The Two Sample T Test is used to test if the means of two samples, each assumed to follow a normal distribution, are equal.

To apply this test, the standard deviations of the two populations do not need to be known. If the standard deviations of the two populations are provided, then the Two Sample Z Test can be used.

If the two samples are paired, then the Two Sample Paired T Test can be used.

Requirements for using Two Sample T Test:

The two populations studied are assumed to be normally distributed.

The two standard deviations of the two populations are unknown.

The formula is:

where $X$ is the sample drawn from the first population, $Y$ is the sample drawn from the second population, $\mathrm{\beta }$ is the test value of the difference, $\mathrm{s1}$: The sample standard deviation of $X$, $\mathrm{s2}$ is the sample standard deviation of $Y$, $\mathrm{N1}$ is the sample size of $X$, and $\mathrm{N2}$ is the sample size of $Y$.

$T$ follows Student's T distribution with degree of freedom computed as follows: $\mathrm{DF}=\frac{{\left(\frac{{\mathrm{s1}}^2}{\mathrm{N1}}+\frac{{\mathrm{s2}}^2}{\mathrm{N2}}\right)}^2}{\frac{{\mathrm{s1}}^4}{{\mathrm{N1}}^2\left(\mathrm{N1}-1\right)}+\frac{{\mathrm{s2}}^4}{{\mathrm{N2}}^2\left(\mathrm{N2}-1\right)}}$   

Determine the null hypothesis:

Null Hypothesis: On average, men are 20 kg heavier than women ($\mathrm{\beta }$ is 20 for this case)

Substitute the information into the formula:

$\mathrm{Mean}\left(X\right)=75.6,\mathrm{Mean}\left(Y\right)=57.25,\mathrm{\beta }=20,\mathrm{s1}=7.18331,\mathrm{s2}=6.75595,\mathrm{N1}=10,\mathrm{N2}=8$ 

$t=\frac{\left(\left(75.6-57.25\right)-20\right)}{\sqrt{\left(\frac{{7.18331}^2}{10}+\frac{{6.75595}^2}{8}\right)}}=-0.500567$

$\mathrm{DF}=\frac{{\left(\frac{{7.18331}^2}{10}+\frac{{6.75595}^2}{8}\right)}^2}{\frac{{\left(\frac{{7.18331}^2}{10}\right)}^2}{9}+\frac{{\left(\frac{{6.75595}^2}{8}\right)}^2}{7}}=15.51621$ 

Compute the p-value:

$p-\mathrm{value}=\mathrm{Probability}\left(\right|T|>|-0.500567\left|\right)=\mathrm{Probability}\left(T<-0.500567\right)+\mathrm{Probability}\left(T>0.500567\right)=0.623702$, $T˜\mathrm{StudentT}\left(15.51621\right)$

Draw the conclusion:

This statistical test does not provide enough evidence to conclude that the null hypothesis is false, so we fail to reject the null hypothesis.