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Question
A psychologist selected a random sample of 22 students. He grouped them in 11 pairs so that the students in each pair have nearly equal scores in an intelligence test. In each pair, one student was taught by method A and the other by method B and examined after the course. The marks obtained by them after the course are as follows:
| Pairs | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| Methods A | 24 | 29 | 19 | 14 | 30 | 19 | 27 | 30 | 20 | 28 | 11 |
| Methods B | 37 | 35 | 16 | 26 | 23 | 27 | 19 | 20 | 16 | 11 | 21 |
Calculate Spearman’s Rank correlation.
Solution
| Pairs | Methods A (X) | Rx | Methods B (Y) | Ry | (Rx-Ry)2 = D2 |
| 1 | 24 | 5 | 37 | 10 | 25 |
| 2 | 29 | 8 | 35 | 9 | 1 |
| 3 | 19 | 3 | 16 | 2 | 1 |
| 4 | 14 | 2 | 26 | 7 | 25 |
| 5 | 30 | 9 | 23 | 6 | 9 |
| 6 | 19 | 3 | 27 | 8 | 25 |
| 7 | 27 | 6 | 19 | 3 | 9 |
| 8 | 30 | 9 | 20 | 4 | 25 |
| 9 | 20 | 4 | 16 | 2 | 4 |
| 10 | 28 | 7 | 11 | 1 | 36 |
| 11 | 11 | 1 | 21 | 5 | 16 |
| N = 11 | Σ D2 = 176 |
Spearman's Rank correlation :
R = 1 - `(6Σ"D"^2)/("N"^3 - "N")`
= 1 - `(6 xx 176)/((11)^3 - 11) = 1 - (1056)/(1320) = 1 - 0.8 = 0.20`
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