A LevelBiologyCambridge (CIE)Revision Notes17. Selection & EvolutionVariationVariation: t-test Worked Example

Variation: t-test Worked Example (Cambridge (CIE) A Level Biology): Revision Note

Exam code: 9700

Phil

Written by: Phil

Reviewed by: Alistair Marjot

Updated on

Variation: t-test worked example

Worked Example

The ear lengths of two populations of rabbits were measured.

Ear lengths of population A (mm):

62, 60, 59, 61, 60, 58, 59, 60, 57, 56, 59, 58, 60, 59, 57

Ear lengths of population B (mm):

58, 59, 57, 59, 59, 57, 55, 60, 57, 58, 59, 58, 57, 58, 59

Use the t-test to determine whether there is a significant difference in ear length between the two populations.

Solution

  • Null hypothesis: There is no significant difference between the ear lengths of the rabbits in populations A and B

  • Sample sizes:

    • Population A: n1 = 15

    • Population B: n2 = 15

    Step 1: Calculate the mean for each data set:

Mean for population A  x̅1 = 885 ÷ 15 = 59 mm

Mean for population B  x̅2 = 870 ÷ 15 = 58 mm

Step 2: Calculate the standard deviation (s) for each set of data

Population A

Population B

Difference between value and mean

(x - x̄)

Difference between value and mean

(x - x̄)2

Difference between value and mean

(x - x̄)

Difference between value and mean

(x - x̄)2

62 - 59 = 3

9

58 - 58 = 0

0

60 - 59 = 1

1

59 - 58 = 1

1

59 - 59 = 0

0

57 - 58 = -1

1

61 - 59 = 2

4

59 - 58 = 1

1

60 - 59 = 1

1

59 - 58 = 1

1

58 - 59 = -1

1

57 - 58 = -1

1

59 - 59 = 0

0

55 - 58 = -3

9

60 - 59 = 1

1

60 - 58 = 2

4

57 - 59 = -2

4

57 - 58 = -1

1

56 - 59 = -3

9

58 - 58 = 0

0

59 - 59 = 0

0

59 - 58 = 1

1

58 - 59 = -1

1

58 - 58 = 0

0

60 - 59 = 1

1

57 - 58 = -1

1

59 - 59 = 0

0

58 - 58 = 0

0

57 - 59 = -2

4

59 - 58 = 1

1

Total ∑(x - x̄)2

36

Total ∑(x - x̄)2

22

To find the standard deviations divide the sum of each square by n - 1 for each data set, and take the square root of each value

Population A (n1 = 15)

Population B (n2 = 15)

n1 - 1 = 14

n2 - 1 = 14

∑(x - x̄)2 = 36 

so 36 ÷ 14 = 2.57

∑(x - x̄)2 = 22 

so 36 ÷ 22 = 1.57

square root of 2.57 end root equals 1.60

square root of 1.57 end root equals 1.25

straight s subscript 1 equals square root of fraction numerator sum for blank of open parentheses straight x minus straight x with bar on top close parentheses squared over denominator straight n minus 1 end fraction end root equals 1.60

straight s subscript 2 equals square root of fraction numerator sum for blank of open parentheses straight x minus straight x with bar on top close parentheses squared over denominator straight n minus 1 end fraction end root equals 1.25

Step 3: Square the standard deviation and divide by n (the number of observations) in each sample, for both samples:

Step 4: Add the values from step 3 together and find the square root

Step 5: Divide the difference between the two means by the value from step 4

Population A

Population B

x̄1 = 59

x̄2 = 58

s1 = 1.60

s2 = 1.25

n1 = 15

n2 = 15

Equation for t-value with calculations: numerator is 59-58, denominator includes squared standard deviations. Result is 1.90. Note: Standard deviations must be squared.

Step 6: Calculate the degrees of freedom (v) for all the data:

 straight v space equals space left parenthesis straight n subscript 1 space minus space 1 right parenthesis space plus space left parenthesis straight n subscript 2 space minus space 1 right parenthesis space equals space 14 space plus space 14 space equals space 28

Step 7: Look at a table that relates t values to the probability that the difference between data sets is due to chance to find where the t value of 1.90 for 28 degrees of freedom (v) calculated lies

Degrees of freedom

Value of t

28

1.70

2.05

2.76

3.67

Probability that chance would have produced this value of t

0.1

0.05

0.01

0.001

Step 8: Draw a conclusion about the statistical relevance of the data

We are considering a confidence level of 0.05, corresponding to a critical value of at least 2.05. However, our t value is 1.90, less than the critical value. So the null hypothesis should be accepted

This means the null hypothesis should be accepted, as there are no significant differences between the two sets of results (any differences between the means of the ear length of rabbits in the two populations are due to chance)

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    Author
    Reviewer
    Phil

    Author: Phil

    Expertise: Biology Content Creator

    Phil has a BSc in Biochemistry from the University of Birmingham, followed by an MBA from Manchester Business School. He has 15 years of teaching and tutoring experience, teaching Biology in schools before becoming director of a growing tuition agency. He has also examined Biology for one of the leading UK exam boards. Phil has a particular passion for empowering students to overcome their fear of numbers in a scientific context.

    Alistair Marjot

    Reviewer: Alistair Marjot

    Expertise: Environmental Systems and Societies & Biology Content Creator

    Alistair graduated from Oxford University with a degree in Biological Sciences. He has taught GCSE/IGCSE Biology, as well as Biology and Environmental Systems & Societies for the International Baccalaureate Diploma Programme. While teaching in Oxford, Alistair completed his MA Education as Head of Department for Environmental Systems & Societies. Alistair has continued to pursue his interests in ecology and environmental science, recently gaining an MSc in Wildlife Biology & Conservation with Edinburgh Napier University.