BUS 308 - Discussion 2
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7
Repeated Measures Designs for Interval Data
Learning Objectives
After reading this chapter, you should be able to:
• Explain the advantages and drawbacks of using data from non-independent groups.
• Complete a paired-samples t-test.
• Complete a within-subjects F.
• Describe “power” as it relates to statistical testing.
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CHAPTER 7Introduction
Chapter Outline
7.1 Dependent Groups Designs Reconsidering the t and F ratios An Example A Matched Pairs Example Comparing the Paired-Samples t-Test to the Independent Samples t-Test The Power of the Dependent Groups Test The Dependent Groups t-Test on Excel The Alternate Approaches to Dependent t-Tests
7.2 The Within-Subjects F Managing Error Variance in the Within-Subjects F A Within-Subjects F Example Calculating the Within-Subjects F Understanding the Result Comparing the Within-Subjects F and the One-Way ANOVA Another Within-Subjects F Example A Within-Subjects F in Excel
Chapter Summary
Introduction
Some of the most critical questions in management relate to change over time. For exam-ple, managers are deeply interested in assessing sales growth, shifts in shopping trends, improvements in employee attitudes, increases in employee performance, and decreases in absenteeism or turnover. They are also often keen to find out the influence of various managerial decisions and business strategies on these and many other change-oriented outcomes. However, none of the analyses completed to this point address these change- related questions, because these analyses do not accommodate repeated measures of the same variables within the same group of subjects over time. For instance, the t-tests and ANOVAs discussed so far compared independent groups, groups that have completely separate subjects. Each subject was only measured once on each variable of interest. The same group of subjects was not measured repeatedly on the same variables to assess change over time.
Another important issue is that independent samples t-tests and ANOVAs assume that the groups being compared are equivalent on most aspects to begin with, except for the independent (grouping or treatment) variable being investigated. When groups are large and individuals are randomly selected, this is usually a reasonable assumption, because any differences between groups tend to be relatively unimportant. The logic behind ran- dom selection is that when groups are randomly drawn from the same population they will differ only by chance—the larger the random sample, the lower the probability of a substantial pre-existing difference. However, when groups are relatively small it can be difficult to determine whether a difference in the measures of the dependent variable occurred because the independent variable had a different impact on the different groups or because there were differences between the groups to begin with.
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CHAPTER 7Section 7.1 Dependent Groups Designs
When variables not included in the analysis prompt differ- ences in dependent variable scores, the result is inflated error variance. Reducing error variance was one of the benefits of including multiple independent variables in factorial ANOVA. However, sometimes an analyst cannot know all the potential variables that can influence the dependent variable, or if they are known, it isn’t feasible to include them in the same analy- sis. Measuring the same subjects repeatedly and comparing their scores over time, especially after introducing important treatments or interventions, makes initial equivalence less of an issue from a statistical standpoint. This is because it is the same group of subjects being measured repeatedly. All of their other characteristics, except for the independent variable (being introduced as a treatment or intervention), are still the same, so in essence they are held constant, controlled, or accounted for.
A third issue is that even if comparing groups is the goal, many of the questions manag- ers are interested in relate to groups that are not entirely independent. Independent sam- ples t-tests and ANOVAs assume that the groups being compared are independent. For example, while it would be appropriate to use an independent samples t-test to compare holiday spending dollar amounts of male and female shoppers, it would be inaccurate to use an independent samples t-test if the question is whether males or females within house- holds are responsible for more holiday spending. This is because in that case the male and female within each household are interdependent. The holiday spending on one partner influences that of the other.
In such situations, whether data are collected on the same dependent variable from the same subjects more than once, or collected from different but related subjects, the goal is to account for this interdependence and control error variance due to initial between- groups differences. This is one of the primary purposes of the dependent groups designs discussed in this chapter.
7.1 Dependent Groups Designs
Dependent groups designs are statistical procedures in which the groups are related. There are three types of dependent group designs: • In the repeated measures design, multiple measures are taken of the same
group of participants. For example, an organization may implement a new rewards system and measure employee satisfaction before and after the new system is implemented to assess its effectiveness in increasing employee satisfaction.
• In dependent samples design, each participant in a particular group is related to a participant in the other group(s) on characteristics relevant to the analysis. The same-household male and female partners’ holiday spending example lends itself to a dependent samples design.
• In the matched pairs design, separate groups are used, but each individual in one group is matched with someone in the other groups who has the same initial characteristics, which makes the groups separate but not independent. For example, employees from two different branches can be matched on their levels of job satisfaction and demographic characteristics such as age, gender,
Review Question A: How do dependent groups tests manage the error variance that comes from compar- ing nonequivalent groups?
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CHAPTER 7Section 7.1 Dependent Groups Designs
ethnic background, education, and pay grade. If a reward system is then implemented in one of the branches, any differences in job satisfaction between the two branches after the implementation will probably not be due to the characteristics on which the subjects were matched, since they are the same for subjects in each group. The groups were made equivalent through the matching process.
The differences between the three designs above are conceptual, not mathematical, since they are all cal- culated the same way. The three approaches have the same statistical purpose, to control the error variance that comes from using nonequivalent groups, which should more accurately reflect the impact of the independent variable. Managers should choose the design that best fits the questions they are interested in answering and the data sources that are available to them.
Reconsidering the t and F ratios
Recall that the t and F values produced in the independent t–test and the one-way ANOVA are ratios. The denominators in both the t and F ratios are measures of how much scores vary within the groups involved in the analysis. If a plant manager compares the pro- ductivity of day and night shift workers in an independent t-test, the denominator in the t-ratio measures how much variability there is among the dayshift workers plus how much variability there is among the night shift workers, SEd 5 "1SEM12 1 SEM22 2. Score variability within the two groups increases the standard error of the difference. The larger the standard error of the difference becomes, the larger the M1 2 M2 difference in the numerator must be for t to be statistically significant.
The point of all of this is that the value of t in the independent t-test—and it’s the same for F in a one-way or factorial ANOVA—is greatly affected by the amount of variability within the groups. Substantial variability within the groups translates into diminished val- ues of t and F. Differences within groups reflect differences in the way individuals in the same samples react to a treatment. If a service manager in an automobile dealership offers bonuses to workers to keep them from leaving to work for competitors, there will still be differences in how long individual employees remain with the dealership because factors besides money are involved. This will particularly be a problem when the groups that are used in the comparison are not equivalent to begin with.
