sample size, the higher the power. Since sample size is typically under an experimenter's control, increasing sample size is one way to increase power. However, it is sometimes difficult and/or expensive to use a large sample size.
Standard Deviation
Figure $1$ also shows that power is higher when the standard deviation is small than when it is large. For all values of $N$, power is higher for the standard deviation of $10$ than for the standard deviation of $15$ (except, of course, when $N = 0$). Experimenters can sometimes control the standard deviation by sampling from a homogeneous population of subjects, by reducing random measurement error, and/or by making sure the experimental procedures are applied very consistently.
Difference between Hypothesized and True Mean
Naturally, the larger the effect size, the more likely it is that an experiment would find a significant effect. Figure $2$ shows the effect of increasing the difference between the mean specified by the null hypothesis ($75$) and the population mean $\mu$ for standard deviations of $10$ and $15$.
Significance Level
There is a trade-off between the significance level and power: the more stringent (lower) the significance level, the lower the power. Figure $3$ shows that power is lower for the $0.01$ level than it is for the $0.05$ level. Naturally, the stronger the evidence needed to reject the null hypothesis, the lower the chance that the null hypothesis will be rejected.
One- versus Two-Tailed Tests
Power is higher with a one-tailed test than with a two-tailed test as long as the hypothesized direction is correct. A one-tailed test at the $0.05$ level has the same power as a two-tailed test at the $0.10$ level. A one-tailed test, in effect, raises the significance level.
13.06: Statistical Literacy
Learning Objectives
• Design of Studies for Alzheimer's Drug
A research design to compare three drugs for the treatment of Alzheimer's disease is described here. For the first two years of the study, researchers will follow the subjects with scans and memory tests.
Example \(1\): what do you think?
The data could be analyzed as a between-subjects design or as a within-subjects design. What type of analysis would be done for each type of design and how would the choice of designs affect power?
Solution
For a between-subjects design, the subjects in the different conditions would be compared after two years. For a within-subjects design, the change in subjects' scores in the different conditions would be compared. The latter would be more powerful.
13.E: Power (Exercises)
General Questions
Q1
Define power in your own words.
Q2
List $3$ measures one can take to increase the power of an experiment. Explain why your measures result in greater power.
Q3
$\text{Population 1 mean} = 36$
$\text{Population 2 mean} = 45$
Both population standard deviations are $10$.
Sample size (per group) $16$.
What is the probability that a $t$ test will find a significant difference between means at the $0.05$ level? Give results for both one- and two-tailed tests. Hint: the power of a one-tailed test at $0.05$ level is the power of a two-tailed test at $0.10$.
Q4
Rank order the following in terms of power. $n$ is the sample size per group.
Population 1 Mean
n
Population 2 Mean
Standard Deviation
a 29 20 43 12
b 34 15 40 6
c 105 24 50 27
d 170 2 120 10
Q5
Alan, while snooping around his grandmother's basement stumbled upon a shiny object protruding from under a stack of boxes . When he reached for the object a genie miraculously materialized and stated: "You have found my magic coin. If you flip this coin an infinite number of times you will notice that heads will show $60\%$ of the time." Soon after the genie's declaration he vanished, never to be seen again. Alan, excited about his new magical discovery, approached his friend Ken and told him about what he had found. Ken was skeptical of his friend's story, however, he told Alan to flip the coin $100$ times and to record how many flips resulted with heads.
1. What is the probability that Alan will be able convince Ken that his coin has special powers by finding a $p$ value below $0.05$ (one tailed). Use the Binomial Calculator (and some trial and error).
2. If Ken told Alan to flip the coin only $20$ times, what is the probability that Alan will not be able to convince Ken (by failing to reject the null hypothesis at the $0.05$ level)? | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/13%3A_Power/13.05%3A_Factors_Affecting_Power.txt |
Statisticians are often called upon to develop methods to predict one variable from other variables. For example, one might want to predict college grade point average from high school grade point average. Or, one might want to predict income from the number of years of education.
14: Regression
Learning Objectives
• Identify errors of prediction in a scatter plot with a regression line
In simple linear regression, we predict scores on one variable from the scores on a second variable. The variable we are predicting is called the criterion variable and is referred to as $Y$. The variable we are basing our predictions on is called the predictor variable and is referred to as $X$. When there is only one predictor variable, the prediction method is called simple regression. In simple linear regression, the topic of this section, the predictions of $Y$ when plotted as a function of $X$ form a straight line.
The example data in Table $1$ are plotted in Figure $1$. You can see that there is a positive relationship between $X$ and $Y$. If you were going to predict $Y$ from $X$, the higher the value of $X$, the higher your prediction of $Y$.
Table $1$: Example data
X Y
1.00 1.00
2.00 2.00
3.00 1.30
4.00 3.75
5.00 2.25
Linear regression consists of finding the best-fitting straight line through the points. The best-fitting line is called a regression line. The black diagonal line in Figure $2$ is the regression line and consists of the predicted score on $Y$ for each possible value of $X$. The vertical lines from the points to the regression line represent the errors of prediction. As you can see, the red point is very near the regression line; its error of prediction is small. By contrast, the yellow point is much higher than the regression line and therefore its error of prediction is large.
The error of prediction for a point is the value of the point minus the predicted value (the value on the line). Table $2$ shows the predicted values ($Y'$) and the errors of prediction ($Y-Y'$). For example, the first point has a $Y$ of $1.00$ and a predicted $Y$ (called $Y'$) of $1.21$. Therefore, its error of prediction is $-0.21$.
Table $2$: Example data
X Y Y' Y-Y' (Y-Y')2
1.00 1.00 1.210 -0.210 0.044
2.00 2.00 1.635 0.365 0.133
3.00 1.30 2.060 -0.760 0.578
4.00 3.75 2.485 1.265 1.600
5.00 2.25 2.910 -0.660 0.436
You may have noticed that we did not specify what is meant by "best-fitting line." By far, the most commonly-used criterion for the best-fitting line is the line that minimizes the sum of the squared errors of prediction. That is the criterion that was used to find the line in Figure $2$. The last column in Table $2$ shows the squared errors of prediction. The sum of the squared errors of prediction shown in Table $2$ is lower than it would be for any other regression line.
The formula for a regression line is
$Y' = bX + A$
where $Y'$ is the predicted score, $b$ is the slope of the line, and $A$ is the $Y$ intercept. The equation for the line in Figure $2$ is
$Y' = 0.425X + 0.785$
For $X = 1$,
$Y' = (0.425)(1) + 0.785 = 1.21$
For $X = 2$,
$Y' = (0.425)(2) + 0.785 = 1.64$
Computing the Regression Line
In the age of computers, the regression line is typically computed with statistical software. However, the calculations are relatively easy, and are given here for anyone who is interested. The calculations are based on the statistics shown in Table $3$. $M_X$ is the mean of $X$, $M_Y$ is the mean of $Y$, $s_X$ is the standard deviation of $X$, $s_Y$ is the standard deviation of $Y$, and $r$ is the correlation between $X$ and $Y$.
Formula for standard deviation
Formula for correlation
Table $3$: Statistics for computing the regression line
MX MY sX sY r
3 2.06 1.581 1.072 0.627
The slope ($b$) can be calculated as follows:
$b = r \frac{s_Y}{s_X}$
and the intercept ($A$) can be calculated as
$A = M_Y - bM_X$
For these data,
$b = \frac{(0.627)(1.072)}{1.581} = 0.425$
$A = 2.06 - (0.425)(3) = 0.785$
Note that the calculations have all been shown in terms of sample statistics rather than population parameters. The formulas are the same; simply use the parameter values for means, standard deviations, and the correlation.
Standardized Variables
The regression equation is simpler if variables are standardized so that their means are equal to $0$ and standard deviations are equal to $1$, for then $b = r$ and $A = 0$. This makes the regression line:
$Z_{Y'} = (r)(Z_X)$
where $Z_{Y'}$ is the predicted standard score for $Y$, $r$ is the correlation, and $Z_X$ is the standardized score for $X$. Note that the slope of the regression equation for standardized variables is $r$.
A Real Example
The case study "SAT and College GPA" contains high school and university grades for $105$ computer science majors at a local state school. We now consider how we could predict a student's university GPA if we knew his or her high school GPA.
Figure $3$ shows a scatter plot of University GPA as a function of High School GPA. You can see from the figure that there is a strong positive relationship. The correlation is $0.78$. The regression equation is:
$\text{University GPA'} = (0.675)(\text{High School GPA}) + 1.097$
Therefore, a student with a high school GPA of $3$ would be predicted to have a university GPA of
$\text{University GPA'} = (0.675)(3) + 1.097 = 3.12$
Assumptions
It may surprise you, but the calculations shown in this section are assumption-free. Of course, if the relationship between $X$ and $Y$ were not linear, a different shaped function could fit the data better. Inferential statistics in regression are based on several assumptions, and these assumptions are presented in a later section of this chapter. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/14%3A_Regression/14.01%3A_Introduction_to_Linear_Regression.txt |
Learning Objectives
• State the relationship between $MSE$ and fit
Instructions
This demonstration allows you to explore fitting data with linear functions. When the demonstration begins, five points are plotted in the graph. The $X$ axis ranges from $1$ to $5$ and the $Y$ axis ranges from $0$ to $5$. The five points are plotted in different colors; next to each point is the $Y$ value of that point. For example, the red point has the value $1.00$ next to it. A vertical black line is drawn with the $Y$ value of $3.0$; this line consists of the predicted values for $Y$. (It is clear that this line does not contain the best predictions.) This line is called the "regression line." The equation for the regression line is $Y' = 0X +3$ where $Y'$ is the predicted value for $Y$. Since the slope is $0$, the same prediction of $3$ is made for all values of $X$.
The error of prediction for each point is represented by a vertical line between the point and the regression line. For the point with a value of $1$ on the $X$ axis, the line goes from the point $(1,1)$ to the point $(1,3)$ on the line. The length of the vertical line is $2$. This means the error of prediction is $2$ and the squared error of prediction is $2 \times 2 = 4$. This error is depicted graphically by the graph on the right. The height of the red square is the error of prediction and the area of the red square is the squared error of prediction.
The errors associated with the other points are plotted in a similar way. Therefore the height of the stacked squares is the sum of the errors of prediction (the lengths of the lines are used, so all errors are positive) and the area of all squares is the total sum of squared errors.
This demonstration allows you to change the regression line and examine the effects on the errors of prediction. If you click and drag an end of the line, the slope of the line will change. If you click and drag the middle of the line, the intercept will change. As the line changes, the errors and squared errors are updated automatically.
You can also change the values of the points by clicking on them and dragging. You can only change the $Y$ values.
You can get a good feeling for the regression line and error by changing the points and the slope and intercept of the line and observing the results.
To see the line that minimizes the squared errors of prediction click the "OK" button.
1. Notice that the total deviation is $6$. Compute this by summing the $5$ separate absolute deviations ($|1-3| + |2-3|$ etc.)
2. The total area is $10$. Compute this by summing the squared deviations.
3. Click middle of the black line and drag it so that it is at $4$. Has the error increased or decreased?
4. Click the left end of the line and drag it until the intercept is about $1$. How has this affected the error?
5. Drag the line further so it has an intercept of about $0$. Then drag the upper portion of the line so that it goes through the point at $5.0$. Within rounding error, the error should be $0$ and the equation for the line should be $Y' = 1X + 0$.
6. Click on the green point at $3.0$ and drag it down so that its value is about $2.0$. Notice the error now is all based on this one point.
7. Adjust the line to see if you can make the absolute error any smaller.
8. Adjust the line to see of you can make the squared error (area) as small as you can and note equation of the line and the size of the area. Press the OK button to see the smallest possible area. How does the line compare with the one you chose?
9. Move all the points around and then try again to fine the line that makes the area smallest. Compare the area to the line that gives the smallest area.
10. After you have found the line that gives the smallest area (smallest squared error) see if you can change the line so that the absolute error is lower than for this line that gives the smallest area.
Illustrated Instructions
Video Demo
The demonstration starts by dragging each of the $5$ points to different locations on the $Y$ axis. Notice how these changes influence the total deviation and area. The video continues by repositioning the regression line by dragging either end as well as the middle. Finally the regression line that minimizes the squared errors is found by clicking the "OK" button. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/14%3A_Regression/14.02%3A_Linear_Fit_Demo.txt |
Learning Objectives
• Compute the sum of squares $Y$
• Convert raw scores to deviation scores
• Compute predicted scores from a regression equation
• Partition sum of squares $Y$ into sum of squares predicted and sum of squares error
• Define $r^2$ in terms of sum of squares explained and sum of squares $Y$
One useful aspect of regression is that it can divide the variation in $Y$ into two parts: the variation of the predicted scores and the variation of the errors of prediction. The variation of $Y$ is called the sum of squares $Y$ and is defined as the sum of the squared deviations of $Y$ from the mean of $Y$. In the population, the formula is
$SSY = \sum (Y-\mu_Y)^2$
where $SSY$ is the sum of squares $Y$, $Y$ is an individual value of $Y$, and $μ_y$ is the mean of $Y$. A simple example is given in Table $1$. The mean of $Y$ is $2.06$ and $SSY$ is the sum of the values in the third column and is equal to $4.597$.
Table $1$: Example of SSY
Y Y-My (Y-My)2
1.00 -1.06 1.1236
2.00 -0.06 0.0036
1.30 -0.76 0.5776
3.75 1.69 2.8561
2.25 0.19 0.0361
When computed in a sample, you should use the sample mean, $M$, in place of the population mean:
$SSY = \sum (Y-M_Y)^2$
It is sometimes convenient to use formulas that use deviation scores rather than raw scores. Deviation scores are simply deviations from the mean. By convention, small letters rather than capitals are used for deviation scores. Therefore the score $y$ indicates the difference between $Y$ and the mean of $Y$. Table $2$ shows the use of this notation. The numbers are the same as in Table $1$.
Table $2$: Example of $SSY$ using Deviation Scores
Y y y2
1.00 -1.06 1.1236
2.00 -0.06 0.0036
1.30 -0.76 0.5776
3.75 1.69 2.8561
2.25 0.19 0.0361
The data in Table $3$ are reproduced from the introductory section. The column $X$ has the values of the predictor variable and the column $Y$ has the criterion variable. The third column, $y$, contains the differences between the column $Y$ and the mean of $Y$.
Table $4$: Example data (The last row contains column sums)
X Y y y2 Y' y' y'2 Y-Y' (Y-Y')2
1.00 1.00 -1.06 1.1236 1.210 -0.850 0.7225 -0.210 0.044
2.00 2.00 -0.06 0.0036 1.635 -0.425 0.1806 0.365 0.133
3.00 1.30 -0.76 0.5776 2.060 0.000 0.0000 -0.760 0.578
4.00 3.75 1.69 2.8561 2.485 0.425 0.1806 1.265 1.600
5.00 2.25 0.19 0.0361 2.910 0.850 0.7225 -0.660 0.436
Sums
15.00 10.30 0.00 4.597 10.300 0.000 1.806 0.000 2.791
The fourth column, $y^2$, is simply the square of the $y$ column. The column $Y'$ contains the predicted values of $Y$. In the introductory section, it was shown that the equation for the regression line for these data is
$Y' = 0.425X + 0.785.$
The values of $Y'$ were computed according to this equation. The column $y'$ contains deviations of $Y'$ from the mean of $Y'$ and $y'^2$ is the square of this column. The next-to-last column, $Y-Y'$, contains the actual scores ($Y$) minus the predicted scores ($Y'$). The last column contains the squares of these errors of prediction.
We are now in a position to see how the $SSY$ is partitioned. Recall that $SSY$ is the sum of the squared deviations from the mean. It is therefore the sum of the $y^2$ column and is equal to $4.597$. $SSY$ can be partitioned into two parts: the sum of squares predicted ($SSY'$) and the sum of squares error ($SSE$). The sum of squares predicted is the sum of the squared deviations of the predicted scores from the mean predicted score. In other words, it is the sum of the $y'^2$ column and is equal to $1.806$. The sum of squares error is the sum of the squared errors of prediction. It is therefore the sum of the $(Y-Y')^2$ column and is equal to $2.791$. This can be summed up as:
$SSY = SSY' + SSE$
$4.597 = 1.806 + 2.791$
There are several other notable features about Table $3$. First, notice that the sum of $y$ and the sum of $y'$ are both zero. This will always be the case because these variables were created by subtracting their respective means from each value. Also, notice that the mean of $Y-Y'$ is $0$. This indicates that although some $Y$ values are higher than their respective predicted $Y$ values and some are lower, the average difference is zero.
The $SSY$ is the total variation, the $SSY'$ is the variation explained, and the $SSE$ is the variation unexplained. Therefore, the proportion of variation explained can be computed as:
$\text{Proportion explained} = \dfrac{SSY'}{SSY}$
Similarly, the proportion not explained is:
$\text{Proportion not explained} = \dfrac{SSE}{SSY}$
There is an important relationship between the proportion of variation explained and Pearson's correlation: $r^2$ is the proportion of variation explained. Therefore, if $r = 1$, then, naturally, the proportion of variation explained is $1$; if $r = 0$, then the proportion explained is $0$. One last example: for $r = 0.4$, the proportion of variation explained is $0.16$.
Since the variance is computed by dividing the variation by $N$ (for a population) or $N-1$ (for a sample), the relationships spelled out above in terms of variation also hold for variance. For example,
$\sigma_{total}^2 = \sigma_{Y'}^2 + \sigma_e^2$
where the first term is the variance total, the second term is the variance of $Y'$, and the last term is the variance of the errors of prediction ($Y-Y'$). Similarly, $r^2$ is the proportion of variance explained as well as the proportion of variation explained.
Summary Table
It is often convenient to summarize the partitioning of the data in a table. The degrees of freedom column ($df$) shows the degrees of freedom for each source of variation. The degrees of freedom for the sum of squares explained is equal to the number of predictor variables. This will always be $1$ in simple regression. The error degrees of freedom is equal to the total number of observations minus $2$. In this example, it is $5 - 2 = 3$. The total degrees of freedom is the total number of observations minus $1$.
Table $4$: Summary Table for Example Data
Source Sum of Squares df Mean Square
Explained 1.806 1 1.806
Error 2.791 3 0.930
Total 4.597 4
Contributor
• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/14%3A_Regression/14.03%3A_Partitioning_Sums_of_Squares.txt |
Learning Objectives
• Make judgments about the size of the standard error of the estimate from a scatter plot
• Compute the standard error of the estimate based on errors of prediction
• Compute the standard error using Pearson's correlation
• Estimate the standard error of the estimate based on a sample
Figure $1$ shows two regression examples. You can see that in $\text{Graph A}$, the points are closer to the line than they are in $\text{Graph B}$. Therefore, the predictions in $\text{Graph A}$ are more accurate than in $\text{Graph B}$.
The standard error of the estimate is a measure of the accuracy of predictions. Recall that the regression line is the line that minimizes the sum of squared deviations of prediction (also called the sum of squares error). The standard error of the estimate is closely related to this quantity and is defined below:
$\sigma _{est}=\sqrt{\frac{\sum (Y-Y')^2}{N}}$
where $\sigma _{est}$ is the standard error of the estimate, $Y$ is an actual score, $Y'$ is a predicted score, and $N$ is the number of pairs of scores. The numerator is the sum of squared differences between the actual scores and the predicted scores.
Note the similarity of the formula for $\sigma _{est}$ to the formula for σ.  It turns out that σest is the standard deviation of the errors of prediction (each $Y - Y'$ is an error of prediction).
Assume the data in Table $1$ are the data from a population of five $X$, $Y$ pairs.
Table $1$: Example data
X Y Y' Y-Y' (Y-Y')2
1.00 1.00 1.210 -0.210 0.044
2.00 2.00 1.635 0.365 0.133
3.00 1.30 2.060 -0.760 0.578
4.00 3.75 2.485 1.265 1.600
5.00 2.25 2.910 -0.660 0.436
Sum 15.00 10.30 10.30 0.000 2.791
The last column shows that the sum of the squared errors of prediction is $2.791$. Therefore, the standard error of the estimate is
$\sigma _{est}=\sqrt{\frac{2.791}{5}}=0.747$
There is a version of the formula for the standard error in terms of Pearson's correlation:
$\sigma _{est}=\sqrt{\frac{(1-\rho )^2SSY}{N}}$
where $ρ$ is the population value of Pearson's correlation and $SSY$ is
$SSY=\sum (Y-\mu _Y)^2$
For the data in Table $1$, $μ_Y = 2.06$, $SSY = 4.597$ and $ρ= 0.6268$. Therefore,
$\sigma _{est}=\sqrt{\frac{(1-0.6268^2)(4.597)}{5}}=\sqrt{\frac{2.791}{5}}=0.747$
which is the same value computed previously.
Similar formulas are used when the standard error of the estimate is computed from a sample rather than a population. The only difference is that the denominator is $N-2$ rather than $N$. The reason $N-2$ is used rather than $N-1$ is that two parameters (the slope and the intercept) were estimated in order to estimate the sum of squares. Formulas for a sample comparable to the ones for a population are shown below.
