Ten competitors in a debate contest are ranked by three judges in the following order.

Competitors A B C D E F G H I J

Ranks By 1^st Judge 7 4 10 5 9 8 6 2 1 3

Ranks By 2^nd Judge 4 1 9 10 7 3 2 5 6 8

Ranks By 3^rd Judge 10 2 8 5 7 6 9 1 4 3

Using the ranking correlation method and state which pair of judges have the nearest approach.

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Solution

Competitors R₁ R₂ R₃ D₁= R₁ - R₂
$D_{1}^{2}$ D₂= R₁ - R₃ $D_{2}^{2}$ D₃= R₂ - R₃ $D_{3}^{2}$

A
B
C
D
E
F
G
H
I
J 7
4
10
5
9
8
6
2
1
3 4
1
9
10
7
3
2
5
6
8 10
2
8
5
7
6
9
1
4
3 3
3
1
−5
2
5
4
−3
−5
−5 9
9
1
25
4
25
16
9
25
25 −3
2
2
0
2
2
−3
1
−3
0 9
4
4
0
4
4
9
1
9
0 −6
−1
1
5
0
−3
−7
4
2
5 36
1
1
25
0
9
49
16
4
25

$Σ {D_{1}}^{2}$ =148 $Σ {D_{2}}^{2}$ =44 $Σ {D_{3}}^{2}$ =166

N = 10

Rank Correlation between Judge 1 and Judge 2
${r_{k}}_{1, 2} = 1 - \frac{6 \sum_{}^{} D_{1}^{2}}{N^{3} - N} = 1 - \frac{6 \times 148}{1000 - 10} or, {r_{k}}_{1, 2} = 1 - 0.896 \Rightarrow {r_{k}}_{1, 2} = 0.103$
Rank Correlation between Judge 1 and Judge 3

${r_{k}}_{1, 3} = 1 - \frac{6 \sum_{}^{} D_{2}^{2}}{N^{3} - N} = 1 - \frac{6 \times 44}{1000 - 10} or, {r_{k}}_{1, 3} = 1 - 0.266 \Rightarrow {r_{k}}_{1, 3} = 0.73$
Rank Correlation between Judge 2 and Judge 3

${r_{k}}_{2, 3} = 1 - \frac{6 \sum_{}^{} D_{3}^{2}}{N^{3} - N} = 1 - \frac{6 \times 166}{1000 - 10} or, {r_{k}}_{2, 3} = 1 - 1.006 \Rightarrow {r_{k}}_{2, 3} = 0.006$

As the rank correlation coefficient between Judge 1 and Judge 3 is highest, thus Judge 1 and Judge 3 has the nearest approach.

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Similar questions

In a singing competition, two judges rank the seven contestants as follows:

$\begin{matrix} Judge 1 & 5 & 4 & 7 & 3 & 1 & 2 & 6 Judge 2 & 6 & 5 & 2 & 1 & 3 & 4 & 7 \end{matrix}$

Calculate the coefficient of rank correlation.

X	1	2	3	4	5	6	7	8	9	10	11	12
Y	12	9	6	10	3	5	4	7	8	2	11	1

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