Chapter 11.3, Problem 8E

(a)

To determine

To find:

The difference of given data.

Answer to Problem 8E

The difference is

Sample   $1$	Sample   $2$	Difference
$28$	$34$	$-6$
$29$	$30$	$-1$
$22$	$31$	$-9$
$25$	$26$	$-1$
$26$	$31$	$-5$
$29$	$30$	$-1$
$27$	$31$	$-4$
$24$	$32$	$-8$
$27$	$29$	$-2$
$28$	$37$	$-9$

Explanation of Solution

Given Information:

Following is a sample of ten matched pairs.

Sample   $1$	$28$	$29$	$22$	$25$	$26$	$29$	$27$	$24$	$27$	$28$
Sample   $2$	$34$	$30$	$31$	$26$	$31$	$30$	$31$	$32$	$29$	$37$

Formula used:

  $\mu 1\text{ and }\mu 2$ Represent the population means and let $\mu d=\mu 1-\mu 2$ .A test will be made of the hypotheses $H0:\mu d=0$ versus $H1:\mu d\ne 0$ .

Calculation:

The difference is

Sample   $1$	Sample   $2$	Difference
$28$	$34$	$-6$
$29$	$30$	$-1$
$22$	$31$	$-9$
$25$	$26$	$-1$
$26$	$31$	$-5$
$29$	$30$	$-1$
$27$	$31$	$-4$
$24$	$32$	$-8$
$27$	$29$	$-2$
$28$	$37$	$-9$

(b)

To determine

To find:

The test statistic.

Answer to Problem 8E

The $p-$ value for this test is $0.002$

Explanation of Solution

Given Information:

To test the hypothesis is that the population mean difference is different $\text{0 at 5\%}$ significance level.

The null and alternative hypothesis is,

  $H_0:{\mu }_d=0$

  $H_1:{\mu }_d\ne 0$

Formula used:

By using MINITAB, find $t-$

test statistics with the help of following steps:

1. Import the data.
2. Select the Stat and choose the Basic Statistics option.
3. Select the Paired $t$ and choose variable option and put First sample and Second sample
4. Click option button choose Confidence level, Test difference and alternative hypothesis.
5. Click 0k.

Calculation:

Paired T-Test and $\text{CI}$ : Sample $\text{1}$ , sample $\text{2}$ .

Paired $T$ for $\text{Sample 1}-\text{Sample}\text{2}$ .

  $\begin{array}{l}\text{ N Mean StDev SE Mean}\\ \text{Sample 1 10 26}\text{.500 2}\text{.273 0}\text{.719}\\ \text{Sample 2 10 31}\text{.100 2}\text{.923 0}\text{.924}\\ \text{Difference 10 }-\text{4}\text{.60 3}\text{.31 }\text{1}\text{.05}\end{array}$

  $\text{95\%}$ CI for mean difference: $\left(-6.97,-2.23\right)$

T-Test of mean difference $=0\left(\text{vs not}=0\right):\text{T-Value}=-4.40\text{ P-Value}=0.002$

  $-4.40$

From the MINITAB output, the $t$ test statistics is,

The $p-$ value for this test is $0.002$

(c)

To determine

To find:

Whether to reject the null hypothesis.

Answer to Problem 8E

The result is statistically significant.

Explanation of Solution

Given Information:

  $\alpha =0.05$

Formula used:

The hypothesis is that the two population proportions are different at $5\%$ significance level.

Calculation:

The conclusion is that $p-$ value in this context is less than $\text{0}\text{.05 which is 0}\text{.002 }$ , so the null hypothesis is rejected at $5\%$ level of significance. There is sufficient evidence to indicate that the population mean difference is different from $0$ . The result is statistically significant.

(d)

To determine

To find:

Whether to reject the null hypothesis.

Answer to Problem 8E

The result is statistically significant.

Explanation of Solution

Given Information:

  $\alpha =0.01$

Formula used:

The hypothesis is that the two population proportions are different at $5\%$ significance level.

Calculation:

The conclusion is that $p-$ value in this context is less than $\text{0}\text{.01 which is 0}\text{.002}$ , so the null hypothesis is rejected at $1\%$ level of significance. There is insufficient evidence to indicate that the population mean difference is different $\text{0}$ . The result is statistically significant.