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

Car Crash Measurements Use the following listed chest deceleration measurements (in g, where g is the force of gravity) from samples of small, midsize, and large cars. (These values are from Data Set 19 “Car Crash Tests” in Appendix B.) Use a 0.05 significance level to test the claim that the different size categories have the same median chest deceleration in the standard crash test. Do the data suggest that larger cars are safer?

Small	44	39	37	54	39	44	42
Midsize	36	53	43	42	52	49	41
Large	32	45	41	38	37	38	33

Three samples are given on the chest deceleration measurements for three car sizes.

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 small car size is computed as follows:

\(\begin{array}{c}{R_1} = 15.5 + 8.5 + 4.5 + .... + 12.5\\ = 86\end{array}\)

The sum of the ranks corresponding to the medium car size is computed as follows:

\(\begin{array}{c}{R_2} = 3 + 20 + 14 + .... + 10.5\\ = 97\end{array}\)

The sum of the ranks corresponding to the large car size is computed as follows:

\(\begin{array}{c}{R_3} = 1 + 17 + 10.5 + .... + 2\\ = 48\end{array}\)

Here, the sample sizes for all samples are given as:

\({n_1} = {n_2} = {n_3} = 7\)

The total size (N) is given as:

\(\begin{array}{c}N = 7 + 7 + 7\\ = 21\end{array}\)

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

\(\begin{array}{c}H = \frac{{12}}{{N\left( {N + 1} \right)}}\left( {\frac{{{R_1}^2}}{{{n_1}}} + \frac{{{R_2}^2}}{{{n_2}}} + \frac{{{R_3}^2}}{{{n_3}}}} \right) - 3\left( {N + 1} \right)\\ = \frac{{12}}{{21\left( {22} \right)}}\left( {\frac{{{{86}^2}}}{7} + \frac{{{{97}^2}}}{7} + \frac{{{{48}^2}}}{7}} \right) - 3\left( {22} \right)\\ = 4.905\end{array}\)

Let k be the number of samples.

Here, k=3.

The degrees of freedom are computed as follows:

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

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

Since the absolute 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.

Since the three samples have equal medians, it can be said that larger cars are not safer than the other cars.