ove.4.8 Twitter users and news, Part I:A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news onTwitter (Pew, 2013). The standard error for this estimate was 2.4%, and a normal distribution naybe used to model the sample proportion. Construct a 99% confidence interval for the fraction of U.S.adult Twitter users who get some news on Twitter, and interpret the confidence interval in context.(please round all percentages to 2 decimal places)The 99% confidence interval is%.%)Interpret the meaning of your confidence interval:O 99% of confidence intervals will be contain the correct sample proportion of U.S. adult twitterusers who get some news on TwitterO We are 99% confident that the true proportion of U.S. adult twitter users who get some newson Twitter is within our confidence intervalO There is a 99% probability that the proportion of U.S. adult twitter users who get some newson Twitter is within our confidence intervalCheck All PartsType here to search1O

Given data: P = 0.52 Confidence level = 99% Standard error(SE) = 2.4 % To find: 99% confidence…

Answered: ove. 4.8 Twitter users and news, Part…

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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Five students took a self-confidence test. Their deviation scores were –4, +3, +1, –2, and +2. Assuming that these scores com form a normaldistribution, which deviation represents the highest raw score?
A study of 420,048 cell phone users found that 137 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0256% for those not using cell phones. Complete parts (a) and (b). a. Use the sample data to construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. enter your response here%<p<enter your response here% (Round to three decimal places as needed.) b. Do cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate of such cancer among those not using cell phones? Why or why not? A. No, because 0.0256% is included in the confidence interval. B. Yes, because 0.0256% is included in the confidence interval. C. Yes, because 0.0256% is not included in the confidence interval. D. No, because 0.0256% is not…
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.
Tourism is extremely important to the economy of Florida. Hotel occupancy is an often-reported measure of visitor volume and visitor activity (Orlando Sentinel, May 19, 2018). Hotel occupancy data for February in two consecutive years are as follows, current year previous year occupied rooms 1470 1458 total rooms 1750 1800 a)What is the 95% confidence interval estimate of the change in occupancy for the one-year period? b)Do you think officials will be pleased with the results? c) Using a 0.05 level of significance, test the hypothesis that there has been an increase in the proportion of rooms occupied over the one-year period.
15. Pain after surgery: In a random sample of 50 patients undergoing a standard surgical procedure, 15 required medication for postoperative pain. In a random sample of 90 patients undergoing a new procedure, only 16 required pain medication. a. Construct a 95% confidence interval for the difference in the proportions of patients needing pain medication between the old and new procedures. b. A physician claims that the proportion of patients who need pain medication is the same for both procedures. Does the confidence interval contradict this claim? 16
Suppose we wish to estimate the average number of hours per week that Americans watch TV. Use the following data from an SRS to find a 90% confidence interval: 30 64 49 40 21 58 9 43 39 43
A survey is taken of 250 students, and a P of 0.45 is found. The same survey is repeated with 1,000 students, and the same value is found. Compare the two 95% confidence intervals. What is the relationship? Is the margin of error for the second one four times smaller? How much smaller is it?
Early-childhood-development studies indicate that the more often a child is read to from birth, the earlier the child begins to read. A local parents’ group wants to test this theory and samples families with young children. They find the following results. Construct a 99% confidence interval to estimate the true difference between the proportions of children who read at an early age when they are read to frequently compared to those who were read to less often, as described in the table of results. Let Population 1 be the children who were read to frequently and Population 2 be the children who were read to less often. Round the endpoints of the interval to three decimal places, if necessary. Ages when Children Begin to Read Read to at Least Three Times per Week Read to Fewer than Three Times per Week Started Reading by age 4 60 35 Started Reading after age 4 36 51 Lower end point: Upper end point:
8.2 Question 3). According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, Arkansas had 1,601 complaints of identity theft out of 3,482 consumer complaints ("Consumer fraud and," 2008). Calculate a 90% confidence interval for the proportion of identity theft in Arkansas.

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