sol7 - Stat 201A HW7 Solution 1. (a) N = 14828, n = 290, nk...

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Stat 201A HW7 Solution 1. (a) N = 14828 , n = 290 , n k = 143, so ˆ p k = 34 / 143 = 23 . 8% and se = ± (1 - n/N p k (1 - ˆ p k ) / ( n k - 1) = 3 . 54% The estimated proportion is 23.8% with a standard error of 3.54%. (b) N k = 7526 , n k = 143 , ˆ p k = 34 /n k , so ˆ τ k = N k ˆ p k = 1789 . 4 and se = N k ± (1 - n k /N k p k (1 - ˆ p k ) / ( n k - 1) = 266 . 3 The estimated number is 1789 with a standard error of 266. 2. See the table for the 10 samples and various estimates for each sample. sample s data y ( s ) ¯ y s 2 se y ) 95% CI 1 1 2 3 2 3 0 1.67 2.33 0.56 - 0.73 4.07 2 1 2 4 2 3 4 3 1 0.37 1.43 4.57 3 1 2 5 2 3 6 3.67 4.33 0.76 0.4 6.94 4 1 3 4 2 0 4 2 4 0.73 - 1.14 5.14 5 1 3 5 2 0 6 2.67 9.33 1.12 - 2.13 7.47 6 1 4 5 2 4 6 4 4 0.73 0.86 7.14 7 2 3 4 3 0 4 2.33 4.33 0.76 - 0.94 5.6 8 2 3 5 3 0 6 3 9 1.1 - 1.71 7.71 9 2 4 5 3 4 6 4.33 2.33 0.56 1.93 6.73 10 3 4 5 0 4 6 3.33 9.33 1.12 - 1.47 8.13 Sum 30 50 (a) μ = 3, τ = 15, σ 2 = 5. There are 10 samples in total so P ( s ) = 0 . 1 for each s . (b) E y ) = 30 / 10 = 3 = μ , and E ( s 2 ) = 50 / 10 = 5 = σ 2 . (c) var y ) = y - 3) 2 / 10 = [(1
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This note was uploaded on 11/24/2010 for the course STAT 201a taught by Professor Wu during the Spring '10 term at Pasadena City College.

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