Chapter 11.1, Problem 2E

a.

To determine

Compute the 90% confidence interval estimate of the population variance.

Answer to Problem 2E

The 90% confidence interval for ${\sigma }^2$ is $\left(15.76,46.95\right)$.

Explanation of Solution

Calculation:

The given information is that there is a sample of 20 items with a sample standard deviation of 5.

The confidence interval for population variance ${\sigma }^2$ is given by:

$\frac{\left(n-1\right)\times s^2}{{\chi }_{\left(\frac{\alpha }{2}\right)}^2}\le {\sigma }^2\le \frac{\left(n-1\right)\times s^2}{{\chi }_{\left(1-\frac{\alpha }{2}\right)}^2}$

Here, the significance level is $\alpha =0.10$.

Degrees of freedom:

$\begin{array}{c}n-1=20-1\\ =19\end{array}$

Critical value for ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2$:

$\begin{array}{c}{\chi }_{1-\left(\frac{\alpha }{2}\right)}^2={\chi }_{1-\left(\frac{0.10}{2}\right)}^2\\ ={\chi }_{0.95}^2\end{array}$

Procedure:

Step by step procedure to obtain ${\chi }^2{}_{0.95}$ value using Table 11.1 is given below:

• Locate the value 19 in the left column of the table.
• Go through the row corresponding to the value 19 and column corresponding to the value ${\chi }^2{}_{0.95}$ of the table.

Thus, the value of ${\chi }^2{}_{0.95}$ with 19 degrees of freedom is 10.117. That is, $\underset{\_}{{\chi }^2{}_{0.95}=10.117}$.

Critical value for ${\chi }_{\frac{\alpha }{2}}^2$:

$\begin{array}{c}{\chi }_{\left(\frac{\alpha }{2}\right)}^2={\chi }_{\left(\frac{0.10}{2}\right)}^2\\ ={\chi }_{0.05}^2\end{array}$

Procedure:

Step by step procedure to obtain ${\chi }^2{}_{0.05}$ value using Table 11.1 is given below:

• Locate the value 19 in the left column of the table.
• Go through the row corresponding to the value 19 and column corresponding to the value ${\chi }^2{}_{0.05}$ of the table.

Thus, the value of ${\chi }_{0.05}^2$ with 19 degrees of freedom is 30.144. That is, $\underset{\_}{{\chi }^2{}_{0.05}=30.144}$.

Substitute $n=20,s^2=25$, ${\chi }_{\left(\frac{\alpha }{2}\right)}^2=30.144$ and ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2=10.117$ in confidence interval formula.

$\begin{array}{c}\left(\frac{\left(20-1\right)\times 25}{30.144}\text{,}\frac{\left(20-1\right)\times 25}{10.117}\right)=\left(\frac{19\times 25}{30.144}\text{,}\frac{19\times 25}{10.117}\right)\\ =\left(\frac{475}{30.144}\text{,}\frac{475}{10.117}\right)\\ =\left(15.76,46.95\right)\end{array}$

Thus, the 90% confidence interval for population variance is $\left(15.76,46.95\right)$.

b.

To determine

Compute the 95% confidence interval estimate of the population variance.

Answer to Problem 2E

The 95% confidence interval for ${\sigma }^2$ is $\left(14.46,53.33\right)$.

Explanation of Solution

Calculation:

For a 95% confidence interval, the confidence level is $0.95$ and $\alpha =0.05$.

Here, the degrees of freedom is,

$\begin{array}{c}n-1=20-1\\ =19\end{array}$

Critical value for ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2$:

$\begin{array}{c}{\chi }_{\left(1-\frac{\alpha }{2}\right)}^2={\chi }_{\left(1-\frac{0.05}{2}\right)}^2\\ ={\chi }_{0.975}^2\end{array}$

Procedure:

Step by step procedure to obtain ${\chi }_{0.975}^2$ value using Table 11.1 is given below:

• Locate the value 19 in the left column of the table.
• Go through the row corresponding to the value 19 and column corresponding to the value ${\chi }^2{}_{0.95}$ of the table.

Thus, the value of ${\chi }^2{}_{0.975}$ with 19 degrees of freedom is 8.907. That is, $\underset{\_}{{\chi }^2{}_{0.975}=8.907}$.

Critical value for ${\chi }_{\frac{\alpha }{2}}^2$:

$\begin{array}{c}{\chi }_{\left(\frac{\alpha }{2}\right)}^2={\chi }_{\left(\frac{0.05}{2}\right)}^2\\ ={\chi }_{0.025}^2\end{array}$

Procedure:

Step by step procedure to obtain ${\chi }^2{}_{0.025}$ value using Table 11.1 is given below:

• Locate the value 19 in the left column of the table.
• Go through the row corresponding to the value 19 and column corresponding to the value ${\chi }^2{}_{0.025}$ of the table.

Thus, the value of ${\chi }^2{}_{0.025}$ with 19 degrees of freedom is 32.852. That is, $\underset{\_}{{\chi }^2{}_{0.025}=38.852}$.

Substitute $n=20,s^2=25$, ${\chi }_{\left(\frac{\alpha }{2}\right)}^2=32.852$ and ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2=8.907$ in confidence interval formula.

$\begin{array}{c}\left(\frac{\left(20-1\right)\times 25}{32.852}\text{,}\frac{\left(20-1\right)\times 25}{8.907}\right)=\left(\frac{19\times 25}{32.852}\text{,}\frac{19\times 25}{8.907}\right)\\ =\left(\frac{475}{32.852}\text{,}\frac{475}{8.907}\right)\\ =\left(14.46,53.33\right)\end{array}$

Thus, the 95% confidence interval for population variance is $\left(14.46,53.33\right)$.

c.

To determine

Compute the 95% confidence interval estimate of the population standard deviation.

Answer to Problem 2E

The 95% confidence interval for population standard deviation is $\left(3.8,7.3\right)$.

Explanation of Solution

Calculation:

The confidence interval for population standard deviation $\sigma$ is given by:

$\sqrt{\frac{\left(n-1\right)\times s^2}{{\chi }_{\left(\frac{\alpha }{2}\right)}^2}}\le \sigma \le \sqrt{\frac{\left(n-1\right)\times s^2}{{\chi }_{1-\left(\frac{\alpha }{2}\right)}^2}}$

From part (b), it is clear that 95% confidence interval for population variance is $\left(14.46,53.33\right)$.

The confidence interval for population standard deviation is,

$\left(\sqrt{14.46},\sqrt{53.33}\right)=\left(3.8,7.3\right)$.

Therefore, the 95% confidence interval for population standard deviation is $\left(3.8,7.3\right)$.