Kendall's Coefficient of Concordance/Examples/Arbitrary Example 2

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Example of Kendall's Coefficient of Concordance

Consider the $4$ competitors $\text {Wilhelmina}$, $\text {Xanthippe}$, $\text {Yondalla}$ and $\text {Zena}$, who are demonstrating their skills to the $4$ judges $\text {Ariadne}$, $\text {Boudicca}$, $\text {Constantine}$ and $\text {Donald}$.

The following table shows the ranking afforded the competitors by each judge, with the row summations per competitor:

  $\textit {Judge}$
$\textit {Competitor}$ $\begin{array} {r {{|}} cccc {{|}} c} & \text A & \text B & \text C & \text D & s_i \\ \hline \text X & 1 & 1 & 4 & 4 & 10 \\ \text Y & 2 & 2 & 3 & 3 & 10 \\ \text Z & 3 & 3 & 2 & 2 & 10 \\ \text Z & 4 & 4 & 1 & 1 & 10 \\ \hline \end{array}$

Kendall's coefficient of concordance is $0$.


Proof

The mean $M$ is given by:

$M = \dfrac 1 2 \times 4 \times \paren {4 + 1} = 10$

The $S$ value is given by:

\(\ds S\) \(=\) \(\ds \sum_{i \mathop = 1}^n \paren {s_i - M}^2\)
\(\ds \) \(=\) \(\ds \paren {10 - 10}^2 + \paren {10 - 10}^2 + \paren {10 - 10}^2 + \paren {10 - 10}^2\)
\(\ds \) \(=\) \(\ds 0\)


Hence Kendall's coefficient of concordance is shown to be:

\(\ds W\) \(=\) \(\ds \dfrac {12 S} {m^2 n \paren {n^2 - 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {12 \times 0} {4^2 \times 4 \times \paren {4^2 - 1} }\)
\(\ds \) \(=\) \(\ds 0\)

$\blacksquare$


Sources

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