$s _{est}=\sqrt{\frac{\sum (Y-Y')^2}{N-2}}$
$s _{est}=\sqrt{\frac{2.791}{3}}=0.964$
$s _{est}=\sqrt{\frac{(1-r)^2SSY}{N-2}}$
14.05: Inferential Statistics for b and r
Learning Objectives
• State the assumptions that inferential statistics in regression are based upon
• Identify heteroscedasticity in a scatter plot
• Compute the standard error of a slope
• Test a slope for significance
• Construct a confidence interval on a slope
• Test a correlation for significance
This section shows how to conduct significance tests and compute confidence intervals for the regression slope and Pearson's correlation. As you will see, if the regression slope is significantly different from zero, then the correlation coefficient is also significantly different from zero.
Assumptions
Although no assumptions were needed to determine the best-fitting straight line, assumptions are made in the calculation of inferential statistics. Naturally, these assumptions refer to the population, not the sample.
1. Linearity: The relationship between the two variables is linear.
2. Homoscedasticity: The variance around the regression line is the same for all values of $X$. A clear violation of this assumption is shown in Figure $1$. Notice that the predictions for students with high high-school GPAs are very good, whereas the predictions for students with low high-school GPAs are not very good. In other words, the points for students with high high-school GPAs are close to the regression line, whereas the points for low high-school GPA students are not.
1. The errors of prediction are distributed normally. This means that the deviations from the regression line are normally distributed. It does not mean that $X$ or $Y$ is normally distributed.
Significance Test for the Slope (b)
Recall the general formula for a t test:
$t=\frac{\text{statistic-hypothesized value}}{\text{estimated standard error of the statistic}}$
As applied here, the statistic is the sample value of the slope ($b$) and the hypothesized value is $0$. The number of degrees of freedom for this test is:
$df = N-2$
where $N$ is the number of pairs of scores.
The estimated standard error of b is computed using the following formula:
$s_b=\frac{s_{est}}{\sqrt{SSX}}$
where $s_b$ is the estimated standard error of $b$, $s_{est}$ is the standard error of the estimate, and $SSX$ is the sum of squared deviations of $X$ from the mean of $X$. $SSX$ is calculated as
$SSX=\sum (X-M_X)^2$
where $M_x$ is the mean of $X$. As shown previously, the standard error of the estimate can be calculated as
$s_{est}=\sqrt{\frac{(1-r^2)SSY}{N-2}}$
These formulas are illustrated with the data shown in Table $1$. These data are reproduced from the introductory section. The column $X$ has the values of the predictor variable and the column $Y$ has the values of the criterion variable. The third column, $x$, contains the differences between the values of column $X$ and the mean of $X$. The fourth column, $x_2$, is the square of the $x$ column. The fifth column, $y$, contains the differences between the values of column $Y$ and the mean of $Y$. The last column, $y_2$, is simply square of the $y$ column.
Table $1$: Example data
X Y x x2 y y2
1.00 1.00 -2.00 4 -1.06 1.1236
2.00 2.00 -1.00 1 -0.06 0.0036
3.00 1.30 0.00 0 -0.76 0.5776
4.00 3.75 1.00 1 1.69 2.8561
5.00 2.25 2.00 4 0.19 0.0361
Sum 15.00 10.30 0.00 10.00 0.00 4.5970
The computation of the standard error of the estimate ($s_{est}$) for these data is shown in the section on the standard error of the estimate. It is equal to $0.964$.
$s_{est} = 0.964$
$SSX$ is the sum of squared deviations from the mean of $X$. It is, therefore, equal to the sum of the $x^2$ column and is equal to $10$.
$SSX = 10.00$
We now have all the information to compute the standard error of $b$:
$s_b=\frac{0.964}{\sqrt{10}}=0.305$
As shown previously, the slope ($b$) is $0.425$. Therefore,
$t=\frac{0.425}{0.305}=1.39$
$df = N-2 = 5-2 = 3$
The $p$ value for a two-tailed $t$ test is $0.26$. Therefore, the slope is not significantly different from $0$.
Confidence Interval for the Slope
The method for computing a confidence interval for the population slope is very similar to methods for computing other confidence intervals. For the $95\%$ confidence interval, the formula is:
$\text{lower limit}: b - (t_{0.95})(s_b)$
$\text{upper limit}: b + (t_{0.95})(s_b)$
where $t_{0.95}$ is the value of $t$ to use for the $95\%$ confidence interval.
The values of $t$ to be used in a confidence interval can be looked up in a table of the t distribution. A small version of such a table is shown in Table $2$. The first column, $df$, stands for degrees of freedom.
Table $2$: Abbreviated $t$ table
df 0.95 0.99
2 4.303 9.925
3 3.182 5.841
4 2.776 4.604
5 2.571 4.032
8 2.306 3.355
10 2.228 3.169
20 2.086 2.845
50 2.009 2.678
100 1.984 2.626
You can also use the "inverse t distribution" calculator to find the $t$ values to use in a confidence interval.
Applying these formulas to the example data,
$\text{lower limit}: 0.425 - (3.182)(0.305) = -0.55$
$\text{upper limit}: 0.425 + (3.182)(0.305) = 1.40$
Significance Test for the Correlation
The formula for a significance test of Pearson's correlation is shown below:
$t=\frac{r\sqrt{N-2}}{\sqrt{1-r^2}}$
where $N$ is the number of pairs of scores. For the example data,
$t=\frac{0.627\sqrt{5-2}}{\sqrt{1-0.627^2}}=1.39$
Notice that this is the same $t$ value obtained in the $t$ test of $b$. As in that test, the degrees of freedom is $N - 2 = 5 -2 = 3$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/14%3A_Regression/14.04%3A_Standard_Error_of_the_Estimate.txt |
Learning Objectives
• Describe what makes a point influential
• Define "distance"
It is possible for a single observation to have a great influence on the results of a regression analysis. It is therefore important to be alert to the possibility of influential observations and to take them into consideration when interpreting the results.
Influence
The influence of an observation can be thought of in terms of how much the predicted scores for other observations would differ if the observation in question were not included. Cook's D is a good measure of the influence of an observation and is proportional to the sum of the squared differences between predictions made with all observations in the analysis and predictions made leaving out the observation in question. If the predictions are the same with or without the observation in question, then the observation has no influence on the regression model. If the predictions differ greatly when the observation is not included in the analysis, then the observation is influential.
A common rule of thumb is that an observation with a value of Cook's D over $1.0$ has too much influence. As with all rules of thumb, this rule should be applied judiciously and not thoughtlessly.
An observation's influence is a function of two factors:
1. How much the observation's value on the predictor variable differs from the mean of the predictor variable and
2. The difference between the predicted score for the observation and its actual score.
The former factor is called the observation's leverage. The latter factor is called the observation's distance.
Calculation of Cook's D (Optional)
The first step in calculating the value of Cook's D for an observation is to predict all the scores in the data once using a regression equation based on all the observations and once using all the observations except the observation in question. The second step is to compute the sum of the squared differences between these two sets of predictions. The final step is to divide this result by $2$ times the $MSE$ (see the section on partitioning the variance).
Leverage
The leverage of an observation is based on how much the observation's value on the predictor variable differs from the mean of the predictor variable. The greater an observation's leverage, the more potential it has to be an influential observation. For example, an observation with a value equal to the mean on the predictor variable has no influence on the slope of the regression line regardless of its value on the criterion variable. On the other hand, an observation that is extreme on the predictor variable has the potential to affect the slope greatly.
Calculation of Leverage (h)
The first step is to standardize the predictor variable so that it has a mean of $0$ and a standard deviation of $1$. Then, the leverage ($h$) is computed by squaring the observation's value on the standardized predictor variable, adding $1$, and dividing by the number of observations.
Distance
The distance of an observation is based on the error of prediction for the observation: The greater the error of prediction, the greater the distance. The most commonly used measure of distance is the $\textit{studentized residual}$. The $\textit{studentized residual}$ for an observation is closely related to the error of prediction for that observation divided by the standard deviation of the errors of prediction. However, the predicted score is derived from a regression equation in which the observation in question is not counted. The details of the computation of a $\textit{studentized residual}$ are a bit complex and are beyond the scope of this work.
Even an observation with a large distance will not have that much influence if its leverage is low. It is the combination of an observation's leverage and distance that determines its influence.
Example $1$
Table $1$ shows the leverage, $\textit{studentized residual}$, and influence for each of the five observations in a small dataset.
Table $1$: Example Data
ID X Y h R D
A 1 2 0.39 -1.02 0.40
B 2 3 0.27 -0.56 0.06
C 3 5 0.21 0.89 0.11
D 4 6 0.20 1.22 0.19
E 8 7 0.73 -1.68 8.86
In the above table, $h$ is the leverage, $R$ is the $\textit{studentized residual}$, and $D$ is Cook's measure of influence.
$\text{Observation A}$ has fairly high leverage, a relatively high residual, and moderately high influence.
$\text{Observation B}$ has small leverage and a relatively small residual. It has very little influence.
$\text{Observation C}$ has small leverage and a relatively high residual. The influence is relatively low.
$\text{Observation D}$ has the lowest leverage and the second highest residual. Although its residual is much higher than $\text{Observation A }$, its influence is much less because of its low leverage.
$\text{Observation E}$ has by far the largest leverage and the largest residual. This combination of high leverage and high residual makes this observation extremely influential.
Figure $1$ shows the regression line for the whole dataset (blue) and the regression line if the observation in question is not included (red) for all observations. The observation in question is circled. Naturally, the regression line for the whole dataset is the same in all panels. The residual is calculated relative to the line for which the observation in question is not included in the analysis. The most influential observation is $\text{Observation E}$ for which the two regression lines are very different. This indicates the influence of this observation. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/14%3A_Regression/14.06%3A_Influential_Observations.txt |
Learning Objectives
• To study Regression toward the mean
Regression toward the mean involves outcomes that are at least partly due to chance. We begin with an example of a task that is entirely chance: Imagine an experiment in which a group of $25$ people each predicted the outcomes of flips of a fair coin. For each subject in the experiment, a coin is flipped $12$ times and the subject predicts the outcome of each flip. Figure $1$ shows the results of a simulation of this "experiment." Although most subjects were correct from $5$ to $8$ times out of $12$, one simulated subject was correct $10$ times. Clearly, this subject was very lucky and probably would not do as well if he or she performed the task a second time. In fact, the best prediction of the number of times this subject would be correct on the retest is $6$, since the probability of being correct on a given trial is $0.5$ and there are $12$ trials.
More technically, the best prediction for the subject's result on the retest is the mean of the binomial distribution with $N = 12$ and $p = 0.50$. This distribution is shown in Figure $2$ and has a mean of $6$.
The point here is that no matter how many coin flips a subject predicted correctly, the best prediction of their score on a retest is $6$.
Now we consider a test we will call "Test A" that is partly chance and partly skill: Instead of predicting the outcomes of $12$ coin flips, each subject predicts the outcomes of $6$ coin flips and answers $6$ true/false questions about world history. Assume that the mean score on the $6$ history questions is $4$. A subject's score on Test A has a large chance component but also depends on history knowledge. If a subject scored very high on this test (such as a score of $10/12$), it is likely that they did well on both the history questions and the coin flips. If this subject is then given a second test (Test B) that also included coin predictions and history questions, their knowledge of history would be helpful and they would again be expected to score above the mean. However, since their high performance on the coin portion of Test A would not be predictive of their coin performance on Test B, they would not be expected to fare as well on Test B as on Test A. Therefore, the best prediction of their score on Test B would be somewhere between their score on Test A and the mean of Test B. This tendency of subjects with high values on a measure that includes chance and skill to score closer to the mean on a retest is called "$\textit{regression toward the mean}$."
The essence of the regression-toward-the-mean phenomenon is that people with high scores tend to be above average in skill and in luck and that only the skill portion is relevant to future performance. Similarly, people with low scores tend to be below average in skill and luck, and their bad luck is not relevant to future performance. This does not mean that all people who score high have above average luck. However, on average they do.
Almost every measure of behavior has a chance and a skill component to it. Take a student's grade on a final exam as an example. Certainly, the student's knowledge of the subject will be a major determinant of his or her grade. However, there are aspects of performance that are due to chance. The exam cannot cover everything in the course and therefore must represent a subset of the material. Maybe the student was lucky in that the one aspect of the course the student did not understand well was not well-represented on the test. Or, maybe the student was not sure which of two approaches to a problem would be better but, more or less by chance, chose the right one. Other chance elements come into play as well. Perhaps the student was awakened early in the morning by a random phone call, resulting in fatigue and lower performance. And, of course, guessing on multiple choice questions is another source of randomness in test scores.
There will be regression toward the mean in a test-retest situation whenever there is less than a perfect ($r = 1$) relationship between the test and the retest. This follows from the formula for a regression line with standardized variables shown below:
$Z_{Y'} = (r)(Z_X)$
From this equation it is clear that if the absolute value of $r$ is less than $1$, then the predicted value of $Z_Y$ will be closer to $0$, the mean for standardized scores, than is $Z_X$. Also, note that if the correlation between $X$ and $Y$ is $0$, as it would be for a task that is all luck, the predicted standard score for $Y$ is its mean, $0$, regardless of the score on $X$.
Figure $3$ shows a scatter plot with the regression line predicting the standardized Verbal SAT from the standardized Math SAT. Note that the slope of the line is equal to the correlation of $0.835$ between these variables.
The point represented by a blue diamond has a value of $1.6$ on the standardized Math SAT. This means that this student scored $1.6$ standard deviations above the mean on Math SAT. The predicted score is $(r)(1.6) = (0.835)(1.6) = 1.34$. The horizontal line on the graph shows the value of the predicted score. The key point is that although this student scored $1.6$ standard deviations above the mean on Math SAT, he or she is only predicted to score $1.34$ standard deviations above the mean on Verbal SAT. Thus, the prediction is that the Verbal SAT score will be closer to the mean of $0$ than is the Math SAT score. Similarly, a student scoring far below the mean on Math SAT will be predicted to score higher on Verbal SAT.
Regression toward the mean occurs in any situation in which observations are selected on the basis of performance on a task that has a random component. If you choose people on the basis of their performance on such a task, you will be choosing people partly on the basis of their skill and partly on the basis of their luck on the task. Since their luck cannot be expected to be maintained from trial to trial, the best prediction of a person's performance on a second trial will be somewhere between their performance on the first trial and the mean performance on the first trial. The degree to which the score is expected to "regress toward the mean" in this manner depends on the relative contributions of chance and skill to the task: the greater the role of chance, the more the regression toward the mean.
Errors Resulting From Failure to Understand Regression Toward the Mean
Failure to appreciate regression toward the mean is common and often leads to incorrect interpretations and conclusions. One of the best examples is provided by Nobel Laureate Daniel Kahneman in his autobiography. Dr. Kahneman was attempting to teach flight instructors that praise is more effective than punishment. He was challenged by one of the instructors who relayed that in his experience praising a cadet for executing a clean maneuver is typically followed by a lesser performance, whereas screaming at a cadet for bad execution is typically followed by improved performance. This, of course, is exactly what would be expected based on regression toward the mean. A pilot's performance, although based on considerable skill, will vary randomly from maneuver to maneuver. When a pilot executes an extremely clean maneuver, it is likely that he or she had a bit of luck in their favor in addition to their considerable skill. After the praise but not because of it, the luck component will probably disappear and the performance will be lower. Similarly, a poor performance is likely to be partly due to bad luck. After the criticism but not because of it, the next performance will likely be better. To drive this point home, Kahneman had each instructor perform a task in which a coin was tossed at a target twice. He demonstrated that the performance of those who had done the best the first time deteriorated, whereas the performance of those who had done the worst improved.
Regression toward the mean is frequently present in sports performance. A good example is provided by Schall and Smith ($2000$), who analyzed many aspects of baseball statistics including the batting averages of players in $1998$. They chose the $10$ players with the highest batting averages (BAs) in $1998$ and checked to see how well they did in $1999$. According to what would be expected based on regression toward the mean, these players should, on average, have lower batting averages in $1999$ than they did in $1998$. As can be seen in Table $1$, $7/10$ of the players had lower batting averages in $1999$ than they did in $1998$. Moreover, those who had higher averages in $1999$ were only slightly higher, whereas those who were lower were much lower. The average decrease from $1998$ to $1999$ was $33$ points. Even so, most of these players had excellent batting averages in $1999$ indicating that skill was an important component of their $1998$ averages.
Table $1$: How the Ten Players with the Highest BAs in $1998$ did in $1999$
1998 1999 Difference
363 379 16
354 298 -56
339 342 3
337 281 -56
336 249 -87
331 298 -33
328 297 -31
328 303 -25
327 257 -70
327 332 5
Figure $4$ shows the batting averages of the two years. The decline from $1998$ to $1999$ is clear. Note that although the mean decreased from $1998$, some players increased their batting averages. This illustrates that regression toward the mean does not occur for every individual. Although the predicted scores for every individual will be lower, some of the predictions will be wrong.
Regression toward the mean plays a role in the so-called "Sophomore Slump," a good example of which is that a player who wins "rookie of the year" typically does less well in his second season. A related phenomenon is called the Sports Illustrated Cover Jinx.
An experiment without a control group can confound regression effects with real effects. For example, consider a hypothetical experiment to evaluate a reading-improvement program. All first graders in a school district were given a reading achievement test and the $50$ lowest-scoring readers were enrolled in the program. The students were retested following the program and the mean improvement was large. Does this necessarily mean the program was effective? No, it could be that the initial poor performance of the students was due, in part, to bad luck. Their luck would be expected to improve in the retest, which would increase their scores with or without the treatment program.
For a real example, consider an experiment that sought to determine whether the drug propranolol would increase the SAT scores of students thought to have test anxiety. Propranolol was given to $25$ high school students chosen because IQ tests and other academic performance indicated that they had not done as well as expected on the SAT. On a retest taken after receiving propranolol, students improved their SAT scores an average of $120$ points. This was a significantly greater increase than the $38$ points expected simply on the basis of having taken the test before. The problem with the study is that the method of selecting students likely resulted in a disproportionate number of students who had bad luck when they first took the SAT. Consequently, these students would likely have increased their scores on a retest with or without the propranolol. This is not to say that propranolol had no effect. However, since possible propranolol effects and regression effects were confounded, no firm conclusions should be drawn.
Randomly assigning students to either the propranolol group or a control group would have improved the experimental design. Since the regression effects would then not have been systematically different for the two groups, a significant difference would have provided good evidence for a propranolol effect.
New York Times Article on the Study
Schall, T., & Smith, G. (2000) Do Baseball Players Regress Toward the Mean? The American Statistician, 54, 231-235. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/14%3A_Regression/14.07%3A_Regression_Toward_the_Mean.txt |
Learning Objectives
• State the regression equation
• Define "regression coefficient"
• Define "beta weight"
• Explain what $R$ is and how it is related to $r$
• Explain why a regression weight is called a "partial slope"
• Explain why the sum of squares explained in a multiple regression model is usually less than the sum of the sums of squares in simple regression
• Define $R^2$ in terms of proportion explained
• Test $R^2$ for significance
• Test the difference between a complete and reduced model for significance
• State the assumptions of multiple regression and specify which aspects of the analysis require assumptions
In simple linear regression, a criterion variable is predicted from one predictor variable. In multiple regression, the criterion is predicted by two or more variables. For example, in the SAT case study, you might want to predict a student's university grade point average on the basis of their High-School GPA ($HSGPA$) and their total SAT score (verbal + math). The basic idea is to find a linear combination of $HSGPA$ and $SAT$ that best predicts University GPA ($UGPA$). That is, the problem is to find the values of $b_1$ and $b_2$ in the equation shown below that give the best predictions of $UGPA$. As in the case of simple linear regression, we define the best predictions as the predictions that minimize the squared errors of prediction.
$UGPA' = b_1HSGPA + b_2SAT + A$
where $UGPA'$ is the predicted value of University GPA and $A$ is a constant. For these data, the best prediction equation is shown below:
$UGPA' = 0.541 \times HSGPA + 0.008 \times SAT + 0.540$
In other words, to compute the prediction of a student's University GPA, you add up their High-School GPA multiplied by $0.541$, their $SAT$ multiplied by $0.008$, and $0.540$. Table $1$ shows the data and predictions for the first five students in the dataset.
Table $1$: Data and Predictions
HSGPA SAT UGPA'
3.45 1232 3.38
2.78 1070 2.89
2.52 1086 2.76
3.67 1287 3.55
3.24 1130 3.19
The values of $b$ ($b_1$ and $b_2$) are sometimes called "regression coefficients" and sometimes called "regression weights." These two terms are synonymous.
The multiple correlation ($R$) is equal to the correlation between the predicted scores and the actual scores. In this example, it is the correlation between $UGPA'$ and $UGPA$, which turns out to be $0.79$. That is, $R = 0.79$. Note that $R$ will never be negative since if there are negative correlations between the predictor variables and the criterion, the regression weights will be negative so that the correlation between the predicted and actual scores will be positive.
