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Homework_9_Solution

Homework_9_Solution - Stat 414 Homework 9 Fall 2008(45...

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Unformatted text preview: Stat 414 Homework 9 Fall 2008 (45 points) 1. 3.4 – 16 (5 points) By using the table IV in the page 685, we have P(X > 30.14) = .05 where X denotes a single observation. Since 10 observations are independent and are identically distributed from the chi-square(df =19), the random variable W (the number out of 10 observations that exceed 30.14) is Bin(n=10,p=.05). Therefore, P(W = 2) = .9885 - .9139 = .0746. 2. 3.5 – 9 (9 points – 3 points for each part) (a) If we set θ 1 = 0 or θ 2 = 1, ∞ < < ∞ − = − = − − x e e x f and e x F x x e x e , ) ( 1 ) ( y = e x so y > 0 and x = g(y) = lny. g’(y) = 1/y ∞ < < = = = − − − y e y ye y e e y f y y e y y , 1 1 ) ( ln ln Therefore Y has an exponential distribution with the parameter θ = 1. (b) If θ 1 ≠ 0 or θ 2 ≠ 1, ∞ < < ∞ − = − = − − − − − x e e x f and e x F x x e x e , 1 ) ( 1 ) ( 2 / ) 1 ( 2 1 2 / ) 1 ( / ) ( 2 θ θ θ θ θ θ θ ∞ < < = = = = + = = > = − − − − y e y e y y ye y f y...
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Homework_9_Solution - Stat 414 Homework 9 Fall 2008(45...

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