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9-5 TESTS ON A POPULATION PROPORTION 313

whereas if the alternative is H : p p ,
0
1
p
p z 1p 0 11
p 0 2
n
0
a b (9-36)
1p11
p2
n
These equations can be solved to find the approximate sample size n that gives a test of level
that has a specified risk. The sample size equations are

z
2 1p 11
p 2 z 1p11
p2 2
0

0
n c d (9-37)
p
p 0

for the two-sided alternative and

z 1p 11
p 2 z 1p11
p2 2

0
0
n c d (9-38)
p
p 0
for a one-sided alternative.

EXAMPLE 9-11 Consider the semiconductor manufacturer from Example 9-10. Suppose that its process fall-
out is really p 0.03. What is the -error for a test of process capability that uses n 200
and 0.05?
The -error can be computed using Equation 9-35 as follows:

0.05
0.03
11.6452 10.0510.952
200
1
c d 1
1
0.442 0.67
10.0311
0.032
200

Thus, the probability is about 0.7 that the semiconductor manufacturer will fail to con-
clude that the process is capable if the true process fraction defective is p 0.03 (3%). That
is, the power of the test against this particular alternative is only about 0.3. This appears to be
a large -error (or small power), but the difference between p 0.05 and p 0.03 is fairly
small, and the sample size n 200 is not particularly large.
Suppose that the semiconductor manufacturer was willing to accept a -error as large as
0.10 if the true value of the process fraction defective was p 0.03. If the manufacturer con-
tinues to use 0.05, what sample size would be required?
The required sample size can be computed from Equation 9-38 as follows:

1.64510.0510.952 1.2810.0310.972 2
n c d
0.03
0.05
832

where we have used p 0.03 in Equation 9-38. Note that n 832 is a very large sample size.
However, we are trying to detect a fairly small deviation from the null value p 0.05.
0