Interpretation of Regression Coefficients
A regression coefficient in multiple regression is the slope of the linear relationship between the criterion variable and the part of a predictor variable that is independent of all other predictor variables. In this example, the regression coefficient for $HSGPA$ can be computed by first predicting $HSGPA$ from $SAT$ and saving the errors of prediction (the differences between $HSGPA$ and $HSGPA'$). These errors of prediction are called "residuals" since they are what is left over in $HSGPA$ after the predictions from $SAT$ are subtracted, and represent the part of $HSGPA$ that is independent of $SAT$. These residuals are referred to as $HSGPA.SAT$, which means they are the residuals in $HSGPA$ after having been predicted by $SAT$. The correlation between $HSGPA.SAT$ and $SAT$ is necessarily $0$.
The final step in computing the regression coefficient is to find the slope of the relationship between these residuals and $UGPA$. This slope is the regression coefficient for $HSGPA$. The following equation is used to predict $HSGPA$ from $SAT$:
$HSGPA' = -1.314 + 0.0036 \times SAT$
The residuals are then computed as:
$HSGPA - HSGPA'$
The linear regression equation for the prediction of $UGPA$ by the residuals is
$UGPA' = 0.541 \times HSGPA.SAT + 3.173$
Notice that the slope ($0.541$) is the same value given previously for $b_1$ in the multiple regression equation.
This means that the regression coefficient for $HSGPA$ is the slope of the relationship between the criterion variable and the part of $HSGPA$ that is independent of (uncorrelated with) the other predictor variables. It represents the change in the criterion variable associated with a change of one in the predictor variable when all other predictor variables are held constant. Since the regression coefficient for $HSGPA$ is $0.54$, this means that, holding $SAT$ constant, a change of one in $HSGPA$ is associated with a change of $0.54$ in $UGPA'$. If two students had the same $SAT$ and differed in $HSGPA$ by $2$, then you would predict they would differ in $UGPA$ by $(2)(0.54) = 1.08$. Similarly, if they differed by $0.5$, then you would predict they would differ by $(0.50)(0.54) = 0.27$.
The slope of the relationship between the part of a predictor variable independent of other predictor variables and the criterion is its partial slope. Thus the regression coefficient of $0.541$ for $HSGPA$ and the regression coefficient of $0.008$ for $SAT$ are partial slopes. Each partial slope represents the relationship between the predictor variable and the criterion holding constant all of the other predictor variables.
It is difficult to compare the coefficients for different variables directly because they are measured on different scales. A difference of $1$ in $HSGPA$ is a fairly large difference, whereas a difference of $1$ on the $SAT$ is negligible. Therefore, it can be advantageous to transform the variables so that they are on the same scale. The most straightforward approach is to standardize the variables so that they each have a standard deviation of $1$. A regression weight for standardized variables is called a "beta weight" and is designated by the Greek letter $β$. For these data, the beta weights are $0.625$ and $0.198$. These values represent the change in the criterion (in standard deviations) associated with a change of one standard deviation on a predictor [holding constant the value(s) on the other predictor(s)]. Clearly, a change of one standard deviation on $HSGPA$ is associated with a larger difference than a change of one standard deviation of $SAT$. In practical terms, this means that if you know a student's $HSGPA$, knowing the student's $SAT$ does not aid the prediction of $UGPA$ much. However, if you do not know the student's $HSGPA$, his or her $SAT$ can aid in the prediction since the $β$ weight in the simple regression predicting $UGPA$ from $SAT$ is $0.68$. For comparison purposes, the $β$ weight in the simple regression predicting $UGPA$ from $HSGPA$ is $0.78$. As is typically the case, the partial slopes are smaller than the slopes in simple regression.
Partitioning the Sums of Squares
Just as in the case of simple linear regression, the sum of squares for the criterion ($UGPA$ in this example) can be partitioned into the sum of squares predicted and the sum of squares error. That is,
$SSY = SSY' + SSE$
which for these data:
$20.798 = 12.961 + 7.837$
The sum of squares predicted is also referred to as the "sum of squares explained." Again, as in the case of simple regression,
$\text{Proportion Explained} = SSY'/SSY$
In simple regression, the proportion of variance explained is equal to $r^2$; in multiple regression, the proportion of variance explained is equal to $R^2$.
In multiple regression, it is often informative to partition the sum of squares explained among the predictor variables. For example, the sum of squares explained for these data is $12.96$. How is this value divided between $HSGPA$ and $SAT$? One approach that, as will be seen, does not work is to predict $UGPA$ in separate simple regressions for $HSGPA$ and $SAT$. As can be seen in Table $2$, the sum of squares in these separate simple regressions is $12.64$ for $HSGPA$ and $9.75$ for $SAT$. If we add these two sums of squares we get $22.39$, a value much larger than the sum of squares explained of $12.96$ in the multiple regression analysis. The explanation is that $HSGPA$ and $SAT$ are highly correlated ($r = 0.78$) and therefore much of the variance in $UGPA$ is confounded between $HSGPA$ and $SAT$. That is, it could be explained by either $HSGPA$ or $SAT$ and is counted twice if the sums of squares for $HSGPA$ and $SAT$ are simply added.
Table $2$: Sums of Squares for Various Predictors
Predictors Sum of Squares
HSGPA 12.64
SAT 9.75
HSGPA and SAT 12.96
Table $3$ shows the partitioning of the sum of squares into the sum of squares uniquely explained by each predictor variable, the sum of squares confounded between the two predictor variables, and the sum of squares error. It is clear from this table that most of the sum of squares explained is confounded between $HSGPA$ and $SAT$. Note that the sum of squares uniquely explained by a predictor variable is analogous to the partial slope of the variable in that both involve the relationship between the variable and the criterion with the other variable(s) controlled.
Table $3$: Partitioning the Sum of Squares
Source Sum of Squares Proportion
HSGPA (unique) 3.21 0.15
SAT (unique) 0.32 0.02
HSGPA and SAT (Confounded) 9.43 0.45
Error 7.84 0.38
Total 20.80 1.00
The sum of squares uniquely attributable to a variable is computed by comparing two regression models: the complete model and a reduced model. The complete model is the multiple regression with all the predictor variables included ($HSGPA$ and $SAT$ in this example). A reduced model is a model that leaves out one of the predictor variables. The sum of squares uniquely attributable to a variable is the sum of squares for the complete model minus the sum of squares for the reduced model in which the variable of interest is omitted. As shown in Table $2$, the sum of squares for the complete model ($HSGPA$ and $SAT$) is $12.96$. The sum of squares for the reduced model in which $HSGPA$ is omitted is simply the sum of squares explained using $SAT$ as the predictor variable and is $9.75$. Therefore, the sum of squares uniquely attributable to $HSGPA$ is $12.96 - 9.75 = 3.21$. Similarly, the sum of squares uniquely attributable to $SAT$ is $12.96 - 12.64 = 0.32$. The confounded sum of squares in this example is computed by subtracting the sum of squares uniquely attributable to the predictor variables from the sum of squares for the complete model: $12.96 - 3.21 - 0.32 = 9.43$. The computation of the confounded sums of squares in analysis with more than two predictors is more complex and beyond the scope of this text.
Since the variance is simply the sum of squares divided by the degrees of freedom, it is possible to refer to the proportion of variance explained in the same way as the proportion of the sum of squares explained. It is slightly more common to refer to the proportion of variance explained than the proportion of the sum of squares explained and, therefore, that terminology will be adopted frequently here.
When variables are highly correlated, the variance explained uniquely by the individual variables can be small even though the variance explained by the variables taken together is large. For example, although the proportions of variance explained uniquely by $HSGPA$ and $SAT$ are only $0.15$ and $0.02$ respectively, together these two variables explain $0.62$ of the variance. Therefore, you could easily underestimate the importance of variables if only the variance explained uniquely by each variable is considered. Consequently, it is often useful to consider a set of related variables. For example, assume you were interested in predicting job performance from a large number of variables some of which reflect cognitive ability. It is likely that these measures of cognitive ability would be highly correlated among themselves and therefore no one of them would explain much of the variance independently of the other variables. However, you could avoid this problem by determining the proportion of variance explained by all of the cognitive ability variables considered together as a set. The variance explained by the set would include all the variance explained uniquely by the variables in the set as well as all the variance confounded among variables in the set. It would not include variance confounded with variables outside the set. In short, you would be computing the variance explained by the set of variables that is independent of the variables not in the set.
Inferential Statistics
We begin by presenting the formula for testing the significance of the contribution of a set of variables. We will then show how special cases of this formula can be used to test the significance of $R^2$ as well as to test the significance of the unique contribution of individual variables.
The first step is to compute two regression analyses:
1. an analysis in which all the predictor variables are included and
2. an analysis in which the variables in the set of variables being tested are excluded.
The former regression model is called the "complete model" and the latter is called the "reduced model." The basic idea is that if the reduced model explains much less than the complete model, then the set of variables excluded from the reduced model is important.
The formula for testing the contribution of a group of variables is:
$F=\cfrac{\cfrac{SSQ_C-SSQ_R}{p_C-p_R}}{\cfrac{SSQ_T-SSQ_C}{N-p_C-1}}=\cfrac{MS_{explained}}{MS_{error}}$
where:
$SSQ_C$ is the sum of squares for the complete model,
$SSQ_R$ is the sum of squares for the reduced model,
$p_C$ is the number of predictors in the complete model,
$p_R$ is the number of predictors in the reduced model,
$SSQ_T$ is the sum of squares total (the sum of squared deviations of the criterion variable from its mean), and
$N$ is the total number of observations
The degrees of freedom for the numerator is $p_C - p_R$ and the degrees of freedom for the denominator is $N - p_C -1$. If the $F$ is significant, then it can be concluded that the variables excluded in the reduced set contribute to the prediction of the criterion variable independently of the other variables.
This formula can be used to test the significance of $R^2$ by defining the reduced model as having no predictor variables. In this application, $SSQ_R$ and $p_R = 0$. The formula is then simplified as follows:
$F=\cfrac{\cfrac{SSQ_C}{p_C}}{\cfrac{SSQ_T-SSQ_C}{N-p_C-1}}=\cfrac{MS_{explained}}{MS_{error}}$
which for this example becomes:
$F=\cfrac{\cfrac{12.96}{2}}{\cfrac{20.80-12.96}{105-2-1}}=\cfrac{6.48}{0.08}=84.35$
The degrees of freedom are $2$ and $102$. The $F$ distribution calculator shows that $p < 0.001$.
F Calculator
The reduced model used to test the variance explained uniquely by a single predictor consists of all the variables except the predictor variable in question. For example, the reduced model for a test of the unique contribution of $HSGPA$ contains only the variable $SAT$. Therefore, the sum of squares for the reduced model is the sum of squares when $UGPA$ is predicted by $SAT$. This sum of squares is $9.75$. The calculations for $F$ are shown below:
$F=\cfrac{\cfrac{12.96-9.75}{2-1}}{\cfrac{20.80-12.96}{105-2-1}}=\cfrac{3.212}{0.077}=41.80$
The degrees of freedom are $1$ and $102$. The $F$ distribution calculator shows that $p < 0.001$.
Similarly, the reduced model in the test for the unique contribution of $SAT$ consists of $HSGPA$.
$F=\cfrac{\cfrac{12.96-12.64}{2-1}}{\cfrac{20.80-12.96}{105-2-1}}=\cfrac{0.322}{0.077}=4.19$
The degrees of freedom are $1$ and $102$. The $F$ distribution calculator shows that $p = 0.0432$.
The significance test of the variance explained uniquely by a variable is identical to a significance test of the regression coefficient for that variable. A regression coefficient and the variance explained uniquely by a variable both reflect the relationship between a variable and the criterion independent of the other variables. If the variance explained uniquely by a variable is not zero, then the regression coefficient cannot be zero. Clearly, a variable with a regression coefficient of zero would explain no variance.
Other inferential statistics associated with multiple regression are beyond the scope of this text. Two of particular importance are:
1. confidence intervals on regression slopes and
2. confidence intervals on predictions for specific observations.
These inferential statistics can be computed by standard statistical analysis packages such as $R$, $SPSS$, $STATA$, $SAS$, and $JMP$.
SPSS Output JMP Output
Assumptions
No assumptions are necessary for computing the regression coefficients or for partitioning the sum of squares. However, there are several assumptions made when interpreting inferential statistics. Moderate violations of Assumptions $1-3$ do not pose a serious problem for testing the significance of predictor variables. However, even small violations of these assumptions pose problems for confidence intervals on predictions for specific observations.
1. Residuals are normally distributed:
As in the case of simple linear regression, the residuals are the errors of prediction. Specifically, they are the differences between the actual scores on the criterion and the predicted scores. A $Q-Q$ plot for the residuals for the example data is shown below. This plot reveals that the actual data values at the lower end of the distribution do not increase as much as would be expected for a normal distribution. It also reveals that the highest value in the data is higher than would be expected for the highest value in a sample of this size from a normal distribution. Nonetheless, the distribution does not deviate greatly from normality.
1. Homoscedasticity:
It is assumed that the variances of the errors of prediction are the same for all predicted values. As can be seen below, this assumption is violated in the example data because the errors of prediction are much larger for observations with low-to-medium predicted scores than for observations with high predicted scores. Clearly, a confidence interval on a low predicted $UGPA$ would underestimate the uncertainty.
1. Linearity:
It is assumed that the relationship between each predictor variable and the criterion variable is linear. If this assumption is not met, then the predictions may systematically overestimate the actual values for one range of values on a predictor variable and underestimate them for another. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/14%3A_Regression/14.08%3A_Introduction_to_Multiple_Regression.txt |
Learning Objectives
• Regression Toward the Mean in American Football
In a discussion about the Dallas Cowboy football team, there was a comment that the quarterback threw far more interceptions in the first two games than is typical (there were two interceptions per game). The author correctly pointed out that, because of regression toward the mean, performance in the future is expected to improve. However, the author defined regression toward the mean as, "In nerd land, that basically means that things tend to even out over the long run."
Example \(1\): what do you think?
Comment on that definition.
Solution
That definition is sort of correct, but it could be stated more precisely. Things don't always tend to even out in the long run. If a great player has an average game, then things wouldn't even out (to the average of all players) but would regress toward that player's high mean performance.
14.E: Regression (Exercises)
General Questions
Q1
What is the equation for a regression line? What does each term in the line refer to? (relevant section)
Q2
The formula for a regression equation based on a sample size of $25$ observations is $Y' = 2X + 9$.
1. What would be the predicted score for a person scoring $6$ on $X$?
2. If someone's predicted score was $14$, what was this person's score on $X$? (relevant section)
Q3
What criterion is used for deciding which regression line fits best? (relevant section)
Q4
What does the standard error of the estimate measure? What is the formula for the standard error of the estimate? (relevant section)
Q5
1. In a regression analysis, the sum of squares for the predicted scores is $100$ and the sum of squares error is $200$, what is $R^2$?
2. In a different regression analysis, $40\%$ of the variance was explained. The sum of squares total is $1000$. What is the sum of squares of the predicted values? (relevant section)
Q6
For the $X,Y$ data below, compute:
1. $r$ and determine if it is significantly different from zero.
2. the slope of the regression line and test if it differs significantly from zero.
3. the $95\%$ confidence interval for the slope.
(relevant section)
X
Y
2
5
4
6
4
7
5
11
6
12
Q7
What assumptions are needed to calculate the various inferential statistics of linear regression? (relevant section)
Q8
The correlation between years of education and salary in a sample of $20$ people from a certain company is $0.4$. Is this correlation statistically significant at the $0.05$ level? (relevant section)
Q9
A sample of $X$ and $Y$ scores is taken, and a regression line is used to predict $Y$ from $X$. If $SSY' = 300$, $SSE = 500$, and $N = 50$, what is: (relevant section relevant section)
1. $SSY$?
2. the standard error of the estimate?
3. $R^2$?
Q10
Using linear regression, find the predicted post-test score for someone with a score of $43$ on the pre-test. (relevant section)
Pre Post
59 56
52 63
44 55
51 50
42 66
42 48
41 58
45 36
27 13
63 50
54 81
44 56
50 64
47 50
55 63
49 57
45 73
57 63
46 46
60 60
65 47
64 73
50 58
74 85
59 44
Q11
The equation for a regression line predicting the number of hours of TV watched by children ($Y$) from the number of hours of TV watched by their parents ($X$) is $Y' = 4 + 1.2X$. The sample size is $12$.
1. If the standard error of $b$ is $0.4$, is the slope statistically significant at the $0.05$ level? (relevant section)
2. If the mean of $X$ is $8$, what is the mean of $Y$? (relevant section)
Q12
Based on the table below, compute the regression line that predicts $Y$ from $X$. (relevant section)
MX
MY
sX sY r
10
12
2.5 3.0 -0.6
Q13
Does $A$ or $B$ have a larger standard error of the estimate? (relevant section)
Q14
True/false: If the slope of a simple linear regression line is statistically significant, then the correlation will also always be significant. (relevant section)
Q15
True/false: If the slope of the relationship between $X$ and $Y$ is larger for $\text{Population 1}$ than for $\text{Population 2}$, the correlation will necessarily be larger in $\text{Population 1}$ than in $\text{Population 1}$. Why or why not? (relevant section)
Q16
True/false: If the correlation is $0.8$, then $40\%$ of the variance is explained. (relevant section)
Q17
True/false: If the actual $Y$ score was $31$, but the predicted score was $28$, then the error of prediction is $3$. (relevant section)
Questions from Case Studies
The following question is from the Angry Moods (AM) case study.
Q18
(AM#23) Find the regression line for predicting Anger-Out from Control-Out.
1. What is the slope?
2. What is the intercept?
3. Is the relationship at least approximately linear?
4. Test to see if the slope is significantly different from $0$.
5. What is the standard error of the estimate?
(relevant section, relevant section, relevant section)
The following question is from the SAT and GPA (SG) case study.
Q19
(SG#3) Find the regression line for predicting the overall university GPA from the high school GPA.
1. What is the slope?
2. What is the $y$-intercept?
3. If someone had a $2.2$ GPA in high school, what is the best estimate of his or her college GPA?
4. If someone had a $4.0$ GPA in high school, what is the best estimate of his or her college GPA?
(relevant section)
The following questions are from the Driving (D) case study.
Q20
(D#5) What is the correlation between age and how often the person chooses to drive in inclement weather? Is this correlation statistically significant at the $0.01$ level? Are older people more or less likely to report that they drive in inclement weather? (relevant section, relevant section )
Q21
(D#8) What is the correlation between how often a person chooses to drive in inclement weather and the percentage of accidents the person believes occur in inclement weather? Is this correlation significantly different from $0$? (relevant section, relevant section )
Q22
(D#10) Use linear regression to predict how often someone rides public transportation in inclement weather from what percentage of accidents that person thinks occur in inclement weather. (Pubtran by Accident)
1. Create a scatter plot of this data and add a regression line.
2. What is the slope?
3. What is the intercept?
4. Is the relationship at least approximately linear?
5. Test if the slope is significantly different from $0$.
6. Comment on possible assumption violations for the test of the slope.
7. What is the standard error of the estimate?
(relevant section, relevant section, relevant section)
Selected Answers
S2
1. $21$
S5
1. $0.33$
S6
1. $b = 1.91$
S9
1. $800$
S12
$a = 19.2$
S18
1. $3.45$
S19
1. $2.6$
S20
$r = 0.43$
S22
1. $0.35$ | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/14%3A_Regression/14.09%3A_Statistical_Literacy.txt |
• 15.1: Introduction to ANOVA
Analysis of Variance (ANOVA) is a statistical method used to test differences between two or more means. It may seem odd that the technique is called "Analysis of Variance" rather than "Analysis of Means." As you will see, the name is appropriate because inferences about means are made by analyzing variance.
• 15.2: ANOVA Designs
There are many types of experimental designs that can be analyzed by ANOVA. This section discusses many of these designs and defines several key terms used.
• 15.3: One-Factor ANOVA
This section shows how ANOVA can be used to analyze a one-factor between-subjects design.
• 15.4: One-Way Demo
• 15.5: Multi-Factor Between-Subjects
• 15.6: Unequal Sample Sizes
Whether by design, accident, or necessity, the number of subjects in each of the conditions in an experiment may not be equal. Since n is used to refer to the sample size of an individual group, designs with unequal sample sizes are sometimes referred to as designs with unequal n. Table 1. Sample Sizes for "Bias Against Associates of the Obese" Study.
• 15.7: Tests Supplementing
The null hypothesis tested in a one-factor ANOVA is that all the population means are equal. When the null hypothesis is rejected, all that can be said is that at least one population mean is different from at least one other population mean.
• 15.8: Within-Subjects
Within-subjects factors involve comparisons of the same subjects under different conditions. A within-subjects factor is sometimes referred to as a repeated-measures factor since repeated measurements are taken on each subject. An experimental design in which the independent variable is a within-subjects factor is called a within-subjects design.
• 15.9: Power of Within-Subjects Designs Demo
This simulation demonstrates the effect of the correlation between measures in a one-way within-subjects ANOVA with two levels. This test is equivalent to a correlated t test.
