Using the Kruskal-Wallis Test. In Exercises 5–8, use the Kruskal-Wallis test.

Speed Dating Use the sample data from Exercise 1 to test the claim that females from the different age brackets give attribute ratings with the same median. Use a 0.05 significance level.

Three samples are given on male attribute ratings by females of three different age groups.

The signficance level is 0.05.

The Kruskal-Wallis test is used to test the difference of medians between three or more samples when the populations of the three samples are not necessarily required to follow normal distribution.

The null hypothesis is as follows:

There is no difference in the medians of the threesamples.

The alternative hypothesis is as follows:

There is a difference in the medians of the threesamples.

This is a two-tailed test.

The ranks of the observations from the three samples are given using the following steps:

• Combine the threesamples and label each observation with the sample name/number it comes from.
• The smallest observation is assigned rank 1; the next smallest observation is assigned rank 2, and so on until the largest value.
• If two observations have the same value, the mean of the ranks is assigned to them.

The following table shows the ranks:

The sum of the ranks corresponding to the age bracket 20-22 is computed as follows:

\(\begin{array}{c}{R_1} = 21 + 25.5 + 6 + .... + 5\\ = 164.5\end{array}\)

The sum of the ranks corresponding to the age bracket 23-26 is computed as follows:

\(\begin{array}{c}{R_2} = 22.5 + 7.5 + 17 + .... + 1\\ = 150\end{array}\)

The sum of the ranks corresponding to the age bracket 27-29 is computed as follows:

\(\begin{array}{c}{R_3} = 17 + 25.5 + 14 + .... + 9.5\\ = 150.5\end{array}\)

Here, the size of sample 1, sample 2, and sample 3 is given as follows:

\({n_1} = {n_2} = {n_3} = 10\)

The total sample size is given as follows:

\(\begin{array}{c}N = 10 + 10 + 10\\ = 30\end{array}\)

The value of the test statistic is computed as shown below:

\(\begin{array}{c}H = \frac{{12}}{{N\left( {N + 1} \right)}}\left( {\frac{{{R_1}^2}}{{{n_1}}} + \frac{{{R_2}^2}}{{{n_2}}} + \frac{{{R_3}^2}}{{{n_3}}}} \right) - 3\left( {N + 1} \right)\\ = \frac{{12}}{{30\left( {31} \right)}}\left( {\frac{{{{164.5}^2}}}{{10}} + \frac{{{{150}^2}}}{{10}} + \frac{{{{150.5}^2}}}{{10}}} \right) - 3\left( {31} \right)\\ = 0.012903\left( {7221.05} \right) - 93\\ = 0.1748\end{array}\)

The degrees of freedom are computed as follows:

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

Let k be the number of samples.

Here, k=3.

The critical value of chi-square for\(\alpha = 0.05\)with 2 degrees of freedom is equal to 5.991.

Since the test statistic value is less than the critical value, so the decision is fail to reject the null hypothesis.

It can be concluded that there is no difference in the medians of the three samples.