Question

a magazine includes a report on the energy costs per year for 32 - inch liquid crystal display (lcd) televisions. the article states that 14 randomly selected 32 - inch lcd televisions have a sample standard deviation of $3.54. assume the sample is taken from a normally distributed population. construct 99% confidence intervals for (a) the population variance ( sigma^{2} ) and (b) the population standard deviation ( sigma ). interpret the results.
(a) the confidence interval for the population variance is (5.46, 45.70).
(round to two decimal places as needed.)
interpret the results. select the correct choice below and fill in the answer box(es) to complete your choice.
(round to two decimal places as needed.)
a. with 99% confidence, you can say that the population variance is between 5.46 and 45.70.
b. with 1% confidence, you can say that the population variance is less than.
c. with 99% confidence, you can say that the population variance is greater than.
d. with 1% confidence, you can say that the population variance is between and.
(b) the confidence interval for the population standard deviation is (, ).
(round to two decimal places as needed )

Explanation:

Step1: Recall the formula for confidence interval of standard deviation

If the confidence interval for variance is \((a,b)\), then for standard deviation it is \((\sqrt{a},\sqrt{b})\)

Step2: Calculate the confidence interval for standard deviation

Given the confidence interval for variance \((5.46,45.70)\)
For the lower - bound of standard deviation: \(\sqrt{5.46}\approx2.34\)
For the upper - bound of standard deviation: \(\sqrt{45.70}\approx6.76\)

Answer:

The confidence interval for the population standard deviation is \((2.34,6.76)\)