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Unformatted text preview: HW 6 Solutions 8.43 a. The distribution function for Y ( n ) is ( 29 n y n y G θ = ) ( , 0 ≤ y ≤ θ, so the distribution function for U is given by = θ = θ ≤ = ≤ = ) ( ) ( ) ( ) ( ) ( u G u Y P u U P u F n n U u , 0 ≤ y ≤ 1. b. (Similar to Example 8.5) We require the value a such that ≤ θ a Y P n ) ( = F U ( a ) = .95. Therefore, a n = .95 so that a = (.95) 1/ n and the lower confidence bound is [ Y ( n ) ](.95) –1/ n . 8.44 a. 2 2 2 2 ) ( 2 ) ( ) ( θ θ = θ θ = ≤ = ∫ y y dt t y Y P y F y Y , 0 < y < θ. b. The distribution of U = Y /θ is given by ) 1 ( 2 2 ) ( ) ( ) ( ) ( 2 u u u u u F u Y P u U P u F Y U = = θ = θ ≤ = ≤ = , 0 < u < 1. Since this distribution does not depend on θ, U = Y /θ is a pivotal quantity. c. Set P ( U ≤ a ) = F Y ( a ) = 2 a (1 – a ) = .9 so that the quadratic expression is solved at a = 1 – 10 . = .6838 and then the 90% lower bound for θ is Y /.6838. 8.47 Note that for all i , the mgf for Y i is 1 ( ) (1 ) Y m t t θ = , t < 1/θ. a. Let ∑ = θ = n i i Y U 1 / 2 . The mgf for U is [ ] 2 / 1 , ) 2 1 ( ) / 2 ( ) ( ) ( < = θ = = t t t m e E t m n n Y tU U . This is the mgf for the chi–square distribution with 2 n degrees of freedom. Thus, U has this distribution, and since the distribution does not depend on θ, U is a pivotal quantity....
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 Spring '08
 EduardoMontoya
 Statistics, Normal Distribution, Standard Deviation, sample variance, mean compartment pressures

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