There are several approaches to calculating the t statistic in two-group dependent groups tests. Whatever their differences, they all take into account the fact that the scores from the two groups are related or matched on some relevant characteristic. One approach is to deal with the relationship directly by calculating the correlation between the two sets of scores and then using the strength of the correlation value as a multiplier in the reduction of the error variance—the higher the correlation between the two sets of scores, the greater the multiplier effect and the lower the resulting error variance. Because correlation is not discussed until Chapter 8, we will use an alternative approach involving what are called “difference scores.” However, whether the t value was calculated with the correlation value or the difference scores, the result will be the same.
Key Terms: Dependent groups designs reduce error variance that comes from using non- equivalent groups. This allows the impact of the independent variable to emerge more readily.
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CHAPTER 7Section 7.1 Dependent Groups Designs
Where
Md 5 the mean of the difference scores
SEMd 5 the standard error of the mean for the difference scores
The steps for the test are as follows:
1. With the scores from the before and after measures listed in two columns, sub- tract the “after” score from the “before” to determine the difference score, d, for each pair.
2. Determine the mean of the d scores, Md. 3. Calculate the standard deviation of the d values, sd. 4. Calculate the standard error of the mean for the difference scores, SEMd,
by dividing the result of step 3 by the square root of the number of pairs of scores, SEMd 5 sd/ "number of pairs .
5. t 5 Md/SEMd.
An Example
A home improvement chain introduced a new training program for its customer service associates. To gauge the effectiveness of the training program, associates were asked to take the same written assessment before and after attending the training. This is also
Formula 7.1 t 5 Md /SEMd
Like the independent samples t, the dependent-samples t is based on the distribution of difference scores. Recall that in the independent samples t-test, the distribution of differ- ence scores indicated how much difference between a pair of sample means (M1 2 M2) could be expected to occur by chance if an infinite number of pairs of sample means were drawn from the same population. In other words, the distribution indicates the point at which the difference between a pair of means is so great that the samples were probably drawn from distinct populations.
The dependent groups tests are based on this same distribution. The difference is that the numerator value in the test statistic is the mean of the differences between each pair of scores, rather than the difference between the means of the independent samples. When the mean of the difference scores varies from the mean of the distribution of differences (which, recall, was 0) by a value as large as the critical value determined by the degrees of freedom for the problem, the t value is statistically significant. The degrees of freedom for a paired- samples t-test are the number of pairs of data, minus 1.
The standard error of the difference in the independent samples t-test “pooled” the variabil- ity within both groups involved in the analysis 1"1SEM12 1 SEM22 2 2 . For the dependent groups test, the denominator is the standard error of the mean for the difference between each pair of scores. The test statistic has this form:
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CHAPTER 7Section 7.1 Dependent Groups Designs
commonly referred to as a before-and-after, or pre-post, design. The goal is to see if the scores of the associates on the assessment significantly increase after attending the training. The assessment scores of the customer service associates before and after training, as well as the solution to the problem, are shown in Figure 7.1.
Figure 7.1: Calculating the before/after t
Before After d
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Before
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After Presentation
Determine the difference between each pair of scores, d by subtraction.
Determine the mean of the difference, the d values (M d )
Calculate the standard deviation of the d values (s d )
Determine standard error of the mean for the difference (SE Md
)
by dividing that result of step 3 by the square root of the number of
pairs, s/ n p
M d = ∑d/10 = –11/10 = –1.1
Verify that S d
= 1.101
Verify that SE Md
= s/ n p = 1.101/ 10 = .348
Divide M d by SE
Md to determine t; t = M
d /SE
Md = −1.1/.348 = –3.161
The degrees of freedom for the critical value of t for this test are the
number of pairs of scores (n p ), –1; t
.05(9) = 2.262
∑d = –11
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CHAPTER 7Section 7.1 Dependent Groups Designs
The calculated value of t exceeds the critical value from the table. The result is statisti- cally significant. The fact that it is negative in this example indicates that the scores were higher after the training than before (because we subtracted the after-training from the before-training scores), which is what is expected if the training is effective. We know the training was effective because the customer service associates’ scores showed a significant increase. Recall that with t-tests, when the question specifies the direction of the difference (did scores increase after the training, as opposed to did they change), the test becomes a one-tailed test and the critical value for one-tailed tests is used.
The mean of the difference scores is just Md 5 21.1. It indicates that there is comparatively little difference between the two sets of scores, and at first glance it might seem surprising that such a minor mean difference can produce a statistically significant result. The expla- nation is in the amount of error variance in this problem. When the error variance is also very small—the standard error of the difference scores is just .348—comparatively small mean differences can result in a statistically significant t-value.
A Matched Pairs Example
Another form of the dependent groups t-test is the “matched pairs” design. In this approach, rather than measure the same people repeatedly, those in the second group of subjects are each paired with someone in the first group. The pairing is based on some quality that would otherwise differ between the groups and increase error variance.
A market analyst wishes to determine whether a new commercial will induce consumers to spend more on a particular product. The analyst selects a group of consumers entering a retail establishment, asks them to view the commercial, and then tracks their expendi- tures. A second group of shoppers also is selected, but they are not asked to view the com- mercial. Wishing to control for age and gender because those characteristics might affect spending for the particular product, the analyst selects people for the second group who match the age and gender characteristics of those in the first group so that each individual from group 1 is matched to someone of the same age and gender in group 2. The expen- ditures in dollars for the members of each group and the solution to the problem are in Figure 7.2.
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CHAPTER 7Section 7.1 Dependent Groups Designs
Figure 7.2: Calculating a matched pairs t-test
The absolute value of t is less than the critical value from the table for df 5 9. The differ- ence is not statistically significant. There are probably several ways to explain the out- come, including the following:
• The most obvious explanation is that the commercial did not work. The shop- pers who viewed the commercial were not induced to spend significantly more than those who did not view it.
