# 005 we fail to reject h 0 we do not have convincing

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= 0.05, we fail to reject H 0 . We do not have convincing evidence that the distribution of drink brand preference is different for the professional and support staffs. (b) The responses of these 6 non-responders might change our conclusion. In the most extreme case if all 6 of them chose Coca-cola the chi-square statistic for the test would change to 7.449, yielding a P- value of 0.0063, which would provide strong evidence that there is a difference in the distribution of drink brand preference between the two groups. Expected Counts for part (a) Professional Staff Support Staff Coca-cola 10.2 33.8 Pepsi 11.8 39.2
© 2011 BFW Publishers The Practice of Statistics, 4/e- Chapter 11 541 Test 11C Part I 1. d We are looking for a difference in the proportion cars choosing each of the three lanes, so the alternative hypothesis is that the population distribution of the categorical variable is not equal for at least one of the three values (the alternative hypothesis includes the possibility that the two proportions are equal to each other, but different from the third). 2. c 3 2 1 2 Observed Expected Expected i 3. d Choice a is about tests of means, choices b and c make reference to observed counts, which are not involved in conditions required for chi-square procedures. 4. c For a chi-square test of homogeneity, the null hypothesis is that the distribution of a categorical variable is the same for two or more different populations. There are three random samples in this setting from three distinct populations. 5. d Degrees of freedom for any chi-square test involving a 2-way table is 1 1 number of rows number of columns . 6. e Expected counts in a two-way table row total column total grand total . 7. a Since the component for midnight/within specifications is a small fraction of the total chi-square statistic (0.090), the observed and expected counts must be close to equal. 8. e Each component of a chi-square statistic is 2 Observed Expected Expected for the given cell. 9. b The 2 x 3 table produces a chi-square statistic with 2 degrees of freedom, thus the P - value is much less that 0.05, providing strong evidence against the null hypothesis that accident type and age are independent. 10. a A chi-square test on a 2 x 2 table is equivalent to a two-sided two-proportion z -test, and the 2 distribution with df = 1 is the square of the standard normal distribution. Thus if 2 2 3.07, 3.07 9.42 z , and the resulting P- value is the same. Part II 11. State: We are testing the hypothesis H o : The proportion of allergy sufferers born during each season of the year matches the proportion of the general population born during each season, against H a : The proportion of allergy sufferers born during each season of the year does not match the proportion of the general population born during each season. We will use a significance level of = 0.05. Plan: The procedure is a chi-square goodness-of-fit test. Conditions: Random : the data come from a simple random sample of 500 people who are allergic to dust mites. Large sample size : Expected counts are: Winter: 150; Spring: 110; Summer: 120; Fall: 120 all are at least 5. Independent : Randomly-selected allergy sufferers should be independent; and there are surely at least 10 x 500 people who suffer from dust mite allergies. Do: 2 2 2 2 2 117 150 105 110 145 120 133 120 14.10 150 110 120 120 , df = 3; P- value = 0.0028.