Ages of students: A simple random sample of 90 U.S. college students had a mean age of 23.06 years. Assume the population standard deviation is a 4.82 years. Construct a 95% confidence interval for the mean age of U.S. college students. Round the answers to two decimal places. A 95% confidence interval for the mean age of U.S. college students is << Credit card debt: In a survey of 1118 U.S. adults conducted in 2012 by the Financial Industry Regulatory Authority, 810 said they always pay their credit cards in full each month. Construct a 90% confidence interval for the proportion of U.S. adults who pay their credit cards in full each month. Round the answers to three decimal places. A 90% confidence interval for the proportion of U.S. adults who pay their credit cards in full each month is <p<0. College tuition: A simple random sample of 35 colleges and universities in the United States has a mean tuition of $17,600 with a standard deviation of $10,200. Construct a 90% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number. A 90% confidence interval for the mean tuition for all colleges and universities is με Call me: A sociologist wants to construct a 99.5% confidence interval for the proportion of children aged 8-10 living in New York who own a cell phone. Part 1 of 3 (a) A survey by the National Consumers League taken in 2012 estimated the nationwide proportion to be 0.41. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.03? A sample of 2118 children aged 8-10 living in New York is needed to obtain a 99.5% confidence interval with a margin of error of 0.03 using the estimate 0.41 for p. Part: 1/3 Part 2 of 3 (b) Estimate the sample size needed if no estimate of p is available. A sample of children aged 8-10 living in New York is needed to obtain a 99.5% confidence interval with a margin of error of 0.03 when no estimate of p is available. Customers, and finds that 75 of them experienced an interruption peed service during the previous month. Part 1 of 3 (a) Find a point estimate for the population proportion of all customers who experienced an interruption. Round the answer to at least three decimal places. The point estimate for the population proportion of all customers who experienced an interruption is 0.136 Part: 1/3 Part 2 of 3 (b) Construct a 98% confidence interval for the proportion of all customers who experienced an interruption. Round the answers to at least three decimal places. A 98% confidence interval for the proportion of all customers who experienced an interruption is <p<0