A 2015 survey of 400 randomly selected drivers reported that 20 percent of the drivers surveyednever use their cruise control. A 95% confidence interval is given by (0.161, 0.239). Which of thefollowing is a correct interpretation of the 95% confidence level?(A) We are 95 percent confident that the true proportion of all drivers who never use their cruisecontrol is between 0.161 and 0.239.(B) There is a 0.95 probability that the true proportion of all drivers who never use their cruisecontrol is between 0.161 and 0.239.(C) 95 percent of all random samples of 400 drivers chosen from the population will result inconfidence intervals which contain 0.20.(D) 95 percent of all random samples of 400 drivers will result in confidence intervals which containthe true proportion of all drivers who never use their cruise control.(E) 95 percent of all random samples of 400 drivers chosen from the population will have sampleproportions between 0.161 and 0.239.

A confidence interval is a range of values, derived from a data set, that is likely to contain the…

Answered: A 2015 survey of 400 randomly selected…

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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Similar questions

In a survey funded by the UW school of medicine, 750 of 1000 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 95% confidence interval estimage of the proportion of adult Seattle residents who don't believe they can contract an STI. (Use a z score of 1.96 for your computations.) (.728, .772) (.723, .777) (.718, .782) (.713, .878) (.665, .835)
In a study of the nicotine patch, 21% of those who used the patch for 2 months reported no smoking incidents in the following year. The 95% confidence interval is (17.4%, 24.8%). Which of the following is an appropriate interpretation of the 95% confidence interval? There is a 95% probability that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%. We can be 95% confident that the proportion of the sample who would report no smoking incidents in the following year is between 17.4% and 24.8%. 95% of samples will have between 17.4% and 24.8% who would report no smoking incidents in the following year. We can be 95% confident that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.
A survey of 2450 golfers showed that 281 of them are left-handed. Construct a 90% confidence interval for the proportion of golfers that are left-handed. Question 10 options: (0.104, 0.125) (0.369, 0.451) (0.100, 0.130) (0.203, 0.293)
An online survey of 385 individuals found that 114 had fully completed the game God of War. Which of the following is a 99% confidence interval for the population proportion of individuals who fully completed this video game? (0.2505, 0.3417) (0.2578, 0.3344) (0.2362, 0.3560) None of the above
The dean of students at a large college is interested in learning about their opinions regarding the percentage of first-year students who should be given parking privileges in the main lot. He sends out an email survey to all students about this issue. A large number of first-year students reply but very few sophomores, juniors, and seniors reply. Based on the responses he receives, he constructs a 90% confidence interval for the true proportion of students who believe first-year students should be given parking privileges in the main lot to be (0.71, 0.79). Which of the following may have an impact on the confidence interval, but is not accounted for by the margin of error? response bias O nonresponse bias O sampling variation O undercoverage bias
I'm having trouble trying to figure out how to work the problem below out. Let’s say that 44 percent of the 1,300 U.S. adults who responded to a poll about marijuana legalization said they support legalization. Let’s also say the margin of error for the results is +/- 3 percentage points. The margin of error tells us there’s a high probability that nationwide support for marijuana legalization falls: The correct interpretation of this confidence interval is:
. In a random sample of 400 private high school students, 320 said they brought a smartphone to school that day, while in a random sample of 400 public high school students, 288 said they brought a smartphone to school that day. Which of the following represents a 98 percent confidence interval estimate for the difference (private high school minus public high school) between the proportions of all private high school and public high school students who bring a smartphone to school? (0.8)(0.2) (0.72)(0.28) (A) (0.8-0.72)±1.96, 400 400 (B) (0.8-0.72)±2.054, + (0.8)(0.2) (0.72)(0.28) 400 400 (C) (0.8-0.72)±2.326 + (0.8)(0.2) (0.72)(0.28) 400 400 1 (D) (0.8-0.72)±2.054, (0.76)(0.24) + 400 400 (E) (0.8-0.72)±2.326, (0.76)(0.24) + I 400 400
Question #8
A professor was curious about her students' grade point averages. She took a random sample of 15 students and found a mean GPA of 3.01 with a standard deviation of 0.534. Which of the following formulas gives a 99% confidence interval for the mean GPA of the professor's students?
If you want to be confident in your estimation of a confidence interval, you should use….
The figure to the right shows the results of a survey in which 2000 college graduates from the year 2016 were asked questions about employment. Construct 99% confidence intervals for the population proportion of college students who gave each response. Employment College students' responses to questions about employment Expect to stay at first employer for 3 or more years Completed an apprenticeship or intemship 70% 66% Employed in field of study 61% Feel underemployed 46% Prefer to work for a large company 15% The 99% confidence interval for the proportion of college students that expect to stay at their first employer for 3 or more years is (Round to three decimal places as needed.)
Listed below are the amounts of net worth (in millions of dollars) of the ten wealthiest celebrities in a country. Construct a 95% confidence interval. What does the result tell us about the population of all celebrities? Do the data appear to be from a normally distributed population as required? 245 210 186 168 166 163 156 156 156 156

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