13. In another pre-election poll, a pollster finds that 438 out of 600 people polled favor Smith. Construct the 95% confidence interval for the proportion (to nearest 0.001) of people in the whole population who
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- In a sample of 460 adults, 281 had children. Construct a 90% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places < p <7. In a random sample of 250 television viewers in a large city, 190 had seen a certain controversial program. Construct a 98% confidence interval for the true proportion of viewers that had seen the program. Express your answer as percentages with 2 decimal places.In a sample of 570 adults, 485 had children. Construct a 95% confidence interval for the true population proportion of adults with children.Give your answers as decimals, to three places < p <A recent survey showed that 42% of US employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 500 companies. Find the 99% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage. (Round to four decimal places)In a preelection poll, a pollster finds that 442 out of 650 people polled favor Smith. Construct the 95% confidence interval for the proportion (nearest to 0.0001) of people in the whole population who favor Smith.5. In a pre-election poll, a pollster finds that 552 out of 1025 people polled favor Smith. Construct the 95% confidence interval for the proportion of people in the whole population who favor Smith.In a sample of 580 adults, 348 had children. Construct a 95% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places10. Suppose we sample 300 UNI students and find the 27 regularly wear a watch. Find a 98% confidence interval for p, the percentage of all UNI students who regularly wear a watch.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 theseOut of 200 people sampled, 76 had kids. Based on this, construct a 90% confidence interval for the truepopulation proportion of people with kids.Give your answers as decimals, to three placesIn a sample of 300 adults, 198 had children. Construct a 90% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places RectanIn a sample of 360 adults, 266 had children. Construct a 95% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three placesSEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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