A poll of 1372 voters asked what thre most significant issue was in the upcoming election. 1056 of those polled answered the economy. Construct a 92% confidence interval for the proportion of voters in the population who feel that the economy is the most significant issue. p' 0.7696793 0.04 .75 Margin of Error: E = .011367 We are 92% confident that the proportion of voters who feel that the economy is the most v and .79 significant issue is between .75

A poll of 1372 voters asked what thre most significant issue was in the upcoming election. 1056 of those polled answered the economy. Construct a 92% confidence interval for the proportion of voters in the population who feel that the economy is the most significant issue. p' 0.7696793 0.04 .75 Margin of Error: E = .011367 We are 92% confident that the proportion of voters who feel that the economy is the most v and .79 significant issue is between .75

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

The confidence interval for a population mean is (7.32, 7.52). What are the sample mean and the margin of error? H = Margin of Error = Give your answers as decimals, to two places.
We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 15 had kids. Based on this, construct a 99% confidence interval for the proportion, p, of adult residents who are parents in this county.
You recently sent out a survey to determine if the percentage of adults who use social media has changed from 62%, which was the percentage of adults who used social media five years ago. Of the 2751 people who responded to survey, 1740 stated that they currently use social media.a) Use the data from this survey to construct a 98% confidence interval estimate of the proportion of adults who use social media. Record the result below in the form of (#,#)(#,#). Round your final answer to four decimal places.b) Can you conclude that the percentage of adults who use social media has changed? Explain. Yes, because the proportion of adults who used social media five years ago is inside of the confidence interval. Yes, because the proportion of adults who used social media five years ago is not inside the confidence interval. No, because the proportion of adults who used social media five years ago is inside the confidence interval. No, because the proportion of adults who used social media…
A Gallup poll of 1487 adults showed that 43% of the respondents have Facebook pages. Construct a 95% confidence interval for the proportion of adults who have a Facebook page.
A random sample of 500 males included 215 that approved of the performance of a President. A random sample of 500 females included 185 that approved of the performance of a President. Determine each of the following parts that will lead you to construct a 95% confidence interval for the overall difference in the percentage of males and females that approve of the President's performance. group one males, gorup 2 females Calculate E, the margin of error, for a 95% level of confidence, Create the final confidence interval for the difference in percentages
If the number of smokers in a sample of 40 students selected from university A is 14 and the number of smokers in a sample of 380 students selected from university B is 84. Then the 99% confidence interval for the difference between proportions of smokers in the two universities (PB- PA) is closest to O a. (-0.1450,-0.1130) O b. (-0.3310,0.0729) O c. None of these o d. (-0.3110,0.0534) e. (-0.1430,-0.1150)
You recently sent out a survey to determine if the percentage of adults who use social media has changed from 56%, which was the percentage of adults who used social media five years ago. Of the 2335 people who responded to survey, 1725 stated that they currently use social media.a) Use the data from this survey to construct a 93% confidence interval estimate of the proportion of adults who use social media. Record the result below in the form of (#,#)(#,#). Round your final answer to four decimal places.
You recently sent out a survey to determine if the percentage of adults who use social media has changed from 62%, which was the percentage of adults who used social media five years ago. Of the 2283 people who responded to survey, 1635 stated that they currently use social media.a) Use the data from this survey to construct a 94% confidence interval estimate of the proportion of adults who use social media. Record the result below in the form of (#,#)(#,#). Round your final answer to four decimal places.
Construct a 96% confidence interval if x=75 and n=200. Round your answers to 3 decimal places.
A poll of 1070 U.S. adults found that 563 U.S. adults, or 52.617% 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. Express all answers as percentages rounded to 2 decimal places. Do not write the percent sign. Margin of Error (as a percentage): Confidence Interval: Margin of Error (as a percentage): Confidence Interval: Now 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. Margin of Error (as a percentage): Confidence Interval: % to O increases O stays roughly the same O decreases % to Finally, 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. and the precision of our estimate O decreases O increases O stays roughly the…
You want to get a 90% confidence interval for the difference between the percentage of females with red hair compared to males with red hair. You sample 76 females and find 34 of them have red hair. You sample 65 males and find 33 have red hair. < p1−p2 <
If the number of smokers in a sample of 40 students selected from university A is 14 and the number of smokers in a sample of 380 students selected from university B is 84. Then the 99% confidence interval for the difference between proportions of smokers in the two universities (PB-PA) is closest to O a. (-0.3110,0.0534) O b. (-0.3310,0.0729) O c. (-0.1430,-0.1150) O d. (-0.1450,-0.1130) Oe. None of these