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Unformatted text preview: STAT 426 HW1 Solutions (provided by TA for your information) 1.3 Let X be the number of defects, = 1. (a) P ( X = 0) = e 1 1 0! = e 1 = 0 . 3679 (b) P ( X = 1) = e 1 1 1 1! = e 1 = 0 . 3679 (c) P ( X 2) = 1 P ( X = 0) P ( X = 1) = 1 2 e 1 = 0 . 2642 1.8 a. Let be the proportion of women reported greater relief with the STAN DARD analgesic H : = 0 . 5 vs H 1 : 6 = 0 . 5 Sample proportion for : p = 40 40+60 = 0 . 4  z  =  p  q (1 ) n =  . 4 . 5  q (0 . 5)(0 . 5) 100 = 2 pvalue= P (  z  2) = 0 . 0456 . 05 Therefore, we reject H at the level of significance 0.05, that is, the probabil ity of greater relief with the standard analgesic is significantly different from that with the new one. b. Let * be the proportion of women reported greater relief with the NEW analgesic Sample proportion for * : p * = 60 40+60 = 0 . 6 95% C.I. for * = p * Z . 025 q p * (1 p * ) n = 0 . 6 1 . 96 q (0 . 6)(1 . 6) 100 = (0 . 504 , . 696)...
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This note was uploaded on 04/29/2010 for the course STAT stat 426 taught by Professor Xe during the Spring '10 term at University of Illinois at Urbana–Champaign.
 Spring '10
 Xe

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