互效应模型 k +τ1+β1+(τB)1+Ei a,J b.k=1 Eik i.i.d N(o, o2 ∑1=0,∑B;=0 ∑(τ阝)n=0,j b ∑(τβ)=0,i=1…a
交互效应模型 ( ) ( ) ( ) ( ) = = = = = = = = = = + + + + j ij i ij i j i j 2 ijk ijk i j ij ijk 0, i 1, a 0, j 1, ,b 0, 0 i.i.d N 0, i 1, ,a, j 1, ,b,k 1, ,m y
方差分析 假设 H =T=0 H 02 2 =β=0 H:(B)2=0对一切i=1…aj=1…,b 偏差平方和的分解
方差分析 假设 偏差平方和的分解 H :( ) 0 i 1, ,a j 1, ,b H : 0 H : 0 03 ij 02 1 2 b 01 1 2 a = = = = = = = = = = = 对一切
S8=∑∑∑↓ k ∑∑∑-y)+∑∑∑-y) k k +∑∑∑-)+∑∑∑n-3-y+y) -sS+ss+ss+ss AB
( ) ( ) ( ) ( ) ( ) A B e AB 2 i j k ij. i.. .j. ... 2 i j k ijk ij. 2 i j k .j. ... 2 i j k i.. ... i j k T ijk ... SS SS SS SS y y y y y y y y y y SS y y = + + + + − + − − + = − + − = −
检验统计量 y.=μ+t+8i.y;=μ+阝+8y=+8 =+τ;+阝1+()+6 ESS =E∑∑∑+E-) bm∑2+a-1)2 ESs=am∑B+(b-1)2
( ) ( ) ( ) ( ) 2 j 2 B j 2 i 2 i i j k 2 i.. ... A i ij. ij. i j ij ... ... .j. .j. j i.. i.. i ESS am b 1 bm a 1 ESS E y y y y = + − = + − = + − = + + + + = + + = + + = + 检验统计量
ESS。=E ∑∑∑(-6n)=a(-12 ESST=E ∑∑∑(+B1+(rB)+Em-) bm∑t2+am∑B+m∑∑()+(abm-1)2 ESSAR=m∑∑(B)+(a-1)b-12
( ) ( ) ( ( ) ) ( ) ( ) ( ) ( )( ) 2 i j 2 AB ij 2 i j 2 ij j 2 j i 2 i i j k 2 ... T i j ij ijk 2 i j k 2 ij. e ijk ESS m a 1 b 1 bm am m abm 1 ESS E ESS E ab m 1 = + − − = + + + − = + + + − = − = −