Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

476 Optimal Tests of Hypotheses

For example, ifn=25andαis 0.05,c 1 =qnorm(0.95,0,1/5) = 0.329,using
R. Hence, the power of the test to detectθ= 1, given in expression (8.1.7), is
computed by1 - pnorm(0.329,1,1/5) = 0.9996.
There is another aspect of this theorem that warrants special mention. It has to
do with the number of parameters that appear in the pdf. Our notation suggests
that there is but one parameter. However, a careful review of the proof reveals
that nowhere was this needed or assumed. The pdf or pmf may depend upon
any finite number of parameters. What is essential is that the hypothesisH 0 and
the alternative hypothesisH 1 be simple, namely, that they completely specify the
distributions. With this in mind, we see that the simple hypothesesH 0 andH 1 do
not need to be hypotheses about the parameters of a distribution, nor, as a matter
of fact, do the random variablesX 1 ,X 2 ,...,Xnneed to be independent. That is, if
H 0 is the simple hypothesis that the joint pdf or pmf isg(x 1 ,x 2 ,...,xn), and ifH 1
is the alternative simple hypothesis that the joint pdf or pmf ish(x 1 ,x 2 ,...,xn),
thenCis a best critical region of sizeαfor testingH 0 againstH 1 if, fork>0,

1.

g(x 1 ,x 2 ,...,xn)

h(x 1 ,x 2 ,...,xn)

≤kfor (x 1 ,x 2 ,...,xn)∈C.

2.

g(x 1 ,x 2 ,...,xn)

h(x 1 ,x 2 ,...,xn)

≥kfor (x 1 ,x 2 ,...,xn)∈Cc.

3.α=PH 0 [(X 1 ,X 2 ,...,Xn)∈C].

Consider the following example.

Example 8.1.3.LetX 1 ,...,Xndenote a random sample onXthat has pmff(x)
with support{ 0 , 1 , 2 ,...}. It is desired to test the simple hypothesis

H 0 :f(x)=

{ e− 1

x! x=0,^1 ,^2 ,...

0elsewhere,

against the alternative simple hypothesis

H 1 :f(x)=

{

(^12 )x+1 x=0, 1 , 2 ,...

0elsewhere.

That is, we want to test whetherXhas a Poisson distribution with meanλ=1
versusXhas a geometric distribution withp=1/2. Here

g(x 1 ,...,xn)

h(x 1 ,...,xn)

=

e−n/(x 1 !x 2 !···xn!)

(^12 )n(^12 )x^1 +x^2 +···+xn

=

(2e−^1 )n 2

Px

i

∏n

1

(xi!)

.

Ifk>0, the set of points (x 1 ,x 2 ,...,xn) such that

(n

∑

1

xi

)

log 2−log

[n

∏

1

(xi!)

]

≤logk−nlog(2e−^1 )=c