Question

Coaching and SAT scores $(8.3,10.2)$Let’s first ask if students who are coached increased their scores significantly.

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

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

(c) Construct and interpret a $99\%$ confidence interval for the mean gain of all students who are coached

Short answer

a). Paired $t$test.

b). There is sufficient evidence to support the claim that the coaching increased their scores.

c). The confidence interval is $(21.501,36.499)$.

Step-by-step solution

Part (a)  Step 1: Given Information

The following table is given:

Part (a) Step 2: Explanation

If every value in one sample has a corresponding value in the other sample, then we use the paired $t$ test, else we use the two-sample $t$ test.

Paired $t$ test

because each subject is in both samples.

Part (b) Step 1: Given Information

The following table is given:

Part (b) Step 2: Explanation

Consider the data for the coached students

The given sample size $=427$

The difference in the mean ${\overline{x}}_d=29$

The difference in standard deviation $s_d=59$

The given sample is random sample and the sample size is $427$.

Therefore, it is a large sample.

Use hypothesis testing for the given data.

Describe the null hypothesis and alternate hypothesis.

Let the null hypothesis $H_0:\mu =0$

Let the alternate hypothesis be $H_1:\mu >0$

Part (b) Step 3: Explanation

Write the test statistics as follows.

Formula used:

$t=\frac{{\overline{x}}_d-\mu }{\frac{s_d}{\sqrt{n}}}$

Substitute the values

Calculations:

$t=\frac{29-0}{\frac{59}{\sqrt{427}}}$

$=\frac{29}{\frac{59}{20.66}}$

$=10.15$

The value of test statistics $t=10.15$.

Use the test statistics to find the P-value. The P-value gives the evidence whether to accept or reject the null hypothesis.

The degree of freedom $=n-1$

Therefore, the degree of freedom is $427-1=426$ as larger values are not in the table.

For calculation use degree of freedom as 100

$P<0.0005$

Take the value of level of significance $\alpha =0.05$

Part (b) Step 4: Explanation

Interpretation:

Reject null hypothesis if the level of significance is more than P-value.

Here level of significance $\alpha =0.05$ and P- value is $0.0005$.

Reject null hypothesis $H_0:\mu =0$.

Therefore, alternate hypothesis is true $H_1:\mu >0$.

This implies that there is convincing evidence to prove that coached students perform better.

Part (c) Step 1: Given Information

${\overline{x}}_d=29$

$s_d=59$

$n=427$

$c=99\%$

Part (c) Step 2: Explanation

Determine the $t_c$using table $B$with $df=n-1=426-1=425>100$and $c=99\%$:

$t_c=2.626$

The endpoints of the confidence interval are then:

localid="1650393262102" $$\begin{gathered} \overline{d}-t_{\alpha /2}·\frac{s_d}{\sqrt{n}}=29-2.626·\frac{59}{\sqrt{427}} \\ \approx 21.502 \end{gathered}$$

localid="1650393275625" $$\begin{gathered} \overline{d}+t_{\alpha /2}·\frac{s_d}{\sqrt{n}}=29+2.626·\frac{59}{\sqrt{427}} \\ \approx 36.498 \end{gathered}$$

We are $99\%$confident that the true population mean difference is between $21.502$and $36.498$.