• Sample size may also have been a problem. Recall that small samples tend to be more variable than larger samples and within group variability is what the denominator in the t-ratio reflects as we noted earlier. Perhaps if this had been a larger sample, the SEMd would have had a smaller value and the t would have been significant.
• As an alternative explanation, perhaps age and gender are not related to how much people spend shopping for the particular product. Perhaps the shop- per’s level of income is the most important characteristic. Shoppers were not matched on that characteristic, which means that it was left uncontrolled.
Viewed Didn’t View d
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Verify that M d
= .825
S d
= 2.167
/ np = 2.167/ 10 = .685
t = M
d /SE
Md = .825/.685 = 1.204
3.25
t .05(9)
= 2.262
SE Md
= S
–1
–1
–1
–1.5
4
4
.5
0
1
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CHAPTER 7Section 7.1 Dependent Groups Designs
The last explanation points out the disadvantage of matched pairs designs compared to repeated measures designs. The matching must neutralize the characteristic(s) most likely to cloud the relationship between the treatment—the commercial in the example just above—and the dependent variable. Otherwise it is impossible to know whether a non-significant outcome reflects an inadequate match or a treatment that does not affect the DV.
Comparing the Dependent Samples t-Test to the Independent t-Test
A good way to compare the dependent samples t-test with the independent samples t is to apply both tests to the same data. We will do that here, but first a word of caution. The factor that should dictate the choice of analytical procedure is the independence of the groups from which the data were gathered. When the groups are independent, the independent samples t-test is the relevant test. When the groups are not indepen- dent, the procedure based on lack of independence between groups is the correct choice. Once the data are gathered, there is no situation where both tests might be appropriate.
Either the groups are independent or they are not. The prob- lem below is an academic exercise completed so that the two procedures can be directly contrasted.
A satellite TV company provides a month of free program- ming as an incentive to customers who refer their friends to the company for service. In an effort to increase sales, for a particular period the company adds a free month of a pre- mium movie channel as a bonus. The dependent variable is the number of people customers refer for service. The independent variable is whether the movie channel bonus was offered.
• For the independent samples t-test, the first group is customers who were offered the extra bonus. The second group is those who were not.
• For the dependent groups t-test, those who received the extra bonus were matched with another customer in the same neighborhood who was not offered the additional incentive.
The data and the solutions for both of these procedures are in Figure 7.3.
Review Question B: How is it that a depen- dent groups t-test can have a smaller value of t than an independent samples t and still be more likely to be statistically significant?
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CHAPTER 7Section 7.1 Dependent Groups Designs
Figure 7.3: The dependent samples t versus the independent samples t
For the independent samples t-test, the groups are unrelated, so the error variance is based upon the differences within both groups of scores that are combined. The result is a value of SEd that is large enough that the t value is not significant. On the other hand, there is very little variance in the difference scores from the matched pairs example. This results in a comparatively small standard deviation of difference score, a small standard error of the mean, and a statistically significant value of t.
The Power of the Dependent Groups Test
In Table 4.1, critical values of t decline as degrees of freedom increase. That is the pattern for all of the procedures we have used, and it will hold for those yet to come. Recall that the degrees of freedom in a dependent samples t-test are the number of pairs of scores – 1.
Bonus No Bonus d
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M
s
SE M
As an independent t-test we have,
SE d
=
SE d .553 .553
As a matched pairs t-test the results are,
t = M
1 /SE
Md = 6.50/.211 = 3.081; t
.05(9) = 2.262. The result is significant.
.650
.669
.211
.5
1
–1
1
1
1
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1
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0
2.850
1.001
.317
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1.434
.453
3.50
(SE M1
2 + SE M2
2) = (.4532 + .3172) = .553
t = M
1 – M
2 = 3.50 – 2.850 = .650 = 1.175; t
.05(18) = 2.101. The result is not significant.
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CHAPTER 7Section 7.1 Dependent Groups Designs
In comparison, in an independent samples t-test, the degrees of freedom are the total number of scores 2 2. Thus, independent samples t-tests will tend to have more degrees of freedom and therefore a lower critical value, as you can see in the satellite TV example above (t.05(18) 5 2.101; t.05(9) 5 2.262).
However, in the paired-samples t-test, there is more control of error variance. The advan- tage gained with the dependent groups test is that when each pair of scores comes from the same participant, or from a matched pair of participants, the scores tend to vary simi- larly, which reduces error variance. The small SEMd value in the denominator will make the t value larger, which will usually more than compensate for the larger critical value connected to dependent groups tests. To illustrate, in the satellite TV example above, although the numerators of the t ratios happened to be equal for the independent samples and paired-samples t values (M1 2 M2 and Md are both .650), the denominator for the independent samples t, SEd, is more than twice as large as the denominator for the paired- samples t, SEMd (.533 versus .211).
As mentioned in Chapter 3, in statistical testing, power is defined as the likelihood of detecting a significant difference when it is present. The more powerful a statistical test is, the more readily it will detect a significant difference. As long as the sets of scores are closely related so that error variance is reduced, dependent groups tests tend to be more powerful than their independent groups equivalents.
The Dependent Groups t-Test on Excel
If the problem in Figure 7.3 is completed in Excel as a dependent groups test, the proce- dure is as follows:
• First create the data file in Excel. · Column A is labeled Bonus to indicate those who received the movie chan-
nel bonus, and column B is labeled NoBonus. · Enter the data beginning with cell A2 and then down for the first group and
from cell B2 down for the second group. • Click the Data tab at the top of the page. • At the extreme right, choose Data Analysis. • In the Analysis Tools window select t-test: Paired Two Sample for Means
and click OK. • There are two blanks near the top of the window for Variable 1 Range and
Variable 2 Range. In the first enter A2:A11 indicating that the data for the first (Class) group are in cells A2 to A11. In the second enter B2:B11 for the NoBonus group.
• Indicate that the hypothesized mean difference is 0, which is the mean of the distribution of difference scores.
• Indicate D1 for the output range so that the solution does not over-write the data scores.
• Click OK.
The result is the screen-shot in Figure 7.4.
