Page 325 - Applied Statistics And Probability For Engineers

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8-2.6 Bootstrap Confidence Intervals (CD Only)

In Section 7-2.5 we showed how a technique called the bootstrap could be used to estimate
ˆ

the standard error ˆ , where is an estimate of a parameter . We can also use the bootstrap
to find a confidence interval on the parameter . To illustrate, consider the case where is the
mean of a normal distribution with known. Now the estimator of is X. Also notice that
z 2 1n is the 100(1 /2) percentile of the distribution of X , and z 2 1n is
the 100( 2) percentile of this distribution. Therefore, we can write the probability statement
associated with the 100(1 )% confidence interval as

P11001 22 percentile
X
10011 22 percentile2 1

or

P1X 10011 22 percentile

X 1001 22 percentile2 1

This last probability statement implies that the lower and upper 100(1 )% confidence lim-
its for are

L X 100 11 22 percentile of X X z 2 1n
U X 100 1 22 percentile of X X z 2 1n

We may generalize this to an arbitrary parameter . The 100(1 )% confidence limits
for are

ˆ
ˆ
L 10011 22 percentile of
ˆ
ˆ
U 1001 22 percentile of
ˆ
Unfortunately, the percentiles of may not be as easy to find as in the case of the normal
distribution mean. However, they could be estimated from bootstrap samples. Suppose we
*
*
ˆ
ˆ * ˆ * ˆ * and * and then calculate ,
find B bootstrap samples and calculate , 21 , p , B 1
*
*
ˆ
ˆ
* , p , * . The required percentiles can be obtained directly from the differences.
2
B
For example, if B 200 and a 95% confidence interval on is desired, the fifth smallest and
*
ˆ
fifth largest of the differences * are the estimates of the necessary percentiles.
i
We will illustrate this procedure using the situation first described in Example 7-3,
involving the parameter of an exponential distribution. Following that example, a random
sample of n 8 engine controller modules were tested to failure, and the estimate of
obtained was 0.0462, where 1 X ˆ ˆ is a maximum likelihood estimator. We used 200
bootstrap samples to obtain an estimate of the standard error for . ˆ
*
ˆ

Figure S8-1(a) is a histogram of the 200 bootstrap estimates , i 1, 2, p , 200. Notice
i
that the histogram is not symmetrical and is skewed to the right, indicating that the sam-
ˆ

pling distribution of also has this same shape. We subtracted the sample average of these
*
bootstrap estimates * 0.5013 from each ˆ * i . The histogram of the differences * , i
ˆ
i
1, 2, p , 200, is shown in Figure S8-1(b). Suppose we wish to find a 90% confidence inter-
*
ˆ
val for . Now the fifth percentile of the bootstrap samples * is 0.0228 and the ninety-
i
fifth percentile is 0.03135. Therefore the lower and upper 90% bootstrap confidence limits are
*
ˆ
*
ˆ
L 95 percentile of 0.0462 0.03135 0.0149
i
*
ˆ
ˆ
U 5 percentile of * 0.0462 1 0.02282 0.0690
i
8-1