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Hw2Sol

# Hw2Sol - 6 70 833 1 96 8 √ 6 ² =(64 632 77 234 Since 80...

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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 = 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 ) = 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 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 = 70 . 833

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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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