STAT100B_midtermS

# STAT100B_midtermS - n i =1 x i x i β true 0 ∑ n i =1 x 2...

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STAT 100B: Midterm Solutions Denise Tsai February 24, 2011 Problem 1 Let the output be X , then ¯ X = 0 . 52. Since X Unif [0 , 1] , f ( x ) = 1 1 - 0 = 1 If there is no problem with the random number generator, μ 0 = E ( X ) = Z 1 0 xf ( x ) dx = Z 1 0 xdx = x 2 2 ± ± ± ± 1 0 = 1 2 σ 2 = V ar ( X ) = E ( X 2 ) - [ E ( X )] 2 = Z 1 0 x 2 f ( x ) dx - ² 1 2 ³ 2 = 1 12 Hypothesis testing: Z = ¯ X - μ 0 σ n = 0 . 52 - 1 2 1 12 300 = 1 . 2 p-value = P ( | Z | > 1 . 2) = 2 × (1 - 0 . 8849) = 0 . 2302 Problem 2 P (Type I error) = P (reject H 0 | H 0 is true) = P p > 0 . 6 | p = 1 2 ) = P Z > 0 . 6 - 0 . 5 q 0 . 5(1 - 0 . 5) 100 = P ( Z > 2) = 0 . 0228 P (Type II error) = P (accept H 0 | H 1 is true) = P p 0 . 6 | p = 0 . 64) = P Z 0 . 6 - 0 . 64 q 0 . 64(1 - 0 . 64) 100 = P ( Z ≤ - 0 . 8333) = 0 . 2033 1

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Problem 3 Set the derivative of f with respect to β equal to zero: f 0 ( β ) = 2 n X i =1 ( Y i - βx i )( - x i ) = 0 Then, n X i =1 x i Y i = β n X i =1 ( x i ) 2 ˆ β = n i =1 x i Y i n i =1 x 2 i Expected Value: E ( ˆ β ) = E ±∑ n i =1 x i Y i n i =1 x 2 i ² = n i =1 x i E ( Y i ) n i =1 x 2 i = n i =1 x i E ( x i β true + ± i ) n i =1 x 2 i =
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Unformatted text preview: n i =1 x i ( x i β true + 0) ∑ n i =1 x 2 i (since E ( ± ) = 0) = β true ∑ n i =1 x 2 i ∑ n i =1 x 2 i = β true Variance: V ar ( ˆ β ) = V ar ±∑ n i =1 x i Y i ∑ n i =1 x 2 i ² = ∑ n i =1 x 2 i V ar ( Y i ) ( ∑ n i =1 x 2 i ) 2 = ∑ n i =1 x 2 i V ar ( x i β true + ± i ) ( ∑ n i =1 x 2 i ) 2 = ∑ n i =1 x 2 i (0 + σ 2 ) ( ∑ n i =1 x 2 i ) 2 (since V ar ( ± ) = σ 2 ) = σ 2 ∑ n i =1 x 2 i ( ∑ n i =1 x 2 i ) 2 = σ 2 ∑ n i =1 x 2 i 2...
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## This note was uploaded on 09/20/2011 for the course STAT 100B taught by Professor Wu during the Winter '11 term at UCLA.

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STAT100B_midtermS - n i =1 x i x i β true 0 ∑ n i =1 x 2...

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