Example:
In an art competition, two judges accorded following ranks to the 10 participants:
Judge X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
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Judge Y | 6 | 2 | 9 | 7 | 1 | 4 | 8 | 3 | 10 | 5 |
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Calculate coefficient of rank correlation.
Solution:
Judge X (R1) | Judge Y (R2) | D = R1 - R2 | D2 |
---|
1 | 6 | -5 | 25 |
2 | 2 | 0 | 0 |
3 | 9 | -6 | 36 |
4 | 7 | -3 | 9 |
5 | 1 | 4 | 16 |
6 | 4 | 2 | 4 |
7 | 8 | -1 | 1 |
8 | 3 | 5 | 25 |
9 | 10 | -1 | 1 |
10 | 5 | 5 | 25 |
N = 10 | | | ∑D2 = 142 |
r_k = 1 - \frac{6\sum{D^2}}{N^3 - N}
= 1 - \frac{6\times{142}}{10^3 - 10}
= 1 - \frac{852}{990}
= 1 - 0.860
= 0.14
Coefficient of Correlation (rk) = 0.14
As the rank correlation is positive and closer to 0, it means that the association between the ranks of the two judges is weaker.
Example:
Calculate the Spearman's Rank Correlation for the following data.
Mathematics | 14 | 15 | 17 | 12 | 16 | 11 | 18 | 9 | 10 |
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Accountancy | 4 | 12 | 8 | 10 | 2 | 5 | 9 | 3 | 7 |
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Solution:
In the given case, there are 9 values, and the ranking for both X and Y or Mathematics and Accountancy is done by giving the highest rank to the highest value and the lowest rank to the lowest value. Therefore, 1st rank is given to 9 in the X series and 2 in the Y series. Similarly, the 9th rank is given to 18 in the X series and 12 in the Y series.
Mathematics (X) | Rank R1 | Accountancy (Y) | Rank R2 | D = R1 - R2 | D2 |
---|
14 | 5 | 4 | 3 | 2 | 4 |
15 | 6 | 12 | 9 | -3 | 9 |
17 | 8 | 8 | 6 | 2 | 4 |
12 | 4 | 10 | 8 | -4 | 16 |
16 | 7 | 2 | 1 | 6 | 36 |
11 | 3 | 5 | 4 | -1 | 1 |
18 | 9 | 9 | 7 | 2 | 4 |
9 | 1 | 3 | 2 | -1 | 1 |
10 | 2 | 7 | 5 | -3 | 9 |
N = 9 | | | | | ∑D2 = 84 |
r_k = 1 - \frac{6\sum{D^2}}{N^3 - N}
= 1 - \frac{6\times84}{9^3 - 9}
= 1 - \frac{504}{720}
= 1 - 0.7
= 0.3
Coefficient of Correlation (rk) = 0.3
It means that there is a positive rank correlation of a moderate degree of 0.3.
Example:
Calculate the coefficient of rank correlation of the scores obtained by 7 students in an essay writing competition by two judges, X and Y.
X | 15 | 12 | 20 | 16 | 18 | 20 | 26 |
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Y | 10 | 15 | 11 | 11 | 25 | 18 | 30 |
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Solution:
In the given case, there are 7 values or students, and ranks have been given as highest rank to the highest score and lowest rank to the lowest score. For instance, for scores given by Judge X, 1st rank is given to the score of 26 and for the scores given by Judge Y, 1st rank is given to the score of 30.
X | Rank R1 | Y | Rank R2 | D | D2 |
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15 | 6 | 10 | 7 | -1 | 1 |
12 | 7 | 15 | 4 | 3 | 9 |
20 | 2.5 | 11 | 5.5 | -3 | 9 |
16 | 5 | 11 | 5.5 | -0.5 | 0.25 |
18 | 4 | 25 | 2 | 2 | 4 |
20 | 2.5 | 18 | 3 | -0.5 | 0.25 |
26 | 1 | 30 | 1 | 0 | 0 |
| | | | | ∑D2 = 23.5 |
Judge X has given 20 scores to two students who are in the place of 2nd and 3rd rank. Therefore, the average of both ranks, i.e., (2+3)/2 = 2.5 rank has been given to both students.
Similarly, Judge Y has given 11 scores to two students who are in the place of 5th and 6th rank. Therefore, the average of both ranks, i.e., (5+6)/2 = 5.5 has been given to both students.
Also, in series X, the number 20 is repeated twice, and in Y series, the number 11 is repeated twice. Therefore, m for series X or m1 is 2 and m for series Y or m2 is 2.
r_k = 1 - \frac{6[\sum D^2 + \frac{1}{12}(m_1^3 - m_1) + \frac{1}{12}(m_2^3 - m_2) + ...]}{N^3 - N}
= 1 - \frac{6[23.5 + \frac{1}{12}(2^3 - 2) + \frac{1}{12}(2^3 - 2)]}{7^3 - 7}
= 1 - \frac{6[23.5 + \frac{1}{2} + \frac{1}{2}]}{336}
= 1 - 6[\frac{24.5}{336}]
= 1 - \frac{147}{336}
= 1 - 0.4375
= 0.5625
Coefficient of Correlation = 0.56
The positive correlation coefficient of 0.56 means that around 25% of the variation is related.