A survey was conducted that asked 1016 people how many books they had read in the past year. Results indicated thatx= 13.7 books and s= 16.2 books. Construct a 95% confidence interval for the mean number of books people read.Interpret the interval.ICCEEDConstruct a 95% confidence interval for the mean number of books people read and interpret the result. Select thecorrect choice below and fill in the answer boxes to complete your choice.(Use ascending order. Round to two decimal places as needed.).OA. There is 95% confidence that the population mean number of books read is betweenB. If repeated samples are taken, 95% of them will have a sample mean betweenOC. There is a 95% probability that the true mean number of books read is betweenandandand

Answered: A survey was conducted that asked 1016…

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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A survey was conducted that asked 1016 people how many books they had read in the past year. Results indicated that x = 13.3 books and s= 16.6 books. Construct a 90% confidence interval for the mean number of books people read. Interpret the interval. Click the icon to view the table of critical t-values. Construct a 90% confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to two decimal places as needed.) OA. If repeated samples are taken, 90% of them will have a sample mean between and OB. There is 90% confidence that the population mean number of books read is between and OC. There is a 90% probability that the true mean number of books read is between and
Use the confidence interval to find the margin of error and the sample mean. (1.77, 2.01) The margin of error is
A poll of 2368 U.S. adults found that 40% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 99% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % The more error we allow, the less precise our estimate. Therefore, as the confidence level increases, the precision of our…
Use the given degree of confidence and sample data to construct a confidence interval for the population mean u. Round your final answers to the nearest hundredth. Test scores: n = 104, x = 95.3, o = 6.5; 99% confidence You will need to calculate the margin of error, followed by the lower and upper end of the confidence interval. Margin of Error: choose your answer.. Confidence Interval: choose your answer... レ choose your answer...
Use the confidence interval to find the margin of error and the sample mean. (0.636, 0.760)
A poll of 2338 U.S. adults found that 71% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to %
Determine whether the statement below makes sense or does not make sense. Explain clearly. Based on our sample, the 95% confidence interval for the mean amount of television watched by adults in a nation is 3.5 to 4.1 hours per day. Therefore, there is 95% chance that the mean for all adults in the nation will fall somewhere in this range and a 5% chance that it will not. Choose the correct answer below. O A. The statement makes sense. There is 95% probability that the confidence interval limits actually contain the true value of the population mean, so the probability it does not fall in this range is 100% - 95% = 5%. O B. The statement does not make sense. The probability the population mean is greater than the upper limit is 5% and the probability it is less than the lower limit is 5%, so the probability it does not is 5% + 5% = 10%. O C. The statement makes sense. There is 5% probability that the confidence interval limits do not contain the true value of the sample mean, so the…
You are interested in finding a 95% 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.
Find the confidence level and α for a. a 90% confidence interval. b. a 99% confidence interval.
find an explain a study that uses confidence intervals. Share and explain what the study was about and how they use confidence intervals in the study to estimate the population parameter.
For the same sample statistics, which level of confidence would produce the widest confidence interval? Explain your reasoning. Choose the correct answer below. 99%, because as the level of confidence increases, ze decreases. 99%, because as the level of confidence increases, Z increases. O 90%, because as the level of confidence decreases, ze increases. 90%, because as the level of confidence decreases, Zc decreases.
Directions: Solve the following. Show your complete solution. Fifty randomly selected jeepneys were stopped, and the tread depth of their right front tires were measured. The mean was 0.32 inch and the standard deviation was 0.08 inch. Find the 99% confidence interval of the mean depth. Assume that the variable is approximately normally distributed.

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