College tuition: A simple random sample of 40 colleges and universities in the United States has a mean tuition of $19,200 with a standard deviation of $10,800. Construct a 98% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number. A 98% confidence interval for the mean tuition for all colleges and universities is S X 8E V ote
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- What is meant by the sample space of an experiment?Does this diet help? A group of 75 people enrolled in a weight-loss program that involved adhering to a special diet and to a daily exercise program. After six months, their mean weight loss was 26 pounds, with a sample standard deviation of 10 pounds. A second group of 39 people went on the diet but didn't exercise. After six months, their mean weight loss was 12 pounds, with a sample standard deviation of 7 pounds. Construct a 99.9 % confidence interval for the mean difference in weight losses. Let μ1 denote the mean weight loss of the group with daily exercise. Use the TI-84 Plus calculator and round the answers to the nearest integer. A 99.9% confidence interval for the mean difference in weight loss, in pounds, is <−μ1-μ2<1. A recent Gallop poll showed that 672 of 1019 adults nationwide think that marijuana should be legalized. a) Find the margin of error that corresponds to a 95% confidence interval. b) Find the 95% confidence interval estimate of the proportion of adults nationwide who think marijuana should be legalized
- In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.03 margin of error and use a confidence level 90%.The Chartered Financial Analyst (CFA) designation is fast becoming a requirement for serious investment professionals. Although it requires a successful completion of three levels of grueling exams, it also entails promising careers with lucrative salaries. A student of finance is curious about the average salary of a CFA® charterholder. He takes a random sample of 25 recent charterholders and computes a mean salary of $195,000 with a standard deviation of $40,000. Use this sample information to determine the 90% confidence interval for the average salary of a CFA charterholder. Assume that salaries are normally distributed. (You may find it useful to reference the t table. Round intermediate calculations to at least 4 decimal places. Round "t" value to 3 decimal places and final answers to the nearest whole number.)The Chartered Financial Analyst (CFA) designation is fast becoming a requirement for serious investment professionals. Although it requires a successful completion of three levels of grueling exams, it also entails promising careers with lucrative salaries. A student of finance is curious about the average salary of a CFA charterholder. He takes a random sample of 36 recent charterholders and computes a mean salary of $129,000 with a standard deviation of $22,000. Use this sample information to determine the 99% confidence interval for the average salary of a CFA charterholder. (You may find it useful to reference the t table. Round final answers to the nearest whole number.)
- The Chartered Financial Analyst (CFA) designation is fast becoming a requirement for serious investment professionals. Although it requires a successful completion of three levels of grueling exams, the designation often results in a promising career with a lucrative salary. A student of finance is curious about the average salary of a CFA charterholder. He takes a random sample of 45 recent charterholders and computes a mean salary of $155,000 with a standard deviation of $18,000. Use this sample information to determine the 95% confidence interval for the average salary of a CFA charterholder. Note: Round your final answers to the nearest whole number.In a test of weight loss programs, 40 adults used the Atkins weight loss program. After 12 months, their mean weight loss was found to be 2.1 lb., with a standard deviation of 4.8 lb. a. Construct a 95% confidence interval estimate of the mean weight loss for all subjects. Does the Atkins program appear to be effective? b.If a sample of 100 adults, construct a 95% confidence interval estimate of the mean weight loss for all subjects. Does the Atkins program appear to be effective? c. Based on the results on a) and b), explain the effect of sample size on the confidence interval estimate and its practical implication.A manufacturing company produces water filters for home refrigerators. Thethe process has typically produced some defective liters. A random sample of 300 liters yielded 12 defects. Construct and interpret a 90% confidence interval for the true proportion of defective water liters.
- A marketing study with a random sample of 400 results in 160 subjects reacting positively to the proposed product. Determine the 95% confidence interval for the population proportion.A doctor claims that the mean number of hours of sleep that seniors in high school get per night differs from the mean number of hours of sleep college seniors get per night. To investigate, he selects a random sample of 50 high school seniors from all high schools in his county. He also selects a random sample of 50 seniors from the colleges in his county. He constructs a 95% confidence interval for the true mean difference in the number of hours of sleep for seniors in high school and seniors in college. The resulting interval is (0.57, 1.25). Based upon the interval, can the doctor conclude that mean number of hours of sleep that seniors in high school get per night differs from the mean number of hours of sleep college seniors get per night? yes because 1 is in the confidence interval yes because 0 is not in the confidence interval no because 1 is in the confidence interval no because 0 is not in the confidence intervalIdentify the y-score for a95% confidence interval if the sample size is 50.