91Ó°ÊÓ

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:

$$\begin{gathered} H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2 \\ {H}_1:{\mathrm{μ}}_1≠{\mathrm{μ}}_2 \end{gathered}$$

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 $\mathrm{Î\pm }=0.05$, 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