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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 = ) ( , 0 y , so the distribution function for U is given by = = = = ) ( ) ( ) ( ) ( ) ( 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 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 ) ( ) ( - = - = = y y dt t y Y P y F y Y , 0 < y < . b. The distribution of U = Y / is given by ) 1 ( 2 2 ) ( ) ( ) ( ) ( 2 u u u u u F u Y P u U P u F Y U- =- = = = = , 0 < u < 1. Since this distribution does not depend on , U = Y / 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 is Y /.6838. 8.47 Note that for all i , the mgf for Y i is 1 ( ) (1 ) Y m t t - =- , t < 1/. a. Let = = n i i Y U 1 / 2 . The mgf for U is [ ] 2 / 1 , ) 2 1 ( ) / 2 ( ) ( ) ( <- = = =- t t t m e E t m n n Y tU U . This is the mgf for the chisquare distribution with 2 n degrees of freedom. Thus, U has this distribution, and since the distribution does not depend on , U is a pivotal quantity....
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This note was uploaded on 10/02/2009 for the course PSTAT 5E taught by Professor Eduardomontoya during the Spring '08 term at UCSB.
- Spring '08