• 15.10: Statistical Literacy
• 15.E: Analysis of Variance (Exercises)
15: Analysis of Variance
Learning Objectives
• What null hypothesis is tested by ANOVA
• Describe the uses of ANOVA
Analysis of Variance (ANOVA) is a statistical method used to test differences between two or more means. It may seem odd that the technique is called "Analysis of Variance" rather than "Analysis of Means." As you will see, the name is appropriate because inferences about means are made by analyzing variance.
ANOVA is used to test general rather than specific differences among means. This can be seen best by example. In the case study "Smiles and Leniency," the effect of different types of smiles on the leniency shown to a person was investigated. Four different types of smiles (neutral, false, felt, miserable) were investigated. The chapter "All Pairwise Comparisons among Means" showed how to test differences among means. The results from the Tukey HSD test are shown in Table $1$.
Table $1$: Six Pairwise Comparisons
Comparison Mi-Mj Q p
False - Felt 0.46 1.65 0.649
False - Miserable 0.46 1.65 0.649
False - Neutral 1.25 4.48 0.010
Felt - Miserable 0.00 0.00 1.000
Felt - Neutral 0.79 2.83 0.193
Miserable - Neutral 0.79 2.83 0.193
Notice that the only significant difference is between the False and Neutral conditions.
ANOVA tests the non-specific null hypothesis that all four population means are equal. That is,
$\mu _{false} = \mu _{felt} = \mu _{miserable} = \mu _{neutral}$
This non-specific null hypothesis is sometimes called the omnibus null hypothesis. When the omnibus null hypothesis is rejected, the conclusion is that at least one population mean is different from at least one other mean. However, since the ANOVA does not reveal which means are different from which, it offers less specific information than the Tukey HSD test. The Tukey HSD is therefore preferable to ANOVA in this situation. Some textbooks introduce the Tukey test only as a follow-up to an ANOVA. However, there is no logical or statistical reason why you should not use the Tukey test even if you do not compute an ANOVA.
You might be wondering why you should learn about ANOVA when the Tukey test is better. One reason is that there are complex types of analysis that can be done with ANOVA and not with the Tukey test. A second is that ANOVA is by far the most commonly-used technique for comparing means, and it is important to understand ANOVA in order to understand research reports. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/15%3A_Analysis_of_Variance/15.01%3A_Introduction_to_ANOVA.txt |
Learning Objectives
• Determine whether a factor is a between-subjects or a within-subjects factor
• Define factorial design
There are many types of experimental designs that can be analyzed by ANOVA. This section discusses many of these designs and defines several key terms used.
Factors and Levels
The section on variables defined an independent variable as a variable manipulated by the experimenter. In the case study "Smiles and Leniency," the effect of different types of smiles on the leniency shown to a person was investigated. Four different types of smiles (neutral, false, felt, and miserable) were shown. In this experiment, "Type of Smile" is the independent variable. In describing an ANOVA design, the term factor is a synonym of independent variable. Therefore, "Type of Smile" is the factor in this experiment. Since four types of smiles were compared, the factor "Type of Smile" has four levels.
An ANOVA conducted on a design in which there is only one factor is called a one-way ANOVA. If an experiment has two factors, then the ANOVA is called a two-way ANOVA. For example, suppose an experiment on the effects of age and gender on reading speed were conducted using three age groups ($8$ years, $10$ years, and $12$ years) and the two genders (male and female). The factors would be age and gender. Age would have three levels and gender would have two levels.
Between- and Within-Subjects Factors
In the "Smiles and Leniency" study, the four levels of the factor "Type of Smile" were represented by four separate groups of subjects. When different subjects are used for the levels of a factor, the factor is called a between-subjects factor or a between-subjects variable. The term "between subjects" reflects the fact that comparisons are between different groups of subjects.
In the "ADHD Treatment" study, every subject was tested with each of four dosage levels ($0, 0.15, 0.30, 0.60$ mg/kg) of a drug. Therefore there was only one group of subjects, and comparisons were not between different groups of subjects but between conditions within the same subjects. When the same subjects are used for the levels of a factor, the factor is called a within-subjects factor or a within-subjects variable. Within-subjects variables are sometimes referred to as repeated-measures variables since there are repeated measurements of the same subjects.
Multi-Factor Designs
It is common for designs to have more than one factor. For example, consider a hypothetical study of the effects of age and gender on reading speed in which males and females from the age levels of $8$ years, $10$ years, and $12$ years are tested. There would be a total of six different groups as shown in Table $1$.
Table $1$: Gender $x$ Age Design
Group Gender Age
1 Female 8
2 Female 10
3 Female 12
4 Male 8
5 Male 10
6 Male 12
This design has two factors: age and gender. Age has three levels and gender has two levels. When all combinations of the levels are included (as they are here), the design is called a factorial design. A concise way of describing this design is as a $Gender (2) \times Age (3)$ factorial design where the numbers in parentheses indicate the number of levels. Complex designs frequently have more than two factors and may have combinations of between- and within-subjects factors. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/15%3A_Analysis_of_Variance/15.02%3A_ANOVA_Designs.txt |
Learning Objectives
• State what the Mean Square Between ($MSB$) estimates when the null hypothesis is true and when the null hypothesis is false
• Compute $MSE$
• Compute $F$ and its two degrees of freedom parameters
• Explain why ANOVA is best thought of as a two-tailed test even though literally only one tail of the distribution is used
• Partition the sums of squares into condition and error
• Format data to be used with a computer statistics program
This section shows how ANOVA can be used to analyze a one-factor between-subjects design. We will use as our main example the "Smiles and Leniency" case study. In this study there were four conditions with $34$ subjects in each condition. There was one score per subject. The null hypothesis tested by ANOVA is that the population means for all conditions are the same. This can be expressed as follows:
$H_0: \mu _1 = \mu _2 = ... = \mu _k$
where $H_0$ is the null hypothesis and $k$ is the number of conditions. In the "Smiles and Leniency" study, $k = 4$ and the null hypothesis is
$H_0: \mu _{false} = \mu _{felt} = \mu _{miserable} = \mu _{neutral}$
If the null hypothesis is rejected, then it can be concluded that at least one of the population means is different from at least one other population mean.
Analysis of variance is a method for testing differences among means by analyzing variance. The test is based on two estimates of the population variance ($\sigma ^2$). One estimate is called the mean square error ($MSE$) and is based on differences among scores within the groups. $MSE$ estimates $\sigma ^2$ regardless of whether the null hypothesis is true (the population means are equal). The second estimate is called the mean square between ($MSB$) and is based on differences among the sample means. $MSB$ only estimates $\sigma ^2$ if the population means are equal. If the population means are not equal, then $MSB$ estimates a quantity larger than $\sigma ^2$. Therefore, if the $MSB$ is much larger than the $MSE$, then the population means are unlikely to be equal. On the other hand, if the $MSB$ is about the same as $MSE$, then the data are consistent with the null hypothesis that the population means are equal.
Before proceeding with the calculation of $MSE$ and $MSB$, it is important to consider the assumptions made by ANOVA:
1. The populations have the same variance. This assumption is called the assumption of $\textit{homogeneity of variance}$.
2. The populations are normally distributed.
3. Each value is sampled independently from each other value. This assumption requires that each subject provide only one value. If a subject provides two scores, then the values are not independent. The analysis of data with two scores per subject is shown in the section on within-subjects ANOVA later in this chapter.
These assumptions are the same as for a t test of differences between groups except that they apply to two or more groups, not just to two groups.
The means and variances of the four groups in the "Smiles and Leniency" case study are shown in Table $1$. Note that there are $34$ subjects in each of the four conditions (False, Felt, Miserable, and Neutral).
Table $1$: Means and Variances from the "Smiles and Leniency" Study
Condition Mean Variance
False 5.3676 3.3380
Felt 4.9118 2.8253
Miserable 4.9118 2.1132
Neutral 4.1176 2.3191
Sample Sizes
The first calculations in this section all assume that there is an equal number of observations in each group. Unequal sample size calculations are shown here. We will refer to the number of observations in each group as $n$ and the total number of observations as $N$. For these data there are four groups of $34$ observations. Therefore, $n = 34$ and $N = 136$.
Computing MSE
Recall that the assumption of homogeneity of variance states that the variance within each of the populations ($\sigma ^2$) is the same. This variance, $\sigma ^2$, is the quantity estimated by $MSE$ and is computed as the mean of the sample variances. For these data, the $MSE$ is equal to $2.6489$.
Computing MSB
The formula for $MSB$ is based on the fact that the variance of the sampling distribution of the mean is
$\sigma _{M}^{2}=\frac{\sigma ^2}{n}$
where $n$ is the sample size of each group. Rearranging this formula, we have
$\sigma ^2=n\sigma _{M}^{2}$
Therefore, if we knew the variance of the sampling distribution of the mean, we could compute $\sigma ^2$ by multiplying it by $n$. Although we do not know the variance of the sampling distribution of the mean, we can estimate it with the variance of the sample means. For the leniency data, the variance of the four sample means is $0.270$. To estimate $\sigma ^2$, we multiply the variance of the sample means ($0.270$) by $n$ (the number of observations in each group, which is $34$). We find that $MSB = 9.179$.
To sum up these steps:
1. Compute the means.
2. Compute the variance of the means.
3. Multiply the variance of the means by $n$.
Recap
If the population means are equal, then both $MSE$ and $MSB$ are estimates of $\sigma ^2$ and should therefore be about the same. Naturally, they will not be exactly the same since they are just estimates and are based on different aspects of the data: The $MSB$ is computed from the sample means and the $MSE$ is computed from the sample variances.
If the population means are not equal, then $MSE$ will still estimate $\sigma ^2$ because differences in population means do not affect variances. However, differences in population means affect $MSB$ since differences among population means are associated with differences among sample means. It follows that the larger the differences among sample means, the larger the $MSB$.
Note
In short, $MSE$ estimates $\sigma ^2$ whether or not the population means are equal, whereas $MSB$ estimates $\sigma ^2$ only when the population means are equal and estimates a larger quantity when they are not equal.
Comparing MSE and MSB
The critical step in an ANOVA is comparing $MSE$ and $MSB$. Since $MSB$ estimates a larger quantity than $MSE$ only when the population means are not equal, a finding of a larger $MSB$ than an $MSE$ is a sign that the population means are not equal. But since $MSB$ could be larger than $MSE$ by chance even if the population means are equal, $MSB$ must be much larger than $MSE$ in order to justify the conclusion that the population means differ. But how much larger must $MSB$ be? For the "Smiles and Leniency" data, the $MSB$ and $MSE$ are $9.179$ and $2.649$, respectively. Is that difference big enough? To answer, we would need to know the probability of getting that big a difference or a bigger difference if the population means were all equal. The mathematics necessary to answer this question were worked out by the statistician R. Fisher. Although Fisher's original formulation took a slightly different form, the standard method for determining the probability is based on the ratio of $MSB$ to $MSE$. This ratio is named after Fisher and is called the $F$ ratio.
For these data, the $F$ ratio is
$F = \frac{9.179}{2.649} = 3.465$
Therefore, the $MSB$ is $3.465$ times higher than $MSE$. Would this have been likely to happen if all the population means were equal? That depends on the sample size. With a small sample size, it would not be too surprising because results from small samples are unstable. However, with a very large sample, the $MSB$ and $MSE$ are almost always about the same, and an $F$ ratio of $3.465$ or larger would be very unusual. Figure $1$ shows the sampling distribution of $F$ for the sample size in the "Smiles and Leniency" study. As you can see, it has a positive skew.
From Figure $1$, you can see that $F$ ratios of $3.465$ or above are unusual occurrences. The area to the right of $3.465$ represents the probability of an $F$ that large or larger and is equal to $0.018$. In other words, given the null hypothesis that all the population means are equal, the probability value is $0.018$ and therefore the null hypothesis can be rejected. The conclusion that at least one of the population means is different from at least one of the others is justified.
The shape of the $F$ distribution depends on the sample size. More precisely, it depends on two degrees of freedom ($df$) parameters: one for the numerator ($MSB$) and one for the denominator ($MSE$). Recall that the degrees of freedom for an estimate of variance is equal to the number of observations minus one. Since the $MSB$ is the variance of $k$ means, it has $k - 1$ $df$. The $MSE$ is an average of $k$ variances, each with $n - 1$ $df$. Therefore, the $df$ for $MSE$ is $k(n - 1) = N - k$, where $N$ is the total number of observations, $n$ is the number of observations in each group, and $k$ is the number of groups. To summarize:
$df_{numerator} = k-1$
$df_{denominator} = N-k$
For the "Smiles and Leniency" data,
$df_{numerator} = k-1=4-1=3$
$df_{denominator} = N-k=136-4=132$
$F = 3.465$
The $F$ distribution calculator shows that $p = 0.018$.
One-Tailed or Two?
Is the probability value from an $F$ ratio a one-tailed or a two-tailed probability? In the literal sense, it is a one-tailed probability since, as you can see in Figure $1$, the probability is the area in the right-hand tail of the distribution. However, the $F$ ratio is sensitive to any pattern of differences among means. It is, therefore, a test of a two-tailed hypothesis and is best considered a two-tailed test.
Relationship to the $t$ test
Since an ANOVA and an independent-groups $t$ test can both test the difference between two means, you might be wondering which one to use. Fortunately, it does not matter since the results will always be the same. When there are only two groups, the following relationship between $F$ and $t$ will always hold:
$F(1,dfd) = t^2(df)$
where $dfd$ is the degrees of freedom for the denominator of the $F$ test and $df$ is the degrees of freedom for the $t$ test. $dfd$ will always equal $df$.
Sources of Variation
Why do scores in an experiment differ from one another? Consider the scores of two subjects in the "Smiles and Leniency" study: one from the "False Smile" condition and one from the "Felt Smile" condition. An obvious possible reason that the scores could differ is that the subjects were treated differently (they were in different conditions and saw different stimuli). A second reason is that the two subjects may have differed with regard to their tendency to judge people leniently. A third is that, perhaps, one of the subjects was in a bad mood after receiving a low grade on a test. You can imagine that there are innumerable other reasons why the scores of the two subjects could differ. All of these reasons except the first (subjects were treated differently) are possibilities that were not under experimental investigation and, therefore, all of the differences (variation) due to these possibilities are unexplained. It is traditional to call unexplained variance error even though there is no implication that an error was made. Therefore, the variation in this experiment can be thought of as being either variation due to the condition the subject was in or due to error (the sum total of all reasons the subjects' scores could differ that were not measured).
One of the important characteristics of ANOVA is that it partitions the variation into its various sources. In ANOVA, the term sum of squares ($SSQ$) is used to indicate variation. The total variation is defined as the sum of squared differences between each score and the mean of all subjects. The mean of all subjects is called the grand mean and is designated as GM. (When there is an equal number of subjects in each condition, the grand mean is the mean of the condition means.) The total sum of squares is defined as
$SSQ_{total}=\sum (X-GM)^2$
which means to take each score, subtract the grand mean from it, square the difference, and then sum up these squared values. For the "Smiles and Leniency" study, $SSQ_{total}=377.19$.
The sum of squares condition is calculated as shown below.
$SSQ_{condition}=n\left [ (M_1-GM)^2 + (M_2-GM)^2 + \cdots +(M_k-GM)^2 \right ]$
where $n$ is the number of scores in each group, $k$ is the number of groups, $M_1$ is the mean for $\text{Condition 1}$, $M_2$ is the mean for $\text{Condition 2}$, and $M_k$ is the mean for $\text{Condition k}$. For the Smiles and Leniency study, the values are:
\begin{align*} SSQ_{condition} &= 34\left [ (5.37-4.83)^2 + (4.91-4.83)^2 + (4.91-4.83)^2 + (4.12-4.83)^2\right ]\ &= 27.5 \end{align*}
If there are unequal sample sizes, the only change is that the following formula is used for the sum of squares condition:
$SSQ_{condition}=n_1(M_1-GM)^2 + n_2(M_2-GM)^2 + \cdots + n_k(M_k-GM)^2$
where $n_i$ is the sample size of the $i^{th}$ condition. $SSQ_{total}$ is computed the same way as shown above.
The sum of squares error is the sum of the squared deviations of each score from its group mean. This can be written as
$SSQ_{error}=\sum (X_{i1}-M_1)^2 + \sum (X_{i2}-M_2)^2 + \cdots + \sum (X_{ik}-M_k)^2$
where $X_{i1}$ is the $i^{th}$ score in $\text{group 1}$ and $M_1$ is the mean for $\text{group 1}$, $X_{i2}$ is the $i^{th}$ score in $\text{group 2}$ and $M_2$ is the mean for $\text{group 2}$, etc. For the "Smiles and Leniency" study, the means are: $5.368$, $4.912$, $4.912$, and $4.118$. The $SSQ_{error}$ is therefore:
\begin{align*} SSQ_{error} &= (2.5-5.368)^2 + (5.5-5.368)^2 + ... + (6.5-4.118)^2\ &= 349.65 \end{align*}
The sum of squares error can also be computed by subtraction:
$SSQ_{error} = SSQ_{total} - SSQ_{condition}$
\begin{align*} SSQ_{error} &= 377.189 - 27.535\ &= 349.65 \end{align*}
Therefore, the total sum of squares of $377.19$ can be partitioned into $SSQ_{condition}(27.53)$ and $SSQ_{error} (349.66)$.
Once the sums of squares have been computed, the mean squares ($MSB$ and $MSE$) can be computed easily. The formulas are:
$MSB = \frac{SSQ_{condition}}{dfn}$
where $dfn$ is the degrees of freedom numerator and is equal to $k - 1 = 3$.
$MSB = \frac{27.535}{3}=9.18$
which is the same value of $MSB$ obtained previously (except for rounding error). Similarly,
$MSE = \frac{SSQ_{error}}{dfd}$
where $dfd$ is the degrees of freedom for the denominator and is equal to $N - k$.
$dfd = 136 - 4 = 132$
$MSE = 349.66/132 = 2.65$
which is the same as obtained previously (except for rounding error). Note that the $dfd$ is often called the $dfe$ for degrees of freedom error.
The Analysis of Variance Summary Table shown below is a convenient way to summarize the partitioning of the variance. The rounding errors have been corrected.
Table $1$: ANOVA Summary Table
Source df SSQ MS F p
Condition 3 27.5349 9.1783 3.465 0.0182
Error 132 349.6544 2.6489
Total 135 377.1893
The first column shows the sources of variation, the second column shows the degrees of freedom, the third shows the sums of squares, the fourth shows the mean squares, the fifth shows the $F$ ratio, and the last shows the probability value. Note that the mean squares are always the sums of squares divided by degrees of freedom. The $F$ and $p$ are relevant only to Condition. Although the mean square total could be computed by dividing the sum of squares by the degrees of freedom, it is generally not of much interest and is omitted here.
Formatting Data for Computer Analysis
Most computer programs that compute ANOVAs require your data to be in a specific form. Consider the data in Table $3$.
Table $3$: Example Data
Group 1 Group 2 Group 3
3 2 8
4 4 5
5 6 5
Here there are three groups, each with three observations. To format these data for a computer program, you normally have to use two variables: the first specifies the group the subject is in and the second is the score itself. The reformatted version of the data in Table $3$ is shown in Table $4$.
Table $4$: Reformatted Data
G Y
1 3
1 4
1 5
2 2
2 4
2 6
3 8
3 5
3 5
To use Analysis Lab to do the calculations, you would copy the data and then
1. Click the "Enter/Edit Data" button. (You may be warned that for security reasons you must use the keyboard shortcut for pasting data.)
2. Paste your data.
3. Click "Accept Data."
4. Set the Dependent Variable to $Y$.
5. Set the Grouping Variable to $G$.
6. Click the ANOVA button.
You will find that $F = 1.5$ and $p = 0.296$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/15%3A_Analysis_of_Variance/15.03%3A_One-Factor_ANOVA.txt |
Learning Objectives
• State how the sums of squares are divided among sources of variation
• State the components of an ANOVA summary table and how they relate to each other
Instructions
This simulation demonstrates the partitioning of sums of squares in analysis of variance (ANOVA). Initially, you see a display of the first sample dataset. There are three groups of subjects and four subjects per group. Each score is represented by a small black rectangle; the mean is represented as a red horizontal line. The values range from $0$ to $10$. The $Y$-axis ticks represent values of $0$, $2.25$, $5$, $7.75$, and $10$. The means of the three groups are $2.50$, $5.50$, and $8.50$.
The table in the lower left-hand portion of the window displays, for each group, the sample size, the mean, the sum of squared deviations of individual scores from the mean, and the squared difference between the group mean and the grand mean multiplied by the sample size. The bottom row shows the sums of these quantities. The sum of the last column is the sums of squares between while the sum of the second-to-last column is the sum of squares within. The ANOVA summary table based on these values is shown to the right. The sum of squares between and within are depicted graphically above the ANOVA summary table.
You can choose other datasets using the pop-up menu. You can also enter your own data by clicking on the data display. If you click in a blank area, an new data point is created (if you hold the mouse down after you click you can position the point by moving the mouse). You can modify the data by clicking on a point and moving it to another location. Finally, you can delete a data point by dragging it outside the colored region.
1. Notice how the sum of squares total is divided up into the sum of squared differences of each score from its group mean and the sum of squared differences of the group mean from the grand mean (GM: the mean of all data). Keep in mind that differences of group means from the grand mean have to be multiplied by the sample size.
