同理ESSB=a∑+(b-1)2 ESST=E ∑∑( +阝;+8:-E b∑2+a∑+(ab-1)2 ESS8=(a-1)b-1)2( ) ( ) ( ) ( )( ) 2 e 2 j 2 j i 2 i i j 2 .. T i j ij 2 j 2 B j ESS a 1 b 1 b a ab 1 ESS E ESS a b 1  = − −  =  +  + −        =  +  +  −  =  + −     同理 