Chapter 18, Problem 26CAP

To determine

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,

$H=\frac{12}{N\left(N+1\right)}\left({\displaystyle \sum \frac{R^2}{n}}\right)-3\left(N+1\right)$

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, $df=k-1$

In formula k denotes the number of groups in the study.

Null hypothesis:

$H_0:$ The productivity levels not differed using the Kruskal-Wallis H test.

Alternative hypothesis:

$H_1:$ The productivity levels differed using the Kruskal-Wallis H test.

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,

$\begin{array}{c}{\displaystyle \sum R_1}=11+12+13+15+7\\ =58\end{array}$

$\begin{array}{c}{\displaystyle \sum R_2}=8+9+10+4+3\\ =34\end{array}$

$\begin{array}{c}{\displaystyle \sum R_3}=5+14+2+1+6\\ =28\end{array}$

Substitute, $N=15,{\displaystyle \sum R_1}=58,{\displaystyle \sum R_2}=34,{\displaystyle \sum R_3}=28,n=5$ in the normal approximation formula,

$\begin{array}{c}H=\frac{12}{15\left(15+1\right)}\left(\frac{{58}^2}{5}+\frac{{34}^2}{5}+\frac{{28}^2}{5}\right)-3\left(15+1\right)\\ =\frac{12}{15\times 16}\left(\frac{3,364}{5}+\frac{1,156}{5}+\frac{784}{5}\right)-3\left(16\right)\\ =\frac{12}{240}\left(672.8+231.2+156.8\right)-48\\ =\left(0.05\right)\left(1,060.8\right)-48\end{array}$

$\begin{array}{c}=53.04-48\\ =5.04\end{array}$

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 $k=3$ in the degrees of freedom formula.

$\begin{array}{c}df=k-1\\ =3-1\\ =2\end{array}$

Critical value:

The given significance level is $\alpha =0.05$ and the degrees of freedom are 5.

From the Appendix B: Table B.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 $df=2$ at a 0.05 level of significance is 5.99.

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.