7Repeated Measures Designs for Interval DataLearnin.docx

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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tan81004_07_c07_163-192.indd 163 2/22/13 3:41 PM
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.
tan81004_07_c07_163-192.indd 164 2/22/13 3:41 PM
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?
tan81004_07_c07_163-192.indd 165 2/22/13 3:41 PM
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.
tan81004_07_c07_163-192.indd 166 2/22/13 3:41 PM
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:
tan81004_07_c07_163-192.indd 167 2/22/13 3:41 PM
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
1.
2.
3.
4.
5.

6.
7.
8.
9.
10.
1
0
3
0
2
1
3
2
1
2
3
2
4

0
3
1
5
4
3
1
–2
–2
–1
0
–1
0
–2
–2
–2
1
Before

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
1
0
3
0
2
1
3
2

1
2
3
2
4
0
3
1
5
4
3
1
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
tan81004_07_c07_163-192.indd 168 2/22/13 3:41 PM
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.
tan81004_07_c07_163-192.indd 169 2/22/13 3:41 PM
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
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.

1.5
4
3
0
2
4.5
6
0
5.25
2
3
0
2
0
3
4
2
1

2
3
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
tan81004_07_c07_163-192.indd 170 2/22/13 3:41 PM
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?
tan81004_07_c07_163-192.indd 171 2/22/13 3:41 PM
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
1.
2.
3.
4.
5.
6.
7.
8.
9.

10.
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
1
1
1
0
2.850
1.001
.317
3
2
2

2
2.5
3
2
4
5
3
4
3
3
2
3
4
1
5
6
4
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.
tan81004_07_c07_163-192.indd 172 2/22/13 3:41 PM
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.
tan81004_07_c07_163-192.indd 173 2/22/13 3:41 PM
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.
tan81004_07_c07_163-192.indd 174 2/22/13 3:41 PM
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.
tan81004_07_c07_163-192.indd 175 2/22/13 3:42 PM
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.
tan81004_07_c07_163-192.indd 176 2/22/13 3:42 PM
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.
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
tan81004_07_c07_163-192.indd 177 2/22/13 3:42 PM
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:

tan81004_07_c07_163-192.indd 178 2/22/13 3:42 PM
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
tan81004_07_c07_163-192.indd 179 2/22/13 3:42 PM
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.
tan81004_07_c07_163-192.indd 180 2/22/13 3:42 PM
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
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:
tan81004_07_c07_163-192.indd 181 2/22/13 3:42 PM
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
tan81004_07_c07_163-192.indd 182 2/22/13 3:42 PM
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
tan81004_07_c07_163-192.indd 183 2/22/13 3:42 PM
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.
tan81004_07_c07_163-192.indd 184 2/22/13 3:42 PM

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-
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.
tan81004_07_c07_163-192.indd 185 2/22/13 3:42 PM
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.
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.
tan81004_07_c07_163-192.indd 186 2/22/13 3:42 PM
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).
tan81004_07_c07_163-192.indd 187 2/22/13 3:42 PM

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.
tan81004_07_c07_163-192.indd 188 2/22/13 3:42 PM
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
tan81004_07_c07_163-192.indd 189 2/22/13 3:42 PM
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
tan81004_07_c07_163-192.indd 190 2/22/13 3:42 PM
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
tan81004_07_c07_163-192.indd 191 2/22/13 3:42 PM
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.
tan81004_07_c07_163-192.indd 192 2/22/13 3:42 PM

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