Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

8.1. Most Powerful Tests 477

is a best critical regionC. Consider the case ofk=1andn= 1. The preceding
inequality may be written 2x^1 /x 1 !≤e/2. This inequality is satisfied by all points
in the setC={x 1 :x 1 =0, 3 , 4 , 5 ,...}. Using R, the level of significance is

PH 0 (X 1 ∈C)=1−PH 0 (X 1 =1,2) = 1−dpois(1,1)−dpois(2,1) = 0. 4482.

The power of the test to detectH 1 is computed as

PH 1 (X 1 ∈C)=1−PH 1 (X 1 =1,2) = 1−(^14 +^18 )=0. 625.

Note that these results are consistent with Corollary 8.1.1.

Remark 8.1.2.In the notation of this section, sayCis a critical region such that

α=

∫

C

L(θ′)andβ=

∫

Cc

L(θ′′),

whereαandβequal the respective probabilities of the Type I and Type II errors
associated withC.Letd 1 andd 2 be two given positive constants. Consider a certain
linear function ofαandβ,namely,

d 1

∫

C

L(θ′)+d 2

∫

Cc

L(θ′′)=d 1

∫

C

L(θ′)+d 2

[

1 −

∫

C

L(θ′′)

]

= d 2 +

∫

C

[d 1 L(θ′)−d 2 L(θ′′)].

If we wished to minimize this expression, we would selectCto be the set of all
(x 1 ,x 2 ,...,xn) such that

d 1 L(θ′)−d 2 L(θ′′)< 0

or, equivalently,

L(θ′)

L(θ′′)

<

d 2

d 1

, for all (x 1 ,x 2 ,...,xn)∈C,

which according to the Neyman–Pearson theorem provides a best critical region
withk=d 2 /d 1. That is, this critical regionCis one that minimizesd 1 α+d 2 β.
There could be others, including points on whichL(θ′)/L(θ′′)=d 2 /d 1 , but these
would still be best critical regions according to the Neyman–Pearson theorem.