91Ó°ÊÓ

Given that,

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

The given claim is that there is a disparity in means.

Now we must determine the most appropriate hypotheses for a significance test.

As a result, either the null hypothesis or the alternative hypothesis is the claim. According to the null hypothesis, the population proportions are equal. If the claim is the null hypothesis, the alternative hypothesis is the polar opposite of the null hypothesis.

The appropriate hypotheses for this are:

$$\begin{gathered} H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2 \\ {H}_a:{\mathrm{μ}}_{1}notequalto{\mathrm{μ}}_2 \end{gathered}$$

Where we have,

${\mathrm{μ}}_1$is the true daily average number of words spoken by all male students at this university.

${\mathrm{μ}}_2$is the true average number of words spoken by all female students at this university on a given day.

Locate the following test statistics:

$$\begin{gathered} =\frac{16569-16177-0}{\sqrt{\frac{91082}{56}+\frac{{7520}^2}{56}}} \\ =0.248 \end{gathered}$$

The degree of liberty will now be:

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

On the other hand using the calculator command: $2×tcdf(0.248,1\mathrm{E}99,55)$which results in the $P$-values as $0.80506$
And we know that the null hypothesis is rejected if the P-value is less than or equal to the significance level.

$P>0.05⇒\text{Fail to Reject}{H}_0$

Therefore, We conclude that there is no convincing evidence that the average number of words spoken per day by all male and female students at this university differs.