Comparing means using student’s t-test

The student’s t-test is a statistical test used to determine if there is a significant difference between the mean values of a particular variable across two populations.

The conditions for using student's t-test are:

• The data must be continuous and normally distributed.
• The variances of the populations should be equal.
• The samples must be independent of each other.

Note: variance is the standard deviation squared. You don't need to learn the formula, but it is:

$\text{variance }=\frac{\mathrm{\Sigma }{(x-\bar{x})}^2}{n-1}$

Conducting a t-test

Step 1 - State the null hypothesis.

This assumes there is no significant difference between the means of the datasets.

Step 2 - Calculate the t statistic using the t-test formula (this will be provided in exams).

$t=\frac{({\bar{x}}_1-{\bar{x}}_2)}{\sqrt{(\frac{{\sigma }_1^2}{n_1})+(\frac{{\sigma }_2^2}{n_2})}}$

Where:

• ${\bar{x}}_1=\text{ mean of the first dataset}$
• ${\bar{x}}_2=\text{ mean of the second dataset}$
• ${\sigma }_1=\text{ standard deviation of the first dataset}$
• ${\sigma }_2=\text{ standard deviation of the second dataset}$
• $n_1=\text{ sample size of the first dataset}$
• $n_2=\text{ sample size of the second dataset}$

Step 3 - Calculate the degrees of freedom.

$\text{degrees of freedom (df) in a t-test = }{n}_1+n_2-2$

Step 4 - Compare the calculated t statistic against a critical value.

The critical value is determined by the degrees of freedom and the chosen significance level (usually p = 0.05). You will be provided with a table of critical values in an exam.

If the t statistic is greater than the critical value:

• Reject the null hypothesis.
• This suggests that the means are significantly different.

If the t statistic is less than the critical value:

• Accept the null hypothesis.
• This suggests that there is no significant difference between the means, and any difference is just due to chance.

Worked example - Comparison of turtle egg hatch rates using student's t-test

Given the following data, use the student's t-test to determine if there is a significant difference between the mean number of turtle eggs hatched in high and low temperature conditions.

Step 1: Equation

$t=\frac{({\bar{x}}_1-{\bar{x}}_2)}{\sqrt{(\frac{{\sigma }_1^2}{n_1})+(\frac{{\sigma }_2^2}{n_2})}}$

Step 2: Substitution and correct evaluation

$t=\frac{(80-70)}{\sqrt{(\frac{5^2}{15})+(\frac{5^2}{15})}}$

$t=\frac{(10)}{\sqrt{(\frac{25}{15})+(\frac{25}{15})}}$

$t=\frac{10}{\sqrt{\mathrm{1.66...}+\mathrm{1.66...}}}$

$t=\frac{10}{\sqrt{\mathrm{3.33...}}}$

$t=\frac{10}{\mathrm{1.82...}}$

$t=5.48\text{ (to 3 s.f.)}$

Step 3: Calculate degrees of freedom (df)

$\text{df in a t-test = }{n}_1+n_2-2$

$df=(15+15)-2=28$

Step 4: Determine significance

with a t value of 5.48 and df of 28, we compare this to the critical t value at a significance level of 0.05

the critical value for the t-test statistic at p = 0.05 and 28 degrees of freedom is 2.05

as our calculated t value (5.48) is greater than this critical value, we reject the null hypothesis and conclude that there is a significant difference between the hatch rates of turtle eggs in high and low temperatures