2. Add a few data points by clicking and note the effects on the sums of squares. Notice that if the points are far from the group mean then the sum of squares within increases greatly.
3. Choose $\text{dataset 2}$. Notice that the means are very different and the data points are all near their group mean. This results in a large sum of squares between and a msall sum of squares within.
4. Look at $\text{dataset 4}$ which is similar but has more subjects.
5. Look at $\text{dataset 6}$ for which the group means are all the same. Note the value of the sum of squares between.
6. Choose "blank dataset" and enter your own data.
Illustrated Instructions
Video Demo
The video demonstration increases the mean of $\text{group 1}$ by dragging the individual points in the group. Notice how the statistics update as the points are moved. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/15%3A_Analysis_of_Variance/15.04%3A_One-Way_Demo.txt |
Learning Objectives
• Define main effect, simple effect, interaction, and marginal mean
• State the relationship between simple effects and interaction
• Compute the source of variation and df for each effect in a factorial design
• Plot the means for an interaction
• Define three-way interaction
Basic Concepts and Terms
In the "Bias Against Associates of the Obese" case study, the researchers were interested in whether the weight of a companion of a job applicant would affect judgments of a male applicant's qualifications for a job. Two independent variables were investigated:
1. whether the companion was obese or of typical weight and
2. whether the companion was a girlfriend or just an acquaintance.
One approach could have been to conduct two separate studies, one with each independent variable. However, it is more efficient to conduct one study that includes both independent variables. Moreover, there is a much bigger advantage than efficiency for including two variables in the same study: it allows a test of the interaction between the variables. There is an interaction when the effect of one variable differs depending on the level of a second variable. For example, it is possible that the effect of having an obese companion would differ depending on the relationship to the companion. Perhaps there is more prejudice against a person with an obese companion if the companion is a girlfriend than if she is just an acquaintance. If so, there would be an interaction between the obesity factor and the relationship factor.
There are three effects of interest in this experiment:
1. Weight: Are applicants judged differently depending on the weight of their companion?
2. Relationship: Are applicants judged differently depending on their relationship with their companion?
3. Weight x Relationship Interaction: Does the effect of weight differ depending on the relationship with the companion?
The first two effects (Weight and Relationship) are both main effects. A main effect of an independent variable is the effect of the variable averaging over the levels of the other variable(s). It is convenient to talk about main effects in terms of marginal means. A marginal mean for a level of a variable is the mean of the means of all levels of the other variable. For example, the marginal mean for the level "Obese" is the mean of "Girlfriend Obese" and "Acquaintance Obese." Table $1$ shows that this marginal mean is equal to the mean of $5.65$ and $6.15$, which is $5.90$. Similarly, the marginal mean for the level "Typical" is the mean of $6.19$ and $6.59$, which is $6.39$. The main effect of Weight is based on a comparison of these two marginal means. Similarly, the marginal means for "Girlfriend" and "Acquaintance" are $5.92$ and $6.37$.
Table $1$: Means for All Four Conditions
Companion Weight
Obese Typical Marginal Mean
Relationship Girlfriend 5.65 6.19 5.92
Acquaintance 6.15 6.59 6.37
Marginal Mean 5.90 6.39
In contrast to a main effect, which is the effect of a variable averaged across levels of another variable, the simple effect of a variable is the effect of the variable at a single level of another variable. The simple effect of Weight at the level of "Girlfriend" is the difference between the "Girlfriend Typical" and the "Girlfriend Obese" conditions. The difference is $6.19-5.65 = 0.54$. Similarly, the simple effect of Weight at the level of "Acquaintance" is the difference between the "Acquaintance Typical" and the "Acquaintance Obese" conditions. The difference is $6.59-6.15 = 0.44$.
Recall that there is an interaction when the effect of one variable differs depending on the level of another variable. This is equivalent to saying that there is an interaction when the simple effects differ. In this example, the simple effects of weight are $0.54$ and $0.44$. As shown below, these simple effects are not significantly different.
Tests of Significance
The important questions are not whether there are main effects and interactions in the sample data. Instead, what is important is what the sample data allow you to conclude about the population. This is where Analysis of Variance comes in. ANOVA tests main effects and interactions for significance. An ANOVA Summary Table for these data is shown in Table $2$.
Table $2$: ANOVA Summary Table
Source df SSQ MS F p
Weight 1 10.4673 10.4673 6.214 0.0136
Relation 1 8.8144 8.8144 5.233 0.0234
W x R 1 0.1038 0.1038 0.062 0.8043
Error 172 289.7132 1.6844
Total 175 310.1818
Consider first the effect of "Weight." The degrees of freedom ($df$) for "Weight" is $1$. The degrees of freedom for a main effect is always equal to the number of levels of the variable minus one. Since there are two levels of the "Weight" variable (typical and obese), the $df$ is $2 - 1 = 1$. We skip the calculation of the sum of squares ($SSQ$) not because it is difficult, but because it is so much easier to rely on computer programs to compute it. The mean square ($MS$) is the sum of squares divided by the $df$. The $F$ ratio is computed by dividing the $MS$ for the effect by the $MS$ for error ($MSE$). For the effect of "Weight," $F = 10.4673/1.6844 = 6.214$. The last column, $p$, is the probability of getting an $F$ of $6.214$ or larger given that there is no effect of weight in the population. The $p$ value is $0.0136$ and therefore the null hypothesis of no main effect of "Weight" is rejected. The conclusion is that being accompanied by an obese companion lowers judgments of qualifications.
The effect "Relation" is interpreted the same way. The conclusion is that being accompanied by a girlfriend leads to lower ratings than being accompanied by an acquaintance.
The $df$ for an interaction is the product of the $df's$ of variables in the interaction. For the "Weight x Relation" interaction ($W \times R$), the $df = 1$ since both Weight and Relation have one $df: 1 \times 1 = 1$. The $p$ value for the interaction is $0.8043$, which is the probability of getting an interaction as big or bigger than the one obtained in the experiment if there were no interaction in the population. Therefore, these data provide no evidence for an interaction. Always keep in mind that the lack of evidence for an effect does not justify the conclusion that there is no effect. In other words, you do not accept the null hypothesis just because you do not reject it.
For "Error," the degrees of freedom is equal to the total number of observations minus the total number of groups. The sample sizes of the four conditions in this experiment are shown in Table $3$. The total number of observations is $40 + 42 + 40 + 54 = 176$. Since there are four groups, $dfe = 176 - 4 = 172$.
Table $3$: Sample Sizes for All Four Conditions
Companion Weight
Obese Typical
Relationship Girlfriend 40 42
Acquaintance 40 54
The final row in the ANOVA Summary Table is "Total." The degrees of freedom total is equal to the sum of all degrees of freedom. It is also equal to the $\textit{number of observations minus 1}$, or $176 - 1 = 175$. When there are equal sample sizes, the sum of squares total will equal the sum of all other sums of squares. However, when there are unequal sample sizes, as there are here, this will not generally be true. The reasons for this are complex and are discussed in the section Unequal Sample Sizes.
Plotting Means
Although the plot shown in Figure $1$ illustrates the main effects as well as the interaction (or lack of an interaction), it is called an interaction plot. It is important to consider the components of this plot carefully. First, the dependent variable is on the $Y$-axis. Second, one of the independent variables is on the $X$-axis. In this case, it is the variable "Weight." Finally, a separate line is drawn for each level of the other independent variable. It is better to label the lines right on the graph, as shown here, than with a legend.
If you have three or more levels on the $X$-axis, you should not use lines unless there is some numeric ordering to the levels. If your variable on the $X$-axis is a qualitative variable, you can use a plot such as the one in Figure $2$. However, as discussed in the section on bar charts, it would be better to replace each bar with a box plot.
Figure $3$ shows such a plot. Notice how it contains information about the medians, quantiles, and minimums and maximums not contained in Figure $2$. Most important, you get an idea about how much the distributions overlap from Figure $3$ which you do not get from Figure $2$.
Line graphs are a good option when there are more than two levels of a numeric variable. Figure $4$ shows an example. A line graph has the advantage of showing the pattern of interaction clearly. Its disadvantage is that it does not convey the distributional information contained in box plots.
An Example with Interaction
The following example was presented in the section on specific comparisons among means. It is also relevant here.
This example uses the made-up data from a hypothetical experiment shown in Table $4$. Twelve subjects were selected from a population of high-self-esteem subjects and an additional $12$ subjects were selected from a population of low-self-esteem subjects. Subjects then performed on a task and (independent of how well they really did) half in each esteem category were told they succeeded and the other half were told they failed. Therefore, there were six subjects in each of the four esteem/outcome combinations and $24$ subjects in all.
After the task, subjects were asked to rate (on a $10$-point scale) how much of their outcome (success or failure) they attributed to themselves as opposed to being due to the nature of the task.
Table $4$: Data from Hypothetical Experiment on Attribution
Esteem
High Low
Outcome Success 7 6
8 5
7 7
8 4
9 5
5 6
Failure 4 9
6 8
5 9
4 8
7 7
3 6
The ANOVA Summary Table for these data is shown in Table $5$.
Table $5$: ANOVA Summary Table for Made-Up Data
Source df SSQ MS F p
Outcome 1 0.0417 0.0417 0.0256 0.8744
Esteem 1 2.0417 2.0417 1.2564 0.2756
O x E 1 35.0417 35.0417 21.5641 0.0002
Error 20 32.5000 1.6250
Total 23 69.6250
As you can see, the only significant effect is the $Outcome \times Esteem$ ($O \times E$) interaction. The form of the interaction can be seen in Figure $5$.
Clearly the effect of "Outcome" is different for the two levels of "Esteem": For subjects high in self-esteem, failure led to less attribution to oneself than did success. By contrast, for subjects low in self-esteem, failure led to more attribution to oneself than did success. Notice that the two lines in the graph are not parallel. Nonparallel lines indicate interaction. The significance test for the interaction determines whether it is justified to conclude that the lines in the population are not parallel. Lines do not have to cross for there to be an interaction.
Three-Factor Designs
Three-factor designs are analyzed in much the same way as two-factor designs. Table $6$ shows the analysis of a study described by Franklin and Cooley investigating three factors on the strength of industrial fans:
1. Hole Shape (Hex or Round)
2. Assembly Method (Staked or Spun), and
3. Barrel Surface (Knurled or Smooth).
The dependent variable, Breaking Torque, was measured in foot-pounds. There were eight observations in each of the eight combinations of the three factors.
As you can see in Table $6$, there are three main effects, three two-way interactions, and one three-way interaction. The degrees of freedom for the main effects are, as in a two-factor design, equal to the number of levels of the factor minus one. Since all the factors here have two levels, all the main effects have one degree of freedom. The interaction degrees of freedom is always equal to the product of the degrees of freedom of the component parts. This holds for the three-factor interaction as well as for the two-factor interactions. The error degrees of freedom is equal to the number of observations ($64$) minus the number of groups ($8$) and equals $56$.
Table $6$: ANOVA Summary Table for Fan Data
Source df SSQ MS F p
Hole 1 8258.27 8258.27 266.68 <0.0001
Assembly 1 13369.14 13369.14 431.73 <0.0001
H x A 1 2848.89 2848.89 92.00 <0.0001
Barrel 1 35.0417 35.0417 21.5641 <0.0001
H x B 1 594.14 594.14 19.1865 <0.0001
A x B 1 135.14 135.14 4.36 0.0413
H x A x B 1 1396.89 1396.89 45.11 <0.0001
Error 56 1734.12 30.97
Total 63 221386.91
A three-way interaction means that the two-way interactions differ as a function of the level of the third variable. The usual way to portray a three-way interaction is to plot the two-way interactions separately. Figure $6$ shows the $\text{Barrel (Knurled or Smooth)} \times \text{Assembly (Staked or Spun)}$ separately for the two levels of Hole Shape (Hex or Round). For the Hex Shape, there is very little interaction with the lines being close to parallel with a very slight tendency for the effect of Barrel to be bigger for Staked than for Spun. The two-way interaction for the Round Shape is different: The effect of Barrel is bigger for Spun than for Staked. The finding of a significant three-way interaction indicates that this difference in two-way interactions is significant.
Formatting Data for Computer Analysis
The data in Table $4$ have been reformatted in Table $7$. Note how there is one column to indicate the level of outcome and one column to indicate the level of esteem. The coding is as follows:
High self-esteem: $1$
Low self-esteem: $2$
Success: $1$
Failure: $2$
Table $7$: Attribution Data Reformatted
outcome esteem attrib
1 1 7
1 1 8
1 1 7
1 1 8
1 1 9
1 1 5
1 2 6
1 2 5
1 2 7
1 2 4
1 2 5
1 2 6
2 1 4
2 1 6
2 1 5
2 1 4
2 1 7
2 1 3
2 2 9
2 2 8
2 2 9
2 2 8
2 2 7
2 2 6
To use Analysis Lab to do the calculations, you would copy the data and then
1. Click the "Enter/Edit Data" button. (You may be warned that for security reasons you must use the keyboard shortcut for pasting data.)
2. Paste your data.
3. Click "Accept Data."
4. Click the "Advanced" button next to the "ANOVA" button.
5. Select "attrib" as the dependent variable and both "outcome" and "esteem" as "group" variables.
6. Click the "Do ANOVA" button. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/15%3A_Analysis_of_Variance/15.05%3A_Multi-Factor_Between-Subjects.txt |
Learning Objectives
• State why unequal $n$ can be a problem
• Define confounding
• Compute weighted and unweighted means
• Distinguish between Type I and Type III sums of squares
• Describe why the cause of the unequal sample sizes makes a difference in the interpretation
The Problem of Confounding
Whether by design, accident, or necessity, the number of subjects in each of the conditions in an experiment may not be equal. For example, the sample sizes for the "Bias Against Associates of the Obese" case study are shown in Table $1$. Although the sample sizes were approximately equal, the "Acquaintance Typical" condition had the most subjects. Since $n$ is used to refer to the sample size of an individual group, designs with unequal sample sizes are sometimes referred to as designs with unequal $n$.
Table $1$: Sample Sizes for "Bias Against Associates of the Obese" Study
Companion Weight
Obese Typical
Relationship Girlfriend 40 42
Acquaintance 40 54
We consider an absurd design to illustrate the main problem caused by unequal $n$. Suppose an experimenter were interested in the effects of diet and exercise on cholesterol. The sample sizes are shown in Table $2$.
Table $2$: Sample Sizes for "Diet and Exercise" Example
Exercise
Moderate None
Diet Low Fat 5 0
High Fat 0 5
What makes this example absurd is that there are no subjects in either the "Low-Fat No-Exercise" condition or the "High-Fat Moderate-Exercise" condition. The hypothetical data showing change in cholesterol are shown in Table $3$.
Table $3$: Data for "Diet and Exercise" Example
Exercise
Moderate None Mean
Diet Low Fat -20 -25
-25
-30
-35
-15
High Fat -20 -5
6
-10
-6
5
Mean -25 -5 -15
The last column shows the mean change in cholesterol for the two Diet conditions, whereas the last row shows the mean change in cholesterol for the two Exercise conditions. The value of $-15$ in the lower-right-most cell in the table is the mean of all subjects.
We see from the last column that those on the low-fat diet lowered their cholesterol an average of $25$ units, whereas those on the high-fat diet lowered theirs by only an average of $5$ units. However, there is no way of knowing whether the difference is due to diet or to exercise since every subject in the low-fat condition was in the moderate-exercise condition and every subject in the high-fat condition was in the no-exercise condition. Therefore, Diet and Exercise are completely confounded. The problem with unequal $n$ is that it causes confounding.
Weighted and Unweighted Means
The difference between weighted and unweighted means is a difference critical for understanding how to deal with the confounding resulting from unequal $n$.
Weighted and unweighted means will be explained using the data shown in Table $4$. Here, Diet and Exercise are confounded because $80\%$ of the subjects in the low-fat condition exercised as compared to $20\%$ of those in the high-fat condition. However, there is not complete confounding as there was with the data in Table $3$.
The weighted mean for "Low Fat" is computed as the mean of the "Low-Fat Moderate-Exercise" mean and the "Low-Fat No-Exercise" mean, weighted in accordance with sample size. To compute a weighted mean, you multiply each mean by its sample size and divide by $N$, the total number of observations. Since there are four subjects in the "Low-Fat Moderate-Exercise" condition and one subject in the "Low-Fat No-Exercise" condition, the means are weighted by factors of $4$ and $1$ as shown below, where $M_W$ is the weighted mean.
$M_W=\frac{(4)(-27.5)+(1)(-20)}{5}=-26$
The weighted mean for the low-fat condition is also the mean of all five scores in this condition. Thus if you ignore the factor "Exercise," you are implicitly computing weighted means.
The unweighted mean for the low-fat condition ($M_U$) is simply the mean of the two means.
$M_U=\frac{-27.5-20}{2}=-23.75$
Table $4$: Data for Diet and Exercise with Partial Confounding Example
Exercise
Moderate None Weighted Mean Unweighted Mean
Diet Low Fat -20 -20 -26 -23.750
-25
-30
-35
M=-27.5 M=-20.0
High Fat -15 6 -4 -8.125
-6
5
-10
M=-15.0 M=-1.25
Weighted Mean -25 -5
Unweighted Mean -21.25 -10.625
One way to evaluate the main effect of Diet is to compare the weighted mean for the low-fat diet ($-26$) with the weighted mean for the high-fat diet ($-4$). This difference of $-22$ is called "the effect of diet ignoring exercise" and is misleading since most of the low-fat subjects exercised and most of the high-fat subjects did not. However, the difference between the unweighted means of $-15.625$ ($(-23.750)-(-8.125)$) is not affected by this confounding and is therefore a better measure of the main effect. In short, weighted means ignore the effects of other variables (exercise in this example) and result in confounding; unweighted means control for the effect of other variables and therefore eliminate the confounding.
Statistical analysis programs use different terms for means that are computed controlling for other effects. SPSS calls them estimated marginal means, whereas SAS and SAS JMP call them least squares means.
Types of Sums of Squares
The section on Multi-Factor ANOVA stated that when there are unequal sample sizes, the sum of squares total is not equal to the sum of the sums of squares for all the other sources of variation. This is because the confounded sums of squares are not apportioned to any source of variation. For the data in Table $4$, the sum of squares for Diet is $390.625$, the sum of squares for Exercise is $180.625$, and the sum of squares confounded between these two factors is $819.375$ (the calculation of this value is beyond the scope of this introductory text). In the ANOVA Summary Table shown in Table $5$, this large portion of the sums of squares is not apportioned to any source of variation and represents the "missing" sums of squares. That is, if you add up the sums of squares for Diet, Exercise, $D \times E$, and Error, you get $902.625$. If you add the confounded sum of squares of $819.375$ to this value, you get the total sum of squares of $1722.000$. When confounded sums of squares are not apportioned to any source of variation, the sums of squares are called Type III sums of squares. Type III sums of squares are, by far, the most common and if sums of squares are not otherwise labeled, it can safely be assumed that they are Type III.
Table $5$: ANOVA Summary Table for Type III SSQ
Source df SSQ MS F p
Diet 1 390.625 390.625 7.42 0.034
Exercise 1 180.625 180.625 3.43 0.113
D x E 1 15.625 15.625 0.30 0.605
Error 6 315.750 52.625
Total 9 1722.000
When all confounded sums of squares are apportioned to sources of variation, the sums of squares are called Type I sums of squares. The order in which the confounded sums of squares are apportioned is determined by the order in which the effects are listed. The first effect gets any sums of squares confounded between it and any of the other effects. The second gets the sums of squares confounded between it and subsequent effects, but not confounded with the first effect, etc. The Type I sums of squares are shown in Table $6$. As you can see, with Type I sums of squares, the sum of all sums of squares is the total sum of squares.
Table $6$: ANOVA Summary Table for Type I SSQ
Source df SSQ MS F p
Diet 1 1210.000 1210.000 22.99 0.003
Exercise 1 180.625 180.625 3.43 0.113
D x E 1 15.625 15.625 0.30 0.605
Error 6 315.750 52.625
Total 9 1722.000
In Type II sums of squares, sums of squares confounded between main effects are not apportioned to any source of variation, whereas sums of squares confounded between main effects and interactions are apportioned to the main effects. In our example, there is no confounding between the $D \times E$ interaction and either of the main effects. Therefore, the Type II sums of squares are equal to the Type III sums of squares.
Which Type of Sums of Squares to Use (optional)
Type I sums of squares allow the variance confounded between two main effects to be apportioned to one of the main effects. Unless there is a strong argument for how the confounded variance should be apportioned (which is rarely, if ever, the case), Type I sums of squares are not recommended.
There is not a consensus about whether Type II or Type III sums of squares is to be preferred. On the one hand, if there is no interaction, then Type II sums of squares will be more powerful for two reasons:
1. variance confounded between the main effect and interaction is properly assigned to the main effect and
2. weighting the means by sample sizes gives better estimates of the effects.
