100BHWS6

100BHWS6 - STAT HW6 Solution March 18, 2008 Problem 1 X1 ,...

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STAT HW6 Solution March 18, 2008 Problem 1 X 1 , ··· ,X n N ( μ 1 2 ) ,Y 1 , ··· ,Y n N ( μ 2 2 ) 1. H 0 : μ 1 = μ 2 ,H 1 : μ 1 6 = μ 2 Λ = L μ 1 , ˆ μ 2 , ˆ σ 2 ) /L μ, ˆ σ 2 0 ) = (ˆ σ 2 0 / ˆ σ 2 ) n + m/ 2 2log Λ = ( n + m )(log ˆ σ 2 0 - log ˆ σ 2 ) where ˆ μ 1 = ¯ X, ˆ μ 2 = ¯ Y , ˆ σ 2 = ( x i - ¯ X ) 2 + ( y j - ¯ Y ) 2 n + m ˆ μ = n ¯ X + m ¯ Y n + m , ˆ σ 2 0 = ( x i - ˆ μ ) 2 + ( y j - ˆ μ ) 2 n + m 2. 2log Λ = 0 . 4336 ,p - value = Pr ( χ 2 1 > 0 . 4336) = 0 . 5102 We have no reason to reject the H 0 . Problem 2 Y i N ( a + bx i 2 ) Λ = L a, ˆ b, ˆ σ 2 ) /L a 0 , ˆ σ 2 0 ) = (ˆ σ 2 0 / ˆ σ 2 ) n + m/ 2 2log Λ = ( n + m )(log ˆ σ 2 0 - log ˆ σ 2 ) where ˆ a = ¯ Y - ˆ b ¯ X, ˆ b = ( x i - ¯ X )( y i - ¯ Y ) ( x i - ¯ X ) 2 , ˆ σ 2 = ( y i - ˆ a - ˆ bx i ) 2 /n ˆ a 0 = ¯ Y , ˆ σ 2 0 = ( y i - ¯ Y ) 2 /n Problem 3 Roll a die 1. L ( p 1 ,p 2 , ··· ,p 5 ) = p n 1 1 p n 2 2 p n 3 3 p n 4 4 p n 5 5 (1 - p 1 - p 2 - p 3 - p 4 - p 5 )) n 6 Λ = L p 1 , ˆ p 2 , ··· , ˆ p 5 ) /L (1 / 6 , 1 / 6 , ··· , 1 / 6) = (6ˆ p 1 ) n 1 (6ˆ p 2 ) n 2 (6ˆ p 3 ) n 3 (6ˆ p 4
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This note was uploaded on 03/30/2011 for the course STAT 100B taught by Professor Wu during the Winter '11 term at UCLA.

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