Chapter 10.3, Problem 102E

(a)

To determine

To explain how much more do coached student gain on average compared to un-coached students and construct and interpret a $99\%$ confidence interval.

Answer to Problem 102E

We are $99\%$ confident that the mean gain for coached students is between $(0.0603,15.9397)$ greater than the mean gain for un-coached students.

Explanation of Solution

It is given:

  $\begin{array}{l}{\overline{x}}_1=29\\ {\overline{x}}_2=21\\ n_1=427\\ n_2=2733\\ s_1=59\\ s_2=52\\ c=0.99\end{array}$

Since all the three conditions: Random, Independent and Normal conditions are satisfied then it is appropriate to conduct the hypothesis test.

The degree of freedom will be as:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (427-1,2733-1)\\ =426\end{array}$

We will check the $df=100$ instead. Hus, the t -value will be:

  $t_{\alpha /2}=2.262$

The confidence interval is:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(29-21)-2.626\times \sqrt{\frac{{59}^2}{427}+\frac{{52}^2}{2733}}\\ =0.0603\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(29-21)+2.626\times \sqrt{\frac{{59}^2}{427}+\frac{{52}^2}{2733}}\\ =15.9397\end{array}$

Thus we conclude that we are $99\%$ confident that the mean gain for coached students is between $(0.0603,15.9397)$ greater than the mean gain for un-coached students.

(b)

To determine

To explain does the interval in part (a) gives the convincing evidence that the coached students gain more on average than un-coached students or not.

Answer to Problem 102E

There is convincing evidence that the coached students gain more on average than un-coached students.

Explanation of Solution

It is given:

  $\begin{array}{l}{\overline{x}}_1=29\\ {\overline{x}}_2=21\\ n_1=427\\ n_2=2733\\ s_1=59\\ s_2=52\\ c=0.99\end{array}$

Since all the three conditions: Random, Independent and Normal conditions are satisfied then it is appropriate to conduct the hypothesis test.

The given claim is: mean difference is positive. So, the claim is either null hypothesis or alternative hypothesis. That is,

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

The test statistics value will be:

  $\begin{array}{l}t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{29-21-0}{\sqrt{\frac{{59}^2}{427}+\frac{{52}^2}{2733}}}\\ =2.646\end{array}$

The degree of freedom will be as:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (427-1,2733-1)\\ =426\end{array}$

We will check the $df=100$ instead. Hus, the P -value will be:

  $0.0025<P<0.005$

If the P -values is less than the significance level we reject the null hypothesis, then we have,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is convincing evidence that the coached students gain more on average than un-coached students.

(c)

To determine

To explain what is your opinion: do you think coaching courses are worth paying for.

Answer to Problem 102E

No.

Explanation of Solution

In the part (a), we have that,

  $(0.0603,15.9397)$

The confidence interval lies very close to zero and thus there does not seem to be a large gain for the coaching. Then coaching courses do not seem worth paying for. Thus, this our opinion as it depends on how well the coaching course explain the subjects to their students.