Chapter 11, Problem 30SE

A sample standard deviation for the number of passengers taking a particular airline flight is 8. A 95% confidence interval estimate of the population standard deviation is 5.86 passengers to 12.62 passengers.

1. a. Was a sample size of 10 or 15 used in the statistical analysis?
2. b. Suppose the sample standard deviation of s = 8 was based on a sample of 25 flights. What change would you expect in the confidence interval for the population standard deviation? Compute a 95% confidence interval estimate of α with a sample size of 25.

To determine

Check whether a sample size of 10 or 15 was used in the statistical analysis.

Answer to Problem 30SE

The sample size 15 is used in the statistical analysis.

Explanation of Solution

Calculation:

It is given that the sample standard deviation is 8. The 95% confidence interval for population standard deviation is (5.86, 12.62).

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

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

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

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

For $n=15$, the degrees of freedom is calculated as follows:

$\begin{array}{c}n-1=15-1\\ =14\end{array}$

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

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

Procedure:

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

• Locate the value 14 in the left column of the table.
• Find the row corresponding to the value 14 and column corresponding to the value ${\chi }^2{}_{0.975}$ of the table.

Thus, the value of ${\chi }^2{}_{0.975}$ with 14 degrees of freedom is 5.629. That is, $\underset{\_}{{\chi }^2{}_{0.975}=5.629}$.

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 14 in the left column of the table.
• Find the row corresponding to the value 14 and column corresponding to the value ${\chi }^2{}_{0.025}$ of the table.

Thus, the value of ${\chi }^2{}_{0.025}$ with 14 degrees of freedom is 26.119. That is, $\underset{\_}{{\chi }^2{}_{0.025}=26.119}$.

Substitute $n=15$, $s^2=64$, ${\chi }_{\left(\frac{\alpha }{2}\right)}^2=26.119$, and ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2=5.629$ in the confidence interval.

$\begin{array}{c}\left(\frac{\left(15-1\right)\times 64}{26.119}\text{,}\frac{\left(15-1\right)\times 64}{5.629}\right)=\left(\frac{14\times 64}{26.119}\text{,}\frac{14\times 64}{5.629}\right)\\ =\left(\frac{896}{26.119}\text{,}\frac{896}{5.629}\right)\\ =\left(34.3,159.2\right)\end{array}$

Thus, the 95% confidence interval for population variance is $\left(34.3,159.2\right)$.

The 95% confidence interval for population standard deviation is given below:

$\left(\sqrt{34.3},\sqrt{159.2}\right)=\left(5.86,12.62\right)$

Therefore, the 95% confidence interval for population standard deviation is $\left(5.86,12.62\right)$.

From the given information, the 95% confidence interval for population standard deviation is 5.86 passengers to 12.62 passengers. For the sample size $n=15$, the 95% confidence interval for population standard deviation is $\left(5.86,12.62\right)$. Hence, the sample size 15 is used in the statistical analysis.

To determine

Compute the 95% confidence interval estimate of $\sigma$ with a sample size of 25. Explain the changes expected in the confidence interval.

Answer to Problem 30SE

The 95% confidence interval estimate of $\sigma$ with a sample size of 25 is $\left(6.25,11.13\right)$.

The change expected in the confidence interval is that the width of the confidence interval is smaller for $n=25$ compared to $n=15$.

Explanation of Solution

Calculation:

For $n=25$, the degrees of freedom is calculated as follows:

$\begin{array}{c}n-1=25-1\\ =24\end{array}$

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

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

Procedure:

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

• Locate the value 24 in the left column of the table.
• Find the row corresponding to the value 24 and column corresponding to the value ${\chi }^2{}_{0.975}$ of the table.

Thus, the value of ${\chi }^2{}_{0.975}$ with 24 degrees of freedom is 12.401. That is, $\underset{\_}{{\chi }^2{}_{0.975}=12.401}$.

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 24 in the left column of the table.
• Find the row corresponding to the value 24 and column corresponding to the value ${\chi }^2{}_{0.025}$ of the table.

Thus, the value of ${\chi }^2{}_{0.025}$ with 24 degrees of freedom is 39.364. That is, $\underset{\_}{{\chi }^2{}_{0.025}=39.364}$.

Substitute $n=25$, $s^2=64$, ${\chi }_{\left(\frac{\alpha }{2}\right)}^2=39.364$, and ${\chi }_{\left(1-\frac{\alpha }{2}\right)}^2=12.401$ in the confidence interval.

$\begin{array}{c}\left(\frac{\left(25-1\right)\times 64}{39.364}\text{,}\frac{\left(25-1\right)\times 64}{12.401}\right)=\left(\frac{24\times 64}{39.364}\text{,}\frac{24\times 64}{12.401}\right)\\ =\left(\frac{\text{1,536}}{39.364}\text{,}\frac{\text{1,536}}{12.401}\right)\\ =\left(39.02,126.86\right)\end{array}$

Thus, the 95% confidence interval for population variance is $\left(39.02,126.86\right)$.

The 95% confidence interval for population standard deviation is as follows:

$\left(\sqrt{39.02},\sqrt{126.86}\right)=\left(6.25,11.13\right)$

Therefore, the 95% confidence interval for population standard deviation is $\left(6.25,11.13\right)$.

The 95% confidence interval for population standard deviation when $n=25$ is $\left(6.25,11.13\right)$ and from part (a), the 95% confidence interval for population standard deviation when $n=15$ is $\left(5.86,12.62\right)$.

The interval width is 4.88$\left(=11.3-6.25\right)$ for $n=25$ and the interval width is 6.76$\left(=12.62-5.86\right)$ for $n=15$. Hence, the interval width is decreased for $n=25$.