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

Correcting the H Test Statistic for Ties In using the Kruskal-Wallis test, there is a correction factor that should be applied whenever there are many ties: Divide H by

\(1 - \frac{{\sum T }}{{{N^3} - N}}\)

First combine all of the sample data into one list, and then, in that combined list, identify the different groups of sample values that are tied. For each individual group of tied observations, identify the number of sample values that are tied and designate that number as t, then calculate\(T = {t^3} - t\). Next, add the T values to get\(\sum T \). The value of N is the total number of observations in all samples combined. Use this procedure to find the corrected value of H for Example 1 in this section on page 628. Does the corrected value of H differ substantially from the value found in Example 1?

Three samples showing the performance IQ scores of subjects with different blood lead levels are given.

Refer to Example 1 of Section 13-5.The following table shows the ranks:

Under the Kruskal-Wallis test, the value of the test statistic (H) is computed without accounting for the number of ties.

Ties represent sample values that occur multiple times in three samples.

If there are ties present, a correction is applied to the value of H.

The correction factor has the following formula:

\(1 - \frac{{\sum T }}{{{N^3} - N}}\)

where

\(T = {t^3} - t\)

Here, t is the number of sample values tied to each tied group.

Referring to Example 1 of Section 13-5, the given table shows the sample values that are tied, the corresponding value of t, and the corresponding value of T.

Here, the sample size of the three groups is given as:

\(\begin{array}{c}{n_1} = 8\\{n_2} = 6\\{n_3} = 5\\\end{array}\)

The total number of observations is given as:

\(\begin{array}{c}N = 8 + 6 + 5\\ = 19\end{array}\)

The correction factor is computed as shown:

\(\begin{array}{c}1 - \frac{{\sum T }}{{{N^3} - N}} = 1 - \frac{{84}}{{6859 - 19}}\\ = 0.987719\end{array}\)

Thus, the correction factor is equal to 0.987719.

The sum of the ranks corresponding to the low blood lead level is computed as follows:

\(\begin{array}{c}{R_1} = 6.5 + 8.5 + 18.5 + .... + 1\\ = 86\end{array}\)

The sum of the ranks corresponding tothe medium blood lead levelis computed as follows:

\(\begin{array}{c}{R_2} = 2 + 12.5 + 18.5 + .... + 5\\ = 50.5\end{array}\)

The sum of the ranks corresponding to the high blood lead level is computed as follows:

\(\begin{array}{c}{R_3} = 10 + 15.5 + 12.5 + .... + 12.5\\ = 53.5\end{array}\)

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}}{{19\left( {20} \right)}}\left( {\frac{{{{86}^2}}}{8} + \frac{{{{50.5}^2}}}{6} + \frac{{{{53.5}^2}}}{5}} \right) - 3\left( {20} \right)\\ = 0.694\end{array}\)

The ordinary value of H is equal to 0.694.

The corrected H can be computed as:

\(\begin{array}{c}{\rm{Corrected}}\;H = \frac{H}{{{\rm{Correction}}\;{\rm{factor}}}}\\ = \frac{{0.694}}{{0.987719}}\\ = 0.703\end{array}\)

Thus, the corrected value of H is equal to 0.703.

The corrected value of H(0.703) does not differ significantly from the H value (0.694). Thus, it can be said that a large number of ties does not seem to have a considerable amount of effect on the H value.