Chapter 10.2, Problem 59E

To determine

To explain do these data provide convincing evidence at the $\alpha =0.05$ significance level of a difference in the average number of words spoken in a day by all male and all female students at this university.

Answer to Problem 59E

There is no convincing evidence of a difference in the average number of words spoken in a day by all male and all female students at this university.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=16569\\ {\overline{x}}_2=16177\\ n_1=56\\ n_2=56\\ s_1=9108\\ s_2=7520\end{array}$

  $\alpha =0.05$

The given claim that: difference in the means.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1\ne {\mu }_2\end{array}$

Where we have,

  ${\mu }_1=$ the true mean number of words spoken in a day by all male students at this university.

  ${\mu }_2=$ the true mean number of words spoken in a day by all female students at this university.

Now, find the test statistics:

  $\begin{array}{l}t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{16569-16177-0}{\sqrt{\frac{{9108}^2}{56}+\frac{{7520}^2}{56}}}\\ =0.248\end{array}$

Now, the degree of freedom will be:

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

Since student’s T distribution table in the appendix does not contains the value of $df=55$ so we will take the nearest value $df=50$ . So the P -value will be:

  $P>2(0.25)=0.50$

On the other hand by using the calculator command: $2\times \text{tcdf(0}\text{.248,1E99,55)}$ which results in the P -values as $0.80506$ .

And we know that if the P -value is less than or equal to the significance level then the null hypothesis is rejected, then,

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we conclude that there is no convincing evidence of a difference in the average number of words spoken in a day by all male and all female students at this university.