In using the Kruskal-Wallis test, there is a correction factor that should be applied whenever there are many ties: Divide H by $$1 - \frac { \sum \mathrm { T } } { \mathrm { N } ^ { 3 } - \mathrm { N } }$$ For each group of tied observations, calculate $\mathrm { T } = \mathrm { t } ^ { 3 } - \mathrm { t }$, where $\mathrm { t }$ is the number of observations that are tied withi the individual group. Find $\mathrm { t }$ for each group of tied values, then compute the value of $\mathrm { T }$ for each group, then add the T values to get $\sum \mathrm { T }$. The total number of observations in all samples combined is N. Find the corrected value of $\mathrm { H }$ for the data below which represents test scores for three different groups.
$\begin{array}{lllllll}
\text { Group 1: } & 20 & 18 & 18 & 18 & 20 & 20 \
\text { Group 2: } & 18 & 12 & 18 & 18 & 20 & \
\text { Group 3: } & 16 & 18 & 13 & 18 & 18 &
\end{array}$
A) $3.69$
B) $- 0.19$
C) $4.91$
D) $3.78$

Question

Quizplus · Accepted Answer

The correction factor for ties is:$$1 - \frac { \sum \mathrm { T } } { \mathrm { N } ^ { 3 } - \mathrm { N } }$$For Group 1, there are 3 ties, so $t = 3$ and $T = t^3 - t = 27 - 3 = 24$.For Group 2, there are 2 ties, so $t = 2$ and $T = t^3 - t = 8 - 2 = 6$.For Group 3, there are 3 ties, so $t = 3$ and $T = t^3 - t = 27 - 3 = 24$.Therefore, $\sum \mathrm { T } = 24 + 6 + 24 = 54$.The total number of observations is $N = 15$.Therefore, the correction factor is:$$1 - \frac { 54 } { 15 ^ { 3 } - 15 } = 1 - \frac { 54 } { 3375 } = 0.984$$The corrected value of $H$ is:$$H = \frac { 12 } { N ( N + 1 ) } \sum \frac { R _ { j } ^ { 2 } } { n _ { j } } = \frac { 12 } { 15 ( 15 + 1 ) } \left ( \frac { 108 } { 6 } + \frac { 72 } { 6 } + \frac { 90 } { 6 } \right ) = 3.78$$Therefore, the correct answer is D.

In Using the Kruskal-Wallis Test, There Is a Correction Factor 1−∑TN3−N1 - \frac { \sum \mathrm { T } } { \mathrm { N } ^ { 3 } - \mathrm { N } }1−N3−N∑T​

In Using the Kruskal-Wallis Test, There Is a Correction Factor $1 - \frac { \sum \mathrm { T } } { \mathrm { N } ^ { 3 } - \mathrm { N } }$