To take advantage of the greater power of Type II sums of squares, some have suggested that if the interaction is not significant, then Type II sums of squares should be used. Maxwell and Delaney (2003) caution that such an approach could result in a Type II error in the test of the interaction. That is, it could lead to the conclusion that there is no interaction in the population when there really is one. This, in turn, would increase the Type I error rate for the test of the main effect. As a result, their general recommendation is to use Type III sums of squares.
Maxwell and Delaney (2003) recognized that some researchers prefer Type II sums of squares when there are strong theoretical reasons to suspect a lack of interaction and the p value is much higher than the typical $α$ level of $0.05$. However, this argument for the use of Type II sums of squares is not entirely convincing. As Tukey (1991) and others have argued, it is doubtful that any effect, whether a main effect or an interaction, is exactly $0$ in the population. Incidentally, Tukey argued that the role of significance testing is to determine whether a confident conclusion can be made about the direction of an effect, not simply to conclude that an effect is not exactly $0$.
Finally, if one assumes that there is no interaction, then an ANOVA model with no interaction term should be used rather than Type II sums of squares in a model that includes an interaction term. (Models without interaction terms are not covered in this book).
There are situations in which Type II sums of squares are justified even if there is strong interaction. This is the case because the hypotheses tested by Type II and Type III sums of squares are different, and the choice of which to use should be guided by which hypothesis is of interest. Recall that Type II sums of squares weight cells based on their sample sizes whereas Type III sums of squares weight all cells the same. Consider Figure $1$ which shows data from a hypothetical $A(2) \times B(2)$design. The sample sizes are shown numerically and are represented graphically by the areas of the endpoints.
First, let's consider the hypothesis for the main effect of $B$ tested by the Type III sums of squares. Type III sums of squares weight the means equally and, for these data, the marginal means for $b_1$ and $b_2$ are equal:
For $b_1:(b_1a_1 + b_1a_2)/2 = (7 + 9)/2 = 8$
For $b_2:(b_2a_1 + b_2a_2)/2 = (14+2)/2 = 8$
Thus, there is no main effect of $B$ when tested using Type III sums of squares. For Type II sums of squares, the means are weighted by sample size.
For $b_1: (4 \times b_1a_1 + 8 \times b_1a_2)/12 = (4 \times 7 + 8 \times 9)/12 = 8.33$
For $b_2: (12 \times b_2a_1 + 8 \times b_2a_2)/20 = (12 \times 14 + 8 \times 2)/20 = 9.2$
Since the weighted marginal mean for $b_2$ is larger than the weighted marginal mean for $b_1$, there is a main effect of $B$ when tested using Type II sums of squares.
The Type II and Type III analysis are testing different hypotheses. First, let's consider the case in which the differences in sample sizes arise because in the sampling of intact groups, the sample cell sizes reflect the population cell sizes (at least approximately). In this case, it makes sense to weight some means more than others and conclude that there is a main effect of $B$. This is the result obtained with Type II sums of squares. However, if the sample size differences arose from random assignment, and there just happened to be more observations in some cells than others, then one would want to estimate what the main effects would have been with equal sample sizes and, therefore, weight the means equally. With the means weighted equally, there is no main effect of $B$, the result obtained with Type III sums of squares.
Unweighted Means Analysis
Type III sums of squares are tests of differences in unweighted means. However, there is an alternative method to testing the same hypotheses tested using Type III sums of squares. This method, unweighted means analysis, is computationally simpler than the standard method but is an approximate test rather than an exact test. It is, however, a very good approximation in all but extreme cases. Moreover, it is exactly the same as the traditional test for effects with one degree of freedom. The Analysis Lab uses unweighted means analysis and therefore may not match the results of other computer programs exactly when there is unequal n and the df are greater than one.
Causes of Unequal Sample Sizes
None of the methods for dealing with unequal sample sizes are valid if the experimental treatment is the source of the unequal sample sizes. Imagine an experiment seeking to determine whether publicly performing an embarrassing act would affect one's anxiety about public speaking. In this imaginary experiment, the experimental group is asked to reveal to a group of people the most embarrassing thing they have ever done. The control group is asked to describe what they had at their last meal. Twenty subjects are recruited for the experiment and randomly divided into two equal groups of $10$, one for the experimental treatment and one for the control. Following their descriptions, subjects are given an attitude survey concerning public speaking. This seems like a valid experimental design. However, of the $10$ subjects in the experimental group, four withdrew from the experiment because they did not wish to publicly describe an embarrassing situation. None of the subjects in the control group withdrew. Even if the data analysis were to show a significant effect, it would not be valid to conclude that the treatment had an effect because a likely alternative explanation cannot be ruled out; namely, subjects who were willing to describe an embarrassing situation differed from those who were not. Thus, the differential dropout rate destroyed the random assignment of subjects to conditions, a critical feature of the experimental design. No amount of statistical adjustment can compensate for this flaw.
1. Maxwell, S. E., & Delaney, H. D. (2003) Designing Experiments and Analyzing Data: A Model Comparison Perspective, Second Edition, Lawrence Erlbaum Associates, Mahwah, New Jersey.
2. Tukey, J. W. (1991) The philosophy of multiple comparisons, Statistical Science, 6, 110-116. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/15%3A_Analysis_of_Variance/15.06%3A_Unequal_Sample_Sizes.txt |
Learning Objectives
• Compute Tukey HSD test
• Describe an interaction in words
• Describe why one might want to compute simple effect tests following a significant interaction
The null hypothesis tested in a one-factor ANOVA is that all the population means are equal. Stated more formally,
$H_0: \mu _1 = \mu _2 = \cdots = \mu _k$
where $H_0$ is the null hypothesis and $k$ is the number of conditions. When the null hypothesis is rejected, all that can be said is that at least one population mean is different from at least one other population mean. The methods for doing more specific tests described All Pairwise Comparisons among Means and in Specific Comparisons apply here. Keep in mind that these tests are valid whether or not they are preceded by an ANOVA.
Main Effects
As shown below, significant main effects in multi-factor designs can be followed up in the same way as significant effects in one-way designs. Table $1$ shows the data from an imaginary experiment with three levels of $\text{Factor A}$ and two levels of $\text{Factor B}$.
Table $1$: Made-Up Example Data
A1 A2 A3 Marginal Means
B1 5 9 7 7.08
4 8 9
6 7 9
5 8 8
Mean = 5 Mean = 8 Mean = 8.25
B2 4 8 8 6.50
3 6 9
6 8 7
8 5 6
Mean = 5.25 Mean = 6.75 Mean = 7.50
Marginal Means 5.125 7.375 7.875 6.79
Table $2$ shows the ANOVA Summary Table for these data. The significant main effect of $A$ indicates that, in the population, at least one of the marginal means for $A$ is different from at least one of the others.
Table $2$: ANOVA Summary Table for Made-Up Example Data
Source df SSQ MS F p
A 2 34.333 17.167 9.29 0.0017
B 1 2.042 2.042 1.10 0.3070
A x B 2 2.333 1.167 0.63 0.5431
Error 18 33.250 1.847
Total 23 71.958
The Tukey HSD test can be used to test all pairwise comparisons among means in a one-factor ANOVA as well as comparisons among marginal means in a multi-factor ANOVA. The formula for the equal-sample-size case is shown below.
$Q = \cfrac{M_i-M_j}{\sqrt{\cfrac{MSE}{n}}}$
where $M_i$ and $M_j$ are marginal means, $MSE$ is the mean square error from the ANOVA, and n is the number of scores each mean is based upon. For this example, $MSE = 1.847$ and $n = 8$ because there are eight scores at each level of $A$. The probability value can be computed using the Studentized Range Calculator. The degrees of freedom is equal to the degrees of freedom error. For this example, $df = 18$. The results of the Tukey HSD test are shown in Table $3$. The mean for $A_1$ is significantly lower than the mean for $A_2$ and the mean for $A_3$. The means for $A_2$ and $A_3$ are not significantly different.
Table $3$: Pairwise Comparisons Among Marginal Means for A
Comparison Mi - Mj Q p
A1 - A2 -2.25 -4.68 0.010
A1 - A3 -2.75 -5.72 0.002
A2 - A3 -0.50 -1.04 0.746
Specific comparisons among means are also carried out much the same way as shown in the relevant section on testing means. The formula for $L$ is
$L = \sum c_i M_i$
where $c_i$ is the coefficient for the $i^{th}$ marginal mean and $M_i$ is the $i^{th}$ marginal mean. For example, to compare $A_1$ with the average of $A_2$ and $A_3$, the coefficients would be $1$, $-0.5$, $-0.5$. Therefore,
$L = (1)(5.125) + (-0.5)(7.375) + (-0.5)(7.875) = -2.5$
To compute $t$, use:
\begin{align*} t &= \cfrac{L}{\sqrt{\cfrac{\sum c_{i}^{2}MSE}{n}}}\ &= -4.25 \end{align*}
where $MSE$ is the mean square error from the ANOVA and n is the number of scores each marginal mean is based on (eight in this example). The degrees of freedom is the degrees of freedom error from the ANOVA and is equal to $18$. Using the Online Calculator, we find that the two-tailed probability value is $0.0005$. Therefore, the difference between $A_1$ and the average of $A_2$ and $A_3$ is significant.
t distribution Calculator
Important issues concerning multiple comparisons and orthogonal comparisons are discussed in the Specific Comparisons section in the Testing Means chapter.
Interactions
The presence of a significant interaction makes the interpretation of the results more complicated. Since an interaction means that the simple effects are different, the main effect as the mean of the simple effects does not tell the whole story. This section discusses how to describe interactions, proper and improper uses of simple effects tests, and how to test components of interactions.
Describing Interactions
A crucial first step in understanding a significant interaction is constructing an interaction plot. Figure $1$ shows an interaction plot from data presented in the section on Multi-Factor ANOVA.
The second step is to describe the interaction in a clear and understandable way. This is often done by describing how the simple effects differed. Since this should be done using as little jargon as possible, the expression "simple effect" need not appear in the description. An example is as follows:
The effect of Outcome differed depending on the subject's self-esteem. The difference between the attribution to self following success and the attribution to self following failure was larger for high-self-esteem subjects (mean difference = $2.50$) than for low-self-esteem subjects (mean difference = $-2.33$).
No further analyses are helpful in understanding the interaction since the interaction means only that the simple effects differ. The interaction's significance indicates that the simple effects differ from each other, but provides no information about whether they differ from zero.
Simple Effect Tests
It is not necessary to know whether the simple effects differ from zero in order to understand an interaction because the question of whether simple effects differ from zero has nothing to do with interaction except that if they are both zero there is no interaction. It is not uncommon to see research articles in which the authors report that they analyzed simple effects in order to explain the interaction. However, this is not a valid approach since an interaction does not depend on the analysis of the simple effects.
However, there is a reason to test simple effects following a significant interaction. Since an interaction indicates that simple effects differ, it means that the main effects are not general. In the made-up example, the main effect of Outcome is not very informative, and the effect of outcome should be considered separately for high- and low-self-esteem subjects.
As will be seen, the simple effects of Outcome are significant and in opposite directions: Success significantly increases attribution to self for high-self-esteem subjects and significantly lowers attribution to self for low-self-esteem subjects. This is a very easy result to interpret.
What would the interpretation have been if neither simple effect had been significant? On the surface, this seems impossible: How can the simple effects both be zero if they differ from each other significantly as tested by the interaction? The answer is that a non-significant simple effect does not mean that the simple effect is zero: the null hypothesis should not be accepted just because it is not rejected.
(See section on Interpreting Non-Significant Results)
If neither simple effect is significant, the conclusion should be that the simple effects differ, and that at least one of them is not zero. However, no conclusion should be drawn about which simple effect(s) is/are not zero.
Another error that can be made by mistakenly accepting the null hypothesis is to conclude that two simple effects are different because one is significant and the other is not. Consider the results of an imaginary experiment in which the researcher hypothesized that addicted people would show a larger increase in brain activity following some treatment than would non-addicted people. In other words, the researcher hypothesized that addiction status and treatment would interact. The results shown in Figure $2$ are very much in line with the hypothesis. However, the test of the interaction resulted in a probability value of $0.08$, a value not quite low enough to be significant at the conventional $0.05$ level. The proper conclusion is that the experiment supports the researcher's hypothesis, but not strongly enough to allow a confident conclusion.
Unfortunately, the researcher was not satisfied with such a weak conclusion and went on to test the simple effects. It turned out that the effect of Treatment was significant for the Addicted group ($p = 0.02$) but not significant for the Non-Addicted group ($p = 0.09$). The researcher then went on to conclude that since there is an effect of Treatment for the Addicted group but not for the Non-Addicted group, the hypothesis of a greater effect for the former than for the latter group is demonstrated. This is faulty logic, however, since it is based on accepting the null hypothesis that the simple effect of Treatment is zero for the Non-Addicted group just because it is not significant.
Components of Interaction (optional)
Figure $3$ shows the results of an imaginary experiment on diet and weight loss. A control group and two diets were used for both overweight teens and overweight adults.
The difference between $\text{Diet A}$ and the Control diet was essentially the same for teens and adults, whereas the difference between $\text{Diet B}$ and $\text{Diet A}$ was much larger for the teens than it was for the adults. Over one portion of the graph the lines are parallel whereas over another portion they are not. It is possible to test these portions or components of interactions using the method of specific comparisons discussed previously. The test of the difference between Teens and Adults on the difference between Diets $A$ and $B$ could be tested with the coefficients shown in Table $4$. Naturally, the same considerations regarding multiple comparisons and orthogonal comparisons that apply to other comparisons among means also apply to comparisons involving components of interactions.
Table $4$: Coefficients for a Component of the Interaction
Age Group Diet Coefficient
Teen Control 0
Teen A 1
Teen B -1
Adult Control 0
Adult A -1
Adult B 1 | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/15%3A_Analysis_of_Variance/15.07%3A_Tests_Supplementing.txt |
Learning Objectives
• Explain why a within-subjects design can be expected to have more power than a between-subjects design
• Explain error in terms of interaction
• Be able to create the Source and df columns of an ANOVA summary table for a design with one between-subjects and one within-subjects variable
• Describe the consequences of violating the assumption of sphericity
• Discuss courses of action that can be taken if sphericity is violated
Within-subjects factors involve comparisons of the same subjects under different conditions. For example, in the "ADHD Treatment" study, each child's performance was measured four times, once after being on each of four drug doses for a week.Therefore, each subject's performance was measured at each of the four levels of the factor "Dose." Note the difference from between-subjects factors for which each subject's performance is measured only once and the comparisons are among different groups of subjects. A within-subjects factor is sometimes referred to as a repeated-measures factor since repeated measurements are taken on each subject. An experimental design in which the independent variable is a within-subjects factor is called a within-subjects design.
Advantage of Within-Subjects Designs
One-Factor Designs
Let's consider how to analyze the data from the "ADHD Treatment" case study. These data consist of the scores of $24$ children with ADHD on a delay of gratification (DOG) task. Each child was tested under four dosage levels. For now, we will be concerned only with testing the difference between the mean in the placebo condition (the lowest dosage, $D0$) and the mean in the highest dosage condition ($D60$). The details of the computations are relatively unimportant since they are almost universally done by computers. Therefore we jump right to the ANOVA Summary table shown in Table $1$.
Table $1$: ANOVA Summary Table
Source df SSQ MS F p
Subjects 23 5781.98 251.39
Dosage 1 295.02 295.02 10.38 0.004
Error 23 653.48 28.41
Total 47 6730.48
The first source of variation, "Subjects," refers to the differences among subjects. If all the subjects had exactly the same mean (across the two dosages), then the sum of squares for subjects would be zero; the more subjects differ from each other, the larger the sum of squares subjects.
Dosage refers to the differences between the two dosage levels. If the means for the two dosage levels were equal, the sum of squares would be zero. The larger the difference between means, the larger the sum of squares.
The error reflects the degree to which the effect of dosage is different for different subjects. If subjects all responded very similarly to the drug, then the error would be very low. For example, if all subjects performed moderately better with the high dose than they did with the placebo, then the error would be low. On the other hand, if some subjects did better with the placebo while others did better with the high dose, then the error would be high. It should make intuitive sense that the less consistent the effect of dosage, the larger the dosage effect would have to be in order to be significant. The degree to which the effect of dosage differs depending on the subject is the $Subjects \times Dosage$interaction. Recall that an interaction occurs when the effect of one variable differs depending on the level of another variable. In this case, the size of the error term is the extent to which the effect of the variable "Dosage" differs depending on the level of the variable "Subjects." Note that each subject is a different level of the variable "Subjects."
Other portions of the summary table have the same meaning as in between-subjects ANOVA. The $F$ for dosage is the mean square for dosage divided by the mean square error. For these data, the $F$ is significant with $p = 0.004$. Notice that this $F$ test is equivalent to the t test for correlated pairs, with $F = t^2$.
Table $2$ shows the ANOVA Summary Table when all four doses are included in the analysis. Since there are now four dosage levels rather than two, the $df$ for dosage is three rather than one. Since the error is the $Subjects \times Dosage$ interaction, the $df$ for error is the $df$ for "Subjects" ($23$) times the $df$ for Dosage ($3$) and is equal to $69$.
Table $2$: ANOVA Summary Table
Source df SSQ MS F p
Subjects 23 9065.49 394.15
Dosage 3 557.61 185.87 5.18 0.003
Error 69 2476.64 35.89
Total 95 12099.74
Carryover Effects
Often performing in one condition affects performance in a subsequent condition in such a way as to make a within-subjects design impractical. For example, consider an experiment with two conditions. In both conditions subjects are presented with pairs of words. In $\text{Condition A}$, subjects are asked to judge whether the words have similar meaning whereas in $\text{Condition B}$, subjects are asked to judge whether they sound similar. In both conditions, subjects are given a surprise memory test at the end of the presentation. If $\text{Condition}$ were a within-subjects variable, then there would be no surprise after the second presentation and it is likely that the subjects would have been trying to memorize the words.
Not all carryover effects cause such serious problems. For example, if subjects get fatigued by performing a task, then they would be expected to do worse on the second condition they were in. However, as long as the order of presentation is counterbalanced so that half of the subjects are in $\text{Condition A}$ first and $\text{Condition B}$second, the fatigue effect itself would not invalidate the results, although it would add noise and reduce power. The carryover effect is symmetric in that having $\text{Condition A}$ first affects performance in $\text{Condition B}$ to the same degree that having $\text{Condition B}$ first affects performance in $\text{Condition A}$.
Asymmetric carryover effects cause more serious problems. For example, suppose performance in $\text{ConditionB }$ were much better if preceded by $\text{Condition A}$, whereas performance in $\text{Condition A}$ was approximately the same regardless of whether it was preceded by $\text{Condition B}$. With this kind of carryover effect, it is probably better to use a between-subjects design.
One between- and one within-subjects factor
In the "Stroop Interference" case study, subjects performed three tasks: naming colors, reading color words, and naming the ink color of color words. Some of the subjects were males and some were females. Therefore, this design had two factors: gender and task. The ANOVA Summary Table for this design is shown in Table $3$.
Table $3$: ANOVA Summary Table for Stroop Experiment
Source df SSQ MS F p
Gender 1 83.32 83.32 1.99 0.165
Error 45 1880.56 41.79
Task 2 9525.97 4762.99 228.06 <0.001
Gender x Task 2 55.85 27.92 1.34 0.268
Error 90 1879.67 20.89
The computations for the sums of squares will not be covered since computations are normally done by software. However, there are some important things to learn from the summary table. First, notice that there are two error terms: one for the between-subjects variable Gender and one for both the within-subjects variable Task and the interaction of the between-subjects variable and the within-subjects variable. Typically, the mean square error for the between-subjects variable will be higher than the other mean square error. In this example, the mean square error for Gender is about twice as large as the other mean square error.
The degrees of freedom for the between-subjects variable is equal to the number of levels of the between-subjects variable minus one. In this example, it is one since there are two levels of gender. Similarly, the degrees of freedom for the within-subjects variable is equal to the number of levels of the variable minus one. In this example, it is two since there are three tasks. The degrees of freedom for the interaction is the product of the degrees of freedom for the two variables. For the $Gender \times Task$ interaction, the degrees of freedom is the product of degrees of freedom Gender (which is $1$) and the degrees of freedom Task (which is $2$) and is equal to $2$.
Assumption of Sphericity
Within-subjects ANOVA makes a restrictive assumption about the variances and the correlations among the dependent variables. Although the details of the assumption are beyond the scope of this book, it is approximately correct to say that it is assumed that all the correlations are equal and all the variances are equal. Table $4$shows the correlations among the three dependent variables in the "Stroop Interference" case study.
Table $4$: Correlations Among Dependent Variables
word reading color naming interference
word reading 1 0.7013 0.1583
color naming 0.7013 1 0.2382
interference 0.1583 0.2382 1
Note that the correlation between the word reading and the color naming variables of $0.7013$ is much higher than the correlation between either of these variables with the interference variable. Moreover, as shown in Table $5$, the variances among the variables differ greatly.
Table $5$: Variances
Variable Variance
word reading 15.77
color naming 13.92
interference 55.07
Naturally the assumption of sphericity, like all assumptions, refers to populations not samples. However, it is clear from these sample data that the assumption is not met in the population.
