hw13ans - Statistics 500 Fall 2009 Solutions to Homework 13...

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Statistics 500 – Fall 2009 Solutions to Homework 13 1 1. Boy/girl ratios (a) () 38 . 6 200 , 43 5 . 0 1 5 . 0 5 . 0 4846 . 0 1 0 0 0 = = = n p z π , p-value is less than 0.0001. There is strong evidence that π is not 0.5. (b) Approximate 95% confidence interval for π is 0.975 1 0.4846 1 0.4846 0.4846 1.96 0.480,0.489 43,200 pp pz n −− ±= ± = where 4846 . 0 200 , 43 937 , 20 = = p . We are 95% confident that the proportion of females is between 0.480 and 0.489. (c) ( ) 016 . 0 , 009 . 344 , 7 4812 . 0 1 4812 . 0 200 , 43 4846 . 0 1 4846 . 0 96 . 1 4812 . 0 4846 . 0 = + ± (d) No. The tests and confidence intervals assume that events (male or female child) are independent. Alternatively, each child has the same probability of being female. If that probability varies between families, the events aren’t independent. 2. Distribution of counts of boys and girls (a) Let 485 . 0 , 6 ~ = = n Binomial X . Expected # with 0 boys = expected # with 6 girls = (6!/(6! 0!) π 6 (1- π ) 0 = 93.71 if you use π =0.485. (b) 0 : proposed model ( 6, 0.485) is correct H Binomial n = = . I give you the contributions to Chi- square for most of the cells. For the 6 girl cell, (E-O) 2 /E = (93.71 – 113) 2 / 93.71 = 3.971. I gave you the sum for the rest of the cells, so Chi-square = 3.971 + 12.894 = 16.866. This has 6 d.f., so ( ) 0098 . 0 866 . 16 2 6 = > = χ P value p There is strong evidence to reject the null hypothesis, the data do not follow the proposed binomial distribution.
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This note was uploaded on 02/11/2012 for the course STAT 500 taught by Professor Staff during the Fall '08 term at Iowa State.

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hw13ans - Statistics 500 Fall 2009 Solutions to Homework 13...

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