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.
population variance is greater than
population variance is between
and
(b) the confidence interval for the population standard deviation is (2.34, 6.76).
(round to two decimal places as needed.)
interpret the results. select the correct ch\ice 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
b. with 1% confidence, you can say that the
population standard deviation is less than
population standard deviation is between and
dollars per year.
dollars per year.
c. with 1% confidence, you can say that the
d. with 99% confidence, you can say that the
population standard deviation is greater than
population standard deviation is between and
dollars per year.
dollars per year.

Explanation:

Step1: Recall the concept of confidence interval

A confidence interval gives a range of values within which the population parameter is likely to lie. For a \(99\%\) confidence interval of the population standard deviation \(\sigma\), we use the formula for the confidence interval of the standard deviation based on the chi - square distribution. The confidence interval \((a,b)\) for \(\sigma\) is constructed such that we are \(99\%\) confident that the true population standard deviation lies within this interval.

Step2: Interpret the confidence interval

Since the confidence interval for the population standard deviation is \((2.34,6.76)\), and the confidence level is \(99\%\). This means that if we were to take many samples and construct confidence intervals in the same way, approximately \(99\%\) of those intervals would contain the true population standard deviation.

Answer:

D. With \(99\%\) confidence, you can say that the population standard deviation is between \(2.34\) and \(6.76\) dollars per year.