Statistical Methods for Psychology

available, and I can first administer the test without saying anything about Asian students
typically performing better, and then I can readminister it in the next week’s lab with the
threatening instructions. (You might do well to consider how this study could be improved
to minimize carryover effects and other contaminants.) Let’s assume that we expect the
scores to go down in the threatening condition, but that because of the fact that the test was
previously given to these same people in the first week, the drop will be from 9.58 down to
only 7.55. Assume that the standard deviation will stay the same at 3.10. To solve for the
standard error of the difference between means we need the correlation between the two
sets of exam scores, but here we are in luck. Aronson’s math questions were taken from a
practice exam for the Graduate Record Exam, and the correlation we seek is estimated sim-
ply by the test-retest reliability of that exam. We have a pretty good idea that the reliability
of that exam will be somewhere around .92. Then

Notice that I have a smaller effect size than in my first lab exercise, because I tried to
be honest and estimate that the difference in means would be reduced because of the
experimental procedures. However, my power is far greater than it was in my original
example because of the added power of matched-sample designs.
Suppose, on the other hand, that we had used a less reliable test, for which 5 .40. We
will assume that sremains unchanged and that we are expecting a 2.03-unit difference
between the means. Then

We see that as drops, so does power. (It is still substantial in this example, but much
less than it was.) When 5 0, our two variables are not correlated and thus the matched-
sample case has been reduced to very nearly the independent-sample case. The important
point here is that for practical purposes the minimum power for the matched-sample case
occurs when 5 0 and we have independent samples. Thus, for all situations in which we
are even remotely likely to use matched samples (when we expect a positive correlation
between and ), the matched-sample design is more powerful than the corresponding
independent-groups design. This illustrates one of the main advantages of designs using
matched samples, and was my primary reason for taking you through these calculations.
Remember that we are using an approximation procedure to calculate power. Essen-
tially, we are assuming the sample sizes are sufficiently large that the t distribution is
closely approximated by z. If this is not the case, then we have to take account of the fact
that a matched-sample t has only one-half as many dfas the corresponding independent-
sample t, and the power of the two designs will not be quite equal when 5 0. This is not
usually a serious problem.

r

X 1 X 2

r

r

r

Power=.91

d=0.60 230 =3.29

d=

m 1 2m 2

sX 12 X 2

=

2.03

3.40

=0.60

sX 12 X 2 =3.10 2 2(1 2 .40)=3.10 2 2(.60)=3.10 2 1.2=3.40

r

Power=.99

d=d 2 n=1.64 230 =8.97

d=

m 1 2m 2

sX 12 X 2

=

9.58 2 7.55

1.24

=1.64

=1.24

sX 12 X 2 =s 2 2(12r)=3.10 2 2(1 2 .92)=3.1 2 2(.08)

Section 8.5 Power Calculations for Matched-Sample t 237