Spearman's rank correlation

The formula

rₛ = 1 - 6Σd² / ( n(n² - 1) )

• d = difference between the ranks of each pair; n = number of pairs.
• rₛ runs from +1 (perfect positive correlation) through 0 (none) to -1 (perfect negative correlation).
• Compare the size of rₛ (ignore the sign) with the critical value for n pairs at p = 0.05. If rₛ ≥ critical value, reject H₀: the correlation is significant.

Critical values of Spearman's rank correlation coefficient, from the Pearson WBI16 specimen paper

Ranking, including ties

1. Rank each variable separately, smallest = 1 (or largest = 1, but the same way for both).

2. Tied values share the mean of the ranks they would have taken: values 3, 5, 5, 8 get ranks 1, 2.5, 2.5, 4.

3. Find d for each pair, square it, add to get Σd².

4. Substitute into the formula.

CP11 example: light intensity and percentage cover in 8 quadrats along a transect

Interpreting

rₛ = 1 - (6 x 16) / (8 x 63) = 0.810. Critical value for n = 8 at p = 0.05 is 0.738.

0.810 > 0.738, so reject H₀: there is a significant positive correlation between light intensity and percentage cover. (At p = 0.01 the critical value 0.881 is not reached.)

A significant correlation does not show light causes the higher cover: soil moisture or trampling may vary along the same transect.

Worked example

Calculate rₛ for 8 quadrats where Σd² = 28, and decide whether the correlation is significant at p = 0.05.

1. rₛ = 1 - (6 x 28) / (8 x (64 - 1))
2. = 1 - 168 / 504 = 0.667
3. Critical value (n = 8) = 0.738; 0.667 < 0.738

Answer: rₛ = 0.667; accept H₀: no significant correlation at p = 0.05.

Check yourself

1. Calculate rₛ for 6 pairs with Σd² = 6. Give your answer to 3 decimal places.
   Show answer

   0.829. 1 - (6 x 6) / (6 x 35) = 1 - 36/210 = 0.829.

2. With n = 6, rₛ = 0.829. Using the Pearson table (critical value 0.886 at p = 0.05), what is the conclusion?
   • Reject the null hypothesis: the correlation is significant
   • There is a significant negative correlation
   • Accept the null hypothesis: the correlation is not significant at p = 0.05
   • There is a significant difference between the means
   Show answer

   Accept the null hypothesis: the correlation is not significant at p = 0.05. 0.829 is less than 0.886; with only 6 pairs a strong correlation is needed.

3. For 10 pairs, Σd² = 24. Calculate rₛ to 3 decimal places.
   Show answer

   0.855. 1 - 144 / 990 = 0.855.

4. Values 4, 7, 7, 7, 9 are ranked from smallest. What rank does each 7 receive?
   Show answer

   3. They take ranks 2, 3 and 4, so each gets 3.

5. From the Pearson table, what is the critical value of rₛ for 9 pairs at p = 0.05?
   Show answer

   0.700. n = 9 gives 0.700.

6. rₛ = -0.83 for 8 pairs (critical value 0.738). What does this show?
   • No significant correlation
   • A significant positive correlation
   • The calculation must be wrong because rₛ is negative
   • A significant negative correlation
   Show answer

   A significant negative correlation. Size 0.83 > 0.738; the minus sign shows one variable falls as the other rises.

7. A significant correlation is found between light intensity and plant cover. Why can you not conclude light causes the change?
   • Correlation tests never give significant results
   • Another factor, such as soil moisture, may vary with light and affect cover
   • Light cannot affect plants
   • rₛ only works with means
   Show answer

   Another factor, such as soil moisture, may vary with light and affect cover. Correlation is not causation.

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Sources

• Pearson Edexcel International Advanced Level in Biology specification (XBI11/YBI11, first teaching 2018): Unit 3, Unit 6 and Appendix 6 mathematical skills
• Pearson WBI16/01 specimen paper (2019) with Spearman's rank critical values (via Physics & Maths Tutor)

Last updated 8 November 2026. Spotted a mistake? Tell us.