Revise Spearman's Rank: Testing Correlation for Biology 6 (IAL) WBI16 (A2 Level) — revision notes and instant AI marking. Free to start.
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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 rs 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
rs = 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.
Compare the calculated rs with the critical value for n at p = 0.05. If rs ≥ critical value, reject the null hypothesis — the correlation is significant. Use the size of rs, ignoring the sign, when comparing with the table.
Easy marks
H₀ for Spearman is: "there is no significant correlation between the two variables." State it, and state the decision in those words.
5. Correlation Is Not Causation
This is the single most examined idea in the topic, and it is worth a mark almost every time.
A significant correlation between light intensity and shoot length
could mean:
light → growth the obvious interpretation
growth → light taller shoots reach more light
X → both soil depth increased along the transect too,
affecting growth AND shading
coincidence with enough variables, some will correlate
Only a controlled experiment, in which you change one variable and hold the rest constant, can establish cause
In fieldwork you are almost always observing, so correlation is the strongest claim available
Name a plausible confounding variable — it shows judgement and often earns the mark
What mark schemes look for
The word "suggests" or "is associated with", not "causes" or "proves" — plus one named confounding variable.
Practice Questions
Practice Question 1
Two values in a set are tied. Explain how they are ranked and what happens to the next value.
Practice Question 2
A student calculates rs = −0.88 for 10 pairs. The critical value at n = 10, p = 0.05 is 0.648. State the conclusion.
Practice Question 3
Explain why a significant correlation between soil moisture and plant abundance does not show that moisture controls abundance.
Practice Question 4
Explain why Spearman's rank is used rather than a test based on the actual measurements.
What to Memorise
Spearman = correlation between paired dataAt least 7 pairsRank each variable separatelyTies share the mean rank, then skiprs = 1 − [6Σd² ÷ n(n² − 1)]+1 perfect positive, −1 perfect negativeCompare the SIZE with the critical valueNon-parametric — no normality assumedCorrelation ≠ causation
Concepts Checklist
Exam Tips
What mark schemes look for
"Suggests an association" rather than "proves" or "causes", plus one named confounding variable. It is close to a guaranteed mark.
The trap
Ranking both variables together in one combined list. They are ranked independently — the test compares the two orderings.
Easy marks
Show the ranking table with the d and d² columns. The intermediate values carry marks even if the final arithmetic slips.
Worth remembering
Ignore the sign when comparing with the table, but always report it in the conclusion — it tells the reader the direction of the relationship.