Spearman's rank correlation

When and why

Use Spearman's rank when you have paired data (two measurements at each site) and want to know if they are related: for example distance from the sea and % vegetation cover, or distance from a city centre and noise. Use at least 10 pairs (fewer makes significance hard to judge).

Formula: rs = 1 - (6 x sum of d squared) / (n cubed - n), where d is the difference between the ranks and n is the number of pairs.

• rs = +1: perfect positive correlation; -1: perfect negative; 0: no correlation.

Method

1. State a null hypothesis: there is no significant relationship between X and Y.

2. Rank each variable separately (rank 1 = highest value). Tied values share the mean of the ranks they would occupy (two values tied for 3rd and 4th both get 3.5).

3. Find d = rank X - rank Y for each pair, and d squared.

4. Add up the d squared values.

5. Put into the formula.

6. Compare rs (ignoring its sign) with the critical value for your n. If it is equal to or greater than the critical value, the result is significant: reject the null hypothesis. Pearson's tables give, for a two-tailed test: n = 10, 0.648 at the 0.05 level and 0.794 at 0.01; n = 12, 0.587 and 0.727; n = 15, 0.521 and 0.654.

7. Interpret: direction, strength, significance, and a geographical explanation. Correlation does not prove causation.

Practice data set

Practice data (not real): distance inland and % vegetation cover on a dune transect.

Worked example

Calculate rs for the practice dune data and decide whether it is significant at the 0.05 level.

1. Sum of d squared = 4; n = 10.
2. rs = 1 - (6 x 4) / (1000 - 10) = 1 - 24 / 990 = 0.976.
3. Critical value for n = 10 at 0.05 (two-tailed) = 0.648.
4. 0.976 > 0.648, so the result is significant: reject the null hypothesis.

Answer: rs = 0.98: a strong, significant positive correlation; vegetation cover increases inland, as succession theory predicts.

Check yourself

1. Calculate rs for the practice dune data. Give your answer to 3 decimal places.
   Show answer

   0.976. rs = 1 - (6 x 4) / 990 = 0.976.

2. Practice data: noise (dB) at the same ten distances from a city centre was 78, 76, 70, 72, 66, 61, 63, 58, 59, 55. Calculate rs between distance and noise. Give your answer to 3 decimal places.
   Show answer

   -0.964. Ranking both and summing d squared gives 324; rs = 1 - (6 x 324) / 990 = -0.964.

3. An investigation with 10 pairs gives rs = 0.61. The critical value at 0.05 is 0.648. What should you conclude?
   • The correlation is significant, so reject the null hypothesis
   • There is a perfect positive correlation
   • The test cannot be used with 10 pairs
   • The correlation is not significant at the 0.05 level, so accept the null hypothesis
   Show answer

   The correlation is not significant at the 0.05 level, so accept the null hypothesis. 0.61 < 0.648, so the relationship could be due to chance.

4. Two values tie for 4th place when ranking. What rank should each be given?
   • 4.5
   • 4
   • 5
   • 9
   Show answer

   4.5. They share the mean of ranks 4 and 5.

5. rs = -0.85 for distance from the sea against % bare sand. This shows
   • a weak positive correlation
   • no relationship
   • a strong negative correlation: bare sand decreases inland
   • that distance causes all changes in sand
   Show answer

   a strong negative correlation: bare sand decreases inland. Negative and close to -1 is strong.

6. Why can a significant Spearman's rank result not prove that distance from the sea causes higher plant cover?
   • Correlation shows a relationship, but other factors such as soil age, humus or trampling may be responsible
   • Spearman's rank is never accurate
   • Plant cover cannot be measured
   • Distance is not a number
   Show answer

   Correlation shows a relationship, but other factors such as soil age, humus or trampling may be responsible. Correlation is not causation.

7. For n = 12, Pearson's critical value at 0.05 (two-tailed) is 0.587. A student calculates rs = 0.590. By how much does it exceed the critical value?
   Show answer

   0.003. 0.590 - 0.587 = 0.003, so it is (just) significant.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Geography specification, Issue 3 (September 2025)
• Pearson WPS03/01 question paper, January 2022 (Spearman's rank and chi-squared critical values tables)
• Pearson WGE02/01 examiners' report, June 2024
• Pearson WGE02/01 mark scheme, January 2024

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