Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

308 Some Elementary Statistical Inferences

cdfĤ.Letφ̂=m(̂θ)andφ̂∗=m(̂θ∗). We have

P

[

θ̂∗≤m−^1 (φ̂−z(1−α/2)σc)

]

= P

[

φ̂∗≤φ̂−z(1−α/2)σc

]

= P

[

φ̂∗−φ̂

σc

≤−z(1−α/2)

]

=α/ 2 ,

similar to (4.9.3). Therefore,m−^1 (φ̂−z(1−α/2)σc)istheα 2 100th percentile of the

cdfĤ. Likewise,m−^1 (φ̂−z(α/2)σc)isthe(1−α 2 )100th percentile of the cdfĤ.

Therefore, in the general case too, the percentiles of the distribution ofĤform the
confidence interval forθ.

4.9.2 BootstrapTestingProcedures..................

Bootstrap procedures can also be used effectively in testing hypotheses. We begin
by discussing these procedures for two-sample problems, which cover many of the
nuances of the use of the bootstrap in testing.
Consider a two-sample location problem; that is,X′ =(X 1 ,X 2 ,...,Xn 1 )is
a random sample from a distribution with cdfF(x)andY′ =(Y 1 ,Y 2 ,...,Yn 2 )
is a random sample from a distribution with the cdfF(x−Δ), where Δ ∈R.
The parameter Δ is the shift in locations between the two samples. Hence Δ can
be written as the difference in location parameters. In particular, assuming that
the meansμY andμXexist, we have Δ =μY−μX. We consider the one-sided
hypotheses given by
H 0 :Δ=0versusH 1 :Δ> 0. (4.9.7)
As our test statistic, we take the difference in sample means, i.e.,

V=Y−X. (4.9.8)

Our decision rule is to rejectH 0 ifV ≥c. As is often done in practice, we base
our decision on thep-value of the test. Recall if the samples result in the values
x 1 ,x 2 ,...,xn 1 andy 1 ,y 2 ,...,yn 2 with realized sample meansxandy, respectively,
then thep-value of the test is

p̂=PH 0 [V≥y−x]. (4.9.9)

Our goal is a bootstrap estimate of thep-value. But, unlike the last section,
the bootstraps here have to be performed whenH 0 is true. An easy way to do this
is to combine the samples into one large sample and then to resample at random
and with replacement the combined sample into two samples, one of sizen 1 (new
xs) and one of sizen 2 (newys). Hence the resampling is performed under one
distribution; i.e.,H 0 is true. LetBbe a positive integer and letv=y−x.Our
bootstrap algorithm is

1. Combine the samples into one sample:z′=(x′,y′).


2. Setj=1.