Explanation The test hypothesis is given as follows: Null hypothesis: H 0 : The durability of the paint was equal for three different conditions. Alternative hypothesis: H 1 : The durability of the paint was different for three different conditions. 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 the critical values of chi-square: Procedure: 1. In the table, first locate 2 in the first z column. 2. Locate the value of 0.01 in the second row. 3. Locate the value, by the intersection of the row and column values give the critical value. The critical value for 2 df for the 0.01 significance level is 9.210. Decision rule: If H > 9.210 , reject the null hypothesis. Otherwise, fail to reject the null hypothesis. In this context, the test statistic for 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 is rank of samples 1, 2,… k, respectively. n 1 , n 2 , ... n k is size of samples 1, 2,… k, respectively. 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: Saltwater Rank 1 Freshwater Rank 2 Freshwater With weeds Rank 3 167

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ISBN: 9781264731657

Author: Lind

Publisher: MCG

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Chapter 16, Problem 19E

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State whether the durability of the paint was different for three different conditions.

Solution Summary: The author states that the durability of the paint was equal for three different conditions. The test statistic H follows the chi-square distribution with degrees of freedom.

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