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Unformatted text preview: Chapter 11 Analysis of Variance and Regression 11.1 a. The first order Taylors series approximation is Var[ g ( Y )] [ g ( )] 2 Var Y = [ g ( )] 2 v ( ) . b. If we choose g ( y ) = g * ( y ) = R y a 1 v ( x ) dx , then dg * ( ) d = d d Z a 1 p v ( x ) dx = 1 p v ( ) , by the Fundamental Theorem of Calculus. Then, for any , Var[ g * ( Y )] 1 p v ( ) ! 2 v ( ) = 1 . 11.2 a. v ( ) = , g * ( y ) = y , dg * ( ) d = 1 2 , Var g * ( Y ) dg * ( ) d 2 v ( ) = 1 / 4, independent of . b. To use the Taylors series approximation, we need to express everything in terms of = E Y = np . Then v ( ) = (1 /n ) and dg * ( ) d 2 = 1 q 1 n 1 2 q n 1 n 2 = 1 4 n (1 /n ) . Therefore Var[ g * ( Y )] dg * ( ) d 2 v ( ) = 1 4 n , independent of , that is, independent of p . c. v ( ) = K 2 , dg * ( ) d = 1 and Var[ g * ( Y )] ( 1 ) 2 K 2 = K , independent of . 11.3 a. g * ( y ) is clearly continuous with the possible exception of = 0. For that value use lH opitals rule to get lim y  1 = lim (log y ) y 1 = log y. b. From Exercise 11.1, we want to find v ( ) that satisfies y  1 = Z y a 1 p v ( x ) dx. Taking derivatives d dy y  1 = y  1 = d dy Z y a 1 p v ( x ) dx = 1 p v ( y ) . 112 Solutions Manual for Statistical Inference Thus v ( y ) = y 2(  1) . From Exercise 11.1, Var y  1 d dy  1 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 Y ij = + i + ij , i = 1 ,...,k, j = 1 ,...,n i , take k = 2. The two parameter configurations ( , 1 , 2 ) = (10 , 5 , 2) ( , 1 , 2 ) = (7 , 8 , 5) , have the same values for + 1 and + 2 , so they give the same distributions for Y 1 and Y 2 . 11.6 a. Under the ANOVA assumptions Y ij = i + ij , where ij independent n(0 , 2 ), so Y ij independent n( i , 2 ). Therefore the sample pdf is k Y i =1 n i Y j =1 (2 2 ) 1 / 2 e ( y ij i ) 2 2 2 = (2 2 ) n i / 2 exp  1 2 2 k X i =1 n i X j =1 ( y ij i ) 2 = (2 2 ) n i / 2 exp 1 2 2 k X i =1 n i 2 i exp  1 2 2 X i X j y 2 ij + 2 2 2 k X i =1 i n i Y i . Therefore, by the Factorization Theorem, Y 1 , Y 2 ,..., Y k , X i X j Y 2 ij is jointly sufficient for ( 1 ,..., k , 2 ) . Since ( Y 1 ,..., Y k ,S 2 p ) is a 1to1 function of this vector, ( Y 1 ,..., Y k ,S 2 p ) is also jointly sufficient....
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This note was uploaded on 04/18/2010 for the course STAT 622 taught by Professor Peruggia,m during the Spring '08 term at Ohio State.
 Spring '08
 Peruggia,M
 Variance

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