BEST CRITICAL REGIONS In this paragraph, let consider the properties a satisfactory test should posses.

Consider the test of the sample null hypothesis H 0 :θ= θ 0 against the simple alternative hypothesis H 1 :θ= θ 1 . Let C be a critical region of size α ; that is, α=P( C; θ 0 ) . Then C is a best critical region of size α if, for every other critical region D of size α=P( D; θ 0 ) , we have that P( C; θ 1 )≥P( D; θ 1 ). That is, when H 1 :θ= θ 1 is true, the probability of rejecting H 0 :θ= θ 0 using the critical region C is at least as great as the corresponding probability using any other critical region D of size α . Thus a best critical region of size α is the critical region that has the greatest power among all critical regions for a best critical region of size α . The Neyman-Pearson lemma gives sufficient conditions for a best critical region of size α .

Neyman-Pearson Lemma Let X 1 , X 2 ,..., X n be a random sample of size n from a distribution with p.d.f. f( x;θ ) , where θ 0 and θ 1 are two possible values of θ . Denote the joint p.d.f. of X 1 , X 2 ,..., X n by the likelihood function L( θ )=L( θ; x 1 , x 2 ,..., x n )=f( x 1 ;θ )f( x 2 ;θ )···f( x n. ;θ ). If there exist a positive constant k and a subset C of the sample space such that P[ ( X 1 , X 2 ,..., X n )∈C; θ 0 ]=α, L( θ 0 ) L( θ 1 ) ≤k for ( x 1 , x 2 ,..., x n )∈C, L( θ 0 ) L( θ 1 ) ≥k for ( x 1 , x 2 ,..., x n )∈C'. Then C is a best critical region of size α for testing the simple null hypothesis H 0 :θ= θ 0 against the simple alternative hypothesis H 1 :θ= θ 1 .

For a realistic application of the Neyman-Pearson lemma, consider the following, in which the test is based on a random sample from a normal distribution. Let X 1 , X 2 ,..., X n be a random sample from a normal distribution N( μ,36 ) . We shall find the best critical region for testing the simple hypothesis H 0 :μ=50 against the simple alternative hypothesis H 1 :μ=55 . Using the ratio of the likelihood functions, namely L( 50 )/L( 55 ) , we shall find those points in the sample space for which this ratio is less than or equal to some constant k. That is, we shall solve the following inequality: L( 50 ) L( 55 ) = ( 72π ) −n/2 exp⁡[ −( 1 72 ) ∑ 1 n ( x i −50 ) 2 ] ( 72π ) −n/2 exp⁡[ −( 1 72 ) ∑ 1 n ( x i −55 ) 2 ] =exp⁡[ −( 1 72 )( 10 ∑ 1 n x i +n 50 2 −n 55 2 ) ]≤k. If we take the natural logarithm of each member of the inequality, we find that −10 ∑ 1 n x i −n 50 2 +n 55 2 ≤( 72 )ln⁡k. Thus, 1 n ∑ 1 n x i ≥− 1 10n [ n 50 2 −n 55 2 +( 72 )ln⁡k ] Or equivalently, x ¯ ≥c, where c=− 1 10n [ n 50 2 −n 55 2 +( 72 )ln⁡k ]. Thus L( 50 )/L( 55 )≤k is equivalent to x ¯ ≥c . A best critical region is, according to the Neyman-Pearson lemma, C={ ( x 1 , x 2 ,..., x n ): x ¯ ≥c }, where c is selected so that the size of the critical region is α . Say n=16 and c=53. Since X ¯ is N( 50,36/16 ) under H 0 we have α=P( X ¯ ≥53;μ=50 )=P[ X ¯ −50 6/4 ≥ 3 6/4 ;μ=50 ]=1−Φ( 2 )=0.0228. The example 1 illustrates what is often true, namely, that the inequality L( θ 0 ) L( θ 1 ) ≤k can be expressed in terms of a function u( x 1 , x 2 ,..., x n ) say, u( x 1 , x 2 ,..., x n )≤ c 1 or u( x 1 , x 2 ,..., x n )≥ c 2 , where c 1 and c 2 is selected so that the size of the critical region is α . Thus the test can be based on the statistic u( X 1 ,..., X n ) . Also, for illustration, if we want α to be a given value, say 0.05, we would then choose our c 1 and c 2 . In example1, with α =0.05, we want 0.05=P( X ¯ ≥c;μ=50 )=P( X ¯ −50 6/4 ≥ c−50 6/4 ;μ=50 )=1−Φ( c−50 6/4 ). Hence it must be true that ( c−50 )/( 3/2 )=1.645 , or equivalently, c=50+ 3 2 ( 1.645 )≈52.47.

