Question

A random sample of 17 values was selected from a population, and the sample standard deviation was

computed to be 150. Based on this sample result, compute a 90% confidence interval estimate for the true
population standard deviation.
State the UPPER bound of this confidence interval. (Hint: Use the formula for
calculating a confidence interval estimate for a population variance. For this
problem, the chi-squared critical value you should use is 7.9616. Also, don't forget,
it is asking you for the confidence interval for the standard deviation. Chapter
11.1)
(Round to two decimal places as needed.)

Answer 1

To compute the upper bound of the 90% confidence interval for the true population standard deviation, you can use the formula:

\[ \sqrt{\frac{(n-1)S^2}{\chi^2_{\alpha/2}}} \]

where:
- \( n \) is the sample size (17 in this case),
- \( S \) is the sample standard deviation (150),
- \( \chi^2_{\alpha/2} \) is the chi-squared critical value (7.9616 for a 90% confidence interval).

Substitute the values into the formula to find the upper bound of the confidence interval. Remember to round to two decimal places.