9 Correlation · 9.1 Correlation
9.1.2 Spearman's rank correlation (with tied ranks) and probable error
Spearman's rank correlation R = 1 − 6Σd² ÷ n(n² − 1) measures agreement between two rankings; the probable error 0.6745 (1 − r²)/√n tells whether a sample r is large enough to be taken seriously.
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- Advanced level
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Core facts
- Spearman's rank correlation coefficient: R = 1 − 6Σd² ÷ n(n² − 1), where d = difference between the two ranks of an item and n = number of pairs.
- Used for qualitative characteristics that can be ranked but not measured: beauty, honesty, merit, taste.
- −1 ≤ R ≤ +1. R = +1: identical rankings; R = −1: exactly reversed rankings.
- Σd is always 0 (a check on the working).
- Tied (repeated) ranks: equal items get the average of the ranks they would occupy (two items tied for 4th and 5th each get 4.5).
- Tie correction: add m(m² − 1)/12 to Σd² for each group of m tied items, in either series.
- Standard error of r: SE = (1 − r²) ÷ √n.
- Probable error of r: PE = 0.6745 × (1 − r²) ÷ √n = 0.6745 × SE.
- Rules: r < PE → not significant; r > 6 PE → significant; in between → nothing definite.
- Limits of the population correlation: r ± PE.
Example 1. Two judges rank 5 entries.
| Entry | Judge 1 | Judge 2 |
|---|---|---|
| A | 1 | 2 |
| B | 2 | 1 |
| C | 3 | 4 |
| D | 4 | 3 |
| E | 5 | 5 |
Solution. d = −1, 1, −1, 1, 0 → Σd² = 4. R = 1 − (6 × 4) ÷ (5 × 24) = 1 − 24/120. Answer: R = 0.8
Example 2. r = 0.6 from a sample of 64 pairs. Find the PE. Solution. (1 − 0.36) ÷ √64 = 0.64/8 = 0.08 (the SE). PE = 0.6745 × 0.08. Answer: PE ≈ 0.054
Explained
1. When actual values are given: rank them first. Rank each series separately (largest = 1, or smallest = 1, but the same rule for both).
Example 3. Marks of 6 students in Economics (x) and Statistics (y):
| Student | x | y |
|---|---|---|
| P | 78 | 84 |
| Q | 36 | 51 |
| R | 98 | 91 |
| S | 25 | 60 |
| T | 75 | 68 |
| U | 82 | 62 |
Working (largest = rank 1):
| Student | Ranks (x, y) | d | d² |
|---|---|---|---|
| P | 3, 2 | 1 | 1 |
| Q | 5, 6 | −1 | 1 |
| R | 1, 1 | 0 | 0 |
| S | 6, 5 | 1 | 1 |
| T | 4, 3 | 1 | 1 |
| U | 2, 4 | −2 | 4 |
Solution. Σd = 0 (check). Σd² = 8. R = 1 − (6 × 8) ÷ (6 × 35) = 1 − 48/210 = 1 − 0.229. Answer: R ≈ 0.771
2. Tied ranks with correction.
Example 4. Find R.
| x | y |
|---|---|
| 48 | 13 |
| 33 | 13 |
| 40 | 24 |
| 9 | 6 |
| 16 | 15 |
| 16 | 4 |
| 65 | 20 |
| 24 | 9 |
| 16 | 6 |
| 57 | 19 |
Solution.
- Ranks of x (largest = 1): 65→1, 57→2, 48→3, 40→4, 33→5, 24→6, the three 16s share places 7, 8, 9 → 8 each, 9→10.
- Ranks of y: 24→1, 20→2, 19→3, 15→4, the two 13s share 5, 6 → 5.5 each, 9→7, the two 6s share 8, 9 → 8.5 each, 4→10.
- In the given order: Rank x = 3, 5, 4, 10, 8, 8, 1, 6, 8, 2; Rank y = 5.5, 5.5, 1, 8.5, 4, 10, 2, 7, 8.5, 3.
- d = −2.5, −0.5, 3, 1.5, 4, −2, −1, −1, −0.5, −1 (Σd = 0).
- Σd² = 6.25 + 0.25 + 9 + 2.25 + 16 + 4 + 1 + 1 + 0.25 + 1 = 41.
- Corrections: x has one tie of m = 3 → 3(9 − 1)/12 = 2. y has two ties of m = 2 → 2(4 − 1)/12 = 0.5 each → 1. Total 3.
- R = 1 − 6(41 + 3) ÷ (10 × 99) = 1 − 264/990.
