Chapter 11, Problem 23SE

Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms.

1. a. What is the point estimate of the population variance?
2. b. Provide a 90% confidence interval estimate of the population variance.
3. c. Provide a 90% confidence interval estimate of the population standard deviation.

To determine

Compute the point estimate of the population variance.

Answer to Problem 23SE

The point estimate of the population variance is 900.

Explanation of Solution

Calculation:

The given information is that the sample mean for a sample of 20 days of operation is 290 rooms occupied per day with a standard deviation of 30 rooms.

The point estimate of population variance is the sample variance.

Therefore,

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

Thus, the point estimate of the population variance is 900.

To determine

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

Answer to Problem 23SE

The 90% confidence interval estimate of the population variance is $\left(567,1,690\right)$.

Explanation of Solution

Calculation:

Here, the sample size is 20.

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

Substitute $n=20,s^2=900$, ${\chi }_{\left(\frac{\alpha }{2}\right)}^2=30.144$ and ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2=10.117$ in the confidence interval formula.

$\begin{array}{c}\left(\frac{\left(20-1\right)\times 900}{30.144}\text{,}\frac{\left(20-1\right)\times 900}{10.117}\right)=\left(\frac{19\times 900}{30.144}\text{,}\frac{19\times 900}{10.117}\right)\\ =\left(\frac{17,100}{30.144}\text{,}\frac{17,100}{10.117}\right)\\ =\left(567,1,690\right)\end{array}$

Thus, the 90% confidence interval for population variance is $\left(567,1,690\right)$.

To determine

Compute the 90% confidence interval estimate of the population standard deviation.

Answer to Problem 23SE

The 90% confidence interval for population standard deviation is $\left(23.8,41.1\right)$.

Explanation of Solution

Calculation:

The confidence interval formula 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 90% confidence interval for population variance is $\left(567,1,690\right)$.

The 90% confidence interval for population standard deviation is,

$\left(\sqrt{567},\sqrt{1690}\right)=\left(23.8,41.1\right)$.

Therefore, the 90% confidence interval for population standard deviation is $\left(23.8,41.1\right)$.