Variation: t-test Worked Example (OCR A Level Biology)

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Variation: t-test worked example

  • The t test method can be used to determine whether the means of two data sets are significantly different

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.

  • 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  1 = 885 ÷ 15 = 59 mm

Mean for population B  2 = 870 ÷ 15 = 58 mm

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

Calculating Standard Deviation

  • Calculate ∑(x - x̄)2
Population A Population B

Difference between value and mean

(x - x̄)

Difference between value and mean squared

(x - x̄)2

Difference between value and mean

(x - x̄)

Difference between value and mean squared

(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

  • Calculate 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
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 22 ÷ 14 = 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

Steps 3-5: use the standard deviations to complete the t test

T-test equation

  Population A Population B
Mean (x̄) 59 58
Standard deviation (s) 1.60 1.25
Sample size (n) 15 15

calculation of t value in t test

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

v = (n1 - 1) + (n2 - 1)

= 14 + 14 

= 28

Step 7: determine where the t value lies in relation to values in a critical values table

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

  • The t value is 1.90, which falls between 1.70 and 2.05 in the critical values table
  • In biology we are looking for a less than 5 % probability that any difference between data sets is due to chance, so we are concerned with whether our t valuer is higher or lower than 2.05

Step 8: draw a conclusion

  • A conclusion should contain:
    • A reference to the t value, the critical value, the degrees of freedom and the probability level
    • Whether or not there is a significant difference between the means of the two data sets
    • Whether the null hypothesis is accepted or rejected

The t value of 1.90 is less than the critical value of 2.05 at 28 df and a probability level of 0.05

There is no significant difference between the data sets and we can accept the null hypothesis. Any difference between the mean ear lengths of population A and population B is due to chance.

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Phil

Author: Phil

Expertise: Biology

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.