Question

In a beauty contest, three judges accorded following ranks to 10 participants:

Judge I	1	6	5	10	3	2	4	9	7	8
Judge II	3	5	8	4	7	10	2	1	6	9
Judge III	6	4	9	8	1	2	3	10	5	7

Find out by Spearman's Rank Difference Method which pair of judges has a common taste in respect of beauty.

Answer 1

R1	R2	R3	D1 = R1 – R2	D2 = R1 – R3	D3 = R2 – R3	D12	D22	D32
1 6 5 10 3 2 4 9 7 8	3 5 8 4 7 10 2 1 6 9	6 4 9 8 1 2 3 10 5 7	–2 1 –3 6 –4 –8 2 8 1 –1	– 5 2 –4 2 2 0 1 –1 2 1	–3 1 –1 –4 6 8 –1 –9 1 8	4 1 9 36 16 64 4 64 1 1	25 4 16 4 4 0 1 1 4 1	9 1 1 16 36 64 1 81 1 64
						${\sum }_{}{D_1}^2=200$	${\sum }_{}{D_2}^2=60$	${\sum }_{}{D_3}^2=214$

N = 10

$$\begin{gathered} \underline{\mathbf{Rank}\mathbf{}\mathbf{Correlation}\mathbf{}\mathbf{between}\mathbf{}\mathbf{Judge}\mathbf{}\mathbf{1}\mathbf{}\mathbf{and}\mathbf{}\mathbf{Judge}\mathbf{}\mathbf{2}} \\ {r}_{k_{1,2}}=1-\frac{6{\sum }_{}{D_1}^2}{N^3-N}=1-\frac{6\times 200}{1000-10}=\frac{990-1200}{990}=-0.212 \\ \underline{\mathbf{Rank}\mathbf{}\mathbf{Correlation}\mathbf{}\mathbf{between}\mathbf{}\mathbf{Judge}\mathbf{}\mathbf{1}\mathbf{}\mathbf{and}\mathbf{}\mathbf{Judge}\mathbf{}\mathbf{3}} \\ {r}_{k_{1,3}}=1-\frac{6{\sum }_{}{D_2}^2}{N^3-N}=1-\frac{6\times 60}{1000-10}=\frac{990-360}{990}=+0.636 \\ \underline{\mathbf{Rank}\mathbf{}\mathbf{Correlation}\mathbf{}\mathbf{between}\mathbf{}\mathbf{Judge}\mathbf{}\mathbf{2}\mathbf{}\mathbf{and}\mathbf{}\mathbf{Judge}\mathbf{}\mathbf{3}} \\ {r}_{k_{2,3}}=1-\frac{6{\sum }_{}{D_3}^2}{N^3-N}=1-\frac{6\times 214}{1000-10}=\frac{990-1284}{990}=-0.296 \end{gathered}$$


Observation and Conclusion:
As the rank correlation coefficient between Judge 1 and Judge 3 is highest and positive, so it can be regarded that they have a common taste in respect of beauty.