Consequences of Violating the Assumption of Sphericity
Although ANOVA is robust to most violations of its assumptions, the assumption of sphericity is an exception: Violating the assumption of sphericity leads to a substantial increase in the Type I error rate. Moreover, this assumption is rarely met in practice. Although violations of this assumption had at one time received little attention, the current consensus of data analysts is that it is no longer considered acceptable to ignore them.
Approaches to Dealing with Violations of Sphericity
If an effect is highly significant, there is a conservative test that can be used to protect against an inflated Type I error rate. This test consists of adjusting the degrees of freedom for all within-subjects variables as follows: The degrees of freedom numerator and denominator are divided by the number of scores per subject minus one. Consider the effect of Task shown in Table $3$. There are three scores per subject and therefore the degrees of freedom should be divided by two. The adjusted degrees of freedom are:
$(2)(1/2) = 1$ for the numerator and
$(90)(1/2) = 45$ for the denominator
The probability value is obtained using the $F$ probability calculator with the new degrees of freedom parameters. The probability of an $F$ of $228.06$ or larger with $1$ and $45$ degrees of freedom is less than $0.001$. Therefore, there is no need to worry about the assumption violation in this case.
Possible violation of sphericity does make a difference in the interpretation of the analysis shown in Table $2$. The probability value of an $F$ of $5.18$ with $1$ and $23$ degrees of freedom is $0.032$, a value that would lead to a more cautious conclusion than the $p$ value of $0.003$ shown in Table $2$.
The correction described above is very conservative and should only be used when, as in Table $3$, the probability value is very low. A better correction, but one that is very complicated to calculate, is to multiply the degrees of freedom by a quantity called $\varepsilon$ (the Greek letter epsilon). There are two methods of calculating $\varepsilon$. The correction called the Huynh-Feldt (or $H-F$) is slightly preferred to the one called the Greenhouse-Geisser (or $G-G$), although both work well. The $G-G$ correction is generally considered a little too conservative.
A final method for dealing with violations of sphericity is to use a multivariate approach to within-subjects variables. This method has much to recommend it, but it is beyond the scope of this text. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/15%3A_Analysis_of_Variance/15.08%3A_Within-Subjects.txt |
Learning Objectives
• State the relationship between variance and power
• State the effect of using a one-tailed test on power
Instructions
This simulation demonstrates the effect of the correlation between measures in a one-way within-subjects ANOVA with two levels. This test is equivalent to a correlated t test. The default values for this demonstration are for an experiment with \(10\) subjects each measured under two conditions. The population difference for the two conditions is \(1.85\) and the variance in each of the conditions is \(4.0\). The graph shows the power of the test as a function of the population correlation between the two scores for the \(0.10\), \(0.05\), and \(0.01\) significance levels. The power of an independent-groups \(t\) test (which assumes the correlation is \(0\)) is shown by the \(x's\).
Experiment with different combinations of the parameters. Is the correlation an important factor in power?
Illustrated Instructions
Video Demo
The video begins by changing the population variance to \(8\) and increases the sample size to \(25\) and then reduces it to \(5\). Notice the impact that these changes have on the relationship between power and population correlation. The video concludes by changing the mean difference to \(4\).
15.10: Statistical Literacy
Learning Objectives
• Testing an Alzheimer's Drug
A research design to compare three drugs for the treatment of Alzheimer's disease is described here. For the first two years of the study, researchers will follow the subjects with scans and memory tests.
Example $1$: what do you think?
Assume the data were analyzed as a two-factor design with pre-post testing as one factor and the three drugs as the second factor. What term in an ANOVA would reflect whether the pre-post change was different for the three drugs?
Solution
It would be the interaction of the two factors since the question is whether the effect of one factor (pre-post) differs as a function of the level of a second factor (drug).
Weight Loss and Cancer Risk
An experiment on risk factors for cancer had four conditions related to weight loss. Quoting the web page description:
"McTiernan and her colleagues randomly assigned 399 overweight and obese postmenopausal women to one of four groups — dieting plus exercise, dieting only, exercise only or a control group not required to do either."
Example $2$: what do you think?
What type of ANOVA design was used?
Solution
There are two between-subjects factors: Diet (yes or no) and Exercise (yes or no) resulting in the four conditions. The design is a $2 \times 2$ factorial design.
15.E: Analysis of Variance (Exercises)
General Questions
Q1
What is the null hypothesis tested by analysis of variance?
Q2
What are the assumptions of between-subjects analysis of variance?
Q3
What is a between-subjects variable?
Q4
Why not just compute $t$-tests among all pairs of means instead computing an analysis of variance?
Q5
What is the difference between "$N$" and "$n$"?
Q6
How is it that estimates of variance can be used to test a hypothesis about means?
Q7
Explain why the variance of the sample means has to be multiplied by "$n$" in the computation of $MSB$.
Q8
What kind of skew does the $F$ distribution have?
Q9
When do $MSB$ and $MSE$ estimate the same quantity?
Q10
If an experiment is conducted with $6$ conditions and $5$ subjects in each condition, what are $dfn$ and $dfe$?
Q11
How is the shape of the $F$ distribution affected by the degrees of freedom?
Q12
What are the two components of the total sum of squares in a one-factor between-subjects design?
Q13
How is the mean square computed from the sum of squares?
Q14
An experimenter is interested in the effects of two independent variables on self esteem. What is better about conducting a factorial experiment than conducting two separate experiements, one for each independent variable?
Q15
An experiment is conducted on the effect of age and treatment condition (experimental versus control) on reading speed. Which statistical term (main effect, simple effect, interaction, specific comparison) applies to each of the descriptions of effects.
1. The effect of the treatment was larger for $15$-year olds than it was for $5$- or $10$-year olds.
2. Overall, subjects in the treatment condition performed faster than subjects in the control condition.
3. The difference between the $10$- and $15$-year olds was significant under the treatment condition.
4. The difference between the $15$- year olds and the average of the $5$- and $10$-year olds was significant.
5. As they grow older, children read faster.
Q16
An $A(3) \times B(4)$ factorial design with $6$ subjects in each group is analyzed. Give the source and degrees of freedom columns of the analysis of variance summary table.
Q17
The following data are from a hypothetical study on the effects of age and time on scores on a test of reading comprehension. Compute the analysis of variance summary table.
12-year olds 16-year olds
30 minutes 66
68
59
72
46
74
71
67
82
76
60 minutes 69
61
69
73
61
95
92
95
98
94
Q18
Define "Three-way interaction"
Q19
Define interaction in terms of simple effects.
Q20
Plot an interaction for an $A(2) \times B(2)$ design in which the effect of $B$ is greater at $A1$ than it is at $A2$. The dependent variable is "Number correct." Make sure to label both axes.
Q21
Following are two graphs of population means for $2 \times 3$ designs. For each graph, indicate which effect(s) ($A$, $B$, or $A \times B$) are nonzero.
Q22
The following data are from an $A(2) \times B(4)$ factorial design.
B1 B2 B3 B4
A1 1
3
4
5
2
2
4
5
3
4
2
6
4
5
6
8
A2 1
1
2
2
2
3
2
4
4
6
7
8
8
9
9
8
1. Compute an analysis of variance.
2. Test differences among the four levels of $B$ using the Bonferroni correction.
3. Test the linear component of trend for the effect of $B$.
4. Plot the interaction.
5. Describe the interaction in words.
Q23
Why are within-subjects designs usually more powerful than between-subjects design?
Q24
What source of variation is found in an ANOVA summary table for a within-subjects design that is not in in an ANOVA summary table for a between-subjects design. What happens to this source of variation in a between-subjects design?
Q25
The following data contain three scores from each of five subjects. The three scores per subject are their scores on three trials of a memory task.
$\begin{matrix} 4 & 6 & 7\ 3 & 7 & 7\ 2 & 8 & 5\ 1 & 4 & 7\ 4 & 6 & 9 \end{matrix}$
1. Compute an ANOVA
2. Test all pairwise differences between means using the Bonferroni test at the $0.01$ level.
3. Test the linear and quadratic components of trend for these data.
Q26
Give the source and df columns of the ANOVA summary table for the following experiments:
1. $22$ subjects are each tested on a simple reaction time task and on a choice reaction time task.
2. $12$ male and $12$ female subjects are each tested under three levels of drug dosage: $0 mg, 10 mg, 20 mg$.
3. $20$ subjects are tested on a motor learning task for $3$ trials a day for $2$ days.
4. An experiment is conducted in which depressed people are either assigned to a drug therapy group, a behavioral therapy group, or a control group. $10$ subjects are assigned to each group. The level of measured once a month for $4$ months.
Questions from Case Studies
The following question is from the Stroop Interference case study.
Q27
The dataset has the scores (times) for males and females on each of three tasks.
1. Do a $Gender (2) \times Task (3)$ analysis of variance.
2. Plot the interaction.
The following question is from the ADHD Treatment case study.
Q28
The data has four scores per subject.
1. Is the design between-subjects or within-subjects?
2. Create an ANOVA summary table.
The following question is from the Angry Moods case study.
Q29
Using the Anger Expression Index as the dependent variable, perform a $2 \times 2$ ANOVA with gender and sports participation as the two factors. Do athletes and non-athletes differ significantly in how much anger they express? Do the genders differ significantly in Anger Expression Index? Is the effect of sports participation significantly different for the two genders?
The following question is from the Weapons and Aggression case study.
Q30
Compute a $2 \times 2$ ANOVA on this data with the following two factors: prime type (was the first word a weapon or not?) and word type (was the second word aggressive or non-aggressive?). Consider carefully whether the variables are between-subject or within-subects variables.
The following question is from the Smiles and Leniency case study.
Q31
Compute the ANOVA summary table. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/15%3A_Analysis_of_Variance/15.09%3A_Power_of_Within-Subjects_Designs_Demo.txt |
In this chapter, we focus on the fact that many statistical procedures work best if individual variables have certain properties. The measurement scale of a variable should be part of the data preparation effort. For example, the correlation coefficient does not require that the variables have a normal shape, but often relationships can be made clearer by re-expressing the variables.
• 16.1: Prelude to Transformations
The introductory chapter covered linear transformations. These transformations normally do not change statistics such as Pearson's r, although they do affect the mean and standard deviation. The first section here is on log transformations which are useful to reduce skew. The second section is on Tukey's ladder of powers. You will see that log transformations are a special case of the ladder of powers. Finally, we cover the relatively advanced topic of the Box-Cox transformation.
• 16.2: Log Transformations
The log transformation can be used to make highly skewed distributions less skewed. This can be valuable both for making patterns in the data more interpretable and for helping to meet the assumptions of inferential statistics.
• 16.3: Tukey Ladder of Powers
Tukey (1977) describes an orderly way of re-expressing variables using a power transformation.
• 16.4: Box-Cox Transformations
The Box-Cox transformation is a particulary useful family of transformations to convert a non-normal behaving data set into an approximately a normal distribution.
• 16.5: Statistical Literacy
• 16.E: Transformations (Exercises)
16: Transformations
The focus of statistics courses is the exposition of appropriate methodology to analyze data to answer the question at hand. Sometimes the data are given to you, while other times the data are collected as part of a carefully-designed experiment. Often the time devoted to statistical analysis is less than \(10\%\) of the time devoted to data collection and preparation. If aspects of the data preparation fail, then the success of the analysis is in jeopardy. Sometimes errors are introduced into the recording of data. Sometimes biases are inadvertently introduced in the selection of subjects or the miscalibration of monitoring equipment.
In this chapter, we focus on the fact that many statistical procedures work best if individual variables have certain properties. The measurement scale of a variable should be part of the data preparation effort. For example, the correlation coefficient does not require that the variables have a normal shape, but often relationships can be made clearer by re-expressing the variables. An economist may choose to analyze the logarithm of prices if the relative price is of interest. A chemist may choose to perform a statistical analysis using the inverse temperature as a variable rather than the temperature itself. But note that the inverse of a temperature will differ depending on whether it is measured in \(^{\circ}F,\; ^{\circ}C,\; or\; ^{\circ}K\).
The introductory chapter covered linear transformations. These transformations normally do not change statistics such as Pearson's \(r\), although they do affect the mean and standard deviation. The first section here is on log transformations which are useful to reduce skew. The second section is on Tukey's ladder of powers. You will see that log transformations are a special case of the ladder of powers. Finally, we cover the relatively advanced topic of the Box-Cox transformation.
16.02: Log Transformations
Learning Objectives
• State how a log transformation can help make a relationship clear
• Describe the relationship between logs and the geometric mean
The log transformation can be used to make highly skewed distributions less skewed. This can be valuable both for making patterns in the data more interpretable and for helping to meet the assumptions of inferential statistics. Figure $1$ shows an example of how a log transformation can make patterns more visible. Both graphs plot the brain weight of animals as a function of their body weight. The raw weights are shown in the upper panel; the log-transformed weights are plotted in the lower panel.
It is hard to discern a pattern in the upper panel whereas the strong relationship is shown clearly in the right panel. The comparison of the means of log-transformed data is actually a comparison of geometric means. This occurs because, as shown below, the anti-log of the arithmetic mean of log-transformed values is the geometric mean.
Table $1$ shows the logs (base $10$) of the numbers $1$, $10$, and $100$. The arithmetic mean of the three logs is
$(0 + 1 + 2)/3 = 1$
The anti-log of this arithmetic mean of $1$ is
$10^1 = 10$
which is the geometric mean:
$(1 \times 10 \times 100)^{0.3333} = 10$
Table $1$: Logarithms
X $\log_{10}(X)$
1 0
10 2
100 3
1,000 4
10,000 5
Therefore, if the arithmetic means of two sets of log-transformed data are equal, then the geometric means are equal. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/16%3A_Transformations/16.01%3A_Prelude_to_Transformations.txt |
Learning Objectives
• Give the Tukey ladder of transformations
• Find a transformation that reveals a linear relationship
• Find a transformation to approximate a normal distribution
Introduction
We assume we have a collection of bivariate data
$(x_1,y_1),(x_2,y_2),...,(x_n,y_n)$
and that we are interested in the relationship between variables $x$ and $y$. Plotting the data on a scatter diagram is the first step. As an example, consider the population of the United States for the $200$ years before the Civil War. Of course, the decennial census began in $1790$. These data are plotted two ways in Figure $1$. Malthus predicted that geometric growth of populations coupled with arithmetic growth of grain production would have catastrophic results. Indeed the US population followed an exponential curve during this period.
Tukey's Transformation Ladder
Tukey ($1977$) describes an orderly way of re-expressing variables using a power transformation. You may be familiar with polynomial regression (a form of multiple regression) in which the simple linear model $y = b_0 + b_1X$ is extended with terms such as $b_2X^2 + b_3X^3 + b_4X^4$. Alternatively, Tukey suggests exploring simple relationships such as
$y = b_0 + b_1X^λ$
or
$y^λ = b_0 + b_1X \label{eq1}$
where $λ$ is a parameter chosen to make the relationship as close to a straight line as possible. Linear relationships are special, and if a transformation of the type $x^\lambda$ or yλworks as in Equation \ref{eq1}, then we should consider changing our measurement scale for the rest of the statistical analysis.
There is no constraint on values of $λ$ that we may consider. Obviously choosing $λ = 1$ leaves the data unchanged. Negative values of $λ$ are also reasonable. For example, the relationship
$y = b_0 + \dfrac{b_1}{x}$
would be represented by $λ = −1$. The value $λ = 0$ has no special value, since $X^0 = 1$, which is just a constant. Tukey ($1977$) suggests that it is convenient to simply define the transformation when $λ = 0$ to be the logarithm function rather than the constant $1$. We shall revisit this convention shortly. The following table gives examples of the Tukey ladder of transformations.
Table $1$: Tukey's Lader of transformation
$\lambda$ $-2$ $-1$ $-1/2$ $0$ $1/2$ $1$ $2$
$y$ $\tfrac{1}{x^2}$ $\tfrac{1}{x}$ $\tfrac{1}{\sqrt{x}}$ $\log x$ $\sqrt{x}$ $x$ $x^2$
If $x$ takes on negative values, then special care must be taken so that the transformations make sense, if possible. We generally limit ourselves to variables where $x > 0$ to avoid these considerations. For some dependent variables such as the number of errors, it is convenient to add $1$ to $x$ before applying the transformation.
Also, if the transformation parameter $λ$ is negative, then the transformed variable $x^\lambda$ is reversed. For example, if $x$ is increasing, then $1/x$ is decreasing. We choose to redefine the Tukey transformation to be $-x^\lambda$ if $λ < 0$ in order to preserve the order of the variable after transformation. Formally, the Tukey transformation is defined as
$y=\left\{\begin{matrix} x^\lambda & if & \lambda >0\ \log x & if & \lambda =0\ -(x^\lambda) & if & \lambda <0 \end{matrix}\right.$
In Table $2$ we reproduce Table $1$ using the modified definition when $λ < 0$.
Table $2$: Modified Tukey's Ladder of Transformations
$\lambda$ $-2$ $-1$ $-1/2$ $0$ $1/2$ $1$ $2$
$y$ $\tfrac{-1}{x^2}$ $\tfrac{-1}{x}$ $\tfrac{-1}{\sqrt{x}}$ $\log x$ $\sqrt{x}$ $x$ $x^2$
The Best Transformation for Linearity
The goal is to find a value of $λ$ that makes the scatter diagram as linear as possible. For the US population, the logarithmic transformation applied to $y$ makes the relationship almost perfectly linear. The red dashed line in the right frame of Figure $1$ has a slope of about $1.35$; that is, the US population grew at a rate of about $35\%$ per decade.
The logarithmic transformation corresponds to the choice $λ = 0$ by Tukey's convention. In Figure $2$, we display the scatter diagram of the US population data for $λ = 0$ as well as for other choices of $λ$.
The raw data are plotted in the bottom right frame of Figure $2$ when $λ = 1$. The logarithmic fit is in the upper right frame when $λ = 0$. Notice how the scatter diagram smoothly morphs from convex to concave as $λ$ increases. Thus intuitively there is a unique best choice of $λ$ corresponding to the "most linear" graph.
One way to make this choice objective is to use an objective function for this purpose. One approach might be to fit a straight line to the transformed points and try to minimize the residuals. However, an easier approach is based on the fact that the correlation coefficient, $r$, is a measure of the linearity of a scatter diagram. In particular, if the points fall on a straight line then their correlation will be $r = 1$. (We need not worry about the case when $r = −1$ since we have defined the Tukey transformed variable $x_\lambda$ to be positively correlated with $x$ itself.)
In Figure $3$, we plot the correlation coefficient of the scatter diagram $(x,y_\lambda )$ as a function of $λ$. It is clear that the logarithmic transformation ($λ = 0$) is nearly optimal by this criterion.
Is the US population still on the same exponential growth pattern? In Figure $4$, we display the US population from $1630$ to $2000$ using the transformation and fit used in the right frame of Figure $1$. Fortunately, the exponential growth (or at least its rate) was not sustained into the Twentieth Century. If it had, the US population in the year $2000$ would have been over $2$ billion ($2.07$ to be exact), larger than the population of China.
We can examine the decennial census population figures of individual states as well. In Figure $5$, we display the population data for the state of New York from $1790$ to $2000$, together with an estimate of the population in $2008$. Clearly something unusual happened starting in $1970$. (This began the period of mass migration to the West and South as the rust belt industries began to shut down.) Thus, we compute the best $λ$ value using the data from $1790$-$1960$ in the middle frame of Figure $5$. The right frame displays the transformed data, together with the linear fit for the $1790-1960$ period. The value of $λ = 0.41$ is not obvious and one might reasonably choose to use $λ = 0.50$ for practical reasons.
If we look at one of the younger states in the West, the picture is different. Arizona has attracted many retirees and immigrants. Figure $6$ summarizes our findings. Indeed, the growth of population in Arizona is logarithmic, and appears to still be logarithmic through $2005$.
Reducing Skew
Many statistical methods such as $t$ tests and the analysis of variance assume normal distributions. Although these methods are relatively robust to violations of normality, transforming the distributions to reduce skew can markedly increase their power.
As an example, the data in the "Stereograms" case study are very skewed. A t test of the difference between the two conditions using the raw data results in a p value of $0.056$, a value not conventionally considered significant. However, after a log transformation ($λ = 0$) that reduces the skew greatly, the $p$ value is $0.023$ which is conventionally considered significant.
The demonstration in Figure $7$ shows distributions of the data from the Stereograms case study as transformed with various values of $λ$. Decreasing $λ$ makes the distribution less positively skewed. Keep in mind that $λ = 1$ is the raw data. Notice that there is a slight positive skew for $λ = 0$ but much less skew than found in the raw data ($λ = 1$). Values of below 0 result in negative skew. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/16%3A_Transformations/16.03%3A_Tukey_Ladder_of_Powers.txt |
Learning Objectives
• To study the Box-Cox transformation
George Box and Sir David Cox collaborated on one paper (Box, $1964$). The story is that while Cox was visiting Box at Wisconsin, they decided they should write a paper together because of the similarity of their names (and that both are British). In fact, Professor Box is married to the daughter of Sir Ronald Fisher.
