Statistical Methods for Psychology

slope quite near 1.0. Such a slope suggests that how much the girl weighed at the begin-

ning of therapy did not seriously influence how much weight she gained or lost by the end

of therapy. (We will discuss regression lines and slopes further in Chapter 9.)

The primary question we wish to ask is whether subjects gained weight as a function of the

therapy sessions. We have an experimental problem here, because it is possible that weight gain

resulted merely from the passage of time, and that therapy had nothing to do with it. However, I

know from other data in Everitt’s experiment that a group that did not receive therapy did not

gain weight over the same period of time, which strongly suggests that the simple passage of

time was not an important variable. If you were to calculate the weight of these girls before and

after therapy, the means would be 83.23 and 90.49 lbs, respectively, which translates to a gain

of a little over 7 pounds. However, we still need to test to see whether this difference is likely to

represent a true difference in population means, or a chance difference. By this I mean that we

need to test the null hypothesis that the mean in the populationof Before scores is equal to the

mean in the populationof After scores. In other words, we are testing H 0 : mA5 mB.

Difference Scores

Although it would seem obvious to view the data as representing two samples of scores, one

set obtained before the therapy program and one after, it is also possible, and very profitable,

to transform the data into one set of scores—the set of differences between X 1 and X 2 for each

subject. These differences are called difference scores, orgain scores,and are shown in the

third row of Table 7.1. They represent the degree of weight gain between one measurement

session and the next—presumably as a result of our intervention. If, in fact, the therapy pro-

gram had noeffect (i.e., if H 0 is true), the average weight would not change from session to

session. By chance some participants would happen to have a higher weight on X 2 than on

X 1 , and some would have a lower weight, but on the averagethere would be no difference.

If we now think of our data as being the set of difference scores, the null hypothesis

becomes the hypothesis that the mean of a population of difference scores (denoted mD)

equals 0. Because it can be shown that mD5 m 1 2 m 2 , we can write H 0 : mD5 m 1 2 m 25 0.

But now we can see that we are testing a hypothesis using onesample of data (the sample of

difference scores), and we already know how to do that.

The tStatistic

We are now at precisely the same place we were in the previous section when we had a

sample of data and a null hypothesis (m 5 0). The only difference is that in this case the

data are difference scores, and the mean and the standard deviation are based on the differ-

ences. Recall that twas defined as the difference between a sample mean and a population

mean, divided by the standard error of the mean. Then we have

where and and are the mean and the standard deviation of the difference scores and N

is the number of difference scores (i.e., the number of pairs,not the number of raw scores).

From Table 7.3 we see that the mean difference score was 7.26, and the standard deviation

of the differences was 7.16. For our data

t=

D 20

sD

=

D 20

sD

1 N

=

7.26 20

7.16

117

=

7.26

1.74

=4.18

D sD

t=

D 20

sD

=

D 20

sD

1 N

Section 7.4 Hypothesis Tests Applied to Means—Two Matched Samples 197

difference scores

gain scores