A random sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that σ is $2.11. Construct the 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.20.1517.5917.9415.2617.8718.0121.7617.3421.4216.2917.6115.7720.7118.7117.2215.71The 90% confidence interval is ($ ☐ $ ).(Round to two decimal places as needed)

The objective of this question is to construct the 90% and 99% confidence intervals for the…

Answered: A random sample of the closing stock…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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In a random sample of five people, the mean driving distance to work was 25.1 miles and the standard deviation was 6.6 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 95% confidence interval for the population mean p. Interpret the results. * *I Identify the margin of error. miles (Round to one decimal place as needed.) rary More Clear all Check answer Get more help - Media View an example ess Help me solve this ESources 58°F A @ S 6. ress e here to search 10/05/17
b. between 2 According to Cardweb.com the mean bank credit card debt for households FE07 was $7868 in 2004. Assume that mean was based on a random sample of 900 households and that the standard deviation of such debts for all households in 2004 was $ 2070. Make a 99% and a 99.22% confidence interval for the 2004 mean bank credit card debt for all households JAn alumni association wants to estimate the mean debt for this year's college graduates. It is known that the population standard deviation of the debts of this years college graduates is $11,800. How large a sample should be selected so that the estimate with a 99% confidence level is within $800 of the population mean How large a sample should be selected so that the estimate with a 94.64% confidence level is within 5900 of the population mean
1. 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 10 randomly selected non-residential college students. Round answers to 3 decimal places where possible. 17 14 23 20 27 27 22 12 7 23 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. Please state what command(s) you typed into your calculator. MIDDLESEX c. If many groups of 10 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced percent of these confidence intervals will from each group. About contain the true population mean number of commute miles and about percent will not contain the true population mean number of
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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.

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