Coaching and SAT scores $(10.2)$ What we really want to know is whether coached students improve more than uncoached students and whether any advantage is large enough to be worth paying for. Use the information above to answer these questions:

(a) Is there good evidence that coached students gained more on average than uncoached students? Carry out a significance test to answer this question.

(b) How much more do coach students gain on the average? Construct and interpret a $99\%$ confidence interval.

(c) Based on your work, what is your opinion: do you think coaching courses are worth paying for?

${\overline{x}}_1=29,{\overline{x}}_2=21$

$s_1=59,s_2=52$

$n_1=427,n_2=2733$

The test statistic formula is:

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

The null and alternative hypotheses for the provided case are:

$H_0:{\mu }_1={\mu }_2$

$H_1:{\mu }_1\ne {\mu }_2$

The test statistic is computed as:

$=\frac{29-21}{\sqrt{\frac{{59}^2}{427}+\frac{{52}^2}{2733}}}$

$=2.646$

The degree of freedom is calculated as:

$$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (427-1,2733-1) \\ =426 \end{gathered}$$

The p-value:

$P-\text{value}=0.0042$

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 $5\%$.

Confidence level is$95\%$.

The $95\%$confidence interval using $\mathrm{Ti}-83$calculator is:

Confidence level is$95\%$.

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.