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CHAPTER 7Section 7.1 Dependent Groups Designs
Figure 7.4: The Excel output for the dependent samples t-test using the data from Figure 7.3
In the Excel solution, t 5 3.074 rather than the 3.081 from the longhand solution. The Excel approach is to calculate the correlation between the two sets of scores, and then to use that value to calculate the standard error of the difference. Note that the Pearson correlation, which is covered in Chapter 8, is indicated to be .909344. It is quite a high correlation, which allows the standard error of the difference to be correspondingly smaller.
In our approach, the correlation between the scores is implicit in the fairly consistent d values from person to person. In any event, the very minor difference of .007 between the Excel solution in Figure 7.4 and the longhand calculation does not alter the outcome. The Excel output also indicates results for one- and two-tailed tests. At p 5 .05, the outcome is statistically significant in either case.
The Alternate Approaches to Dependent t-Tests
The repeated measures and the matched pairs approaches to calculating a dependent groups t-test each have their advantages. The repeated measures design provides the greatest control over extraneous variables that can confound results when there are dif- ferent subjects in each group. As careful as an analyst might be in matching, the chance remains that some other characteristic that affects the dependent variable will be more prevalent in one group than in the other, and error variance inflated as a result.
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CHAPTER 7Section 7.2 The Within-Subjects F
Note that the matched pairs approach also assumes a large sample from which to draw participants for the second group who match those in the first group. As the number of variables on which participants must be matched increases, so must the size of the sample from which to draw in order to find participants with the correct combination of characteristics.
The advantage of the matched pairs design, on the other hand, is that once the groups are formed it takes less time to execute. The treatment group and the control group can be involved simultaneously. Figure 7.5 summarizes the comparison among independent samples, repeated measures, and matched pairs t-tests.
Figure 7.5: Comparing the t-tests
7.2 The Within-Subjects F
Sometimes two measures of the same group are not enough to track changes in the DV. Maybe the analysts in the satellite TV example wish to compare the effect of offering several kinds of bonuses over a lengthier period to determine which provides the best response. The within-subjects F is a dependent groups procedure for two or more groups when the dependent variable is interval or ratio scale. Just as the dependent groups t-test is the repeated measures equivalent of the independent t-test, the within-subjects F is the repeated measures or matched pairs equivalent of the one-way ANOVA. Fisher, who
developed analysis of variance, also developed the within-subjects test, which is why the test statistic is still F.
Although the dependent groups can be formed by either repeatedly measuring the same group or by matching participants from each group, when there are more than two groups, matching becomes untenable. While theoretically possible to match
participants in any number of groups, it becomes increasingly difficult to match partici- pants for more than two or three groups. Multiple measures of the same group are far more common.
Independent Samples Before/After Matched Pairs
The t-tests
Groups
Denominator/
Error Term
Independent groups One group measured
twice
Two groups: each
subject from 1st group
matched to one in the 2nd
Within groups variability
plus between groups
Only within groups
variability
Only within groups
variability
Key Term: The within- subjects F is the dependent groups equivalent to the one- way ANOVA. It typically employs one group measured repeatedly over time.
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CHAPTER 7Section 7.2 The Within-Subjects F
Managing Error Variance in the Within-Subjects F
Recall from Chapter 5 that the variability measure in ANOVA is the sums of squares, SS. In the one-way ANOVA, there were SS values for total variability (SStot), between- groups variability (SSbet), and within-groups variability (SSwith). Although the terminol- ogy changes and there are some conceptual differences between the one-way ANOVA and a within-subjects F, the procedures have a great deal in common.
If a group of participants in a study is measured on a dependent variable at three different intervals and their scores recorded in parallel columns, the data sheet might be as follows:
1st measure 2nd measure 3rd measure
Participant 1
Participant 2
The scores in each column are similar to what the independent groups’ scores were in a one-way ANOVA. Differences from column to column primarily indicate the effect the different levels of the IV have on the subjects’ DV scores. Within any particular column the differences from row to row are the differences from subject to subject that were the within-group differences (SSwith) in the one-way ANOVA. Those differences are also error variance in the within-subjects ANOVA, but the within-subjects F approach is to calcu- late the row-to-row variability and then eliminate it from the analysis. It makes sense to do this because each group involves the same people, and this source of error variance should be the same for each group.
Even after eliminating person-to-person differences, however, factors not included in the analysis still contribute to error variance. Those sources of error are reflected in the residual variance and remain, but the row differences tend to represent a substantial part of the overall error variance. Eliminating it typically results in a more powerful F test.
In the dependent samples t-test the within-subjects variance was managed by rely- ing on the standard deviation of the difference scores, or by reducing the denominator in the t-ratio according to how highly correlated the two sets of measures were (the Excel approach).
In the within-subjects F the variability due to person-to-person differences is calculated and then simply discarded so that it is no longer a part of the analysis. That was not pos- sible in the one-way ANOVA because with different subjects in each group there was no way to separate person-to-person differences from other sources of error variance in the problem.
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CHAPTER 7Section 7.2 The Within-Subjects F
A Within-Subjects F Example
The production department would like to track productivity changes over time in an elec- tronics components assembly facility. Five newly hired workers are selected for the study. The number of components each employee averages per hour is measured at three dif- ferent times: one week, one month, and two months after hire. The question is whether there is a relationship between the length of employment and number of successfully assembled components. The data for the five employees are as follows:
Average Number of Components Assembled per Hour
1 week 1 month 2 months
Diego 2 5 4
John 4 7 7
Ann 3 6 5
Carol 4 5 6
Dan 5 8 9
• The independent variable (IV, or treatment variable) is the time employed. • The dependent variable (DV) is the average number of components assem-
bled per hour. • The issue is whether there are significant differences in the measures from
column to column—differences over time.
The differences over time are equivalent to the SSbet in the one-way ANOVA. For this pro- cedure, that source of variance is called the sum of squares between columns, SScol.
Calculating the Within-Subjects F
As with the one-way ANOVA, we begin by determining all variability from all sources, SStot. It is calculated the same way as before:
1. The sum of squares total. SStot 5 (x 2 MG)
2
a. Subtract each score from the mean of all the scores from all the groups. b. Square the difference. c. Then sum the squared differences.
