401quiz - stat 401 Quiz 3 solution (Spring, 2008) 1 (1) p =...

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stat 401 Quiz 3 solution (Spring, 2008) 1 (1) ˆ p = 4 13 = . 308 , s.e. p ) = q ˆ p (1 - ˆ p ) n = 0 . 128 ˆ p ± Z 0 . 025 × s.e. p ) = 0 . 308 ± 1 . 96 × 0 . 128 (0 . 057 , 0 . 559) Note: we should use Z 0 . 025 ,since the distribution of ˆ p approximates to a normal distribution based on the Central Limit Theorem. (2) n = 4 · Z 2 α/ 2 ˆ p (1 - ˆ p ) L 2 = 4 · 1 . 96 2 · ( . 308)(1 - . 308) . 1 2 = 327 . 3. We need n = 328. (3) If we don’t have information, we use ˆ p = . 5 instead. Therefore n = 1 . 96 2 . 1 2 = 384 . 16.We need n = 385. (4) p ? = ˆ p 2 = 0 . 0946 95% C.I = (0 . 057 2 , 0 . 559 2 ) 2 (1) E ( X ) == 5 + 1 λ So, λ = 1 E ( X ) - 5 ˆ λ = 1 ¯ X - 5 ¯ X = 7 . 201538 ˆ λ = 1 ¯ X - 5 = 1 7 . 201538 - 5 = 0 . 4542279 (2) d P ( X > 8) = exp ± - 3 ˆ λ ² = exp ( - 3 · ( . 4542278)) = . 2559730 3 (1) ˆ β 1 = n · x i y i - ( x i )( y i ) n · x 2 i - ( x i ) 2 = 7 * 718 . 61 - 98 . 8 * 49 . 9 7 * 1411 . 28 - (98 . 8) 2 = . 852. ˆ α 1 = ¯ Y - ˆ β 1 · ¯ X = - 4 . 90 ˆ μ Y | X = x = - 4 . 9 + 0 . 852 x (2) S 2 = 1 n - 2 h y 2 i - ˆ α 1 y i - ˆ β 1 x i y i i = 1 5 [368 . 63 + 4 . 9 * 49 . 9 - 0 . 852 * 718 . 61] = 0 . 176856 b σ = S = . 421 (3) ˆ μ Y | X =7 = - 4 . 9 + 0 . 852 * 7 = 1 . 064 The estimated variance of the estimator is given by d σ ˆ μ Y | X =7 = S ˆ μ Y | X =7 = S q 1 n + n ( x - ¯ X ) 2 n X 2 i - ( X i ) 2 = . 421 q 1 7 + 7(7 - 14 . 11429) 2 7 * 1411 . 28 - (98 . 8) 2 =
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This note was uploaded on 05/08/2008 for the course STAT 401 taught by Professor Akritas during the Spring '00 term at Pennsylvania State University, University Park.

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401quiz - stat 401 Quiz 3 solution (Spring, 2008) 1 (1) p =...

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