Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

4.9. Bootstrap Procedures 311

replacement; see Exercise 4.9.10. Usually, the permutation tests and the bootstrap
tests give very similar solutions; see Efron and Tibshirani (1993) for discussion.
As our second testing situation, consider a one-sample location problem. Sup-
poseX 1 ,X 2 ,...,Xnis a random sample from a continuous cdfF(x) with finite
meanμ. Suppose we want to test the hypotheses

H 0 : μ=μ 0 versusH 1 : μ>μ 0 ,

whereμ 0 is specified. As a test statistic we useXwith the decision rule

RejectH 0 in favor ofH 1 ifXis too large.

Letx 1 ,x 2 ,...,xnbe the realization of the random sample. We base our decision

on thep-value of the test, namely,

̂p=PH 0 [X≥x],

wherexis the realized value of the sample average when the sample is drawn. Our
bootstrap test is to obtain a bootstrap estimate of thisp-value. At first glance, one
might proceed by bootstrapping the statisticX. But note that thep-value must be
estimated underH 0. To assure thatH 0 is true, bootstrap the values:

zi=xi−x+μ 0 ,i=1, 2 ,...,n. (4.9.11)

Our bootstrap procedure is to randomly sample with replacement fromz 1 ,z 2 ,...,zn.
Let (zj,∗ 1 ,...,z∗j, 1 ) denote, say, thejthbootstrap sample. As in expression (4.9.4),
it follows thatE(zj,i∗)=μ 0. Hence, using thezis, the bootstrap resampling is
performed underH 0. Denote the test statistic by the sample meanz∗j. Then the
bootstrapp-value is

̂p∗=

#Bj=1{z∗j≥x}

B

. (4.9.12)

Example 4.9.3.To illustrate the bootstrap test just described, consider the fol-
lowing data set. We generatedn=20observationsXi=10Wi+ 100, whereWi
has a contaminated normal distribution with proportion of contamination 20% and
σc= 4. Suppose we are interested in testing

H 0 :μ=90versusH 1 : μ> 90.

Because the true mean ofXiis 100, the null hypothesis is false. The data generated

are

119.7 104.1 92.8 85.4 108.6 93.4 67.1 88.4 101.0 97.2

95.4 77.2 100.0 114.2 150.3 102.3 105.8 107.5 0.9 94.1

Thesamplemeanofthesevaluesisx=95.27, which exceeds 90, but is it significantly
over 90? As discussed above, we bootstrap the valueszi=xi− 95 .27 + 90. The
R functionboottestonemeanperforms this bootstrap test. For the run we did,
it computed the 3000 valuesz∗j, which are displayed in the histogram in Figure