HW5Sol - men 40-x 60 + x women x 100-x and the table of...

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Homework 5 – solution 2. Since n = 12, we should use no more than 4 bins. Let the bins correspond to the four seasons (with the first bin corresponding to winter). Then we have E i = 3 for i = 1 ,..., 4. O 1 = 3, O 2 = 6, O 3 = 2, and O 4 = 1. The value of the test statistic is χ 2 = 1 3 (0 + 9 + 1 + 4) = 4 . 67 The corresponding critical value is χ 2 0 . 05 , 4 - 0 - 1 = χ 2 0 . 05 , 3 = 7 . 81. Since χ 2 < χ 2 0 . 05 , 3 , we cannot reject the null hypothesis, and conclude that the data is consistent with the uniform birthday hypothesis. 3. Let x be the number of female smokers in the sample. Then the table of observed values looks like smoking nonsmoking
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Unformatted text preview: men 40-x 60 + x women x 100-x 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 = (20-x ) 2 1 20 + 1 80 + 1 20 + 1 80 = (20-x ) 2 8 . Equating it to the critical value, we obtain (20-x ) 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 .

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