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Unformatted text preview: men 40x 60 + x women x 100x and the table of expected values is smoking nonsmoking men 20 80 women 20 80 Computing the value of the test statistic, we obtain 2 = (20x ) 2 1 20 + 1 80 + 1 20 + 1 80 = (20x ) 2 8 . Equating it to the critical value, we obtain (20x ) 2 8 = 2 . 05 , 1 = 3 . 84 . Solving the latter for x , we get x = 20 8 3 . 84 = 14 . 46 . Therefore, the smallest number of female smokers for which the independence hypothesis would not be rejected is 15. 1...
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This note was uploaded on 03/17/2008 for the course IE 121 taught by Professor Perevalov during the Spring '08 term at Lehigh University .
 Spring '08
 Perevalov

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