Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

4.9. Bootstrap Procedures 313

4.9.3.LetX 1 ,X 2 ,...,Xnbe a random sample from a Γ(1,β) distribution.

(a)Show that the confidence interval (2nX/(χ^22 n)(1−(α/2)), 2 nX/(χ^22 n)(α/2))isan

exact (1−α)100% confidence interval forβ.

(b)Using part (a), show that the 90% confidence interval for the data of Example

4.9.1 is (64. 99 , 136 .69).

4.9.4. Consider the situation discussed in Example 4.9.1. Suppose we want to
estimate the median ofXiusing the sample median.

(a)Determine the median for a Γ(1,β) distribution.

(b)The algorithm for the bootstrap percentile confidence intervals is general

and hence can be used for the median. Rewrite the R code in the func-

tionpercentciboot.sso that the median is the estimator. Using the sample

given in the example, obtain a 90% bootstrap percentile confidence interval

for the median. Did it trap the true median in this case?

4.9.5.SupposeX 1 ,X 2 ,...,Xnis a random sample drawn from aN(μ, σ^2 ) distri-

bution. As discussed in Example 4.2.1, the pivot random variable for a confidence

interval is

t=

X−μ

S/

√

n

, (4.9.13)

whereXandSare the sample mean and standard deviation, respectively. Recall
by Theorem 3.6.1 thatthas a Studentt-distribution withn−1 degrees of freedom;
hence, its distribution is free of all parameters for this normal situation. In the
notation of this section,t(nγ−) 1 denotes theγ100% percentile of at-distribution with
n−1 degrees of freedom. Using this notation, show that a (1−α)100% confidence
interval forμis
(
x−t(1−α/2)

s

√

n

,x−t(α/2)

s

√

n

)

. (4.9.14)

4.9.6.Frequently, the bootstrap percentile confidence interval can be improved if
the estimator̂θis standardized by an estimate of scale. To illustrate this, consider a
bootstrap for a confidence interval for the mean. Letx∗ 1 ,x∗ 2 ,...,x∗nbe a bootstrap
sample drawn from the samplex 1 ,x 2 ,...,xn. Consider the bootstrap pivot [analog
of (4.9.13)]:

t∗=

x∗−x

s∗/

√

n

, (4.9.15)

wherex∗=n−^1

∑n

i=1x

∗

iand

s∗^2 =(n−1)−^1

∑n

i=1

(x∗i−x∗)^2.