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

Arsenic in Rice Listed below are amounts of arsenic in samples of brown rice from three different states. The amounts are in micrograms of arsenic and all samples have the same serving size. The data are from the Food and Drug Administration. Use a 0.01 significance level to test the claim that the three samples are from populations with the same median.

Arkansas	4.8	4.9	5	5.4	5.4	5.4	5.6	5.6	5.6	5.9	6	6.1
California	1.5	3.7	4	4.5	4.9	5.1	5.3	5.4	5.4	5.5	5.6	5.6
Texas	5.6	5.8	6.6	6.9	6.9	6.9	7.1	7.3	7.5	7.6	7.7	7.7

Three samples are given showing the amount of arsenic present in brown rice servings of the three different states.

The significance level is 0.01.

The Kruskal-Wallis test is used to test the difference of medians between the given threesamples.

The null hypothesis is as follows:

The three samples are obtained from populations with the same median.

The alternative hypothesis is as follows:

The three samples are obatined from populations with different medians.

This is a two-tailed test.

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

• Combine the three samples and tag the sample name against each observation.
• Ranks are assigned to all observations. The smallest observation is assigned a rank of 1; the next smallest observation is assigned a rank of 2, and so on.
• 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 Arkansas is computed as follows:

\(\begin{array}{c}{R_1} = 5 + 6.5 + 8 + .... + 26\\ = 192\end{array}\)

The sum of the ranks corresponding to California is computed as follows:

\(\begin{array}{c}{R_2} = 1 + 2 + 3 + .... + 19.5\\ = 116.5\end{array}\)

The sum of the ranks corresponding toTexas is computed as follows:

\(\begin{array}{c}{R_3} = 19.5 + 23 + 27 + .... + 35.5\\ = 357.5\end{array}\)

Here,

\({n_1} = {n_2} = {n_3} = 12\)

and

\(\begin{array}{c}N = 12 + 12 + 12\\ = 36\end{array}\)

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

\(\begin{array}{c}H = \frac{{12}}{{N\left( {N + 1} \right)}}\left( {\frac{{{R_1}^2}}{{{n_1}}} + \frac{{{R_2}^2}}{{{n_2}}} + \frac{{{R_3}^2}}{{{n_3}}}} \right) - 3\left( {N + 1} \right)\\ = \frac{{12}}{{36\left( {37} \right)}}\left( {\frac{{{{192}^2}}}{{12}} + \frac{{{{116.5}^2}}}{{12}} + \frac{{{{357.5}^2}}}{{12}}} \right) - 3\left( {37} \right)\\ = 22.816\end{array}\)

Let k be the number of samples.

Thus, 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\(\alpha = 0.01\)with 2 degrees of freedom is equal to 9.21.

Since the test statistic value is greater than the critical value, the null hypothesis is rejected.

It can be concluded that the three samples are obtained from populations with different medians.