Question

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

a. How much more do coached students gain, on average, compared to uncoached students? Construct and interpret a $99\%$confidence interval.

b. Does the interval in part (a) give convincing evidence that coached students gain more, on average, than uncoached students? Explain your answer.

c. Based on your work, what is your opinion: Do you think coaching courses are worth paying for?

Short answer

Part(a) We are $99\%$ confident that coached students gain more on average, compared to uncoached students.

Part(b) There is convincing evidence that coached students gain more, on average, than uncoached students.

Part(c) No, coaching courses are not worth paying for.

Step-by-step solution

Part(a) Step 1 : Given information

We need to find how more do coached students gain, on average, compared to uncoached students.

Part(a) Step 2 : Simplify

It is given:
$$\begin{gathered} \overline{x_1}=29 \\ \overline{x_2}=21 \\ {n}_1=427 \\ {n}_2=2733 \\ {s}_1=59 \\ {s}_2=52 \\ c=0.99 \end{gathered}$$
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:
$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(427-1,2733-1) \\ =426 \end{gathered}$$
Now, check the $df=100$instead. Therefore,, the t-value will be:

$t_{\alpha /2}=2.262$
The confidence interval is:
$$\begin{gathered} (\overline{x_1}-\overline{x_2})-t_{\frac{\alpha }{2}}\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =(29-21)-2.626\times \sqrt{\frac{{59}^2}{427}+\frac{{52}^2}{2733}} \\ =0.0603 \\ (\overline{x_1}-\overline{x_2})+t_{\frac{\alpha }{2}}\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =(29-21)+2.626\times \sqrt{\frac{{59}^2}{427}+\frac{{52}^2}{2733}} \\ =15.9397 \end{gathered}$$
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.

Part(b) Step 1 : Given information

We need to find convincing evidence that coached students gain more, on average, than uncoached students.

Part(b) Step 2 : Simplify

As given,

$$\begin{gathered} \overline{x_1}=29 \\ \overline{x_2}=21 \\ {n}_1=427 \\ {n}_2=2733 \\ {s}_1=59 \\ {s}_2=52 \\ c=0.99 \end{gathered}$$

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,

$$\begin{gathered} H_0:{\mu }_1={\mu }_2 \\ {H}_a:{\mu }_1>{\mu }_2 \end{gathered}$$
The test statistics value will be:
$$\begin{gathered} t=\frac{\left(\overline{x_1}-\overline{x_2}\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{{\displaystyle \frac{{s_1}^2}{n_1}}+\frac{{s_2}^2}{n_2}}} \\ =\frac{29-21}{\sqrt{{\displaystyle \frac{{59}^2}{427}}+{\displaystyle \frac{{52}^2}{2733}}}} \\ =2.646 \end{gathered}$$
The degree of freedom will be as:
$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(427-12733-1) \\ =426 \end{gathered}$$
Now check $df=100$, P-value will be

$0.0025<P<0.005$

P-value is less than significance value.

Therefore, We conclude that convincing evidence that coached students gain more, on average, than uncoached students

Part(c) Step 1 : Given information

We need to check weather coaching courses are worth paying for or not.

Part(c) Step 2 : Simplify

No,

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.