2. The sum of squares between columns (SScol), is similar to the SSbet in the one- way ANOVA. For columns 1, 2 through “k,” the formula is:
Formula 7.2 SScol 5 (Mcol1 2 MG) 2ncol1 1 (Mcol2 2 MG)
2ncol2 1 . . . 1 (Mcolk 2 MG) 2ncolk
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CHAPTER 7Section 7.2 The Within-Subjects F
a. Take the variance from all sources, SStot. b. Subtract the variability due to the IV, which is SScol. c. Then subtract the person-to-person differences, SSrows, to produce the SSresid.
Understanding the Result
The ANOVA table is completed with the following degrees of freedom values:
• df total 5 N 2 1 • df columns 5 number of columns 2 1 • df rows 5 number of rows 2 1 • df residual 5 df columns times df rows
Like all ANOVA problems, the mean square values are calculated by dividing the sums of squares by their degrees of freedom. The only MS values required are the MScol, which includes the treatment effect, and the MSresid, which is the error term. The MS is not neces- sary for the total or the rows. The ratio of treatment effect to residual error is F:
Formula 7.4 SSresid 5 SStot 2 SScol 2 SSrows
a. Calculate the mean for each column of scores. b. Subtract the mean for all the data (MG) from each column mean. c. Square the result. d. Multiply the squared value by the number of scores in the column.
3. The sum of squares between rows. Here the scores for each row are treated as a separate group. For rows 1, 2 through “i”:
Formula 7.5 F 5 MScol/MSresid
Formula 7.3 SSrows 5 (Mr1 2 MG) 2nr1 1 (Mr2 2 MG)
2nr2 1 . . . (Mri 2 MG) 2nri
a. Calculate the mean for each row of scores. b. Subtract the mean for all the data from each row mean. c. Square the result. d. Multiply the squared value by the number of scores in the row.
4. The residual sum of squares is the error term in the within-subjects F and is used the same way that SSwith was used in the one-way ANOVA. It is deter- mined by subtraction as follows:
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CHAPTER 7Section 7.2 The Within-Subjects F
Figure 7.6 shows the calculations and the table for the “average components assembled per hour” problem.
Figure 7.6: A within-subjects F example
The calculated value of F exceeds the critical value of F from the table. The length of employment is significantly related to the number of components assembled per hour.
1 week 1 month 2 months Row Means
Average Components Assembled per Hour
Diego
John
Ann
Carol
Dan
Column Means
Grand Mean (M G )
2
4
3
4
5
3.60
5
7
6
5
8
6.20
4
7
5
6
9
6.20
3.667
6.0
4.667
5.0
7.333
1. SS tot
= ∑(x – M G )2
(2 – 5.333)2 + (4 – 5.333)2 + . . . + (9 – 5.333)2 = 49.333
2. SS col
= (M col1
– M G )2n
col1 + (M
col2 – M
G )2n
col2 + . . . + (M
col1 – M
G )2n
colk
(3.6 – 5.333)2 5 + (6.2 – 5.333)2 5 + (6.2 – 5.333)2 5 = 22.533
3. SS rows
= (M r1 – M
G )2n
r1 + (M
r2 – M
G )2n
r2 + . . . + (M
ri – M
G )2n
ri
(3.667 – 5.333)2 3 + (6.0 – 5.333)2 3 + (4.667 – 5.333)2 3 + (5.0 – 5.333)2 3
+ (7.333 – 5.333)2 3 = 23.325
4. The residual sum of squares:
SS res
= SS tot
– SS col
– SS rows
= 49.333 – 22.533 – 23.325 = 3.475
SS df MS F FcritSource
The ANOVA table
Total
Columns
Rows
Residual
49.333
22.533
23.325
14
2
4
11.267 25.961 4.46
.4343.475 8
5.333
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CHAPTER 7Section 7.2 The Within-Subjects F
The Post Hoc Test
Ordinarily, the calculation of F leaves unanswered the question of which set of measures is significantly different from which. In this particular problem, however, there is only one possibility. Since the two later groups of measures have the same mean (M 5 6.20), they must both be significantly different from the only other group of measures in the problem, the one-week-on-the-job column for which M 5 3.6. As a demonstration, Tukey’s HSD is completed anyway. The HSD error term is now MSresid. Substituting MSresid for MSwith in the formula provides:
Formula 7.6 HSD 5 x"1MSresid /n2
Where
x is a table value (Table 5.4) based on the number of means, which is the same as the number of sets of measures, 3; the df for MSresid are 8 n 5 the number of scores for any one measure, 5 in this example
For the number of products components per hour study,
HSD 5 4.04 "1.434/52 5 1.190—a difference between any pair of means 1.190, or greater, is statistically significant.
Using a matrix indicating the difference between each pair of means makes it easier to interpret the HSD value.
1 week (3.6) 1 month (6.2) 2 months (6.2)
1 week (3.6) diff 5 2.6* diff 5 2.6*
1 month (6.2) diff 5 0
2 months
The one-week measures of productivity are significantly different from either of the other two, and of course since the mean values of the one- and two-month measures are the same, neither of the last two measures is significantly different from the other. The largest increase in productivity comes between the first week and the first month of employment.
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CHAPTER 7Section 7.2 The Within-Subjects F
For the problem just completed, h2 5 22.533/49.333 5 .457. About indicates that about 46% of the variance in productivity can be explained by how long employees have been on the job.
Comparing the Within-Subjects F and the One-Way ANOVA
In the one-way ANOVA, within-group variance is different for each group because each group is made up of different participants. That means that the within-group variance is different for each group, and because that variance cannot be distinguished from the bal- ance of the error variance, it remains in the analysis. Eliminating within-group variance from the error term allows relatively small differences between groups to be statistically significant.
This is illustrated by applying one-way ANOVA to the within- subjects F test just completed. As with the t-test comparison earlier, this is for illustration purposes only. Groups are either independent or dependent, and that should be the determin- ing criterion for test selection.
The SStot and the SSbet will be the same as the SStot and the SScol were in the within-subjects problem.
