Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Normal distribution with standard deviation - 100. You read a report that says, "On the basis of a simple random sample of 100 high school seniors that took the SAT-M test this year, a confidence interval for u is found to be 512.00 + 25.76." What was the confidence level used to calculate this confidence interval? A) 90% В) 95% С) 99% D) 99.9%

Suppose you take 4 air quality samples in parts per million: 19.6, 1.88, 12.7, 11.9. Give a 90% confidence interval of the mean air quality? (Also, t(0.95,3) = 2.3534 and t(0.90,4)= 1.5332) (a) (3.45, 16.83) (b) (0, 25) (c) (2.93, 20.10) (d) (5.22, 21.54) (e) None of the above

A study of the career paths of hotel general managers sent questionnaires to an simple random sample of 260 hotels belonging to major U.S. hotel chains. There were 155 responses. The average time these 155 general managers had spent with their current company was 11.85 years. (Take it as known that the standard deviation of time with the company for all general managers is 3.8 years.) Find the margin of error for an 80% confidence interval to estimate the mean time a general manager had spent with their current company. Round your answer to 2 decimal places.

A flu shot is administered during flu season. We look at the effect of the shot on individuals who received the shot versus those who did not. In a sample of 100 patients that received the flu shot, the average recovery time after being infected with the flu was 4 days with a standard deviation of 1 day. What is a 95% confidence interval for the population average recovery time? O (3.99.4.01) (3.80, 4.20)

If a sample of the lives of 18 car batteries has a standard deviation of 0.722 years, what is the LOWER bound for a 80% confidence interval for ? what is the UPPER bound for a 80% confidence interval for ?

A study of the career paths of hotel general managers sent questionnaires to an SRS of 200 hotels belonging to major U.S. hotel chains. There were 99 responses. The average time these 99 general managers had spent with their current company was 11.96 years. (Take it as known that the standard deviation of time with the company for all general managers is 3.8 years.) (a) Find the margin of error for an 85% confidence interval to estimate the mean time a general manager had spent with their current company:     years   (b) Find the margin of error for a 99% confidence interval to estimate the mean time a general manager had spent with their current company:      years

A study of the career paths of hotel general managers sent questionnaires to an SRS of 280 hotels belonging to major U.S. hotel chains. There were 157 responses. The average time these 157 general managers had spent with their current company was 13.28 years. (Take it as known that the standard deviation of time with the company for all general managers is 3 years.) (a) Find the margin of error for an 80% confidence interval to estimate the mean time a general manager had spent with their current company:  years (b) Find the margin of error for a 99% confidence interval to estimate the mean time a general manager had spent with their current company:  years

A study of the career paths of hotel general managers sent questionnaires to an SRS of 210 hotels belonging to major U.S. hotel chains. There were 88 responses. The average time these 88 general managers had spent with their current company was 11.73 years. (Take it as known that the standard deviation of time with the company for all general managers is 3.6 years.)