Cambridge International A Level Biology Revision Guide

Cambridge International A Level Biology
434
Quadrat Percentage cover
C. vulgaris V. myrtillus
1 30 15
2 37 23
3 15 6
4 15 10
5 20 11
6 9 10
7 3 3
8 5 1
9 10 5
10 25 17
11 35 30
Table 18.2 Results from study of composition of vegetation
on moorland.
to make a null hypothesis that there is no correlation
between the percentage cover of the two species.
The equation to calculate the Spearman rank
correlation, r s
, is:
r s
= 1 −
6 × ∑D 2
n 3 − n
where n is the number of pairs of items in the sample
and D is the difference between each pair of ranked
measurements and ∑ is the ‘sum of’.
The next step is to draw a scatter graph to see if it
looks as if there is a correlation between the abundance
of the two species. This can be done very quickly using
the graphing facility of a spreadsheet program. You can
now follow the steps shown on page 503 to calculate the
value for r s
. Again, the quickest way to do this is to set up a
spreadsheet that will do the calculations.
The ecologist calculated the value of r s
to be 0.930.
A correlation coefficient of +0.930 is very close to +1, so we
can conclude that there is a positive correlation between
the two species and that the strength of the association
is very high. The ecologist was also able to reject the null
hypothesis and accept the alternative hypothesis that there
is a correlation between the abundance of C. vulgaris and
V. myrtillus on this reclaimed moorland.
QUESTION
18.8 a Draw a scatter graph to show the data in
Table 18.2.
b Follow the worked example on page 503 to
calculate the Spearman rank correlation
coefficient (r s
) for the data in Table 18.2. Show all
the steps in your calculation.
c Explain why the ecologist was able to reject the
null hypothesis.
d State the conclusion that the ecologist could make
from this investigation.
e The ecologist thinks that the relationship is the
result of habitat preference, that both species
prefer drier soil. Describe an investigation that he
could carry out to test this hypothesis.
Pearson’s linear correlation
In many investigations, the data collected are for two
continuous variables and the data within each variable
show a normal distribution. When this is the case,
Pearson’s correlation coefficient can be used. This also
calculates numbers between +1 and -1 and the result is
interpreted in the same way. The method of calculation
looks more complex, but the test can be done easily with a
spreadsheet.
The first step is to check whether the relationship
between the two continuous variables appears to be linear
by drawing a scatter graph. The correlation coefficient
should not be calculated if the relationship is not linear.
For correlation only purposes, it does not really matter on
which axis the variables are plotted. The nearer the scatter
of points is to a straight line, the higher the strength of
association between the variables. Before using this test
you must also be satisfied that the data for the variables
you are investigating show a normal distribution.
As trees grow older, they tend to get cracks in their
bark. A student measured the width of cracks on many
pine trees in a plantation and found that they varied
considerably. The data she collected showed a normal
distribution. She noticed that the larger, and presumably
older, trees tended to have wider cracks in their bark
than the smaller trees. She wanted to see if there was a
correlation between the size of the trees and the size of
these cracks. She chose to measure the circumference of
each tree as a measure of their overall size. She measured
the width of the cracks in the bark. This means that she
collected continuous data for each of the two variables
– tree circumference and crack width. She investigated
this by selecting twelve trees at random and measuring