Let X 1 , X 2 ,..., X n denote a random sample of size n from a Poisson distribution with mean λ . A best critical region for testing H 0 :λ=2 against H 1 :λ=5 is given by L( 2 ) L( 5 ) = 2 ∑ x i e −2n x 1 ! x 2 !··· x n ! x 1 ! x 2 !··· x n ! 5 ∑ x i e −5n ≤k. The inequality is equivalent to ( 2 5 ) ∑ x i e 3n ≤k and ( ∑ x i )ln⁡( 2 5 )+3n≤ln⁡k. Since ln⁡( 2/5 )<0 , this is the same as ∑ i=1 n x i ≥ ln⁡k−3n ln⁡( 2/5 ) =c. If n=4 and c=13, then α=P( ∑ i=1 4 X i ≥13;λ=2 )=1−0.936=0.064, from the tables, since ∑ i=1 4 X i has a Poisson distribution with mean 8 when λ =2. When H 0 :θ= θ 0 and H 1 :θ= θ 1 are both simple hypotheses, a critical region of size α is a best critical region if the probability of rejecting H 0 when H 1 is true is a maximum when compared with all other critical regions of size α . The test using the best critical region is called a most powerful test because it has the greatest value of the power function at θ= θ 1 when compared with that of other tests of significance level α . If H 1 is a composite hypothesis, the power of a test depends on each simple alternative in H 1 .

A test, defined by a critical region C of size α , is a uniformly most powerful test if it is a most powerful test against each simple alternative in H 1 . The critical region C is called a uniformly most powerful critical region of size α .

Let now consider the example when the alternative is composite. Let X 1 , X 2 ,..., X n be a random sample from N( μ,36 ) . We have seen that when testing H 0 :μ=50 against H 1 :μ=55 , a best critical region C is defined by C={ ( x 1 , x 2 ,..., x n ): x ¯ ≥c }, where c is selected so that the significance level is α . Now consider testing H 0 :μ=50 against the one-sided composite alternative hypothesis H 1 :μ>50 . For each simple hypothesis in H 1 , say μ= μ 1 the quotient of the likelihood functions is L( 50 ) L( μ 1 ) = ( 72π ) −n/2 exp⁡[ −( 1 72 ) ∑ 1 n ( x i −50 ) 2 ] ( 72π ) −n/2 exp⁡[ −( 1 72 ) ∑ 1 n ( x i − μ 1 ) 2 ] =exp⁡[ −( 1 72 ){ 2( μ 1 −50 ) ∑ 1 n x i +n( 50 2 − μ 1 2 ) } ]. Now L( 50 )/L( μ 1 )≤k if and only if x ¯ ≥ ( −72 )ln⁡( k ) 2n( μ 1 −50 ) + 50+ μ 1 2 =c. Thus the best critical region of size α for testing H 0 :μ=50 against H 1 :μ= μ 1 , where μ 1 >50 , is given by C={ ( x 1 , x 2 ,..., x n ): x ¯ ≥c }, where is selected such that P( X ¯ ≥c; H 0 :μ=50 )=α.