midterm11Wsol

midterm11Wsol - Statistics 105 Midterm Exam Solution...

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Statistics 105 Midterm Exam Solution (Winter 2011) 1. (a) ˆ p = . 008, se= ± ( . 008)( . 992) / 1000 = . 0028 (b) . 008 ± (1 . 96)( . 0028) = ( . 0025 , . 0135). (c) n = ( 1 . 96 . 003 ) 2 ( . 008)( . 992) = 3387 . 438. Take n = 3388. 2. (a) ¯ x ± t α/ 2 ,n - 1 s/ n = 102 . 4 ± (2 . 064)(2 . 3 / 25) = (101 . 45 , 103 . 35). (b) ¯ x ± ks = 102 . 4 ± (2 . 208)(2 . 3) = (97 . 32 , 107 . 48). (c) The 95% lower conﬁdence bound for σ 2 is σ 2 ( n - 1) s 2 χ 2 α,n - 1 = (25 - 1)(2 . 3) 2 36 . 42 = 3 . 486 So σ 1 . 867. 3. (a) E ( ¯ X ) = 2 θ + 1 and V ( ¯ X ) = σ 2 /n = 3 θ/ 9 = θ/ 3. (b) The moment estimator is the solution of 2 ˆ θ + 1 = ¯ X , i.e., ˆ θ = ( ¯ X - 1) / 2. Clearly E ( ˆ θ ) = ( E ( ¯ X ) - 1) / 2 = θ so it is unbiased. V ( ˆ θ ) = V (( ¯ X - 1) / 2) = V ( ¯ X ) / 4 = θ/ 12. So MSE = V ( ˆ θ ) + bias 2 = V ( ˆ θ ) = θ/ 12. 4. (a) P ( X i < 9 . 5) = P ( Z < (9 . 5 - 10) / 1) = P ( Z < - 0 . 5) = . 30854 so the answer is ( . 30854) 4 = . 00906. (b) ¯ X N (10 , 1 / 4) so P ( ¯ X < 9 . 5) = P ( Z < (9 .
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This note was uploaded on 04/08/2012 for the course STAT 105 taught by Professor Hansen during the Winter '12 term at UCLA.

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