6. Consider the accompanying 2 x 3 table displaying the sample proportions that fell in the various combinations of categories (e.g., 13% of those in the sample were in the first category of both factors). \begin{tabular}{c|ccc} & 1 & 2 & 3 \\ \hline 1 & .13 & .19 & .28 \\ 2 & .07 & .11 & .22 \\ \end{tabular} What is the smallest sample size $n$ for which these observed proportions would result in rejection of the independence hypothesis? Use $\alpha = .01$.
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We need to determine the expected proportions under the null hypothesis of independence. To do this, we can multiply the marginal proportions for each category. For example, the expected proportion for the first category of both factors would be: 0.25 (proportion Show more…
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Consider the accompanying contingency table displaying the sample proportions that fell in the various combinations of categories (e.g., 13% of those in the sample were in the first category of both factors). a. Suppose the sample consisted of n = 100 people. Use the chi-squared test for independence with significance level .10. b. Repeat the above assuming that the sample size was n = 1000. c. What is the smallest sample size n for which these observed proportions would result in rejection of the independence hypothesis?
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3. If you assume that the observations in the sample are independent, what is the smallest value the sample size could be to meet the conditions for this hypothesis test? A. 10 B. 20 C. 48 D. 47 E. 68 F. None of the above 4. Calculate the test statistic. 5. Calculate the p value. 6. Based on the p value we have evidence that the null model is not a good fit for our observed data. Note: You can earn partial credit on this problem.
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For Exercises 3 through $8,$ the null hypothesis was rejected. Use the Scheffé test when sample sizes are unequal or the Tukey test when sample sizes are equal, to test the differences between the pairs of means. Assume all variables are normally distributed, samples are independent, and the population variances are equal. Exercise 13 in Section $12-1$
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