Question

what is the margin of error and the confidence interval?

Question Help In a random sample of seven people, the mean driving distance to work was 24.7 miles and the standard deviation was 6.6 miles. Assuming the population is normally distributed and using the I-distribution, a 90% confidence interval for the population mean is (15.5, 33.9) (and the margin of error is 9.2). Through research, it has been found that the population standard deviation of driving distances to work is 5.8. Using the standard normal distribution with the appropriate calculations for a standard deviation that is known, find the margin of error and construct a 99% confidence interval for the population mean . Interpret and compare the results Identify the margin of error (Round to one decimal place as needed.)

Answer 1

n=7, X = 24.7, s = 6.6 99% (15.5, 33.9) 6 = 5.8 ggo%. I .: We have touse z :. Zgg% = 2.5758. .. margin of tror = = 2.5758 x 5.8 Margin oferrux 5.6 : 99% cl Lower limit=X - Bit = 24.7-5.6 - 19.1 Upper limit=x+ME = 29.7 +5.6 = 30.3 in99% CI ( 19.1, 30.3 )