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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 />

Answers to self-assessment questions<br />

% cover<br />

C.<br />

vulgaris<br />

below, there are two quadrats for each<br />

species where this applies.<br />

Next, calculate D (the difference between the<br />

ranks). Then calculate D 2 for each species (as<br />

in the χ 2 test this removes the negative signs).<br />

Calculate the sum (Σ) of D 2 . (You could type<br />

the table into a spreadsheet and get it to do<br />

all the calculations for you.)<br />

% cover<br />

V.<br />

myrtillus<br />

Rank<br />

C.<br />

vulgaris<br />

C.<br />

vulgaris<br />

D D 2<br />

30 15 9 8 1 1<br />

37 23 11 10 1 1<br />

15 6 5.5 4 1.5 2.25<br />

15 10 5.5 5.5 0 0<br />

20 11 7 7 0 0<br />

9 10 3 5.5 –2.5 6.25<br />

3 3 1 2 –1 1<br />

5 1 2 1 1 1<br />

10 5 4 3 1 1<br />

25 17 8 9 –1 1<br />

35 30 10 11 –1 1<br />

n = 11 ΣD 2 =<br />

15.5<br />

The figures are now inserted into the above<br />

equation.<br />

r s = 1 – 6 × ΣD2<br />

n 3 – n<br />

r s = 1 – 6 × 15.5<br />

1331 – 11<br />

r s = 1 – 0.070<br />

r s = 0.930<br />

A correlation coefficient of +0.93 is very close to<br />

+1, so we can conclude that there is a positive<br />

correlation between the two species and that<br />

the strength of the association is very high.<br />

Now look up the Spearman’s rank coefficient<br />

in the table of critical values that correspond<br />

to the number of pairs of measurements in<br />

results table (there are 11).<br />

Table showing the critical values of r s at p =<br />

0.05 for different numbers of paired values<br />

Number of pairs of<br />

measurements<br />

Critical value at p = 0.05<br />

(5%)<br />

5 1.00<br />

6 0.89<br />

7 0.79<br />

8 0.74<br />

9 0.68<br />

10 0.65<br />

11 0.60<br />

12 0.59<br />

13 0.56<br />

14 0.54<br />

15 0.52<br />

16 0.51<br />

17 0.49<br />

18 0.48<br />

19 0.45<br />

20 0.44<br />

30 0.36<br />

If the value for r s is greater than the critical<br />

value, then you can reject the null hypothesis.<br />

If the value of r s is less than the critical value<br />

you can accept the null hypothesis.<br />

In this case the value for r s is greater than<br />

the critical value so the ecologist can<br />

reject the null hypothesis and accept the<br />

alternative hypothesis.<br />

d There is a significant correlation between<br />

the abundance (as measured by percentage<br />

cover) of the two species on the moorland.<br />

e There are two variables: soil moisture and<br />

percentage cover (or some other measurement<br />

of abundance, such as species density)<br />

Random sampling or systematic sampling<br />

could be used in this investigation. This is the<br />

method for random sampling:<br />

<strong>Cambridge</strong> <strong>International</strong> AS and A <strong>Level</strong> <strong>Biology</strong> © <strong>Cambridge</strong> University Press 2014

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