Find the Kruskal-Wallis H test at a .05 level of significance.
Determine whether to retain or reject the null hypothesis.
Answer to Problem 26CAP
The Kruskal-Wallis H test at a .05 level of significance is 5.040.
The decision is to retain the null hypothesis.
The productivity levels not differed using the Kruskal-Wallis H test.
Explanation of Solution
Calculation:
The information given that, the study contains three locations in a small business and employees listened to music in one location, listened to news radio in a second location, and did not listen to radio during work in a third location.
Kruskal-Wallis H test:
The Kruskal-Wallis H test is a non-parametric statistical method used to determine the whether the total ranks for two or more groups that are independent would be significantly different or not. This is a test that is used as an alternative for one-way between-subjects ANOVA.
Steps to find the test statistic:
- Combine scores from each group and rank them in numerical order.
- Sum the ranks for each group.
- Compute the test statistic (H).
The formula to find the test statistic is,
In the formula, N is the total number of participants, n is the number of participants per group, and R is the total rank in each group.
Degrees of freedom:
The degrees of freedom for the test is,
In formula k denotes the number of groups in the study.
Null hypothesis:
Alternative hypothesis:
The ranks are,
Productivity | Levels | Ranks |
32 | Music | 11 |
34 | Music | 12 |
36 | Music | 13 |
40 | Music | 15 |
28 | Music | 7 |
29 | News Radio | 8 |
30 | News Radio | 9 |
31 | News Radio | 10 |
25 | News Radio | 4 |
23 | News Radio | 3 |
26 | No Radio | 5 |
38 | No Radio | 14 |
20 | No Radio | 2 |
18 | No Radio | 1 |
27 | No Radio | 6 |
The sum of ranks for each group is,
Substitute,
Hence, the Kruskal-Wallis H test at a .05 level of significance is 5.040.
Decision rules:
- If the test statistic value is greater than critical value then reject the null hypothesis. The test is significant.
- If the test statistic value is less than critical value then retain the null hypothesis. The test is no significant.
Degrees of freedom:
Substitute
Critical value:
The given significance level is
From the Appendix C: Table C.7 Critical values for Chi-square:
- Locate the value 2 in the degrees of freedom (df) column.
- Locate the 0.05 in level of significance row.
- The intersecting value that corresponds to the 2 with level of significance 0.05 is 5.99.
Thus, the critical value for
Conclusion:
The test statistic value is 5.04.
The critical value is 5.99.
The test statistic value is less than the critical value.
The null hypothesis is retained.
Hence, the productivity levels not differed using the Kruskal-Wallis H test.
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Chapter 18 Solutions
Statistics for the Behavioral Sciences
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