ADDITIONAL PP 11.1
Refer to the experiment described in ADDITIONAL PP 10.2. For convenience, the data are
shown again below. We are using  = 0.05 for rejecting the null hypothesis.
Subject
1
2
3
4
5
6
7
8
9
10
11
12
13
14
a.
b.
Experimental
Condition
18
12
8
10
12
5
15
16
10
13
17
15
9
16
Control
Condition
13
14
6
8
15
7
11
14
9
10
16
12
14
13
What is the power of the experiment to detect a moderate effect (Preal = 0.70 or 0.30).
What is the probability of a Type II error?
SOLUTION is on next page.
SOLUTION
a.
Calculation of power involves a two-step process.
1: Assume the null hypothesis is true (Pnull = 0.50) and determine the possible
sample outcomes in the experiment that will allow H0 to be rejected.  = 0.052 tail. With
N = 14, and P = 0.50, from Table B,
STEP
p(0 pluses)
p(1 pluses)
p(2 pluses)
p(12 pluses)
p(13 pluses)
p(14 pluses)
p(0, 1, 2, 12, 13, or14 pluses)
= 0.0001
= 0.0009
= 0.0056
= 0.0056
= 0.0009
= 0.0001
= 0.0132
p(0 pluses)
= 0.0001
p(1pluses)
= 0.0009
p(2 pluses)
= 0.0056
p(3 pluses)
0.0222
p(11 pluses)
0.0222
p(12 pluses)
= 0.0056
p(13 pluses)
= 0.0009
p(14 pluses)
= 0.0001
p(0, 1, 2, 3, 11, 12, 13, or14 pluses) = 0.0576
We can reject H0 if the obtained result is 2 or 12 pluses, but not if it is 3 or 11 pluses.
Therefore, the outcomes that will allow rejection of H0 are 0, 1, 2, 12, 13, or 14 pluses.
STEP 2: For Preal = 0.30, determine the probability of getting any of the above sample
outcomes. This probability is the power of the experiment to detect this
hypothesized real effect. With N = 14 and Preal = 0.30, from Table B,
Power = probability of rejecting H0 if IV has a real effect.
= p(0, 1, 2, 12, 13, or14 pluses) as sample outcomes if Preal = 0.30
= 0.0068 + 0.0407 + 0.1134 +0.0000 +0.0000 +0.0000
= 0.1609
Note that the same answer would result for Preal = 0.70.
b.
Calculation of beta:
 = 1 – Power
= 1 – 0.1609
= 0.8391