Question

use technology to construct the confidence intervals for the population variance $sigma^{2}$ and the population standard deviation $sigma$. assume the sample is taken from a normally distributed population.
$c = 0.99, s^{2}=7.29, n = 29$
the confidence interval for the population variance is $(4.00, 16.38)$.
(round to two decimal places as needed.)
the confidence interval for the population standard deviation is $(square,square)$.
(round to two decimal places as needed.)

Explanation:

Step1: Recall the relationship between variance and standard deviation

The standard deviation \(\sigma\) is the square - root of the variance \(\sigma^{2}\). If the confidence interval for the variance is \((a,b)\), then the confidence interval for the standard deviation is \((\sqrt{a},\sqrt{b})\)

Step2: Calculate the lower bound of the standard deviation interval

Given the lower bound of the variance interval \(a = 4.00\). Using the formula \(x=\sqrt{a}\), we have \(x=\sqrt{4.00}=2.00\)

Step3: Calculate the upper bound of the standard deviation interval

Given the upper bound of the variance interval \(b = 16.38\). Using the formula \(y=\sqrt{b}\), we have \(y=\sqrt{16.38}\approx4.05\)

Answer:

The confidence interval for the population standard deviation is \((2.00,4.05)\)