1( = @ 1( > 2( > 3( 4( > 2( 45 % @ @ > > > @ > > = > 1( 180 2( 162 3 ...

108Z = 2/ 56 (2Z = −256 (4( P(Z < 2 / 33)= 0/99. H 0 H 0 (2. H 1 H 1 (4H 0 H 1H 0 H 0 N( 0,δ2) . Z = 256 (1Z = −2/ 56 , (7(2(4(3. H 0 H 0 (1. ) 2H 1 H 1 (3: (8H 0 H 1H 0 H 0 (1(3X1,X2,...,X n y = n 2∑ Xii=0δ 2 / n (4nδ (3δ 2 (2 (1 n t T (10 T 2 X 2 (n) (4 X2(n −1)(3 F(1,n) (2 F(n,1)(111 θ1 ≠ θ 2 θ 1 = θ 2 (11 . θ1 ≠ θ 2 θ 1 = θ 2. θ1 ≠ θ 2 θ 1 = θ 2. θ 1 = θ 2 (1 (2 (3. θ 1 = θ 2 (4 H 0 : P = 0/5 X ~ bin( 4,P) (12 . { x x = 0,1 } H : P = 0 31 /(9