Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ.
Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.
-Weights of men: 90% confidence; n = 14, $\overline { \mathrm { x } }$ = 161.5 lb, s = 13.7 lb
A) $10.2 \mathrm { lb }

Question

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ.
Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.
-Weights of men: 90% confidence; n = 14, $\overline { \mathrm { x } }$ = 161.5 lb, s = 13.7 lb 
A) $10.2 \mathrm { lb } < \sigma < 19.3 \mathrm { lb }$
B) $11.1 \mathrm { lb } < \sigma < 2.7 \mathrm { lb }$
C) $10.4 \mathrm { lb } < \sigma < 20.3 \mathrm { lb }$
D) $10.8 \mathrm { lb } < \sigma < 17.7 \mathrm { lb }$

Quizplus · Accepted Answer

Using the formula for the confidence interval of the population standard deviation:

$\sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2,n-1}}} < \sigma < \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2,n-1}}}$

Substituting the given values, we get:

$\sqrt{\frac{(14-1)(13.7)^2}{\chi^2_{0.05/2,13}}} < \sigma < \sqrt{\frac{(14-1)(13.7)^2}{\chi^2_{1-0.05/2,13}}}$

Using a chi-square distribution table, we can find that:
$\chi^2_{0.05/2,13} = 5.267$
$\chi^2_{1-0.05/2,13} = 23.685$

Substituting these values, we get:
$\sqrt{\frac{(14-1)(13.7)^2}{5.267}} < \sigma < \sqrt{\frac{(14-1)(13.7)^2}{23.685}}$

Simplifying, we get:
$10.4 \mathrm { lb } < \sigma < 20.3 \mathrm { lb }$

Rounding to the same number of decimal places as the sample standard deviation, we get:
$10.4 \mathrm { lb } < \sigma < 20.3 \mathrm { lb }$

Therefore, the answer is C.

Use the Given Degree of Confidence and Sample Data to Find