Cambridge International A Level Biology Revision Guide

Cambridge International A Level Biology
500
The t-test
The t-test is used to assess whether or not the means of
two sets of data with roughly normal distributions, are
significantly different from one another.
For this example, we will use data from another
investigation.
The corolla (petal) length of two populations of gentian
were measured in mm.
Corolla lengths of population A:
13, 16, 15, 12, 18, 13, 13, 16, 19, 15, 18, 15, 15, 17, 15,
Corolla lengths of population B:
16, 14, 16, 18, 13, 17, 19, 20, 17, 15, 16, 16, 19, 21, 18,
The formula for the t-test is:
− x1 − − x2
t =
s2 1
s2
n
+ 2
1
n 2
− x1 is the mean of sample 1
x − 2
is the mean of sample 2
s 1
is the standard deviation of sample 1
s 2
is the standard deviation of sample 2
n 1
is the number of individual measurements in sample 1
n 2
is the number of individual measurements in sample 2
1 For each set of data, calculate the mean.
2 Calculate the differences from the mean of all
observations in each data set. This is x − x − .
3 Calculate the squares of these. This is (x − x − ) 2 .
4 Calculate the sum of the squares. This is ∑(x − x − ) 2 .
5 Divide this by n 1
− 1 for the first set and n 2
− 1 for the
second set.
6 Take the square root of this. The result is the standard
deviation for each set of data.
For population A, s 1
= 4.24.
For population B, s 2
= 4.86.
7 Square the standard deviation and divide by the
number of observations in that sample, for
both samples.
8 Add these values together for the two samples and take
the square root of this.
9 Divide the difference in the two sample means with the
value from step 8. This is t and, in this case, is 1.93.
10 Calculate the total degrees of freedom for all the data (v).
v = (n 1
− 1) + (n 2
− 1) = 28
11 Refer to the table of t values for 28 degrees of freedom
and a value of t = 1.93 (Table P2.2).
Degrees of
freedom
Value of t
1 6.31 12.7 63.7 63.6
2 2.92 4.30 9.93 31.6
3 2.35 3.18 5.84 12.9
4 2.13 2.78 4.60 8.61
5 2.02 2.57 4.03 6.87
6 1.94 2.45 3.71 5.96
7 1.90 2.37 3.50 5.41
8 1.86 2.31 3.36 5.04
9 1.83 2.26 3.25 4.78
10 1.81 2.23 3.17 4.59
11 1.80 2.20 3.11 4.44
12 1.78 2.18 3.06 4.32
13 1.77 2.16 3.01 4.22
14 1.76 2.15 2.98 4.14
15 1.75 2.13 2.95 4.07
16 1.75 2.12 2.92 4.02
17 1.74 2.11 2.90 3.97
18 1.73 2.10 2.88 3.92
19 1.73 2.09 2.86 3.88
20 1.73 2.09 2.85 3.85
22 1.72 2.07 2.82 3.79
24 1.71 2.06 2.80 3.75
26 1.71 2.06 2.78 3.71
28 1.70 2.05 2.76 3.67
30 1.70 2.04 2.75 3.65
>30 1.64 1.96 2.58 3.29
Probability that
chance could have
produced this
0.10 0.05 0.01 0.001
value of t
Confidence level 10% 5% 1% 0.1%
Table P2.2 Values of t.
Using the table of probabilities in the t-test
In statistical tests that compare samples, it is the
convention to start off by making the assumption that
there is no significant difference between the samples. You
assume that they are just two samples from an identical
population. This is called the null hypothesis. In this case,
the null hypothesis would be:
There is no difference between the corolla length in
population A and population B.