sol8 - PSTAT 120C Solutions to Assignment 8 June 7 2006 1(a...

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Unformatted text preview: PSTAT 120C: Solutions to Assignment # 8 June 7, 2006 1. (a) In the case where all the r ij = r , ¯ y i ·· = 1 br b X j =1 r X m =1 y ijm and ¯ y ··· = 1 bkr b X j =1 k X i =1 r X m =1 y ijm The expected value of each observation is E ( y ijk ) = μ + τ i + β j , and so the expected value of the overall mean is E (¯ y ··· ) = 1 bkr b X j =1 k X i =1 R X m =1 ( μ + τ i + β j ) = 1 bkr b X j =1 k X i =1 r ( μ + τ i + β j ) = 1 bk b X j =1 k X i =1 ( μ + τ i + β j ) = μ + 1 k k X i =1 τ i + 1 b b X j =1 β j = μ because, by definition ∑ k i =1 τ i = ∑ b j =1 β j = 0. Similarly, E (¯ y i ·· ) = 1 br b X j =1 r X m =1 ( μ + τ i + β j ) = 1 b b X j =1 ( μ + τ i + β j ) = μ + τ i + 1 b b X j =1 β j = μ + τ i . Therefore, E (ˆ τ i ) = μ + τ i- μ = τ i , and this estimator is unbiased. 1 (b) When the r ij are different, we can write the estimator as ˆ τ i = 1 ∑ b j =1 r ij b X j =1 r i,j X m =1 y ijk- 1 ∑ k i =1 ∑ b j =1 r ij k X i =1 b X...
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This note was uploaded on 04/09/2008 for the course PSTAT 120c taught by Professor Idk during the Spring '07 term at UCSB.

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sol8 - PSTAT 120C Solutions to Assignment 8 June 7 2006 1(a...

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