SStot 5 49.333
SSbet 5 22.533
SSwith 5 (xa 2 Ma) 2 1 (xb 2 Mb)
2 1 (xc 2 Mc) 2 (Formula 6.3)
5 (2 2 3.60)2 1 (4 2 3.60)2 1 . . . 1 (9 2 6.20)2 5 26.80
Review Question C: What is the equivalent in the one-way ANOVA of the between col- umns variability in the within-subjects F?
Formula 7.7 h2 5 SScol /SStot
Calculating the Effect Size
The other question when F is significant is regarding the practical importance of the result. Eta-squared is perhaps the easiest measure of effect size to calculate. It is adjusted from the independent groups test application by substituting SScol for what was SSbet in the earlier application, which gives it this form:
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CHAPTER 7Section 7.2 The Within-Subjects F
Because participant-to-participant differences cannot be separated from the balance of the error variance, the SSwith in a one-way ANOVA is the same as SSrows 1 SSresid in the within- subjects F. With the SSrows added back in to the error term, note in Figure 7.7 the change that makes to the ANOVA table, and to F in particular.
Figure 7.7: The within-subjects F example repeated as a one-way ANOVA
• The degrees of freedom for “within” change to 12 from the 8 for residual which results in a smaller critical value for the independent groups test, but that adjustment does not compensate for the additional variance in the error term.
• Note that the sum of squares for within becomes 26.800 compared to 3.475 in the within-subjects test.
• Because of the larger error term, the F value is reduced from 25.961 in the within problem to 5.046 in the one-way problem, a factor of about 1/5th.
The calculations illustrate the gains in statistical power from dependent groups designs.
Another Within-Subjects F Example
A human resources specialist is tasked with examining whether employees who have been with the organization for different periods have different patterns of sick leave. Examin- ing the records, the HR specialist determines the number of sick days taken by employees during their first, second, third, and fourth years of employment. The data and the solu- tion are shown in Figure 7.8.
SS df MS F F crit
Source
The ANOVA table
Total
Between
Within
49.333
22.533
26.800
14
2
12
11.267 5.046 3.89
2.233
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CHAPTER 7Section 7.2 The Within-Subjects F
Figure 7.8: Number of sick days during the first, second, third, and fourth years of employment
1st 2nd 3rd 4th Row MeansEmployee
Employment Year
1
2
3
4
5
Column Means
4
5
3
4
2
3.60
3
4
1
2
1
2.20
2
3
1
1
2
1.80
5
4
2
3
3
3.40
3.50
4.0
1.750
2.50
2.0
M G = 2.750
Verify that, 1. SS
tot = ∑(x – M
G )2 = 31.750
2. SS col
= (M col1
– M G )2n
col1 + (M
col2 – M
G )2n
col2 + (M
col3 – M
G )2n
col3 + (M
col4 – M
G )2n
col4
3. SS rows
= (M r1 – M
G )2n
r1 + (M
r2 – M
G )2n
r2 + (M
r3 – M
G )2n
r3 + (M
r4 – M
G )2n
r4 + (M
r5 – M
G )2n
r5
= (3.5 – 2.75)2 4 + (4.0 – 2.75)2 4 + (1.75 – 2.75)2 4 + (2.5 – 2.75)2 4 + (2.0 – 2.75)2 4 = 15.0
4. SS res
= SS tot
– SS col
– SS rows
= 31.75 – 11.75 – 15 = 5.0
SS df MS FSource
Total
Columns
Rows
Residual
19
3
4
12
3.917
.417
9.393 F.05 (3,12) = 3.49. F is sig.
M 1 = 3.6
M 1 = 3.6
M 2 = 2.2
M 2 = 2.2
M 3 = 1.8
M 3 = 1.8
1.4* 1.8*
1.2
1.6*
The post hoc test: HSD = x .05
(MS w /n) = 4.20 (.417/5) = 1.213
M 4 = 3.4
M 4 = 3.4
SS tot
11.75
31.75 37% of the variance in sick days taken is related to the length of employment
31.75
11.75
15.0
5.0
.2
.4
= = = SS
col2
= (3.6 – 2.75)2 5 + (2.2 – 2.75)2 5 + (1.8 – 2.75)2 5 + (3.4 – 2.75)2 5 = 11.750
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CHAPTER 7Section 7.2 The Within-Subjects F
The results (F) indicate that the number of sick days taken depends, to some degree, on the length of employment. The post hoc test indicates that those in their first year of employment take a significantly greater number of sick days (the newest employees had the highest mean number of sick days) than those in their second or third year of employ- ment. Those who have been employed for three years have significantly fewer sick days than those employed for four years. The eta-squared value indicates that about 37% of the variance in number of sick days taken is a function of the length of employment.
A Within-Subjects F in Excel
Dependent groups ANOVA is not one of the options Excel offers in the list of Data Analy- sis Tools. However, like any data analysis task involving a number of repetitive calcula- tions, any business spreadsheet can be a great help. The last problem will be completed in Excel as an example.
1. Begin by setting the data up in four columns just as they are in Figure 7.8, but insert a blank column to the right of each data column. With a row at the top for the labels, data for the first year begin in cell A2. Column B will be blank. The data for the second year will be in column C, and so on.
2. Calculate the row and column means as well as a grand mean as follows: a. For the column means, place the cursor in cell A7 just beneath the last
value in the first column and enter the formula 5average(A2:A6) followed by Enter.
To repeat this for the other columns, left click on the solution that is now in A7, drag the cursor across to G7, and release the mouse but- ton. In the Home tab click Fill and then Right. This will repeat the column means calculations for the other columns. Delete the entries this makes to cells B7, D7, and F7 since there aren’t yet any data in those columns.
b. For the row means, place the cursor in cell I2 and enter the formula 5average(A2, C2, E2, G2) followed by Enter.
To repeat this for the other columns, left-click on the solution that is now in I2, drag the cursor down to I6, and release the mouse button. In the Home tab click Fill and then Down. This will repeat the calcu- lation of means for the other rows.
c. For the grand mean, place the cursor in cell I7 and enter the formula 5average(I2:I6) followed by Enter (the mean of the row means will be the same as the grand mean—the same could have been done with the column means). Note that MG 5 2.75.
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CHAPTER 7Section 7.2 The Within-Subjects F
3. To determine the SStot: a. In cell B2 enter the formula 5(A2-2.75)^2 and press Enter. This will square
the difference between the value in A2 and the grand mean.
