Chapter 11, Problem 23SE

a.

To determine

To find: The point estimate of the variance of the population.

Answer to Problem 23SE

The point estimate is 900.

Explanation of Solution

Given:

Sample size $\left(n\right)=20$

Sample mean $\left(\overline{x}\right)=290$

Sample standard deviation $\left(s\right)=30$

Formula used:

The formula to compute the confidence interval for population standard deviation is:

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

The formula to compute the confidence interval for the population variance is:

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

Calculation:

The point estimate of the variance of the population.

  $\begin{array}{c}\text{Estimate of population variance =}{\left(s\right)}^2\\ ={\left(30\right)}^2\\ =900\end{array}$

Thus, the required estimate is 900.

b.

To determine

To find: The 90% confidence interval for the population variance.

Answer to Problem 23SE

The confidence interval is (567, 1690)

Explanation of Solution

Calculation:

Degrees of freedom $\left(df\right)$ = $\left(n-1\right)=\left(20-1\right)=19$

The critical values can be calculated as follows:

  $\begin{array}{c}{\chi }^2{}_{right}={\chi }^2{}_{\frac{\alpha }{2},df}\\ ={\chi }^2{}_{\frac{0.05}{2},19}\\ ={\chi }^2{}_{0.025,19}\\ =30.144\text{ (Using }{\chi }^2\text{-table})\end{array}$

  $\begin{array}{c}{\chi }^2{}_{left}={\chi }^2{}_{1-\frac{\alpha }{2},df}\\ ={\chi }^2{}_{1-\frac{0.05}{2},19}\\ ={\chi }^2{}_{0.975,19}\\ =10.117\text{ (Using }{\chi }^2\text{-table})\end{array}$

Substitute the values in the confidence interval formula.

  $\begin{array}{c}\frac{\left(n-1\right)s^2}{{\chi }^2{}_{right}}<{\sigma }^2<\frac{\left(n-1\right)s^2}{{\chi }^2{}_{left}}\\ \frac{\left(20-1\right)900}{30.144}<{\sigma }^2<\frac{\left(20-\right)900}{10.117}\\ 567<{\sigma }^2<1690\end{array}$

Thus, the obtained confidence interval is (567, 1690)

c.

To determine

To find: The 90% confidence interval for the population standard deviation

Answer to Problem 23SE

The confidence interval is (23.8, 41.1)

Explanation of Solution

Calculation:

From the above part, it is known that the confidence interval is (567, 1690)

The confidence interval for the population standard deviationis calculated as:

  $\begin{array}{c}567<{\sigma }^2<1690\\ \sqrt{567}<\sqrt{{\sigma }^2}<\sqrt{1690}\\ 23.8<\sigma <41.1\end{array}$

Thus, the required confidence interval is (23.8, 41.1)