Hw2Sol - 6 , 70 . 833 + 1 . 96 8 . √ 6 ² = (64 . 632 ,...

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Homework Solutions Assignment Number 2 Due January 29, 2007 Section 4.3 Problem 23. Let Y = donation collected tomorrow. Given that μ = $20 , 000 and σ = $5000, P ( Y > $30 , 000) = P ± Y - $20 , 000 $5 , 000 > $30 , 000 - $20 , 000 $5 , 000 ² = P ( Z > 2) = 0 . 0228 . Section 5.2 Problem 9. (a) L ( θ ) = ( 1 θ ) n , if 0 y 1 , y 2 , ..., y n θ, and 0 otherwise. Thus θ e = y max , which for these data is 14.2. (b) L ( θ ) = ³ 1 θ 2 - θ 1 ´ n , if θ 1 y 1 , y 2 , ..., y n θ 2 , and 0 otherwise. Thus θ 1 e = y min and θ 2 e = y max . For these data, θ 1 e = 1 . 8, θ 2 e = 14 . 2. Problem 15. For Y uniform on (0 , θ ) , E ( Y ) = θ/ 2 . Set θ/ 2 = y, so the method of moments estimate is θ e = 2 y. For the data given, θ e = 2(50) = 100. The maximum likelihood estimate is y max = 92. Problem 20. E ( Y ) = Z k yθk θ ± 1 y ² θ +1 dy = θk θ Z k y - θ dy = θk θ - 1 Set θk θ - 1 = y , and get θ e = y y - k . Section 5.3 Problem 2. The confidence interval is ± y - z α/ 2 σ n , y + z α/ 2 σ n ² = ± 70 . 833 - 1 . 96 8 . 0
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Unformatted text preview: 6 , 70 . 833 + 1 . 96 8 . √ 6 ² = (64 . 632 , 77 . 234) . Since 80 does not fall within the confidence interval, that men and women metabolize methylmercury at the same rate is not believable. 1 Problem 3. The length of the confidence interval is 2 z α/ 2 σ √ n = 2(1 . 96)(14 . 3) √ n = 56 . 056 √ n . For 56 . 056 √ n ≤ 3 . 06 , n ≥ ± 56 . 056 3 . 06 ² 2 = 335 . 58 , so take n = 336. Problem 4. (a) P (-1 . 64 < Z < 2 . 33) = 0 . 94, a 94% confidence level. (b) P (-∞ < Z < 2 . 58) = 0 . 995, a 99.5% confidence level. (c) P (-1 . 64 < Z < 0) = 0 . 45, a 45% confidence level. 2...
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This note was uploaded on 03/23/2008 for the course MATH 181A taught by Professor Xu,lily during the Spring '07 term at UCSD.

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Hw2Sol - 6 , 70 . 833 + 1 . 96 8 . √ 6 ² = (64 . 632 ,...

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