When conditions change across a habitat, from shade to sun or from wet to dry, species change with them. Transects record that change, and Spearman's rank and Pearson's correlation tell you whether the link between a factor and a species is real or could be chance.
A line transect is a tape laid across a habitat, recording the organisms that touch it at regular intervals; a belt transect places quadrats at intervals along the line, giving abundance as well as presence. Measuring an abiotic factor (light, soil moisture, pH) or a biotic factor at each point lets you test for a correlation with a species' abundance. Spearman's rank correlation, rs = 1 − 6ΣD2 ÷ (n3 − n), works on ranks and suits data that are not normally distributed. Pearson's correlation coefficient r measures how close the points are to a straight line and needs both variables to be normally distributed. Both run from −1 to +1; the value is compared with a critical value at p = 0.05 to accept or reject the null hypothesis of no correlation. A correlation never proves cause.
Fig. 1: a belt transect from woodland into an open field, and the scatter graph of plant numbers against light intensity
Section 01
Line and belt transects.
Press Sample along the transect. Ten quadrats are placed along the white line from the woodland into the open field, and the graph plots dandelions per quadrat against light intensity (about 200 to 2000 lux). Repeat a few times and read Spearman's rs and Pearson's r each time.
Method
How
Gives
line transect
lay a tape across the habitat; record each species touching the line at regular intervals (for example every metre)
presence and distribution along a gradient; quick, but no measure of abundance
belt transect
place a frame quadrat at regular intervals along the tape (continuous or interrupted)
density, frequency or percentage cover at each point, so abundance as well as distribution
When to use a transect: where there is an obvious gradient, such as from a wood into a field, up a seashore or away from a pond. Random sampling would miss the pattern.
Pair the data: at every quadrat, also measure the factor of interest (light meter, soil moisture probe, pH probe, or the number of a competing or prey species).
Repeat: run several parallel transects and use mean values at each distance.
Examiner trap
A transect is systematic, not random sampling: it is used deliberately along a gradient. For estimating the overall diversity of a uniform area, use random quadrats instead.
Stage 1 · Learn
Check what the sim just showed you
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
In the transect tab, how are the ten quadrats arranged?
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The simulator usually gives a Spearman rs well above 0.65 for 10 quadrats. What does this show?
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What abiotic factor does the simulator measure at each quadrat?
Quick check+10 XP
Why is the simulated belt transect more useful here than a line transect?
Section 02
Spearman's rank correlation.
Spearman's test is the one to reach for with field counts, which are rarely normally distributed.
Null hypothesis: there is no correlation between the two variables.
Step 1: rank each variable separately, 1 for the smallest. Tied values share the mean of the ranks they would occupy (two values tying for 3rd and 4th each get 3.5).
Step 2: for each pair, find the difference in ranks, D, and square it.
Step 3: rs = 1 − (6 × ΣD2) ÷ (n3 − n), where n is the number of pairs.
Step 4: compare rs (ignoring its sign) with the critical value for your n at p = 0.05. If it is equal to or greater than the critical value, reject the null hypothesis: the correlation is significant, and there is less than a 5% probability that it arose by chance.
Sign: positive rs means both rise together; negative means one falls as the other rises.
Light / lux
Rank
Plants
Rank
D
D2
120
1
2
1
0
0
300
2
5
3
−1
1
450
3
4
2
1
1
600
4
8
5
−1
1
900
5
7
4
1
1
1100
6
11
6
0
0
1500
7
12
7
0
0
Worked result: ΣD2 = 4, n = 7, so rs = 1 − 24 ÷ 336 = 0.93. The critical value for n = 7 at p = 0.05 is about 0.79, so there is a significant positive correlation.
Examiner trap
A significant correlation does not show that one variable causes the other. Light might affect the plants, or both might depend on a third factor such as soil moisture or trampling.
Section 03
Pearson's linear correlation.
Pearson's coefficient uses the actual values, not ranks, so it asks a stricter question: how close do the points lie to a straight line?
Formula (given in the exam): r = (Σxy − n x̄ ȳ) ÷ ((n − 1) sx sy), where x̄ and ȳ are the means and sx and sy the standard deviations.
Conditions: both variables are measured on a continuous scale, are approximately normally distributed, and a scatter graph suggests a linear relationship. Plot the scatter graph first.
Values: +1 is a perfect positive linear correlation, −1 a perfect negative one, 0 no linear correlation. Values such as 0.8 or −0.8 indicate strong correlation; significance is again judged against a critical value for n.
Choose
When
Spearman's rank
data not normally distributed, ordinal (ranked) data, small samples, or a relationship that rises (or falls) steadily but is not a straight line
Pearson's
both variables continuous and normally distributed, and the scatter graph looks linear
Stage 2 · Exam
Exam-style questions
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Finish the four checks above to unlock the exam questions
Exam style+20 XP
For 8 pairs of data, ΣD2 = 12. Using rs = 1 − 6ΣD2 ÷ (n3 − n), what is Spearman's rs?
Exam style+20 XP
A student finds rs = −0.52 for 10 quadrats. The critical value at p = 0.05 is 0.65. What should the student conclude?
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Which condition must be met before Pearson's correlation coefficient is used?
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A belt transect up a rocky shore shows a significant negative correlation between height above low tide and the number of a seaweed. Which is the best conclusion?
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