Spearman's rank correlation

Business Statistics · Unit IV · Topic 11 of 15 · 5 min read

The rank correlation formula, ranks given and ranks to be assigned, tied ranks with the correction factor, merits and demerits, and the three-judges beauty contest, the accountancy and statistics marks and the tea and coffee prices solved.

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On this page

1. Meaning and formula
2. Worked: three judges (January 2025)
3. Worked: assigning ranks (June 2025)
4. Worked: tied ranks (illustration)
5. Merits and demerits (the August 2024 short note)
How this is examined

Cases checked to exist. Read the judgment before quoting a holding.

Every case cited here has been checked to exist — the citation resolves and names the right parties. The propositions attributed to each case have not been verified line by line against the judgment. Case names link through to the full judgment: read it before quoting a holding in an examination answer, and never carry a proposition from notes into court on trust.

The examinable core. ρ = 1 − 6ΣD² / (n³ − n). The three-judges problem (January 2025) asks which pair has “the nearest approach to common taste”: compute ρ for all three pairs and pick the highest positive value. Check every time that ΣD = 0.

1. Meaning and formula

Charles Spearman (1904) devised a coefficient based on ranks rather than on actual values. It is used for qualitative characteristics that cannot be measured but can be ranked (beauty, honesty, intelligence, taste), and for quantitative data where only the order matters or where extreme values would distort Karl Pearson’s r.

ρ (or R) = 1 − 6ΣD² / [n(n² − 1)], where D = difference between the two ranks of each item and n = number of pairs. Like r, ρ lies between −1 and +1; it is in fact Karl Pearson’s r computed on the ranks.

Three situations:

Ranks given (the judges problem): use them directly.
Values given: assign ranks, highest value = 1 (or lowest = 1, but use the same rule for both series).
Repeated (tied) values: give each tied item the average of the ranks they would occupy, and add a correction factor (m³ − m)/12 to ΣD² for every group of m tied items: ρ = 1 − 6[ΣD² + Σ(m³ − m)/12] / (n³ − n).

2. Worked: three judges (January 2025)

Ten competitors in a beauty contest were ranked by three judges. Which pair has the nearest approach to common taste?

R₁	R₂	R₃	D₁₂	D²	D₂₃	D²	D₁₃	D²
1	3	6	−2	4	−3	9	−5	25
6	5	4	1	1	1	1	2	4
5	8	9	−3	9	−1	1	−4	16
10	4	8	6	36	−4	16	2	4
3	7	1	−4	16	6	36	2	4
2	10	2	−8	64	8	64	0	0
4	2	3	2	4	−1	1	1	1
9	1	10	8	64	−9	81	−1	1
7	6	5	1	1	1	1	2	4
8	9	7	−1	1	2	4	1	1
Total			0	200	0	214	0	60

n = 10, n(n² − 1) = 990.

ρ₁₂ = 1 − (6 × 200)/990 = 1 − 1.212 = −0.212.
ρ₂₃ = 1 − (6 × 214)/990 = 1 − 1.297 = −0.297.
ρ₁₃ = 1 − (6 × 60)/990 = 1 − 0.364 = +0.636.

Conclusion: Judges 1 and 3 have the nearest approach to common taste in beauty (ρ = +0.636). Judge 2’s taste runs mildly opposite to both others.

3. Worked: assigning ranks (June 2025)

Marks in Accountancy (X): 78, 39, 36, 65, 62, 90, 82, 75, 25, 98. Statistics (Y): 84, 47, 51, 53, 58, 86, 62, 68, 60, 91. Rank 1 = highest.

X	R_X	Y	R_Y	D	D²
78	4	84	3	1	1
39	8	47	10	−2	4
36	9	51	9	0	0
65	6	53	8	−2	4
62	7	58	7	0	0
90	2	86	2	0	0
82	3	62	5	−2	4
75	5	68	4	1	1
25	10	60	6	4	16
98	1	91	1	0	0
Total				0	30

ρ = 1 − (6 × 30)/990 = 1 − 0.182 = +0.818: high positive correlation between the two subjects.

Tea and coffee prices (100-mark January 2026). Tea: 75, 88, 95, 70, 60, 80, 81, 50 → ranks 5, 2, 1, 6, 7, 4, 3, 8. Coffee: 120, 134, 150, 115, 110, 140, 142, 100 → ranks 5, 4, 1, 6, 7, 3, 2, 8. D = 0, −2, 0, 0, 0, 1, 1, 0; ΣD² = 6; n = 8, n(n² − 1) = 504. ρ = 1 − 36/504 = 1 − 0.071 = +0.929.

4. Worked: tied ranks (illustration)

X: 10, 15, 12, 15, 18; Y: 8, 12, 10, 14, 11. Ranks of X (highest = 1): 18 → 1, the two 15s share ranks 2 and 3 → 2.5 each, 12 → 4, 10 → 5. Ranks of Y: 14 → 1, 12 → 2, 11 → 3, 10 → 4, 8 → 5.

Pairs (R_X, R_Y): (5, 5), (2.5, 2), (4, 4), (2.5, 1), (1, 3). D = 0, 0.5, 0, 1.5, −2; ΣD² = 0 + 0.25 + 0 + 2.25 + 4 = 6.5. One tie of m = 2: correction = (8 − 2)/12 = 0.5.

ρ = 1 − 6(6.5 + 0.5)/(125 − 5) = 1 − 42/120 = 0.65.

5. Merits and demerits (the August 2024 short note)

Merits: simple to understand and calculate; the only method for qualitative data that can be ranked; not distorted by extreme values; gives the same result as Pearson’s r when there are no ties and values are ranks. Demerits: uses only the order, so some information in the actual values is lost; cannot be used for a bivariate frequency table; laborious when n is large (above about 30) unless ranks are given; it is only an approximation of r for the original values.

Trap. ΣD must be zero. If it is not, a rank has been assigned wrongly. In the three-judges problem, the “nearest approach” is the largest positive ρ; a negative value means opposite tastes.

How this is examined

Three judges: January 2025 (ten marks). Assigning ranks to marks: June 2025 (ten marks). Tea and coffee prices: the 100-mark January 2026 paper (“by Spearman’s rank correlation method only”). “Write a note on Spearman’s coefficient of rank correlation”: August 2024. Show the D column with signs and the ΣD = 0 check.

Now try the real questions: Business Statistics past question papers.