Answer: R ≈ 0.733
3. Probable error: reading the result.
| r compared with PE | Conclusion |
|---|---|
| r < PE | not significant (no evidence of correlation) |
| PE < r < 6 PE | nothing definite |
| r > 6 PE | significant (correlation is real) |
Example 5. r = 0.6, n = 64 (from Example 2, PE ≈ 0.054). Is r significant? Give the limits. Solution. 6 PE ≈ 0.324 < 0.6, so significant. Limits 0.6 ± 0.054. Answer: significant; ρ lies between about 0.546 and 0.654
Example 6. r = 0.5 from 16 pairs. Comment. Solution. PE = 0.6745 × 0.75 ÷ 4 ≈ 0.126; 6 PE ≈ 0.759. r is above PE but below 6 PE. Answer: nothing definite can be said
Deeper
1. Correcting a wrong rank difference.
Example 7. For 10 pairs, R was found to be 0.5. Later one d was found to be 3 instead of the correct 7. Find the correct R. Solution.
- From R = 1 − 6Σd²/990: 0.5 = 1 − 6Σd²/990 → Σd² = 0.5 × 990/6 = 82.5.
- Correct Σd² = 82.5 − 3² + 7² = 82.5 − 9 + 49 = 122.5.
- R = 1 − 6 × 122.5/990 = 1 − 735/990.
Answer: R ≈ 0.258
2. Extreme values of Σd². Σd² = 0 when the rankings are identical (R = 1). For exactly reversed rankings Σd² = n(n² − 1)/3 and R = −1. Check with n = 5: ranks 1–5 against 5–1 give d = 4, 2, 0, −2, −4 and Σd² = 40 = 5 × 24/3.
Example 8. n = 10 and Σd² = 30. Find R. Solution. 1 − 180/990 = 1 − 0.182. Answer: R ≈ 0.818
3. How R relates to Pearson's r. R is Pearson's r worked out on the ranks. With no ties the two give exactly the same value. With ties, the textbook formula with the correction factor is the standard answer in exams; it can differ slightly from Pearson's r computed on the averaged ranks (in Example 4, 0.733 against about 0.747).
4. Merits and limits.
| Spearman's R | Pearson's r |
|---|---|
| works on ranks; qualitative data | needs measured values |
| simple, quick for small n | more work, uses full values |
| little affected by extreme values | pulled by extreme values |
| loses the size of differences | uses all the information |
Spearman's method is not used for a grouped (bivariate frequency) table, and it becomes tedious for large n (say above 30) unless ranks are already given.
5. About the probable error.
- 0.6745 is the value z such that the middle 50% of a normal distribution lies within ±z standard deviations; so r ± PE is the range within which the population correlation is expected to lie with a 50% chance.
- PE is meaningful only when the sample is random, reasonably large and drawn from a roughly normal population.
- PE falls as n rises: r = 0.8 from 100 pairs gives PE = 0.6745 × 0.36/10 ≈ 0.024.
Common traps
- ✗ "R = 1 − 6Σd ÷ n(n² − 1)." ✓ Use the sum of squared differences, Σd²; Σd itself is always 0.
- ✗ "The denominator is n(n − 1)." ✓ It is n(n² − 1), e.g. 10 × 99 = 990 for n = 10.
- ✗ "Tied items get the first rank of the group." ✓ They get the average rank (tied 7th, 8th, 9th → 8 each).
- ✗ "Add one correction per series." ✓ Add m(m² − 1)/12 for each group of ties, in both series.
- ✗ "PE = 0.6745 (1 − r) ÷ √n." ✓ It is (1 − r²): PE = 0.6745 (1 − r²)/√n.
- ✗ "r is significant if r > PE." ✓ Significant only if r > 6 PE; r < PE means not significant.
- ✗ "Ranks must be given from smallest." ✓ Either direction works if both series use the same rule.
Quick revision
- R = 1 − 6Σd²/[n(n² − 1)].
- Σd = 0 always; check it.
- Ties: average rank; add m(m² − 1)/12 per tie group.
- R = 1 identical, R = −1 reversed; reversed gives Σd² = n(n² − 1)/3.
- No ties: Spearman's R equals Pearson's r on the ranks.
- SE = (1 − r²)/√n.
- PE = 0.6745 (1 − r²)/√n.
- r < PE not significant; r > 6 PE significant.
- Limits of population r: r ± PE.
- Use rank correlation for qualitative traits (beauty, merit).
Practise this topic. Questions on Karl Pearson's & Spearman's correlation, probable error, with answers explained.
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