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266 CHAPTER 8 STATISTICAL INTERVALS FOR A SINGLE SAMPLE

Equation 8-23 contain the unknown parameter p. However, as suggested at the end of Section
ˆ
P
8-2.5, a satisfactory solution is to replace p by in the standard error, which results in
ˆ
ˆ
ˆ
ˆ
P11 P2 P11 P2
ˆ
q ˆ
P P z 2 B n p P z 2 B n r 1 (8-24)
˛
This leads to the approximate 100(1 )% confidence interval on p.
Definition
If is the proportion of observations in a random sample of size n that belongs to a
p ˆ
class of interest, an approximate 100(1 )% confidence interval on the proportion
p of the population that belongs to this class is
p ˆ 11 p ˆ 2 p ˆ 11 p ˆ 2
p ˆ z 2 B n p p ˆ z 2 ˛ B n (8-25)

is the upper 2 percentage point of the standard normal distribution.
where z 2

This procedure depends on the adequacy of the normal approximation to the binomial. To
be reasonably conservative, this requires that np and n(1 p) be greater than or equal to 5. In
situations where this approximation is inappropriate, particularly in cases where n is small,
other methods must be used. Tables of the binomial distribution could be used to obtain a con-
fidence interval for p. However, we could also use numerical methods based on the binomial
probability mass function that are implemented in computer programs.

EXAMPLE 8-6 In a random sample of 85 automobile engine crankshaft bearings, 10 have a surface finish that
is rougher than the specifications allow. Therefore, a point estimate of the proportion of bear-
ings in the population that exceeds the roughness specification is p ˆ x n 10 85 0.12.
A 95% two-sided confidence interval for p is computed from Equation 8-25 as

p ˆ 11 p ˆ 2 p ˆ 11 p ˆ 2
p ˆ z p p ˆ z ˛
0.025 B n 0.025 B n
or

0.1210.882 0.1210.882
B 85 B 85
0.12 1.96 p 0.12 1.96
which simplifies to

0.05 p 0.19
Choice of Sample Size
Since P ˆ is the point estimator of p, we can define the error in estimating p by P ˆ as
ˆ
E 0 p P0. Note that we are approximately 100(1 )% confident that this error is less
˛1p11 p2 n. For instance, in Example 8-6, we are 95% confident that the sample
than z 2
proportion p ˆ 0.12 differs from the true proportion p by an amount not exceeding 0.07.
In situations where the sample size can be selected, we may choose n to be 100 (1 )%
confident that the error is less than some specified value E. If we set E z 2 ˛1p11 p2 n
and solve for n, the appropriate sample size is