Chapter 13.1, Problem 6E

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 is (H₁) $\left({\text{μ}}_1-{\text{μ}}_2\right)\ne 0$.

The t table value can be calculated as follows:

$\begin{array}{c}\text{t}>{\text{t}}_{\frac{\alpha }{2},\text{ν}}\\ \text{t}={\text{t}}_{\frac{0.05}{2},22}\\ ={\text{t}}_{.025,22}\\ =2.074\end{array}$

The t table value is 2.074.

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

$\begin{array}{c}\text{t}=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{(74-71)-0}{\sqrt{\left(\frac{(12-1){18}^2+(12-1){16}^2}{12+12-2}\right)\left(\frac{1}{12}+\frac{1}{12}\right)}}\\ =0.43\end{array}$

The calculated t value is 0.43, and the p value is 0.6703. Since the p value is significantly different from 0, the alternate hypothesis is not accepted.

b:

To determine

The test hypothesis with increasing standard deviation.

Explanation of Solution

The t table value can be calculated as follows:

$\begin{array}{c}\text{t}>{\text{t}}_{\frac{\alpha }{2},\text{ν}}\\ \text{t}={\text{t}}_{\frac{0.05}{2},22}\\ ={\text{t}}_{.025,22}\\ =2.074\end{array}$

The t table value is 2.074.

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

$\begin{array}{c}\text{t}=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{(74-71)-0}{\sqrt{\left(\frac{(12-1){210}^2+(12-1){198}^2}{12+12-2}\right)\left(\frac{1}{12}+\frac{1}{12}\right)}}\\ =0.04\end{array}$

The calculated t value is 0.04, and the p value is 0.9716. Since the p value is significantly different from 0, the alternate hypothesis is not accepted.

c:

To determine

The impact of increasing standard deviation.

Explanation of Solution

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

d:

To determine

The test hypothesis with increasing sample.

Explanation of Solution

The t table value can be calculated as follows:

$\begin{array}{c}\text{t}>{\text{t}}_{\frac{\alpha }{2},\text{ν}}\\ \text{t}={\text{t}}_{\frac{0.05}{2},298}\\ ={\text{t}}_{.025,298}\\ =1.96\end{array}$

The t table value is 1.96.

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

$\begin{array}{c}\text{t}=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{(74-71)-0}{\sqrt{\left(\frac{(150-1){18}^2+(150-1){16}^2}{150+150-2}\right)\left(\frac{1}{150}+\frac{1}{150}\right)}}\\ =1.53\end{array}$

The calculated t value is 1.53, and the p value is 0.1282. Since the p value is significantly different from 0, the alternate hypothesis is not accepted.

e:

To determine

The impact of increasing sample.

Explanation of Solution

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

f:

To determine

The test hypothesis with increasing mean value.

Explanation of Solution

The t table value can be calculated as follows:

$\begin{array}{c}\text{t}>{\text{t}}_{\frac{\alpha }{2},\text{ν}}\\ \text{t}={\text{t}}_{\frac{0.05}{2},22}\\ ={\text{t}}_{.025,22}\\ =2.074\end{array}$

The t table value is 2.074.

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

$\begin{array}{c}\text{t}=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{(76-71)-0}{\sqrt{\left(\frac{(12-1){18}^2+(12-1){16}^2}{12+12-2}\right)\left(\frac{1}{12}+\frac{1}{12}\right)}}\\ =0.72\end{array}$

The calculated t value is 0.72, and the p value is 0.4796. Since the p value is significantly different from 0, the alternate hypothesis is not accepted.

g:

To determine

The impact of increasing mean value.

Explanation of Solution

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