Spearman Rank Correlation Coefficient
Laura Armstrong & Joe Wolfensohn
Teachers
Contents
Recall Questions
This topic requires prior knowledge of data collection methods in ecology and an understanding of statistical relationships and correlation. You can test your knowledge on these below.
What is a correlation?
A statistical relationship or association between two variables.
What is meant by an abiotic factor?
A non-living environmental factor, such as light intensity, temperature, or soil pH.
When might a biologist want to measure the strength of a relationship between two variables?
When investigating how one factor (e.g. light) affects the abundance or distribution of a species.
Topic Explainer Video 1
Check out this @JoeDoesBiology video that explains spearman rank correlation coefficient or read the full notes below. Once you've gone through the whole note, try out the practice questions!
Topic Explainer Video 2
If you want to finish off the second part of @JoeDoesBiology's explanation of spearman rank correlation coefficient, check out this video and once you've gone through the whole note, try out the practice questions!
What Is Spearman’s Rank?
A statistical test used to determine whether there is a correlation between two sets of ranked data.
It calculates the strength of the correlation and whether it is positive or negative and can be used to test the significance of the correlation.
Formula:
Where:
rs = Spearman’s rank correlation coefficient
d = difference in ranks between paired values
n = number of pairs
Step-by-Step Calculation Guide
-
Rank each set of data (lowest = 1, highest = n). If the values are the same, the mean rank between them is given.
-
Calculate the difference in ranks (d) for each pair.
-
Square each difference (d²).
-
Sum the d² values.
-
Substitute into the formula to calculate rs.
Interpreting Spearman’s Rank (rs)
|
rs Value |
Interpretation |
|
+1 |
Perfect positive correlation |
|
0 |
No correlation |
|
–1 |
Perfect negative correlation |
-
The closer rs is to +1 or –1, the stronger the correlation.
-
Use a critical value table to determine if the correlation is statistically significant (based on sample size and significance level, e.g. p = 0.05).
Worked Example
A biologist wants to investigate the relationship between soil nitrate concentration and the index of diversity.
Null hypothesis: There is no correlation between soil nitrate concentration and the index of diversity.
|
Site |
NO₂ concentration (ppm) |
Rank |
Simpson's Index of Diversity (D) |
Rank |
d |
d2 |
|
A |
0.012 |
1 |
0.82 |
10 |
9 |
81 |
|
B |
0.032 |
4 |
0.69 |
8 |
4 |
16 |
|
C |
0.025 |
3 |
0.73 |
9 |
6 |
36 |
|
D |
0.037 |
5 |
0.52 |
6 |
1 |
1 |
|
E |
0.018 |
2 |
0.40 |
4 |
2 |
4 |
|
F |
0.053 |
8.5 |
0.60 |
7 |
1.5 |
2.25 |
|
G |
0.041 |
6 |
0.49 |
5 |
1 |
1 |
|
H |
0.049 |
7 |
0.38 |
3 |
4 |
16 |
|
I |
0.053 |
8.5 |
0.30 |
1 |
7.5 |
56.25 |
|
J |
0.065 |
10 |
0.33 |
2 |
8 |
64 |
|
Σ = 277.5 |
-
Rank nitrate and index of diversity values.
-
Calculate d = difference between ranks.
-
Calculate d² and sum.
-
Use formula to calculate rs.
- rs= 1 - ((6 x 277.5) / 10(102-1))
- rs= - 0.682
Result: rs= - 0.682 — moderate negative correlation
Compare to critical value for n = 10 at p = 0.05
This is because we have 10 paired values.
|
n |
p = 0.1 |
p = 0.05 |
p = 0.01 |
|
9 |
0.6 |
0.7 |
0.833 |
|
10 |
0.564 |
0.648 |
0.794 |
|
11 |
0.536 |
0.618 |
0.755 |
Critical value = 0.648
0.682 > 0.648 (we can ignore the negative sign when comparing to the critical value)
As the calculated value exceeds the critical value, we reject the null hypothesis and the correlation is significant.
There is a less than 5% probability that the correlation is due to chance.
Key Terms
-
Correlation: a relationship between two variables.
-
Spearman’s rank: a statistical test to measure the strength and direction of correlation.
No answer provided.
Exam Tips
Make sure data is ranked correctly - lowest to highest.
Clearly state whether your rs value shows a positive or negative correlation and the strength of the correlation.
Use the same phrasing when concluding to ensure you get the marks.
Don’t confuse correlation with causation - correlation does not prove causation, other factors could influence the results.
No answer provided.
A student measured light intensity and % cover of bracken at 10 points along a woodland transect. She calculated a Spearman’s rank correlation coefficient of rs= – 0.43. The critical value at p = 0.05 for n = 10 is 0.648.
Explain what this result shows about the relationship between light intensity and bracken cover. (3 marks)
-
The correlation is negative, suggesting that as light intensity increases, bracken cover decreases.
-
The rs value (–0.43) is less than the critical value of 0.648, so the correlation is not significant.
-
There is a more than 5% probability that the correlation is due to chance.
Practice Question
Try to answer the practice question from the TikTok on your own, then watch the video to see how well you did!