The Student's t-test tests whether there is a significant difference between the means of two sets of data, for example the mean shell height of limpets on an exposed shore and on a sheltered shore. It is used when the data are continuous and approximately normally distributed.t = (x̄1 − x̄2) ÷ √(s12/n1 + s22/n2) where x̄ is the mean, s is the standard deviation and n is the number of measurements in each sample.Steps:
1. Null hypothesis: there is no significant difference between the two means.
2. Calculate the mean and standard deviation of each sample.
3. Calculate t (use its size, ignoring any minus sign).
4. Degrees of freedom = n1 + n2 − 2.
5. Compare t with the critical value at p = 0.05. If t is greater than or equal to the critical value, reject the null hypothesis.Worked example: limpets on an exposed shore (n = 10) have a mean height of 30 mm, s = 4 mm. On a sheltered shore (n = 10) the mean is 26 mm, s = 3 mm.t = (30 − 26) ÷ √(16/10 + 9/10) = 4 ÷ √2.5 = 4 ÷ 1.58 = 2.53 Degrees of freedom = 10 + 10 − 2 = 18, and the critical value at p = 0.05 is 2.10. Since 2.53 > 2.10, reject the null hypothesis: there is a significant difference in mean shell height between the two shores, and the probability that it is due to chance is less than 5%.
1. Null hypothesis: there is no significant difference between the two means.
2. Calculate the mean and standard deviation of each sample.
3. Calculate t (use its size, ignoring any minus sign).
4. Degrees of freedom = n1 + n2 − 2.
5. Compare t with the critical value at p = 0.05. If t is greater than or equal to the critical value, reject the null hypothesis.Worked example: limpets on an exposed shore (n = 10) have a mean height of 30 mm, s = 4 mm. On a sheltered shore (n = 10) the mean is 26 mm, s = 3 mm.