HW10-Solutions - STAT 101 Agresti Homework 10 Solutions...

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STAT 101 - Agresti Homework 10 Solutions 12/2/10 Chapter 12 12.1. The response variable is the number of firefights reported. The explanatory variable was branch of service and deployment venue combination. The null hypothesis is that the population means are the same for each branch of service and deployment venue combination. The alternative hypothesis is that at least two of the population means differ. 12.2. (a) 0 1 2 12 : H μ μ μ = = = L ; a H : at least two of the population means are unequal. (b) If the null hypothesis were true, we would expect F to equals about 1, with relatively large values above 1 giving strong evidence against the null. Since F = 0.61 does not differ from 1 by much, it does not appear that we have strong evidence against 0 H . (c) If all 12 population means were equal (e.g., the mean number of good friends was not associated with astrological sign), we would see results at least as extreme as those observed with probability 0.82. There is not enough evidence to show an association between the mean number of good friends and astrological sign. 12.3. (a) 0 1 2 3 4 5 : H μ μ μ μ μ = = = = ; a H : at least two of the population means are unequal. (b) If the two estimates of the variance were equal, we would expect F to be 1. Since F = 0.80 does not differ from 1 by much, it does not appear that we have strong evidence against 0 H . (c) If all 5 population means were equal (e.g., the mean number of good friends was not associated with marital status), we would see results as extreme as those observed with probability 0.53. There is not enough evidence to show an association between the mean number of good friends and marital status. 12.5. (a) (i) 0 1 2 3 : H μ μ μ = = ; a H : at least two of the population means are unequal. (ii) F = 3.03; (iii) P = 0.049; (iv) We reject the null hypothesis at the 0.05 significance level. (b) Yes. The standard deviations are all much larger than the means, suggesting that the three distributions are very highly skewed to the right, rather than normal. 12.6. (a) 0 1 2 3 : H μ μ μ = = ; a H : at least two of the population means are unequal. F = 2.50, P = 0.18. There is not enough evidence to conclude that the population mean quiz scores differ for the three groups. (b) The F statistic will be smaller, because the between groups variation would not be as large, so the numerator of the F statistic would be smaller. (c) The F statistic will be larger, since the within groups variation is smaller. (d) The F statistic would be larger, since larger sample sizes provide stronger evidence, for a given size of effect. (e) The P -values would be larger for part (b), smaller for part (c), and smaller for part (d). 12.8. (a) Within-groups SS = 12.0; within estimate = 12/3 = 4.0. (b) Between-groups SS = 148.0; between estimate = 148/2 = 74.0. (c) 0 1 2 3 : H μ μ μ = = ; a H : at least two of the population means are unequal. F = 18.5, P = 0.02. There is enough evidence to conclude that the mean damages differ for the three bumper types.
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