A random sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that σ is $2.11. Construct the 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. 20.15 17.59 17.94 15.26 17.87 18.01 21.76 17.34 21.42 16.29 17.61 15.77 20.71 18.71 17.22 15.71 The 90% confidence interval is ($ ☐ $ ). (Round to two decimal places as needed)

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2. The data below represent the amounts of time (in minutes) spent checking email for a random sample of employees at a company. Assume the times are normally distributed. 7.5 2.0 12.1 8.8 9.4 7.3 1.9 2.8 7.0 7.3 a. Find a point estimate for the population mean. b. Construct a 90% confidence interval for the population mean. Interpret the results. C. If we change the level of confidence to 95%, would the confidence interval would be shorter or wider than your answer in (c)? Explain. (Working the problem is not necessary.)

You are interested in finding a 95% confidence interval for the average commute that non- residential students have to their college. The data below show the number of commute miles for 11 randomly selected non-residential college students. Round answers to 3 decimal places where possible. 11 10 17 10 27 15 20 8 10 8 12 a. To compute the confidence interval use a ?v distribution. b. With 95% confidence the population mean commute for non-residential college students is between and miles. c. If many groups of 11 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of commute miles and about percent will not contain the true population mean number of commute miles.

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A laboratory tested 85 chicken eggs and found that the mean amount of cholesterol was 191 milligrams. Assume that o = 16.2 milligrams. Construct a 95% confidence interval for the true mean cholesterol content, µ, of all such eggs. Round to one decimal place. O A. 186.9 to 195.1 milligrams B. 186.5 to 195.5 milligrams C. 188.1 to 193.9 milligrams D. 187.6 to 194.4 milligrams

Use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a recent season, the population standard deviation of the yards per carry for all running backs was 1.36. The yards per carry of 25 randomly selected running backs are shown below. Assume the yards per carry are normally distributed. 2.6 3.5 3.1 4.6 3.4 4.9 7.4 6.7 4.3 5.2 3.9 4.6 2.9 5.6 6.6 7.3 6.5 7.2 96 1.9 5.6 2.7 22 2.8 5.2 4.1 3.6 Interpret the results. Choose the correct answer below. OA. If a large sample of players are taken approximately 95% of them will have yards per carry between the bounds of the confidence interval. OB. With 95% confidence, it can be said that the population mean yards per carry is between the bounds of the confidence interval. OC. It can be said that 95% of players have a yards per carry between the bounds of the confidence interval.…

ou are interested in finding a 95% confidence interval for the average commute that non-residential students have to their college. The data below show the number of commute miles for 11 randomly selected non-residential college students. Round answers to 3 decimal places where possible.   15 12 9 25 20 9 6 10 7 26 27   a. To compute the confidence interval use a distribution. b. With 95% confidence the population mean commute for non-residential college students is between  and   miles. c. If many groups of 11 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of commute miles and about percent will not contain the true population mean number of  commute miles.

You are interested in finding a 90% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 12 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.   7 26 5 10 23 14 14 8 19 6 8 12   a. To compute the confidence interval use a ? z or t  distribution. b. With 90% confidence the population mean number of visits per physical therapy patient is between __and  ___visits. c. If many groups of 12 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About __percent of these confidence intervals will contain the true population mean number of visits per patient and about __percent will not contain the true population mean number of visits per patient.