Spearman's rank correlation coefficient (rs) tests whether there is a significant correlation between two variables, for example light intensity and the percentage cover of a plant along a transect. It can be used with data that are not normally distributed. You need at least 5 pairs of data, ideally 10 or more.rs = 1 − (6Σd2) ÷ (n(n2 − 1)) where d is the difference between the ranks of each pair and n is the number of pairs.Steps:
1. Null hypothesis: there is no significant correlation between the two variables.
2. Rank each variable separately, from smallest (rank 1) to largest. Tied values get the mean of the ranks they would occupy.
3. Find d for each pair, square it and add the squares to get Σd2.
4. Calculate rs. It lies between +1 (perfect positive correlation) and −1 (perfect negative correlation); 0 means no correlation.
5. Compare rs, ignoring its sign, with the critical value for n at p = 0.05.Worked example: 6 quadrats. Light ranks are 1 to 6; the matching cover ranks are 5, 6, 4, 3, 2, 1. So d = −4, −4, −1, 1, 3, 5 and Σd2 = 68. Here n(n2 − 1) = 6 × 35 = 210.rs = 1 − (6 × 68) ÷ 210 = 1 − 1.94 = −0.94 The critical value for n = 6 is 0.886. Since 0.94 > 0.886, reject the null hypothesis: there is a significant negative correlation between light intensity and cover.
1. Null hypothesis: there is no significant correlation between the two variables.
2. Rank each variable separately, from smallest (rank 1) to largest. Tied values get the mean of the ranks they would occupy.
3. Find d for each pair, square it and add the squares to get Σd2.
4. Calculate rs. It lies between +1 (perfect positive correlation) and −1 (perfect negative correlation); 0 means no correlation.
5. Compare rs, ignoring its sign, with the critical value for n at p = 0.05.Worked example: 6 quadrats. Light ranks are 1 to 6; the matching cover ranks are 5, 6, 4, 3, 2, 1. So d = −4, −4, −1, 1, 3, 5 and Σd2 = 68. Here n(n2 − 1) = 6 × 35 = 210.