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HW1_Sol - STAT 426 HW1 Solutions(provided by TA for your...

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STAT 426 HW1 Solutions (provided by TA for your information) 1.3 Let X be the number of defects, λ = 1. (a) P ( X = 0) = e - 1 1 0 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 π be the proportion of women reported greater relief with the STAN- DARD analgesic H 0 : π = 0 . 5 vs H 1 : π = 0 . 5 Sample proportion for π : p = 40 40+60 = 0 . 4 | z | = | p - π | π (1 - π ) n = | 0 . 4 - 0 . 5 | (0 . 5)(0 . 5) 100 = 2 p-value= P ( | z | ≥ 2) = 0 . 0456 0 . 05 Therefore, we reject H 0 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 π * be the proportion of women reported greater relief with the NEW analgesic Sample proportion for π * : p * = 60 40+60 = 0 . 6 95% C.I. for π * = p * ± Z 0 . 025 p * (1 - p * ) n = 0 . 6 ± 1 . 96 (0 . 6)(1 - 0 . 6) 100 = (0 . 504 , 0 . 696) Since this interval contains only positive values and larger than 0 . 5 , we con- clude with 95% confidence that π * 0 . 5. 2.2 Let p 1 be the estimated annual probability that a woman over the age 35 dies of lung cancer for current smokers, and p 2

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