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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 end-of-chapter questions<br />

s = 13 659<br />

39<br />

s = 116.486<br />

39<br />

s = 2.99 [4]<br />

b s M = 2.99<br />

40<br />

s M = 2.99<br />

6.32<br />

s M = 0.47 [2]<br />

c<br />

Mean mass / g<br />

100<br />

80<br />

60<br />

40<br />

20<br />

0<br />

Error bars represent<br />

±2 × the standard deviation<br />

Lemon fruits set 1 Lemon fruits set 2<br />

[5]<br />

d the error bars for the two sets of fruits do<br />

not overlap; it is therefore possible that the<br />

difference between them is significant, but<br />

we cannot be sure about this; we would need<br />

to do a further statistical test, such as a t-test,<br />

to be able to answer this question; [3]<br />

[Total: 14]<br />

2 Formula for the t-test is:<br />

x 1 – x 2<br />

t =<br />

2<br />

s 1<br />

n + s 2<br />

2<br />

1 n 2<br />

All of these values are shown on page 476.<br />

They are:<br />

x 1 = 3.12<br />

x 2 = 3.02<br />

s 1 = 0.17<br />

s 2 = 0.16<br />

n 1 = 21<br />

n 2 = 19<br />

x 1 – x 2 =3.12 – 3.02 = 0.10<br />

s 1 2 = 0.17 2 = 0.0289<br />

so s 1 2 ÷ n 1 = 0.0289 ÷ 21 = 0.00138<br />

s 2 2 = 0.16 2 = 0.0256<br />

so s 2 2 ÷ n 2 = 0.0256 ÷ 19 = 0.00135<br />

s 1 2 ÷ n 1 + s 2 2 ÷ n 2 = 0.00138 + 0.00135 = 0.00273<br />

The square root of this value is 0.0522.<br />

Therefore t = 0.10 ÷ 0.0522 = 1.92<br />

Now work out the number of degrees of<br />

freedom, v:<br />

v = (n 1 – 1) + (n 2 –1) = (21 – 1) + (19 – 1) = 38<br />

In the probability table for t on page 000, look<br />

up the critical value of t at the 5% level for<br />

this number of degrees of freedom. This is<br />

the value in the >30 row, which is 1.96.<br />

Our value of t is smaller than this. We would have<br />

to move to the left in the table, towards the value<br />

for the 10% confidence level, to get this value.<br />

This means that the probability of the difference<br />

between the two sets of data being caused by<br />

chance is more than 0.05, or more than 5%. We<br />

can therefore say that there is no significant<br />

difference between the two sets of data. [10]<br />

3 For species P:<br />

x x – x (x – x) 2<br />

10 –0.2 0.04<br />

9 –1.2 1.44<br />

11 0.8 0.64<br />

7 3.2 10.24<br />

8 2.2 4.84<br />

14 –3.8 14.44<br />

10 –0.2 0.04<br />

12 –1.8 3.24<br />

12 –1.8 3.24<br />

total = 102<br />

9 –1.2 1.44<br />

mean x = 10.2 Σ(x – x) 2 = 39.60<br />

n – 1 = 9<br />

Σ (x – x) 2<br />

s =<br />

n – 1<br />

s = 4.40<br />

s = 2.10<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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