Explanation The test hypotheses are given as follows: Null hypothesis: H 0 : There is no difference in the distribution of scores. Alternate hypothesis: H 1 : There is a difference in the distribution of scores. Here, the test statistic H follows the chi-square distribution with k − 1 degrees of freedom where k is the number of populations and H is the test statistic. In order to formulate the decision rule, the chi-square is used. In this context, the number of populations is 3. Degrees of freedom for populations: k − 1 = 3 − 1 = 2 From Appendix B, Table B.7 Critical Values of Chi-Square: Procedure: 1. In the table, first locate 2 in the first z column. 2. Locate the value of 0.05 in the second row. 3. Locate the value by the intersection of the row and column values, which gives the critical value. The critical value for 2 df for the 0.05 significance level is 5.991. Decision rule: If H > 5.991 , reject the null hypothesis. Otherwise fail to reject the null hypothesis. In this context, the test statistic for the Kruskal–Wallis test is denoted as H . Test statistic, H : H = 12 n ( n + 1 ) [ ( ∑ R 1 ) 2 n 1 + ( ∑ R 2 ) 2 n 2 + ... + ( ∑ R k ) 2 n k ] − 3 ( n + 1 ) . Where k is the number of populations, R 1 , R 2 , ... R k are the ranks of samples 1, 2,… k . n 1 , n 2 , ... n k are the sizes of samples 1, 2,… k . n is the combined number of observations for all samples. In this context, it is necessary to rank all the observations. The following table represents the ranks of all the samples: 8 a.m. Rank 1 10 a.m. Rank 2 1.30 p.m. Rank 3 68 6 59 1.5 67 5 84 20 59 1.5 69 7 75 10.5 63 4 75 10.5 78 15

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Author: Lind

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State whether there is a difference in the distribution of scores.

Solution Summary: The author states whether there is a difference in the distribution of scores. The Kruskal–Wallis test statistic H follows the chi-square distribution with degrees of freedom.

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