(b) Based on your sample, graph the 99% confidence interval for the population mean of all the distances the voters in Vanville County live from their nearest polling station. Enter the values for the lower and upper limits on the graph-to show your confidence interval. For the point (*), enter the claim 7.78 from the article. 0.000 99% confidence interval: 0.000 2.000 4.000 5.000 10.000 6.000 8.000 10.000 (c) Does the 99% confidence interval you constructed contradict the claim made in the article? Choose the best answer from the choices below. X O No, the confidence interval does not contradict the claim. The mean of 7.78 km from the article is inside the 99% confidence interval. O No, the confidence interval does not contradict the claim. The mean of 7.78 km from the article is outside the 99% confidence interval. Yes, the confidence interval contradicts the claim. The mean of 7.78 km from the article is inside the 99% confidence interval. O Yes, the confidence interval contradicts the claim. The mean of 7.78 km from the article is outside the 99% confidence interval. 5 & In anticipation of an upcoming election, officials in Vanville County are looking at the distance from each voter's home to that voter's nearest polling station. Assume that the population of all such distances for voters in Vanville County is approximately normally distributed. An article for the newspaper Frontline Freedom claimed that the mean of this population is 7.78 km. You want to test this claim, so you select a random sample of 17 Vanville County voters, and for each you record the distance the voter lives from their nearest polling station. Follow the steps below to construct a 99% confidence interval for the population mean of all the distances voters in Vanville County live from their nearest polling station. Then state whether the confidence interval you construct contradicts the reporter's claim. (If necessary, consult a list of formulas.) (a) " Click on "Take Sample" to see the results for your random sample. Number of people 17 Sample standard Sample mean deviation 4.179 2.023 5 your 99% Enter the values of the sample size, the point estimate of the mean, the sample standard deviation, and the critical value you need for confidence interval. (Choose the correct critical value from the table of critical values provided.) When you are done, select "Compute". Take Sample Sample size: 1 Point estimate: Standard error: X Sample standard deviation: Critical values Margin of error: 0.005 = 2.921 Critical value: 0.010 = 2.583 99% confidence interval: 50.025 =2.120 Compute 0.050 = 1.746 40.100 = 1.337 8 B

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Z-score for an 88% confidence level

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…

you want to find out the percent of people who use an iPhone so you survey 120 random adults of whom 57 use an iPhone. what is the 95% confidence interval for P the put portion of all adults who use iPhone

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.

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…

In a survey of 500 students, 327 said they were planning to stay in town for the holidays.  What is the point estimate for the proportion of students staying home for the holidays? Construct the 95% confidence interval for the proportion of students staying home for the holidays.  Assume all requirements have been met.

You recently sent out a survey to determine if the percentage of adults who use social media has changed from 58%, which was the percentage of adults who used social media five years ago. Of the 2670 people who responded to survey, 1767 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.