Coaching and SAT scores Let’s first ask if students who are coached increased their scores significantly, on average.

a. You could use the information on the Coached line to carry out either a two-sample t test comparing Try 1 with Try 2 or a paired t test using Gain. Which is the correct test? Why?

b. Carry out the proper test. What do you conclude?

We need to find which test to carry out either a two-sample t test comparing Try 1 with Try 2 or a paired t test using Gain.

The researcher wants to know if the students are coached and if so, whether their scores will improve.

To use the paired t test if every value in one sample has a corresponding value in the other sample; otherwise, use the two-sample t test.
As each subject is dependent, we will use the paired t test in this scenario.

We need to draw conclusion from test.

As given,

$$\begin{gathered} \alpha =0.05 \\ n=427 \\ \overline{x_D}=29 \\ {s}_D=59 \end{gathered}$$

Now, all conditions Random, Independent and Normal are fulfilled then it is right to take hypothesis test.

Mean difference is positive. So, it is either null hypothesis or alternative hypothesis.

$$\begin{gathered} H_0:{\mu }_D=0 \\ {H}_a:{\mu }_D>0 \end{gathered}$$

Statistics of test :

$$\begin{gathered} t=\frac{\overline{x_D}-{\mu }_D}{{\displaystyle \frac{s_D}{\sqrt{n}}}} \\ =\frac{29-0}{{\displaystyle \frac{59}{\sqrt{427}}}} \\ =10.157 \end{gathered}$$

Now,

$$\begin{gathered} df=n-1 \\ =427-1 \\ =426 \end{gathered}$$

Check $df=100$, P-value will be :

$P<0.0005$

P-value is less than significance level then reject null hypothesis,

Then $\mathrm{P}<0.05\Rightarrow \mathrm{Reject}{\mathrm{H}}_0$

Therefore, there is convincing evidence that students who are coached increased their scores significantly.