To repeat this for the other data in the column, left-click the cursor in cell B2 and drag down to cell B6. Click Fill and Down. With the cur- sor in cell B7, click the summation sign () at the upper right of the screen and hit Enter. Repeat these steps for columns C, E, and G.
b. With the cursor in H9 type in SStot5 and click Enter. In cell I9 enter the formula 5Sum(B7,D7,F7,H7) and press Enter. The value will be 31.75, which is the value of SStot.
4. For the SScol: a. In cell A8 enter the formula 5(3.6-2.75)^2*5 and press Enter. This will
square the difference between the column mean and the grand mean and multiply the result by the number of measures in the column, 5.
In cells C8, E8, and G8, repeat this for each of the other columns, substituting the mean for the each column for the 3.60 that was the column 1 mean.
b. With the cursor in H10 type in SScol5 and click Enter. In cell I10 enter the formula 5Sum(A8,C8,E8,G8) and press Enter. The value will be 11.75, which is the sum of squares for the columns.
5. For the SSrows: a. In cell J2 enter the formula 5(I2-2.75)^2*4 and press Enter. Repeat this
in rows J3–J6 by left clicking on what is now J2 and dragging the cursor down to cell J6. Click Fill and Down.
b. With the cursor in H11 type in SSrow5 and click Enter. In cell I11 enter the formula 5Sum(J2:J6) and press Enter. The value will be 15.0, which is the sum of squares for the participants.
6. For the SSresid, in cell H12 enter SSresid5 and click Enter. In cell I12 enter the formula 5I10-I11-I12. The resulting value will be 5.0.
Having used Excel to determine all the sums of squares values, the mean squares are determined by dividing the sums of square for columns and residual by their degrees of freedom:
MScol 5 11.75/3 5 3.917
MSresid 5 5/12 5 .417
F 5 MScol/MSresid 5 3.917/.417 5 9.393, which agrees with the earlier calculations done by hand.
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CHAPTER 7Section 7.2 The Within-Subjects F
To create the ANOVA table:
• Beginning in cell A10, type in Source, in B10 SS, df in C10, MS in D10, F in E10, and Fcrit in F10.
• Beginning in cell A11 and working down type in total, columns, rows, residual.
For the sum of squares values:
• In cell B11 enter 5I9. • In cell B12 enter 5I10. • In cell B13 enter 5I11. • In cell B14 enter 5I12.
For the degrees of freedom:
• In cell C11 enter 19 for total degrees of freedom. • In cell C12 enter 3 for columns degrees of freedom. • In cell C13 enter 4 for rows degrees of freedom. • In cell C14 enter 12 for residual degrees of freedom.
For the mean squares:
• in cell D12 enter 5B12/C12. The result is MScol. • in cell D14 enter 5B14/C14. The result is MSresid.
For the F value in cell E12 enter 5D12/D14.
In cell F12 enter the critical value of F for 3 and 12 degrees of freedom, which is 3.49.
The list of commands looks intimidating, but mostly because every keystroke has been included. With some practice, most of what’s been done here will become second nature. A screenshot of the result of all the above, with some color added to separate sections, is Figure 7.9.
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CHAPTER 7Chapter Summary
Figure 7.9: A screenshot of a within-subjects F problem
Chapter Summary
In any analysis involving groups of subjects, individuals within the same group may still respond to the same stimulus differently. Those differences constitute error variance, which is compounded in independent groups tests where the individuals are different for each group. No matter how carefully a researcher randomly selects the groups to be used in a study, people in the same group are going to respond differently to whatever is measured. The before/after or paired-samples t and the within-subjects F tests eliminate that source of error variance by either using the same people repeatedly or matching sub- jects on the most important characteristics (Objective 1). Controlling error variance in this fashion results in what is ordinarily a more powerful test (Objective 4).
Using the same group repeatedly requires fewer participants for dependent groups designs, but because the same groups are used repeatedly, completing analyses with mul- tiple measures requires more time. One way to respond to the time requirement is to match subjects so that the different levels of the independent variable can be administered concurrently. However, finding matching subjects on all of the relevant characteristics creates its own difficulty. Using the same group multiple times eliminates that difficulty (Objective 1).
Having noted some of the differences between dependent groups designs and their inde- pendent groups equivalents, it is important to note their consistencies as well. Indepen- dent samples t-tests, paired-samples t-tests, one-way ANOVA, and within-subjects F all have a categorical independent variable (nominal scale) and a continuous dependent vari- able (interval or ratio scale). Like the z-test, the t-tests and ANOVAs test for significant differences between means (Objectives 1, 2, and 3).
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CHAPTER 7Chapter Formulas
Answers to Review Questions
A. The dependent groups designs manage the non-equivalence of groups by either using the same group repeatedly or matching subjects in multiple groups on the most relevant characteristics.
B. The key to the dependent groups test’s power is in the magnitude of the error term. Smaller calculated values of t may still be significant because the error term—the denominator in the t-ratio—is relatively small. The error is con- trolled by using the same group repeatedly or by matching subjects.
C. The SSbet (the sum of squares between groups) in the one-way ANOVA gauges the same variance that the SScol (between columns) measures in the within- subjects F.
Chapter Formulas
Formula 7.1 t 5 Md /SEmd This is the formula for either the paired-samples, or matched pairs t-test. The numerator is the mean of the differences between the first and second score, and the denominator is the standard error of the mean for the difference scores.
Formula 7.2 SScol 5 (Mcol1 2 MG) 2ncol1 1 (Mcol2 2 MG)
2ncol2 1 . . . 1 (Mcolk 2 MG) 2ncolk
This is the formula for determining the sum of squares between columns for a within- subjects F. It indicates the treatment effect.
Formula 7.3 SSrows 5 (Mr1 2 MG) 2nr1 1 (Mr2 2 MG)
2nr2 1 . . . (Mri 2 MG) 2nri
This formula determines the person-to-person variance within a group. It is a source of error variance, and since it is common to each group in a repeated measures design, it is calculated to eliminate it from what will be the denominator in the F ratio.
