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Cambridge International A Level Biology Revision Guide

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<strong>Cambridge</strong> <strong>International</strong> A <strong>Level</strong> <strong>Biology</strong><br />

The formula for this test is:<br />

∑xy −nx − y − r =<br />

ns x<br />

s y<br />

where:<br />

r is the correlation coefficient<br />

x is the number of species P in a quadrat<br />

y is the number of species Q in the same quadrat<br />

n is the number of readings (in this case, 10)<br />

−x is the mean number of species P<br />

−y is the mean number of species Q<br />

s x<br />

is the standard deviation for the numbers of P<br />

s y<br />

is the standard deviation for the numbers of Q<br />

Table P2.4 shows you the steps to follow in order to<br />

calculate the value of r, using the data in Table P2.3.<br />

The value of r, 0.81, is the correlation coefficient. This<br />

value should always work out somewhere between −1 and<br />

+1. (If it doesn’t check your calculation!)<br />

A value of +1 means total positive correlation<br />

between your two sets of figures.<br />

A value of −1 means total negative correlation<br />

between your two sets of figures.<br />

A value of 0 means there is no correlation.<br />

Here, we have a value of r that lies quite close to 1. We<br />

can say that there is a positive, linear correlation between<br />

the numbers of species P and the numbers of species Q.<br />

1 Calculate x × y for<br />

each set of values.<br />

502<br />

2 Calculate the means for<br />

each set of figures, − x<br />

and − y.<br />

3 Calculate n − x − y.<br />

Here, n = 10, − x = 10.2 and<br />

− y = 20.4, so<br />

n − x − y = 10 × 10.2 × 20.4<br />

Quadrat<br />

Number of<br />

species P,<br />

x<br />

Number of<br />

species Q,<br />

y<br />

1 10 21 210<br />

2 9 20 180<br />

3 11 22 242<br />

4 7 17 119<br />

5 8 16 128<br />

6 14 23 322<br />

7 10 20 200<br />

8 12 24 288<br />

9 12 22 264<br />

10 9 19 171<br />

mean x − = 10.2 y − = 20.4<br />

∑xy = 2124<br />

nx − y − 10 × 10.2 × 20.4<br />

= 2080.8<br />

standard deviation s x<br />

= 2.10 s y<br />

= 2.55<br />

r = ∑xy – n− xy<br />

−<br />

ns x<br />

s y<br />

= 2124 – (10 × 10.2 × 20.4)<br />

10 × 2.10 × 2.55<br />

= 2124 – 2080.8<br />

53.55<br />

= 43.2<br />

53.55<br />

= 0.81<br />

xy<br />

4 Add up all the values<br />

of xy, to find ∑xy.<br />

5 Now calculate the<br />

standard deviation, s,<br />

for each set of figures.<br />

The method for doing<br />

this is shown in Table<br />

P2.1 on page 498.<br />

6 Now substitute<br />

your numbers into<br />

the formula and<br />

calculate r.<br />

Table P2.4 Calculating Pearson’s linear correlation for the data in Table P2.3.

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