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. population variance is less than population variance is between 5713052 and 47786031. what is the confidence interval for the population standard deviation $\sigma$? (2390, 6913) (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 standard deviation is greater than $ b. with 99% confidence, you can say that the population standard deviation is between $ and $ c. with 1% confidence, you can say that the population standard deviation is between $ and $ d. with 99% confidence, you can say that the population standard deviation is less than $

Explanation:

Step1: Recall the confidence level concept

A 99% confidence interval means that if we were to construct many such intervals from different samples, approximately 99% of them would contain the true population parameter.

Step2: Analyze the options

• Option A: Talks about 1% confidence, which is not relevant as we are using a 99% confidence level.
• Option B: Since we have a 99% confidence interval for the population standard deviation (already given as (2390, 6913)), with 99% confidence, we can say the population standard deviation is between these two values.
• Option C: 1% confidence is incorrect for our 99% confidence level scenario.
• Option D: The interval is a range (not just less than a single value) for the 99% confidence case.

Answer:

B. With 99% confidence, you can say that the population standard deviation is between $2390$ and $6913$.