Formula 7.4 SSresid 5 SStot 2 SScol 2 SSrows
The error term in the within-subjects F is determined by subtracting the column-to-column (treatments) and the row-to-row (participants) differences from all variance. Whatever is left, SSresid, when divided by its degrees of freedom, becomes the error term in the F ratio.
Formula 7.5 F 5 MScol/MSresid The F statistic in dependent groups ANOVA.
Formula 7.6 HSD 5 x"1MSresid /n2 Tukey’s post hoc HSD test in dependent groups ANOVA.
Formula 7.7 h2 5 SScol/SStot Eta-squared as an estimate of effect size, adapted for dependent groups ANOVA.
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CHAPTER 7Management Application Exercises
Management Application Exercises
Unless otherwise stated, use p 5 .05 in all your answers.
1. A dental office wants to gauge patients’ reaction to a new cleaning procedure. Eight patients are asked about their level of anxiety before and after receiving the new procedure.
Before After
1. 5 4
2. 6 4
3. 4 3
4. 9 5
5. 5 6
6. 7 3
7. 4 2
8. 5 5
a. What is the standard deviation of the difference scores? b. What is the standard error of the mean for the difference scores? c. What is the calculated value of t? d. Are the differences statistically significant? e. What was the impact of the new procedure? Should the dental office con-
tinue to use it?
2. A courier service in a large city tracks the number of deliveries it is asked to make by 10 clients before and after it offers a progressive discount for repeat business to assess the effects of the discount.
Before After
1. 0 10
2. 20 20
3. 10 0
4. 25 50
5. 0 0
6. 50 75
7. 10 20
8. 0 20
9. 50 60
10. 25 35
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CHAPTER 7Management Application Exercises
a. What is the most appropriate statistical test in this situation? Why? b. Are there significant differences in the number of deliveries? c. If the goal is to promote repeat business, should the discount be continued?
3. Eight participants attend three consecutive sessions in a business seminar. In the first there is no reinforcement for responding to the session moderator’s questions. In the second, those who respond are provided with positive feedback as rein- forcement. In the third, responders receive cafeteria discount coupons. The num- ber of times the participants responded in each session is provided below.
None Feedback Coupons
1. 2 4 5
2. 3 5 6
3. 3 4 7
4. 4 6 7
5. 6 6 8
6. 2 4 5
7. 1 3 4
8. 2 5 7
a. Did the different reinforcers have significantly different effects on the num- ber of responses? If so, which reinforcers are significantly different from which? Rank the reinforcers from most to least effective.
b. Calculate and explain the effect size. c. If instead of having the same participants attend the three sessions, three
different groups of participants attended one session each, and the table above showed the number of responses in each of those groups, how would your answers to the above two questions have changed? Perform all your calculations again.
d. Why are the F values of the two answers different?
4. Eight college students take summer jobs as door-to-door sales representatives for a cleaning supplies company. Their number of sales made per week during their first four weeks of summer employment are as follows.
Week 1 Week 2 Week 3 Week 4
1. 5 8 9 9
2. 4 7 8 10
3. 4 4 4 5
4. 2 3 5 5
5. 4 6 6 8
6. 3 5 7 9
7. 4 5 5 4
8. 2 3 6 7
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CHAPTER 7Management Application Exercises
a. Are there significant differences among the weeks? b. Which weeks are significantly different from which? c. Is sales success related to experience? d. How much of the variations in sales can be explained by amount of experi-
ence?
5. A business department at a university sponsors a study of the relationship between participation in a summer internship program and students’ grade point average (GPA). Eight students who participate in a summer internship are matched with eight students in the same year who receive no internship. Students’ GPAs at the end of the academic year are compared.
Internship No internship
1. 3.6 3.2
2. 2.8 3.0
3. 3.3 3.0
4. 3.8 3.2
5. 3.2 2.9
6. 3.3 3.1
7. 2.9 2.9
8. 3.1 3.4
a. Although there are two separate groups, an independent samples t-test is not appropriate for this analysis. Why?
b. Are the differences statistically significant? c. Write a paragraph explaining your findings to new students who have not
yet taken any statistics classes.
6. A utility company notes the number of complaints in a particular community about the quality of service provided by service representatives. The company then requires service representatives to attend a quality training class, and then the number of complaints is tracked in a second community serviced by the same group of representatives and where residents have socioeconomic characteristics similar to those in the first community. The data are as follows:
Before training After training
1. 12 5
2. 10 3
3. 5 6
4. 8 5
5. 6 5
6. 12 10
7. 9 8
8. 7 7
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CHAPTER 7Key Terms
a. What is the most appropriate statistical test in this situation? Why? b. Are the differences statistically significant? c. To what extent was the training effective in reducing complaints?
7. A supervisor is monitoring the number of sick days employees take by month. For seven employees reporting to him, the number of sick days are as follows:
Oct Nov Dec
1. 2 4 3
2. 0 0 0
3. 1 5 4
4. 2 5 3
5. 2 7 7
6. 1 3 4
7. 2 3 2
a. What are the independent and dependent variables in this analysis? What is the type of data scale of each?
b. Are the month-to-month differences significant? c. How much of the variance does the month explain? d. If instead of tracking the number of sick days of the same seven employees,
the manager randomly selected seven different employees every month and used their number of sick days, how would your answer have changed? Perform all your calculations again.
e. Why are the F values of the two answers different?
Key Terms
• Dependent groups designs are statistical procedures in which the groups are related, either because multiple measures are taken of the same participants or because each participant in a particular group is matched with participants in the other group or groups according to whichever characteristics are relevant to the analysis.
• Repeated measures design is a type of dependent groups design where multiple measures are taken of the same group of participants.
• Dependent samples design is a type of dependent groups design where each participant in a particular group is related to a participant in the other group(s) on characteristics relevant to the analysis.
• Matched pairs design is a type of dependent groups design where separate groups are used, but with individuals in each group matched with someone in each of the other groups who has the same initial characteristics.
• The within-subjects F is the dependent groups equivalent of the one-way ANOVA. In this procedure, either participants in each group are paired on the relevant char- acteristics with participants in the other groups or one group is measured repeatedly after different levels of the independent variable are introduced.
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