Question

the annual earnings of 14 randomly selected computer software engineers have a sample standard deviation of $3620. assume the sample is from a normally distributed population. construct a confidence interval for the population variance $sigma^{2}$ and the population standard deviation $sigma$. use a 99% level of confidence. interpret the results.
what is the confidence interval for the population variance $sigma^{2}$?
(5713052, 47786031) (round to the nearest integer as needed.)
interpret the results. select the correct choice below and fill in the answer box(es) to complete your choice. (round to the nearest integer as needed.)
○ a. with 1% confidence, you can say that the population variance is between and.
○ b. with 99% confidence, you can say that the population variance is greater than.
○ c. with 1% confidence, you can say that the population variance is less than.
○ d. with 99% confidence, you can say that the population variance is between and.

Explanation:

Step1: Recall the concept of confidence interval

A confidence interval for the population variance gives a range of values within which the true population variance is likely to lie. A 99% confidence level means that if we were to construct many such intervals, 99% of them would contain the true population variance.

Step2: Analyze each option

• Option A: Incorrect. The confidence level is 99% not 1%.
• Option B: Incorrect. A confidence interval for variance gives a range (lower and upper bound), not just a lower bound.
• Option C: Incorrect. The confidence level is 99% not 1% and a confidence interval for variance gives a range (lower and upper bound), not just an upper bound.
• Option D: Correct. Since we have a 99% confidence interval for the population variance \(\sigma^{2}\) as \((5713052,47786031)\), with 99% confidence, we can say that the population variance is between these two values.

Answer:

D. With 99% confidence, you can say that the population variance is between \(5713052\) and \(47786031\)