Cambridge International A Level Biology Revision Guide

Chapter P2: Planning, analysis and evaluation
Spearman’s rank correlation
Spearman's rank correleation is used to find out if there
is a correlation between two sets of variables, when they
are not normally distributed.
As with Pearson’s linear correlation test, the first thing
to do is to plot your data as a scatter graph, and see if they
look as though there may be a correlation. Note that, for
this test, the correlation need not be a straight line – the
correlation need not be linear.
Let’s say that you have counted the numbers of species
R and species S in 10 quadrats. Table P2.5 shows your
results, and Figure P2.9 shows these data plotted as a
scatter graph.
Number of species Number of species
Quadrat
R
S
1 38 24
2 2 5
3 22 8
4 50 31
5 28 27
6 8 4
7 42 36
8 13 6
9 20 11
10 43 30
Table P2.5 Numbers of species R and species S found in 10
quadrats.
Number of individuals of species R
50
40
30
20
10
0
0 5 10 15 20 25 30 35
Number of individuals of species S
Figure P2.9 Scatter graph of the data in Table P2.5.
Now rank each set of data. For example, for the
number of species R, Quadrat 4 has the largest number, so
that is ranked as number 1. This is shown in Table P2.6.
Quadrat
Number
of species
R
Rank for
species
R
Number
of species
S
Rank for
species
S
1 38 7 24 6
2 2 1 5 2
3 22 5 8 4
4 50 10 31 9
5 28 6 27 7
6 8 2 4 1
7 42 8 36 10
8 13 3 6 3
9 20 4 11 5
10 43 9 30 8
Table P2.6 Ranked data from Table 2.5.
Once you have ranked both sets of results, you need
to calculate the differences in rank, D, by subtracting the
rank of species S from the rank of species R. Then square
each of these values. Add them together to find ∑D 2 . This
is shown in Table P2.7.
Quadrat
Rank for
species
R
Rank for
species S
Difference
in rank, D
1 7 6 1 1
2 1 2 –1 1
3 5 4 1 1
4 10 9 1 1
5 6 7 –1 1
6 2 1 1 1
7 8 10 –2 4
8 3 3 0 0
9 4 5 –1 1
10 9 8 1 1
∑D 2 = 12
Table P2.7 Calculating ∑D 2 for the data in Table P2.5.
The formula for calculating Spearman’s rank correlation
coefficient is:
6 × ∑D 2
r s
= 1 −
n 3 − n
where:
r s
is Spearman’s rank coefficient
∑D 2 is the sum of the differences between the ranks
of the two samples
n is the number of samples
D 2
503