Chapter 10.2, Problem 74E

a.

To determine

To find: Whether there is sufficient evidence to conclude that coached students are getting average more score than uncoached students.

Answer to Problem 74E

Yes, there is sufficient evidence to support the claim that coached students are getting more on average than the uncoached students.

Explanation of Solution

Given:

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

Formula used:

Test statistic formula is:

  $t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Calculation:

The null and alternative hypotheses for the provided case are:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_1:{\mu }_1\ne {\mu }_2\end{array}$

The test statistic is computed as:

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

The degree of freedom is calculated as:

  $df=\min \left(n_1-1,n_2-1\right)=\min \left(427-1,2733-1\right)=426$

The p-value:

  $P-value=0.0042$

In this case, $P-value=0.00424>0.05$

The null hypothesis could not be rejected which is not showing the sufficient evidences for the claim at significance level of 5%.

b.

To determine

To find: The 95% confidence interval and interpret it.

Answer to Problem 74E

The confidence interval is (1.0002, 15).

Explanation of Solution

The 95% confidence interval using Ti-83 calculator is:

  

Thus, the confidence interval is (1.0002, 15).

c.

To determine

To explain: If the coaching is worth paying.

Answer to Problem 74E

No, it is not worth paying.

Explanation of Solution

The above computed confidence interval is very close to the value 0. Thus, no larger gain is being earned which implies that it is not worth paying for.