8.1. Most Powerful Tests 471x 0 1 2
f(x;1/2) 1 / 32 5 / 32 10 / 32
f(x;3/4) 1 / 1024 15 / 1024 90 / 1024
f(x;1/2)/f(x;3/4) 32 / 1 32 / 3 32 / 9
x 3 4 5
f(x;1/2) 10/32 5 / 32 1 / 32
f(x;3/4) 270 / 1024 405 / 1024 243 / 1024
f(x;1/2)/f(x;3/4) 32 / 27 32 / 81 32 / 243We shall use one random value ofXto test the simple hypothesisH 0 :θ=^12
against the alternative simple hypothesisH 1 :θ=^34 , and we shall first assign
the significance level of the test to beα= 321. We seek a best critical region of
sizeα= 321 .IfA 1 ={x:x=0}orA 2 ={x:x=5},thenP{θ=1/ 2 }(X ∈
A 1 )=P{θ=1/ 2 }(X∈A 2 )= 321 and there is no other subsetA 3 of the space{x:
x=0, 1 , 2 , 3 , 4 , 5 }such thatP{θ=1/ 2 }(X ∈A 3 )= 321. Then eitherA 1 orA 2 is
the best critical regionCof sizeα= 321 for testingH 0 againstH 1. Wenotethat
P{θ=1/ 2 }(X∈A 1 )= 321 andP{θ=3/ 4 }(X∈A 1 )= 10241 .Thus,ifthesetA 1 is used as
a critical region of sizeα= 321 , we have the intolerable situation that the probability
of rejectingH 0 whenH 1 is true (H 0 is false) is much less than the probability of
rejectingH 0 whenH 0 is true.
On the other hand, if the setA 2 is used as a critical region, thenP{θ=1/ 2 }(X∈
A 2 )= 321 andP{θ=3/ 4 }(X∈A 2 )= 1024243. That is, the probability of rejectingH 0
whenH 1 is true is much greater than the probability of rejectingH 0 whenH 0 is
true. Certainly, this is a more desirable state of affairs, and actuallyA 2 is the best
critical region of sizeα= 321. The latter statement follows from the fact that when
H 0 is true, there are but two subsets,A 1 andA 2 , of the sample space, each of whose
probability measure is 321 and the fact that
243
1024 =P{θ=3/^4 }(X∈A^2 )>P{θ=3/^4 }(X∈A^1 )=1
1024.It should be noted in this problem that the best critical regionC =A 2 of size
α= 321 is found by including inCthe point (or points) at whichf(x;^12 )issmallin
comparison withf(x;^34 ). This is seen to be true once it is observed that the ratio
f(x;^12 )/f(x;^34 )isaminimumatx= 5. Accordingly, the ratiof(x;^12 )/f(x;^34 ), that
is given in the last line of the above tabulation, provides us with a precise tool by
which to find a best critical regionCfor certain given values ofα. To illustrate this,
takeα= 326 .WhenH 0 is true, each of the subsets{x:x=0, 1 },{x:x=0, 4 },
{x:x=1, 5 },{x:x=4, 5 }has probability measure 326. By direct computation it
is found that the best critical region of this size is{x:x=4, 5 }. This reflects the
fact that the ratiof(x;^12 )/f(x;^34 ) has its two smallest values forx=4andx=5.
The power of this test, which hasα= 326 ,is
P{θ=3/ 4 }(X=4,5) = 1024405 + 1024243 = 1024648.The preceding example should make the following theorem, due to Neyman and
Pearson, easier to understand. It is an important theorem because it provides a
systematic method of determining a best critical region.