91Ó°ÊÓ

$$\begin{gathered} {\stackrel{¯}{x}}_1=16569 \\ {\stackrel{¯}{x}}_2=16177 \\ {n}_1=56 \\ {n}_2=56 \\ {s}_1=9108 \\ {s}_2=7520 \\ \mathrm{Î\pm }=0.05 \end{gathered}$$

There are three requirements that must be met:

Because the samples are drawn at random from different populations, it is satisfied.

Independent: It is satisfying because the sample of 56 female students represents less than $10\%$of the total female student population, and the sample of $56$male students represents less than $10\%$of the total male student population.

Normal: It is satisfying because both samples are large, with a sample size of at least $30$for each.

As a result, all of the requirements have been met.

The degree of liberty will now be:

$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(56-1,56-1) \\ =55 \end{gathered}$$

Thus the t value will be:

$t=2.009$

The confidence interval will be calculated as follows:

$$\begin{gathered} ({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2)-t_{\frac{\mathrm{Î\pm }}{2}}×\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =(16177-16569)-2.009×\sqrt{\frac{{9108}^2}{56}+\frac{{7520}^2}{56}} \\ =-3562.90 \end{gathered}$$

$$\begin{gathered} ({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2)+t_{\frac{\mathrm{Î\pm }}{2}}×\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =(16177-16569)+2.009×\sqrt{\frac{{9108}^2}{56}+\frac{{7520}^2}{56}} \\ =2778.90 \end{gathered}$$

Therefore, we conclude that we are $95\%$confident that the mean number of words spoken in a day by all male students at this university is between $3562.90$words lower and $2778.90$words higher than the mean number of words spoken in a day by all female students at this university.

Because the confidence interval gives us a range of possible values for the difference between the true means, whereas the significance test only checks one possible value for the difference between the means, that is, if the means are equal, the confidence interval gives us more information than the significance test.