The Box-Cox transformation of the variable $x$ is also indexed by $λ$, and is defined as
$x' = \dfrac{x^\lambda-1}{\lambda} \label{eq1}$
At first glance, although the formula in Equation \ref{eq1} is a scaled version of the Tukey transformation $x^\lambda$, this transformation does not appear to be the same as the Tukey formula in Equation (2). However, a closer look shows that when $λ < 0$, both $x_\lambda$ and $X_{\lambda }^{'}$ change the sign of $x^\lambda$ to preserve the ordering. Of more interest is the fact that when $λ = 0$, then the Box-Cox variable is the indeterminate form $0/0$. Rewriting the Box-Cox formula as
$X_{\lambda }^{'}=\frac{e^{\lambda \log (x)}-1}{\lambda }\approx \frac{\left ( 1+\lambda \log (x) + \tfrac{1}{2}\lambda ^2\log (x)^2 + \cdots \right )-1}{\lambda }\rightarrow \log (x)$
as $\lambda \rightarrow 0$. This same result may also be obtained using l'Hôpital's rule from your calculus course. This gives a rigorous explanation for Tukey's suggestion that the log transformation (which is not an example of a polynomial transformation) may be inserted at the value $λ = 0$.
Notice with this definition of $X_{\lambda }^{'}$ that $x = 1$ always maps to the point $X_{\lambda }^{'} = 0$ for all values of $λ$. To see how the transformation works, look at the examples in Figure $1$. In the top row, the choice $λ = 1$ simply shifts $x$ to the value $x−1$, which is a straight line. In the bottom row (on a semi-logarithmic scale), the choice $λ = 0$ corresponds to a logarithmic transformation, which is now a straight line. We superimpose a larger collection of transformations on a semi-logarithmic scale in Figure $2$.
Transformation to Normality
Another important use of variable transformation is to eliminate skewness and other distributional features that complicate analysis. Often the goal is to find a simple transformation that leads to normality. In the article on $q-q$ plots, we discuss how to assess the normality of a set of data,
$x_1,x_2, \ldots ,x_n.$
Data that are normal lead to a straight line on the q-q plot. Since the correlation coefficient is maximized when a scatter diagram is linear, we can use the same approach above to find the most normal transformation.
Specifically, we form the $n$ pairs
$\left ( \Phi ^{-1} \left ( \frac{i-0.5}{n} \right ), x_{(i)} \right ),\; for\; i=1,2,\cdots ,n$
where $\Phi ^{-1}$ is the inverse CDF of the normal density and $x_{(i)}$ denotes the $i^{th}$ sorted value of the data set. As an example, consider a large sample of British household incomes taken in $1973$, normalized to have mean equal to one ($n = 7125$). Such data are often strongly skewed, as is clear from Figure $3$. The data were sorted and paired with the $7125$ normal quantiles. The value of $λ$ that gave the greatest correlation ($r = 0.9944$) was $λ = 0.21$.
The kernel density plot of the optimally transformed data is shown in the left frame of Figure $4$. While this figure is much less skewed than in Figure $3$, there is clearly an extra "component" in the distribution that might reflect the poor. Economists often analyze the logarithm of income corresponding to $λ = 0$; see Figure $4$. The correlation is only $r = 0.9901$ in this case, but for convenience, the log-transform probably will be preferred.
Other Applications
Regression analysis is another application where variable transformation is frequently applied. For the model
$y =\beta_o + \beta_1 x_1 + \beta_2 x_2 + \ldots \beta_p x_p + \epsilon$
and fitted model
$\widehat{y}=b_0 + b_1x_1 + b_2x_2 + \cdots + b_px_p$
each of the predictor variables $x_j$ can be transformed. The usual criterion is the variance of the residuals, given by
$\frac{1}{n} \sum_{i=1}^{n} (\widehat{y}_i-y_i)^2$
Occasionally, the response variable y may be transformed. In this case, care must be taken because the variance of the residuals is not comparable as $λ$ varies. Let $\bar{g}_y$ represent the geometric mean of the response variables.
$\bar{g}_y = \left ( \prod_{i-1}^{n} y_i \right )^{1/n}$
Then the transformed response is defined as
$y_{\lambda }^{'} = \frac{y^\lambda -1}{\lambda \cdot \bar{g}_{y}^{\lambda -1}}$
When $λ = 0$ (the logarithmic case),
$y_{0}^{'} = \bar{g}_y \cdot \log (y)$
For more examples and discussions, see Kutner, Nachtsheim, Neter, and Li (2004).
16.05: Statistical Literacy
Learning Objectives
• Stock Appreciation
Many financial web pages give you the option of using a linear or a logarithmic \(Y\)-axis. An example from Google Finance is shown below.
Example \(1\): what do you think?
To get a straight line with the linear option chosen, the price would have to go up the same amount every time period. What would result in a straight line with the logarithmic option chosen?
Solution
The price would have to go up the same proportion every time period. For example, go up \(0.1\%\) every day.
16.E: Transformations (Exercises)
General Questions
Q1
When is a log transformation valuable?
Q2
If the arithmetic mean of \(\log_{10}\) transformed data were \(3\), what would be the geometric mean?
Q3
Using Tukey's ladder of transformation, transform the following data using a \(λ\) of \(0.5: 9, 16, 25\)
Q4
What value of \(λ\) in Tukey's ladder decreases skew the most?
Q5
What value of \(λ\) in Tukey's ladder increases skew the most?
Question from Case Study
Q6
In the ADHD case study, transform the data in the placebo condition (\(D0\)) with \(λ's\) of \(0.5\), \(0\), \(-0.5\), and \(-1\). How does the skew in each of these compare to the skew in the raw data. Which transformation leads to the least skew? | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/16%3A_Transformations/16.04%3A_Box-Cox_Transformations.txt |
Chi Square is a distribution that has proven to be particularly useful in statistics. The first section describes the basics of this distribution. The following two sections cover the most common statistical tests that make use of the Chi Square distribution. The section "One-Way Tables" shows how to use the Chi Square distribution to test the difference between theoretically expected and observed frequencies. The section "Contingency Tables" shows how to use Chi Square to test the association between two nominal variables. This use of Chi Square is so common that it is often referred to as the "Chi Square Test."
17: Chi Square
Learning Objectives
• Define the Chi Square distribution in terms of squared normal deviates
• Describe how the shape of the Chi Square distribution changes as its degrees of freedom increase
A standard normal deviate is a random sample from the standard normal distribution. The Chi Square distribution is the distribution of the sum of squared standard normal deviates. The degrees of freedom of the distribution is equal to the number of standard normal deviates being summed. Therefore, Chi Square with one degree of freedom, written as $\chi ^2(1)$, is simply the distribution of a single normal deviate squared. The area of a Chi Square distribution below $4$ is the same as the area of a standard normal distribution below $2$, since $4$ is $2^2$.
Consider the following example.
Example $1$
You sample two scores from a standard normal distribution, square each score, and sum the squares. What is the probability that the sum of these two squares will be six or higher?
Solution
Since two scores are sampled, the answer can be found using the Chi Square distribution with two degrees of freedom. A Chi Square calculator can be used to find that the probability of a Chi Square (with $2 df$) being six or higher is $0.050$.
The mean of a Chi Square distribution is its degrees of freedom. Chi Square distributions are positively skewed, with the degree of skew decreasing with increasing degrees of freedom. As the degrees of freedom increases, the Chi Square distribution approaches a normal distribution. Figure $1$ shows density functions for three Chi Square distributions. Notice how the skew decreases as the degrees of freedom increases.
The Chi Square distribution is very important because many test statistics are approximately distributed as Chi Square. Two of the more common tests using the Chi Square distribution are tests of deviations of differences between theoretically expected and observed frequencies (one-way tables) and the relationship between categorical variables (contingency tables). Numerous other tests beyond the scope of this work are based on the Chi Square distribution.
17.02: One-Way Tables
Learning Objectives
Describe what it means for there to be theoretically-expected frequencies
• Compute expected frequencies
• Compute Chi Square
• Determine the degrees of freedom
The Chi Square distribution can be used to test whether observed data differ significantly from theoretical expectations. For example, for a fair six-sided die, the probability of any given outcome on a single roll would be $1/6$. The data in Table $1$ were obtained by rolling a six-sided die $36$ times. However, as can be seen in Table $1$, some outcomes occurred more frequently than others. For example, a "$3$" came up nine times, whereas a "$4$" came up only two times. Are these data consistent with the hypothesis that the die is a fair die? Naturally, we do not expect the sample frequencies of the six possible outcomes to be the same since chance differences will occur. So, the finding that the frequencies differ does not mean that the die is not fair. One way to test whether the die is fair is to conduct a significance test. The null hypothesis is that the die is fair. This hypothesis is tested by computing the probability of obtaining frequencies as discrepant or more discrepant from a uniform distribution of frequencies as obtained in the sample. If this probability is sufficiently low, then the null hypothesis that the die is fair can be rejected.
Table $1$: Outcome Frequencies from a Six-Sided Die
Outcome Frequency
1 8
2 5
3 9
4 2
5 7
6 5
The first step in conducting the significance test is to compute the expected frequency for each outcome given that the null hypothesis is true. For example, the expected frequency of a "$1$" is $6$ since the probability of a "$1$" coming up is $1/6$ and there were a total of $36$ rolls of the die.
$\text{Expected frequency} = (1/6)(36) = 6$
Note that the expected frequencies are expected only in a theoretical sense. We do not really "expect" the observed frequencies to match the "expected frequencies" exactly.
The calculation continues as follows. Letting $E$ be the expected frequency of an outcome and $O$ be the observed frequency of that outcome, compute
$\frac{(E-O)^2}{E}$
for each outcome. Table $2$ shows these calculations.
Table $2$: Outcome Frequencies from a Six-Sided Die
Outcome E O
1 6 8 0.667
2 6 5 0.167
3 6 9 1.500
4 6 2 2.667
5 6 7 0.167
6 6 5 0.167
Next we add up all the values in Column 4 of Table $2$.
$\sum \frac{(E-O)^2}{E} = 5.333$
This sampling distribution of $\sum \frac{(E-O)^2}{E}$ is approximately distributed as Chi Square with $k-1$ degrees of freedom, where $k$ is the number of categories. Therefore, for this problem the test statistic is
$\chi _{5}^{2}=5.333$
which means the value of Chi Square with $5$ degrees of freedom is $5.333$.
From a Chi Square calculator it can be determined that the probability of a Chi Square of $5.333$ or larger is $0.377$. Therefore, the null hypothesis that the die is fair cannot be rejected.
This Chi Square test can also be used to test other deviations between expected and observed frequencies. The following example shows a test of whether the variable "University GPA" in the SAT and College GPA case study is normally distributed.
The first column in Table $3$ shows the normal distribution divided into five ranges. The second column shows the proportions of a normal distribution falling in the ranges specified in the first column. The expected frequencies ($E$) are calculated by multiplying the number of scores ($105$) by the proportion. The final column shows the observed number of scores in each range. It is clear that the observed frequencies vary greatly from the expected frequencies. Note that if the distribution were normal, then there would have been only about $35$ scores between $0$ and $1$, whereas $60$ were observed.
Table $3$: Expected and Observed Scores for $105$ University GPA Scores
Range Proportion E O
Above 1 0.159 16.695 9
0 to 1 0.341 35.805 60
-1 to 0 0.341 35.805 17
Below -1 0.159 16.695 19
The test of whether the observed scores deviate significantly from the expected scores is computed using the familiar calculation.
$\chi _{3}^{2} = \sum \frac{(E-O)^2}{E} = 30.09$
The subscript "$3$" means there are three degrees of freedom. As before, the degrees of freedom is the number of outcomes minus $1$, which is $4 - 1 = 3$ in this example. The Chi Square distribution calculator shows that $p < 0.001$ for this Chi Square. Therefore, the null hypothesis that the scores are normally distributed can be rejected. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/17%3A_Chi_Square/17.01%3A_Chi_Square_Distribution.txt |
Learning Objectives
• Develop a basic understanding of the properties of a sampling distribution based on the properties of the population
Instructions
In this simulation, $100$ numbers are either sampled from a normal distribution or a uniform distribution. The frequencies in each of $10$ "bins" is then displayed in the "observed" column. The expected frequencies based on both a normal distribution (on the left) or a uniform distribution (on the right) are shown just to the left of the observed frequencies. For each bin the value $\frac{(E-O)^2}{E}$ is computed where $E$ is the expected frequency and $O$ is the observed frequency. The sum of these quantities is the value of Chi Square shown at the bottom.
1. The default is to sample from a normal distribution. Click the sample button and $100$ values will be sampled from a normal distribution. Compare the observed values in the "From a Normal Distribution" section to the expected values. Is the Chi Square test significant at the $0.05$ level? How often would you expect it to be significant.
2. Compare the observed frequencies from the "From a Uniform Distribution" section to the expected frequencies. In what way are they different? Is the difference significant? If so, then the null hypothesis that the numbers were sampled from a uniform distribution could be rejected. Of course, in this simulation, you know where the numbers were sampled so you know the null hypothesis is false.
3. Simulate several experiments and see if the significance for the test of a uniform distribution is always significant.
4. Make the actual distribution a uniform distribution and do more simulated experiments. Compare the results to when the actual distribution was normal.
Illustrated Instructions
This simulation samples $100$ values from a normal or uniform distribution and calculates the the Chi Square value. As can be seen from the image below, the simulation begins by displaying a table with expected frequencies.
Clicking on the "Sample" button, samples $100$ values from a normal distribution (by default) and displays the observed frequencies as well as the results of the Chi Square tests.
17.04: Contingency Tables
Learning Objectives
• State the null hypothesis tested concerning contingency tables
• Compute expected cell frequencies
• Compute Chi Square and $df$
This section shows how to use Chi Square to test the relationship between nominal variables for significance. For example, Table $1$ shows the data from the Mediterranean Diet and Health case study.
Table $1$: Frequencies for Diet and Health Study
Outcome
Diet Cancers Fatal Heart Disease Non-Fatal Heart Disease Healthy Total
AHA 15 24 25 239 303
Mediterranean 7 14 8 273 302
Total 22 38 33 512 605
The question is whether there is a significant relationship between diet and outcome. The first step is to compute the expected frequency for each cell based on the assumption that there is no relationship. These expected frequencies are computed from the totals as follows. We begin by computing the expected frequency for the AHA Diet/Cancers combination. Note that $22/605$ subjects developed cancer. The proportion who developed cancer is therefore $0.0364$. If there were no relationship between diet and outcome, then we would expect $0.0364$ of those on the AHA diet to develop cancer. Since $303$ subjects were on the AHA diet, we would expect $(0.0364)(303) = 11.02$ cancers on the AHA diet. Similarly, we would expect $(0.0364)(302) = 10.98$ cancers on the Mediterranean diet. In general, the expected frequency for a cell in the $i^{th}$ row and the $j^{th}$ column is equal to
$E_{i,j} = \frac{T_iT_j}{T}$
where $E_{i,j}$ is the expected frequency for cell $i,j$, $T_i$ is the total for the $i^{th}$ row, $T_j$ is the total for the $j^{th}$ column, and $T$ is the total number of observations. For the AHA Diet/Cancers cell, $i = 1$, $j = 1$, $T_i = 303$, $T_j = 22$, and $T = 605$. Table $2$ shows the expected frequencies (in parenthesis) for each cell in the experiment.
Table $2$: Observed and Expected Frequencies for Diet and Health Study
Outcome
Diet Cancers Fatal Heart Disease Non-Fatal Heart Disease Healthy Total
AHA 15
(11.02)
24
(19.03)
25
(16.53)
239
(256.42)
303
Mediterranean 7
(10.98)
14
(18.97)
8
(16.47)
273
(255.58)
302
Total 22 38 33 512 605
The significance test is conducted by computing Chi Square as follows.
$\chi _{3}^{2} = \sum \frac{(E-O)^2}{E} = 16.55$
The degrees of freedom is equal to $(r-1)(c-1)$, where r is the number of rows and $c$ is the number of columns. For this example, the degrees of freedom is $(2-1)(4-1) = 3$. The Chi Square calculator can be used to determine that the probability value for a Chi Square of $16.55$ with three degrees of freedom is equal to $0.0009$. Therefore, the null hypothesis of no relationship between diet and outcome can be rejected.
A key assumption of this Chi Square test is that each subject contributes data to only one cell. Therefore, the sum of all cell frequencies in the table must be the same as the number of subjects in the experiment. Consider an experiment in which each of $16$ subjects attempted two anagram problems. The data are shown in Table $3$.
Table $3$: Anagram Problem Data
Anagram 1 Anagram 2
Solved 10 4
Did not Solve 6 12
It would not be valid to use the Chi Square test on these data since each subject contributed data to two cells: one cell based on their performance on $\text{Anagram 1}$ and one cell based on their performance on $\text{Anagram 2}$. The total of the cell frequencies in the table is $32$, but the total number of subjects is only $16$.
The formula for Chi Square yields a statistic that is only approximately a Chi Square distribution. In order for the approximation to be adequate, the total number of subjects should be at least $20$. Some authors claim that the correction for continuity should be used whenever an expected cell frequency is below $5$. Research in statistics has shown that this practice is not advisable. For example, see:
Bradley, D. R., Bradley, T. D., McGrath, S. G., & Cutcomb, S. D. (1979) Type I error rate of the chi square test of independence in r x c tables that have small expected frequencies. Psychological Bulletin, 86, 1200-1297.
The correction for continuity when applied to $2 \times 2$ contingency tables is called the Yates correction. The simulation 2 x 2 tables lets you explore the accuracy of the approximation and the value of this correction. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/17%3A_Chi_Square/17.03%3A_Testing_Distribution_Demonstration.txt |
Learning Objectives
• Explore the accuracy of the Chi Square test
• Determine the extent to which the Yate's correction affects the accuracy of the test
Instructions
The Chi Square Test is an approximate not an exact test. This simulation allows you to examine the accuracy and power of the Chi Square Test in a variety of situations. The significance of the difference between the proportion who succeed in $\text{Condition 1}$ and the proportion who succeed in $\text{Condition 2}$ is tested.
With the default parameters, the probability of success is the same ($0.60$) in both conditions so the null hypothesis is true. The default sample size is $10$ per condition and the default significance level is $0.05$.
Some authors have suggested that a correction called the "Yates Correction" be done whenever the expected frequency of any cell is below five. A test without this correction and one with the correction (if a cell has an expected frequency bellow five) is conducted for each simulated experiment.
If you click the "Simulate $1$" button, one simulated experiment will be conducted. The observed and expected frequencies are presented as well as the Chi Square Test with and without the Yates correction. A tally of the number of times the tests were significant is shown below.
If you click the "Simulate $1000$" (or $5000$) button then $1000$ (or $5000$) simulated experiments are conducted and the numbers of significant and non-significant tests are shown.
1. Using the default values, click on the "Simulate $1$" button. Note the number of successes and failures in each of the two conditions. Note the value of the Chi Square with and without the Yate's correction. Was either significant? Check the lower section that shows the count of the number significant. Most likely, it will show one non-significant.
2. Test whether the Type I error rate is close to the nominal significance level of $0.05$. Do this by clicking the "Simulate $5000$" button several times. Compare the proportion significant to $0.05$. Look at the results both when the Yates correction is never used and when it is used when an expected cell frequency is less than $5$. Is the test conservative ($\text{proportion significant} < 0.05$) or is it liberal ($\text{proportion significant} > 0.05$)
3. Redo the previous simulations when the probability of success is $0.50$ for each condition. Are the results similar? Try making one of the sample sizes $10$ and the other one $6$.
4. Try to find a set of parameters such that the proportion significant is greater than $0.06$. (Make sure the null hypothesis is always true -- that the probability of success is the same for both conditions.) Could you find such a set of parameters? Are there many circumstances in which the test is that liberal? Do you think the Yates correction is a good procedure?
5. Now consider cases in which the probability of success is different for the two conditions. Here the null hypothesis is false so the higher the proportion significant the better. What is the effect of using the Yates correction on rejecting a false null hypothesis?
Illustrated Instructions
The demo begins by running a single simulation, which turns out to be not statistically significant. The video concludes by running $1,000$ simulations and then $5,000$ simulations.
Video Demo
Try changing the proportions of the conditions?
17.06: Statistical Literacy
Learning Objectives
• A Spice Inhibits Liver Cancer
An experiment was conducted to test whether the spice saffron can inhibit liver cancer. Two groups of rats were tested. Both groups were injected with chemicals known to increase the chance of liver cancer. The experimental group was fed saffron ($n = 24$) whereas the control group was not ($n = 8$). The experiment is described here.
Only $4$ of the $24$ subjects in the saffron group developed cancer as compared to $6$ of the $8$ subjects in the control group.
Example $1$: what do you think?
What method could be used to test whether this difference between the experimental and control groups is statistically significant? Use Analysis Lab to do the test.
Solution
The Chi Square test of contingency tables could be used. It yields a $\chi ^2$ ($df = 1$) of $9.50$ which has an associated $p$ of $0.002$. | textbooks/stats/Introductory_Statistics/Introductory_Statistics_(Lane)/17%3A_Chi_Square/17.05%3A_2_x_2_Simulation.txt |
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