Applied Statistics and Probability for Engineers

9-5 TESTS ON A POPULATION PROPORTION 313

whereas if the alternative is H 1 : pp 0 ,

(9-36)

These equations can be solved to find the approximate sample size nthat gives a test of level

that has a specified risk. The sample size equations are

 a

p 0

 p z    1 p 011

 p 02

 n

1 p 11

 p 2

 n

b

for the two-sided alternative and

for a one-sided alternative.

EXAMPLE 9-11 Consider the semiconductor manufacturer from Example 9-10. Suppose that its process fall-

out is really p0.03. 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:

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 p0.03 (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 p0.05 and p0.03 is fairly

small, and the sample size n200 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 p0.03. 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:

where we have used p0.03 in Equation 9-38. Note that n832 is a very large sample size.

However, we are trying to detect a fairly small deviation from the null value p 0 0.05.

 832

nc

1.645 1 0.05 1 0.95 2 1.28 1 0.03 1 0.97 2

0.03 0.05

d

2

 1 

 c

0.05 0.03 

1 1.645 21 0.05 1 0.95 2

200

1 0.03 11

 0.03 2

200

d 1 

 1

 0.44 2 0.67

nc (9-37)

z   

21 p 011

 p 02 z 1 p 11

 p 2

p p 0 d

2

nc (9-38)

z   1 p 011

 p 02 z 1 p 11

 p 2

p p 0 d

2

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