Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

480 Optimal Tests of Hypotheses

As will be seen presently, uniformly most powerful tests do not always exist.
However, when they do exist, the Neyman–Pearson theorem provides a technique
for finding them. Some illustrative examples are given here.

Example 8.2.2. LetX 1 ,X 2 ,...,Xndenote a random sample from a distribution
that isN(0,θ), where the varianceθis an unknown positive number. It will be
shown that there exists a uniformly most powerful test with significance levelα
for testing the simple hypothesisH 0 :θ=θ′,whereθ′is a fixed positive number,
against the alternative composite hypothesisH 1 :θ>θ′.ThusΩ={θ:θ≥θ′}.
The joint pdf ofX 1 ,X 2 ,...,Xnis

L(θ;x 1 ,x 2 ,...,xn)=

(

1

2 πθ

)n/ 2

exp

{

−

1

2 θ

∑n

i=1

x^2 i

}

.

Letθ′′represent a number greater thanθ′,andletkdenote a positive number. Let

Cbe the set of points where

L(θ′;x 1 ,x 2 ,...,xn)

L(θ′′;x 1 ,x 2 ,...,xn)

≤k,

that is, the set of points where

(

θ′′

θ′

)n/ 2

exp

[

−

(

θ′′−θ′

2 θ′θ′′

)∑n

1

x^2 i

]

≤k

or, equivalently,

∑n

1

x^2 i≥

2 θ′θ′′

θ′′−θ′

[

n

2

log

(

θ′′

θ′

)

−logk

]

=c.

The setC={(x 1 ,x 2 ,...,xn):

∑n
1 x
2
i≥c}is then a best critical region for testing
thesimplehypothesisH 0 :θ=θ′against the simple hypothesisθ=θ′′.Itremains
to determinec, so that this critical region has the desired sizeα.IfH 0 is true, the
random variable

∑n

1 X

2

i/θ

′has a chi-square distribution withndegrees of freedom.

Sinceα=Pθ′(

∑n

1 X

2

i/θ

′ ≥c/θ′),c/θ′ may be computed, for example, by the

Rcodeqchisq(1−α, n). ThenC ={(x 1 ,x 2 ,...,xn):

∑n

1 x

2
i ≥c}is a best
critical region of sizeαfor testingH 0 :θ=θ′ against the hypothesisθ=θ′′.
Moreover, for each numberθ′′greater thanθ′, the foregoing argument holds. That
is,C={(x 1 ,...,xn):

∑n

1 x

2
i ≥c}is a uniformly most powerful critical region
of sizeαfor testingH 0 :θ=θ′againstH 1 :θ>θ′.Ifx 1 ,x 2 ,...,xndenote
the experimental values ofX 1 ,X 2 ,...,Xn,thenH 0 :θ=θ′ is rejected at the
significance levelα,andH 1 :θ>θ′is accepted if

∑n

1 x

2

i≥c;otherwise,H^0 :θ=θ

′

is accepted.
If, in the preceding discussion, we taken= 15,α=0.05, andθ′=3,then
the two hypotheses areH 0 :θ=3andH 1 :θ>3. Using R,c/3 is computed by
qchisq(0.95,15) = 24.996. Hence,c=74.988.