6.18 Is college worth it? Part II. Exercise 6.16 presents the results of a poll where 48% of 331 Americans who decide to not go to college do so because they cannot afford it. (a) Calculate a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it, and interpret the interval in context. (b) Suppose we wanted the margin of error for the 90% confidence level to be about 1.5%. How large of a survey would you recommend?
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- 10. A research group studying the use of Apple Pay asked the question, "Do you ever use Apple Pay to make a payment at a grocery store?" to people selected from two random samples. One sample consisted of older adults, aged 35 years and older, and the other sample consisted of younger adults, ages 18 to 34 years old. The proportion of people who answered yes in each sample was used to create a 95 percent confidence interval of (0.097,0.125) to estimate the difference (younger minus older) between the population of people who would answer yes to the question. Which of the following is the best description of what is meant by 95 percent confidence? a) The probability is 0.95 that the difference in the population proportions of people who would answer yes to the question is between 0.097 and 0.125. b) The probability is 0.95 that the difference in the sample proportion of people who would answer yes to the question is between 0.97 and 0.125. c) In repeated random sampling with the same…Sleep apnea: Sleep apnea is a disorder in which there are pauses in breathing during sleep. People with this condition must wake up frequently to breathe. In a sample of 421 people aged 65 and over, 115 of them had sleep apnea. ((c) In another study, medical researchers concluded that more than 34% of elderly people have sleep apnea. Based on the confidence interval, does it appear that more than 34% of people aged 65 and over have sleep apnea? Explain.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 survey of randomly selected university students found that 91 of the 101 first-year students and 96 of the 136 second-year students surveyed had purchased used textbooks in the past year. (a) Construct a 95% confidence interval for the difference the proportions of first-year and second-year university students who purchased used textbooks. (Round your answers to 4 decimal places, if needed.) Answer: (b) If a 98% confidence interval for the difference in proportion of first-year and second-year university students who purchased used textbooks is (0.0809, 0.3093). Give an interpretation of this confidence interval. O We are 98% confident that the first-year students at this university bought between 8.09% and 30.93% more used textbooks than second-year students. O We know that 98% of the first-year students bought between 8.09% and 30.93% higher than the proportion of second-year students who bought used textbooks. O We are 98% confident that, at this university, the proportion of…2) A random sample of 120 students at a high school was asked whether they would ask their father or mother for help with a homework assignment in science. A second random sample of 150 different students was asked the same question for an assignment in history. Fifty-four students in the first sample and 72 students in the second sample replied that they turned to their mother rather than their father for help. Construct a 99% confidence interval for the difference of the two proportions and express it as . (Round both numbers to the 4th decimal placMales and females were asked what they would do if they received a $100 bill in the mail that was addressed to their neighbor, but had been incorrectly delivered to them. Of the 70 males sampled, 55 said yes they would return it to their neighbor. Of the 130 females sampled, 120 said yes. We want to construct a 90% confidence interval estimate for the difference between the true proportion of males who say they would return the money to their neighbor and the true proportion of females who say they would return the money to their neighbor. Choose the best option for the result of constructing that interval. O -226 < p1 -p2 < +.003 O-226 < p1 -p2 <.-.048 O -243 < p1 -p2 < -.031
- In September 2011, Gallup surveyed 1,004 American adults and asked them whether they blamed Barack Obama a great deal, a moderate amount, not much, or not at all for U.S. economic problems. The results showed that 53% of respondents blamed Barack Obama a great deal or a moderate amount. Calculate the 99% confidence interval for the proportion of all American adults who blame Barack Obama a great deal or a moderate amount for U.S. economic problems. O (0.504, 0.556) O (0.499, 0.561) O (0.489, 0.571) O The confidence interval cannot be determined from the information given.A computer store near campus is stocking its shelves and wants to know the relative proportions of students who use PCs vs Macs. A random sample of 60 students is surveyed, and each student is asked whether they use a PC or Mac more often. 55% select ”PC.” (a) What is the population and what is the sample in this study? (b) Calculate a 95% confidence interval for the proportion of UCI students that are PC users. (c) Provide an interpretation of this confidence interval in the context of this problem. (d) The confidence interval is pretty wide and leaves a lot of uncertainty over the proportion of UCIundergarduates who use PCs. With the goal to estimate a narrower 95% confidence interval, what is a simple change to this study that you could suggest for the next time that a similar survey is conducted?Several years ago, 45% of parents who had children in grades K-12 were satisfied with the quality of education the students receive. A recent poll asked 1,025 parents who have children in grades K-12 if they were satisfied with the quality of education the students receive. Of the 1,025 surveyed, 484 indicated that they were satisfied. Construct a 99% confidence interval to assess whether this represents evidence that parents' attitudes toward the quality of education have changed.
- Credit card ownership varies across age groups. Suppose that the estimated percentage of people who own at least one credit card is 64% in the 18–24 age group, 84% in the 25–34 age group, 75% in the 35–49 age group, and 77% in the 50+ age group. Suppose these estimates are based on 465 randomly selected people from each age group. (a) Construct a 95% confidence interval for the proportion of people in the 18–24 age group who own at least one credit card. (Round your answers to four decimal places.) to Construct a 95% confidence interval for the proportion of people in the 25–34 age group who own at least one credit card. (Round your answers to four decimal places.) to Construct a 95% confidence interval for the proportion of people in the 35–49 age group who own at least one credit card. (Round your answers to four decimal places.) to Construct a 95% confidence interval for the proportion of people in the 50+ age group who own at least one credit card. (Round your answers to four…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:Suppose NBC has asked the same question to a sample of 1296 adults nine months earlier, finding 47% with an optimistic outlook for the year ahead. Develop a 95% confidence interval for the percentage of American adults who had an optimistic outlook for the economy nine months earlier.
