Chapter 13.1, Problem 4E

a:

To determine

The test hypothesis.

Explanation of Solution

The null hypothesis (H₀) is $\left({\text{μ}}_1-{\text{μ}}_2\right)=0$, and the alternate hypothesis (H₁) is $\left({\text{μ}}_1-{\text{μ}}_2\right)>0$.

The unequal variance estimator (v) can be calculated as follows:

$\begin{array}{l}\nu =\frac{{(s_1^2/n_1+s_2^2/n_2)}^2}{\frac{{(s_1^2/n_1)}^2}{n_1-1}+\frac{{(s_2^2/n_2)}^2}{n_2-1}}\\ =200.4\end{array}$

The value of unequal variance is 200.4, which can be round off as 200.

The t table value can be calculated as follows:

$\begin{array}{c}\text{t}>{\text{t}}_{\alpha ,\text{ν}}\\ \text{t}={\text{t}}_{0.05,200}\\ =1.653\end{array}$

The t table value is 1.653.

The t value (t) can be calculated as follows:

$\begin{array}{c}\text{t}=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}}\\ =\frac{(412-405)-0}{\sqrt{\left(\frac{{128}^2}{150}+\frac{{54}^2}{150}\right)}}\\ =0.62\end{array}$

The calculated t value is 0.62, and the p value is 0.2689. Since the p value is significantly different from 0, the alternate hypothesis is not accepted.

b:

To determine

The test hypothesis with decreasing standard deviation.

Explanation of Solution

The unequal variance estimator (v) can be calculated as follows:

$\begin{array}{l}\nu =\frac{{(s_1^2/n_1+s_2^2/n_2)}^2}{\frac{{(s_1^2/n_1)}^2}{n_1-1}+\frac{{(s_2^2/n_2)}^2}{n_2-1}}\\ =223.1\end{array}$

The value of unequal variance is 223.1, which can be round off as 223..

The t table value can be calculated as follows:

$\begin{array}{c}\text{t}>{\text{t}}_{\alpha ,\text{ν}}\\ \text{t}={\text{t}}_{0.05,223}\\ =1.645\end{array}$

The t table value is 1.645.

The t value (t) can be calculated as follows:

$\begin{array}{c}\text{t}=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}}\\ =\frac{(412-405)-0}{\sqrt{\left(\frac{{31}^2}{150}+\frac{{16}^2}{150}\right)}}\\ =2.46\end{array}$

The calculated t value is 2.46, and the p value is 0.0074. Since the p value is nearly equal to 0, the alternate hypothesis is accepted.

c:

To determine

The impact of decreasing standard deviation.

Explanation of Solution

When the standard deviation decreases, it increases the calculated t value and decreases the p value.

d:

To determine

The test hypothesis with decreasing sample.

Explanation of Solution

The unequal variance estimator (v) can be calculated as follows:

$\begin{array}{l}\nu =\frac{{(s_1^2/n_1+s_2^2/n_2)}^2}{\frac{{(s_1^2/n_1)}^2}{n_1-1}+\frac{{(s_2^2/n_2)}^2}{n_2-1}}\\ =25.6\end{array}$

The value of unequal variance is 25.6, which can be round off as 26.

The t table value can be calculated as follows:

$\begin{array}{c}\text{t}>{\text{t}}_{\alpha ,\text{ν}}\\ \text{t}={\text{t}}_{0.05,26}\\ =1.706\end{array}$

The t table value is 1.706.

The t value (t) can be calculated as follows:

$\begin{array}{c}\text{t}=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}}\\ =\frac{(412-405)-0}{\sqrt{\left(\frac{{128}^2}{20}+\frac{{54}^2}{20}\right)}}\\ =0.23\end{array}$

The calculated t value is 0.23, and the p value is 0.4118. Since the p value is significantly different from 0, the alternate hypothesis is not accepted.

e:

To determine

The impact of decreasing sample.

Explanation of Solution

When the sample size decreases, it decreases the t value and increases the p value.

f:

To determine

The test hypothesis with decreasing mean value.

Explanation of Solution

The unequal variance estimator (v) can be calculated as follows:

$\begin{array}{l}\nu =\frac{{(s_1^2/n_1+s_2^2/n_2)}^2}{\frac{{(s_1^2/n_1)}^2}{n_1-1}+\frac{{(s_2^2/n_2)}^2}{n_2-1}}\\ =200.4\end{array}$

The value of unequal variance is 200.4, which can be round off as 200.

The t table value can be calculated as follows:

$\begin{array}{c}\text{t}>{\text{t}}_{\alpha ,\text{ν}}\\ \text{t}={\text{t}}_{0.05,200}\\ =1.653\end{array}$

The t table value is 1.653.

The t value (t) can be calculated as follows:

$\begin{array}{c}\text{t}=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}}\\ =\frac{(409-405)-0}{\sqrt{\left(\frac{{128}^2}{150}+\frac{{54}^2}{150}\right)}}\\ =0.35\end{array}$

The calculated t value is 0.35, and the p value is 0.3624. Since the p value is significantly different from 0, the alternate hypothesis is not accepted.

g:

To determine

The impact of decreasing mean value.

Explanation of Solution

When the mean decreases, it decreases the t value and increases the p value.