Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

8.2. Uniformly Most Powerful Tests 481

Example 8.2.3.LetX 1 ,X 2 ,...,Xndenote a random sample from a distribution
that isN(θ,1), whereθis unknown. It will be shown that there is no uniformly most
powerful test of the simple hypothesisH 0 :θ=θ′,whereθ′is a fixed number against
the alternative composite hypothesisH 1 :θ =θ′.ThusΩ={θ:−∞<θ<∞}.
Letθ′′be a number not equal toθ′.Letkbe a positive number and consider

(1/ 2 π)n/^2 exp

[

−

∑n

1

(xi−θ′)^2 / 2

]

(1/ 2 π)n/^2 exp

[

−

∑n

1

(xi−θ′′)^2 / 2

]≤k.

The preceding inequality may be written as

exp

{

−(θ′′−θ′)

∑n

1

xi+

n

2

[(θ′′)^2 −(θ′)^2 ]

}

≤k

or

(θ′′−θ′)

∑n

1

xi≥

n

2

[(θ′′)^2 −(θ′)^2 ]−logk.

This last inequality is equivalent to

∑n

1

xi≥

n

2

(θ′′+θ′)−

logk

θ′′−θ′

,

provided thatθ′′>θ′, and it is equivalent to

∑n

1

xi≤

n

2

(θ′′+θ′)−

logk

θ′′−θ′

ifθ′′<θ′. The first of these two expressions defines a best critical region for testing
H 0 :θ=θ′against the hypothesisθ=θ′′provided thatθ′′>θ′, while the second
expression defines a best critical region for testingH 0 :θ=θ′against the hypothesis
θ=θ′′provided thatθ′′<θ′. That is, a best critical region for testing the simple
hypothesis against an alternative simple hypothesis, sayθ=θ′+1, does not serve as
a best critical region for testingH 0 :θ=θ′against the alternative simple hypothesis
θ=θ′−1. By definition, then, there is no uniformly most powerful test in the case
under consideration.
It should be noted that had the alternative composite hypothesis been one-sided,
eitherH 1 :θ>θ′orH 1 :θ<θ′, a uniformly most powerful test would exist in each
instance.

Example 8.2.4.In Exercise 8.1.10, the reader was asked to show that if a random

sample of sizen= 10 is taken from a Poisson distribution with meanθ, the critical

region defined by

∑n

1 xi≥3 is a best critical region for testingH^0 :θ=0.1 against