Statistics for the Behavioral Sciences by Frederick J. Gravetter, Larry B. Wallnau ISBN 10: 1305504917 ISBN 13: 9781305504912

256 CHAPTER 8 | Introduction to Hypothesis Testing
Original
Population
Normal with
s 5 80 and
m 5 10
FIGURE 8.10
A demonstration of
measuring power for
a hypothesis test. The
left-hand side shows the
distribution of sample
means that would occur if
the null hypothesis were
true. The critical region
is defined for this distribution.
The right-hand
side shows the distribution
of sample means
that would be obtained
if there were an 8-point
treatment effect. Notice
that if there is an 8-point
effect, then essentially
all of the sample means
would be in the critical
region. Thus, the probability
of rejecting the null
hypothesis (the power of
the test) is nearly 100%
for an 8-point treatment
effect.
Distribution of sample means
for n 5 25 if H 0 is true
Reject
H 0
If H 0 is true (no
treatment effect)
m 5 80 and
s 5 10
21.96 0
s M 5 2
Reject
H 0
76 78 80 82 84 86 88 90 92
11.96 z
With an 8-point
treatment effect
m 5 88 and
s 5 10
Distribution of sample means
for n 5 25 with 8-point effect
s M 5 2
test. Using α = .05, the critical region consists of extreme values in this distribution, specifically
sample means beyond z = 1.96 or z = –1.96. These values are shown in Figure 8.10
and, for both distributions, we have shaded all the sample means located in the critical region.
Now turn your attention to the distribution on the right, which shows all of the possible
sample means if there is an 8-point treatment effect. Notice that most of these sample
means are located beyond the z = 1.96 boundary. This means that, if there is an 8-point
treatment effect, you are almost guaranteed to obtain a sample mean in the critical region
and reject the null hypothesis. Thus, the power of the test (the probability of rejecting H 0
)
is close to 100% if there is an 8-point treatment effect.
To calculate the exact value for the power of the test we must determine what portion of
the distribution on the right-hand side is shaded. Thus, we must locate the exact boundary
for the critical region, then find the probability value in the unit normal table. For the distribution
on the left-hand side, the critical boundary of z = +1.96 corresponds to a location
that is above the mean by 1.96 standard deviations. This distance is equal to
1.96σ M
= 1.96(2) = 3.92 points
Thus, the critical boundary of z = +1.96 corresponds to a sample mean of M = 80 +
3.92 = 83.92. Any sample mean greater than M = 83.92 is in the critical region and would