Page 314 - Applied statistics and probability for engineers

292 Chapter 8/Statistical intervals for a single sample

This may be rearranged as
⎛ p(1 − p) p − p) ⎞
(1
⎜ ˆ − ≤ p ≤ ˆ + ⎟ ≃ 1 − ≠ (8-21)
α
α
P P z / 2 P z / 2
⎝ n n ⎠
The quantity p(1− p n in Equation 8-21 is called the standard error of the point estimator P.
ˆ
/
)
This was discussed in Chapter 7. Unfortunately, the upper and lower limits of the coni dence
interval obtained from Equation 8-21 contain the unknown parameter p. However, as sug-
ˆ
gested at the end of Section 8-1.5, a solution that is often sai sfactory is to replace p by P in
the standard error, which results in
⎛ ˆ − P) ˆ − P) ⎞
ˆ
ˆ
⎜ ˆ − P( 1 ≤ p ≤ ˆ + P( 1 ⎟ − ≠ (8-22)
α
α
⎜ n n ⎟
P P z / 2 P z / 2 ≃ 1
⎝ ⎠

This leads to the approximate 100(1 – α)% conidence interval on p.
Approximate
Coni dence ^
Interval on a If p is the proportion of observations in a random sample of size n that belongs to a

Binomial class of interest, an approximate 100(1 – α)% conidence interval on the proportion
Proportion p of the population that belongs to this class is
ˆ p − ˆ p) ˆ p − ˆ p)
(1
(1
−
+
ˆ p z /α 2 ≤ p ≤ ˆ p z /α 2 (8-23)
n n
where z /α 2 is the upper α / 2 percentage point of the standard normal distribution.
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 when this approximation is inappropriate, particularly in cases when n is small,
other methods must be used. Tables of the binomial distribution could be used to obtain a con-

idence interval for p. However, we could also use numerical methods that are implemented
on the binomial probability mass function in some computer program.

Example 8-8 Crankshaft Bearings In a random sample of 85 automobile engine crankshaft bearings, 10

have a surface inish that is rougher than the speciications allow. Therefore, a point estimate of

the proportion of bearings in the population that exceeds the roughness specii cation is ˆ p = x / n = 10 85 = .12 .
/
0
A 95% two-sided conidence interval for p is computed from Equation 8-23 as

ˆ p − ˆ p) ˆ p − ˆ p)
(1
(1
ˆ p − ≤ p ≤ ˆ p z+
. z 0 025
n . 0 025 n
or
0 12 0 88) 0 12 0 88)
. (
. (
.
.
0 12 − 1 96 ≤ p ≤ 0 12 1 96
+
.
.
.
.
85 85
which simplii es to
0 0509 ≤ p ≤ 0 2243
.
.
Practical Interpretation: This is a wide CI. Although the sample size does not appear to be small (n = 85), the value
of ˆ p is fairly small, which leads to a large standard error for ˆ p contributing to the wide CI.