midtermsolution

midtermsolution - Statistics 105 Midterm Exam Solution...

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Statistics 105 Midterm Exam Solution (Winter 2008) 1. (a) E ( X ) = 2 θ = ¯ X , so ˆ θ = ¯ X/ 2. (b) Yes, because E ( ˆ θ ) = E ( ¯ X/ 2) = E ( ¯ X ) / 2 = (2 θ ) / 2 = θ . 2. (a) ¯ X has N ( μ, σ 2 /n ), i.e., N (24 , 1). (b) P ( ¯ X > 25) = P ( Z > (25 - 24) / 1) = P ( Z > 1 . 0) = 1 - . 8413 = . 1587. (c) z 0 . 05 = 1 . 645 so the 95 percentile is 24 + (1 . 645)(1) = 25 . 645. 3. (a) ˆ p = 140 / 500 = 0 . 28, se p ) = ± ( . 28)( . 72) / 500 = . 020. (b) ˆ p ± z α/ 2 se p ) = . 280 ± (2 . 576)(0 . 020) = . 280 ± . 0515 = ( . 2285 , . 3315). (c) E = . 04 so n = ( z 0 . 005 E ) 2 (0 . 5) 2 = ² 2 . 576 0 . 04 ³ 2 (0 . 5) 2 = 1036 . 84. Take n = 1037. 4. The likelihood function is L ( λ ) = n ´ i =1 f ( x i , λ ) = n ´ i =1 λx i e - λx 2 i / 2 = λ n ( n ´ i =1 x i ) e - λ/ 2 n i =1 x 2 i ln L ( λ ) = n ln λ + ln( n ´ i =1 x i ) - λ 2 ln( n µ i =1 x 2 i ) d ln L ( λ ) = n λ - 1 2 ln( n µ i =1 x 2 i ) = 0 So the MLE is ˆ λ = 2 n n i =1 X 2 i 5. (a) True (b) False (c) False (d) False. If the process of selecting a random sample of size 50 and then computing the cor-
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This note was uploaded on 03/18/2008 for the course STATS 105 taught by Professor Hongquanxu during the Winter '08 term at UCLA.

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