The mean number of English courses taken in a two–year time period by male and female college students is believed
to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an
average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with
a standard deviation of 1.0. Are the means statistically the same?

Find the male and female of English course

Given the table with data:

Assume$\overline{X_1}-\overline{X_2}$ represent that the difference between the mean of English course taken by males and females

Test these hypothesis:

$$\begin{gathered} H_0:{\mu }_1={\mu }_2 \\ {H}_1:{\mu }_1\ne {\mu }_2 \end{gathered}$$

To perform the test, we need to distribute the student.

First, calculate the standard error:

$$\begin{gathered} SE=\sqrt{\frac{s\frac{2}{1}}{n_1}+\frac{s\frac{2}{2}}{n_2}} \\ =\sqrt{\frac{0.8^2}{29}+\frac{1^2}{16}} \\ =\sqrt{0.084569} \\ =0.290807 \end{gathered}$$

Test statistic

$t=\frac{\overline{X_1}-{\overline{X}}_2}{SE}=\frac{3-4}{SE}=-3.438707$

Now we have to find the number of degree of freedom of our test statistic and for that, we are using the formula

$df=\frac{SE}{{\displaystyle \frac{s\frac{4}{1}}{n\frac{2}{1}(n_1-1)}}+{\displaystyle \frac{s\frac{4}{2}}{n\frac{2}{1}(n_2-1)}}}=25.7436$

We get the $p=0.002$

If we take $\alpha =0.05$, then p- value is less than $0.005$, thus we reject the hypothesis $H_0$and conclude that at $5\%$significance level, and there is a sufficient evidence to conclude that the mean number of college English course that male and female takes is different