Question

Five competitors in a beauty contest are ranked by three judges in the following order :

$\begin{array}{cccccc}\text{Rank by Judge A}& 1& 2& 3& 4& 5\\ \text{Rank by Judge B}& 2& 4& 1& 5& 3\\ \text{Rank by Judge C}& 1& 3& 5& 2& 4\end{array}$

Using rank correlation coefficient, determine which pair of judges has the nearest approach to tastes in beauty.

Answer 1

$\begin{array}{ccccccccc}{R}_1& R_2& R_3& D_{12}=R_1-R_2& D_{23}=R_2-R_3& D_{13}=R_1-R_3& D_{12}^2& D_{23}^2& D_{13}^2\\ 1& 2& 1& -1& +1& 0& 1& 1& 0\\ 2& 4& 3& -2& +1& -1& 4& 1& 1\\ 3& 1& 5& +2& -4& -2& 4& 16& 4\\ 4& 5& 2& -1& +3& +2& 1& 9& 4\\ 5& 3& 4& +2& -1& +1& 4& 1& 1\\ N=5& & & \sum D_{12}=0& \sum D_{23}=0& \sum D_{13}=0& \sum D_{12}^2& \sum D_{23}^2& \sum D_{13}^2\\ & & & & & & =14& =28& 10\end{array}$

Applying the formulae,

$R_{12}=\frac{6\sum D_{12}^2}{N^3-N}=1-\frac{6\times 14}{5^3-5}=1-\frac{84}{120}=\frac{36}{120}=0.3R_{23}=\frac{6\sum D_{23}^2}{N^3-N}=1-\frac{6\times 28}{5^3-5}=1-\frac{168}{120}=-0.4R_{13}=\frac{6\sum D_{13}^2}{N^3-N}=1-\frac{6\times 10}{5^3-5}=1-\frac{60}{120}=+0.5$

Since the coefficient of rank correlation is positive and highest in the judgement of the judges A and C, we conclude that they have the similar tastes to common tastes in beauty. Judges B and C have very different tastes.