482 Optimal Tests of HypothesesH 1 :θ=0.5. This critical region is also a uniformly most powerful one for testing
H 0 :θ=0.1 againstH 1 :θ> 0 .1 because, withθ′′> 0 .1,
(0.1)Px
ie−10(0.1)/(x 1 !x 2 !···xn!)
(θ′′)Px
ie−10(θ′′)/(x 1 !x 2 !···xn!) ≤kis equivalent to
(
0. 1
θ′′)Pxi
e−10(0.^1 −θ′′)
≤k.The preceding inequality may be written as
(n
∑1xi)
(log 0. 1 −logθ′′)≤logk+ 10(1−θ′′)or, sinceθ′′> 0 .1, equivalently as∑n1xi≥logk+10− 10 θ′′
log 0. 1 −logθ′′.Of course,
∑n
1 xi≥3 is of the latter form.
Let us make an important observation, although obvious when pointed out. Let
X 1 ,X 2 ,...,Xndenote a random sample from a distribution that has pdff(x;θ),θ∈
Ω. Suppose thatY=u(X 1 ,X 2 ,...,Xn) is a sufficient statistic forθ. In accordance
with the factorization theorem, the joint pdf ofX 1 ,X 2 ,...,Xnmay be written
L(θ;x 1 ,x 2 ,...,xn)=k 1 [u(x 1 ,x 2 ,...,xn);θ]k 2 (x 1 ,x 2 ,...,xn),wherek 2 (x 1 ,x 2 ,...,xn) does not depend uponθ. Consequently, the ratio
L(θ′;x 1 ,x 2 ,...,xn)
L(θ′′;x 1 ,x 2 ,...,xn)=k 1 [u(x 1 ,x 2 ,...,xn);θ′]
k 1 [u(x 1 ,x 2 ,...,xn);θ′′]depends uponx 1 ,x 2 ,...,xnonly throughu(x 1 ,x 2 ,...,xn). Accordingly, if there
is a sufficient statisticY=u(X 1 ,X 2 ,...,Xn)forθand if a best test or a uniformly
most powerful test is desired, there is no need to consider tests that are based upon
any statistic other than the sufficient statistic. This result supports the importance
of sufficiency.
In the above examples, we have presented uniformly most powerful tests. For
some families of pdfs and hypotheses, we can obtain general forms of such tests.
We sketch these results for the general one-sided hypotheses of the form
H 0 :θ≤θ′versusH 1 : θ>θ′. (8.2.1)The other one-sided hypotheses with the null hypothesisH 0 :θ≥θ′, is completely
analogous. Note that the null hypothesis of (8.2.1) is a composite hypothesis. Recall
from Chapter 4 that the level of a test for the hypotheses (8.2.1) is defined by