Dr. Hackney STA Solutions pg 174

# Dr. Hackney STA Solutions pg 174 - 11-2 Solutions Manual...

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Unformatted text preview: 11-2 Solutions Manual for Statistical Inference Thus v(y) = y -2(-1) . From Exercise 11.1, Var y -1 d -1 dy 2 v() = 2(-1) -2(-1) = 1. Note: If = 1/2, v() = , which agrees with Exercise 11.2(a). If = 1 then v() = 2 , which agrees with Exercise 11.2(c). 11.5 For the model Yij = + i + ij , i = 1, . . . , k, j = 1, . . . , ni , take k = 2. The two parameter configurations (, 1 , 2 ) (, 1 , 2 ) = (10, 5, 2) = (7, 8, 5), have the same values for + 1 and + 2 , so they give the same distributions for Y1 and Y2 . 11.6 a. Under the ANOVA assumptions Yij = i + ij , where ij independent n(0, 2 ), so Yij independent n(i , 2 ). Therefore the sample pdf is k ni 1 k ni (y ij -i )2 = (2 2 )-ni /2 exp - 2 (y ij - i )2 (2 2 )-1/2 e- 22 2 i=1 j=1 i=1 j=1 = (2 2 )-ni /2 exp - 1 exp - 2 2 Therefore, by the Factorization Theorem, Y1 , Y2 , . . . , Yk , i j 1 2 2 k 2 ni i i=1 k i j 2 2 yij + 2 2 i ni Yi i=1 . 2 Yij 2 is jointly sufficient for 1 , . . . , k , 2 . Since (Y1 , . . . , Yk , Sp ) is a 1-to-1 function of this 2 1 , . . . , Yk , Sp ) is also jointly sufficient. vector, (Y b. We can write 1 k ni (2 2 )-ni /2 exp - 2 (y ij - i )2 2 i=1 j=1 1 k ni = (2 2 )-ni /2 exp - 2 ([y ij - yi ] + [i - i ])2 y 2 i=1 j=1 k 1 k ni 1 [y ij - yi ]2 exp - 2 = (2 2 )-ni /2 exp - 2 ni [i - i ]2 , y 2 2 i=1 i=1 j=1 so, by the Factorization Theorem, Yi , i = 1, . . . , n, is independent of Yij - Yi , j = 1, . . . , ni , 2 i . so Sp is independent of each Y c. Just identify ni Yi with Xi and redefine i as ni i . ...
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