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294 Chapter 8/Statistical intervals for a single sample
Approximate
One-Sided Coni dence The approximate 100(1 – α)% lower and upper conidence bounds are
Bounds on a Binomial ˆ p) ˆ p)
Proportion ˆ p z− ˆ p(1 − ˆ p z+ ˆ p(1 −
α ≤ p and p ≤ α (8-26)
n n
respectively.
A Different Cofi dence interval on the Binomial Proportion
There is a different way to construct a CI on a binomial proportion than the traditional approach
in Equation 8-23. Starting with Equation 8-22 and replacing the inequalities with an equality
and solving the resulting quadratic equation for p results in
2 p − ˆ p) 2
( ˆ 1
p + z α /2 ± z /α 2 + z α /2 2
ˆ
p = 2 n n 4 n
1 + z α 2 /2 n
This implies that a two-sided CI on a proportion p is as follows:
2 p − ˆ p) 2
( ˆ 1
p + z α /2 + z /2 + z α /2
ˆ
α
UCL = 2 n n 4 n 2
1 + z α 2 /2 n
2 p − ˆ p) 2 (8-27)
( ˆ 1
p + z α /2 − z α/2 + z α /2
ˆ
n
LCL = 2n n 4 n 2
2
1 + z α /2 n
The article by Agresti and Coull in The American Statistician (“Approximate Better Than
‘Exact’ for Interval Estimation of a Binomial Proportion,” 1998, pp. 119–126) reports that the
actual conidence level for the CI in Equation 8-27 is closer to the “advertised” or nominal level
for almost all values of α and p than for the traditional CI in Equation 8-23. The authors also
report that this new interval can be used with nearly all sample sizes. So the requirements that
np ˆ ≥ 5 or 10 or n(1− ˆ p) • 5 or 10 are not too important. If the sample size is large, the quantity
ˆ
2
2
2
z α/2 /(2 n) will be small relative to p , z α/ /(4 a ) will be small relative to ˆ(p 1− p 2 2 n
ˆ) / n, and z α/ /
2
will be small, so as a result the Agresti-Coull CI in Equation 8-27 will reduce to the traditional
CI given in Equation 8-23.
Example 8-10 The Agresti-Coull CI on a Proportion Reconsider the crankshaft bearing data introduced in
Example 8-8. In that example we reported that ˆ p = 0 .12 and n = 85. The traditional 95% CI was
.
.
0 0509 ≤ p ≤ 0 2243
To construct the new Agresti-Coull CI, we use Equation 8-27:
2 p − ˆ p) 2 . 1 96 2 0 12 0 88) 1 96 2
.
.
.
(
( ˆ 1
p + z α /2 + z /α 2 + z α /2 . 0 12 + + 1 96 + 2
.
ˆ
(
UCL = 2 n n 4 n 2 = ( 2 885) 85 4 85 ) = 0 2024
.
+
2
.
1 + z α 2 /2 n 1 ( 1 96 / 85)
2 ( p ˆ 1 − ) p ˆ 2 . 1 96 2 0 . ( . ) . 1 96 2
0
8
8
2
1
ˆ p + z α /2 − z α/2 + z α /2 2 . 0 12 + −1.96 + 2
+
.
( )
LCL = 2 n n 4 n = 2 85 85 ( 4 85 ) = . 0 0654
2
1
1 + z α /2 n 1+ ( .96 2 /85 )
The two CIs would agree more closely if the sample size were larger.