Spearman's Rank: Testing Correlation

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Edexcel IAL Biology • Unit 6: Practical Skills in Biology II

Spearman's Rank: Testing Correlation

📉 Big idea: Spearman's rank asks whether two variables move together — and it does it by comparing ranks rather than raw values, which is why it works on data that is not normally distributed. What it can never do is tell you which variable causes the other.

Summary — What This Topic Covers

When to use Spearman's rank rather than a t-test or chi-squared
Ranking data, and handling tied ranks
The formula and setting the calculation out
Interpreting r_s between −1 and +1
Why correlation is not causation, and what to say instead

1. When to Use It

Use Spearman's rank when

you have pairs of measurements from the same individual, quadrat or site, and you want to know whether the two variables are associated. At least 7 pairs are needed.

✓ light intensity and shoot length at 12 points along a transect ✓ soil moisture and species abundance in 10 quadrats ✓ body length and mass in 15 fish ✗ comparing two group means → t-test ✗ counts in categories → chi-squared ✗ fewer than 7 pairs → too few to test

Why ranks

Ranking discards the actual values and keeps only the order. This makes the test non-parametric: it does not assume the data is normally distributed, which is often untrue of field data.

2. Ranking and Ties

value: 4.2 6.8 6.8 9.1 12.0 rank: 1 2.5 2.5 4 5 the two 6.8 values would occupy ranks 2 and 3 each is given the MEAN of those ranks: (2 + 3) ÷ 2 = 2.5 the next value then takes rank 4 — rank 3 is skipped

Rank each variable separately, from smallest to largest
Tied values share the mean of the ranks they would have occupied
After a tie, continue from the next unused rank — do not renumber
Check: the highest rank should equal n if there are no ties at the top

Common mistake

Ranking the two variables together as one list. They are ranked independently — the whole point is to compare the two orderings.

3. The Calculation

The formula

r_s = 1 − [ 6 Σd² ÷ n(n² − 1) ]

where d is the difference between the two ranks for each pair, and n is the number of pairs.

site light rank shoot rank d d² ────────────────────────────────────────────────── 1 12 1 4.1 1 0 0 2 18 2 5.0 2 0 0 3 25 3 6.2 4 −1 1 4 31 4 5.8 3 1 1 5 40 5 7.4 5 0 0 6 52 6 8.1 6 0 0 7 63 7 9.0 7 0 0 ────────────────────────────────────────────────── Σd² = 2 rs = 1 − [ (6 × 2) ÷ 7(49 − 1) ] = 1 − [ 12 ÷ 336 ] = 1 − 0.0357 = 0.964

4. Interpreting r_s

+1.0 perfect positive correlation — ranks identical +0.8 strong positive +0.4 weak positive 0.0 no correlation −0.4 weak negative −0.8 strong negative −1.0 perfect negative — ranks exactly reversed

The test

Compare the calculated r_s with the critical value for n at p = 0.05. If r_s ≥ critical value, reject the null hypothesis — the correlation is significant. Use the size of r_s, ignoring the sign, when comparing with the table.

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