Chapter 10.2, Problem 72E

(a)

To determine

To find out of which is the correct test of the two given tests.

1. Two sample $t$test comparing Try 1 with Try 2 for coached students
2. A pair test using Gain.

Explanation of Solution

Given Information : The following table is given

		Try1		Try 2		Gain
	$n$	$\overline{x}$	$s_x$	$\overline{x}$	$s_x$	$\overline{x}$	$s_x$
Coached	$427$	$500$	$92$	$529$	$97$	$29$	$59$
Uncoached	$2733$	$506$	$101$	$527$	$101$	$21$	$52$

The correct test to use is paired test using Gain. Paired test will be used because for every value in TRY1 there is corresponding value in Try 2 and Gain.

(b)

To determine

To perform the proper test and conclude the result after performing the proper test.

Explanation of Solution

Given Information The following table is given

		Try1		Try 2		Gain
	$n$	$\overline{x}$	$s_x$	$\overline{x}$	$s_x$	$\overline{x}$	$s_x$
Coached	$427$	$500$	$92$	$529$	$97$	$29$	$59$
Uncoached	$2733$	$506$	$101$	$527$	$101$	$21$	$52$

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 $H_1:\mu >0$

Write the test statistics as follows.

Formula used:

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

Substitute the values

Calculations:

  $\begin{array}{c}t=\frac{29-0}{\frac{59}{\sqrt{427}}}\\ =\frac{29}{\frac{59}{20.66}}\\ =\frac{29}{2.856}\\ =10.15\end{array}$

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 $\alpha =0.05$

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.

(c)

To determine

To find: a $99\%$ confidence interval for the mean gain of all students who are coached. Also interpret the results.

Explanation of Solution

Given Information : The following table is given

		Try1		Try 2		Gain
	$n$	$\overline{x}$	$s_x$	$\overline{x}$	$s_x$	$\overline{x}$	$s_x$
Coached	$427$	$500$	$92$	$529$	$97$	$29$	$59$
Uncoached	$2733$	$506$	$101$	$527$	$101$	$21$	$52$

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 confidence interval c = $99\%$

The degree of freedom = $n-1$

Therefore, the degree of freedom is $427-1=426$

Formula used:

The end points of the confidence interval are given by $\left({\overline{x}}_d-t\cdot \frac{s_d}{\sqrt{n}},{\overline{x}}_d+t\cdot \frac{s_d}{\sqrt{n}}\right)$

For confidence interval at $99\%$ the value of $t=2.626$

Substitute the values

Calculations:

  $\begin{array}{c}\left(29-2.626\cdot \frac{\left(59\right)}{\sqrt{427}},29+2.626\cdot \frac{\left(59\right)}{\sqrt{427}}\right)=\left(29-2.626\cdot \frac{\left(59\right)}{20.66},29+2.626\cdot \frac{\left(59\right)}{20.66}\right)\\ =\left(29-2.626\left(2.856\right),29+2.626\left(2.856\right)\right)\\ =\left(29-7.499,29+7.499\right)\\ =\left(21.501,36.499\right)\end{array}$

The confidence interval is $\left(21.501,36.499\right)$