ࡱ> DFC[ &5bjbj77 4tU\U\&- >>>>>RRRRtRy1R0000000$I351>1>>31m m m j>>0m 0m m 8/|h0G.P5~/0I10y1/66$h06>h0|m 1176y16 > : Comparing Two Population Means (matched pairs and independent samples) The personnel manager of a large retail clothing store suspects a difference in the mean amount of break time taken by workers during the weekday shifts compared to that of the weekend shifts. It is suspected that the weekday workers take longer breaks on the average. A random sample of 46 weekday workers had a mean 1  EMBED Equation.3 = 53 minutes of break time per 8-hour shift and  EMBED Equation.3  = 7.3 minutes. A random sample of 40 weekend workers had a mean  EMBED Equation.3  = 47 minutes and  EMBED Equation.3  = 9.1 minutes. Test the managers suspicion at the 5% level of significance. Two-Sample Z-Test and CI Sample N Mean StDev SE Mean 1 46 53.00 7.30 1.1 2 40 47.00 9.10 1.4 Difference = mu (1) - mu (2) Estimate for difference: 6.00 95% lower bound for difference: 3.01 Z-Test of difference = 0 (vs >): Z-Value = 3.34 P-Value = 0.001 Two growth hormones are being considered. A random sample of 10 rats were given the first hormone and their average weight gain was  EMBED Equation.3  = 2.3 pounds with standard deviation  EMBED Equation.3  = 0.4 pound. For the second hormone, a random sample of 15 rats showed their average weight gain to be  EMBED Equation.3 = 1.9 pounds with standard deviation  EMBED Equation.3  = 0.2 pound. Assume the weight gains follow a normal distribution. Using a 10% level of significance, can we say there is a difference in average weight gains for the two growth hormones? Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 10 2.300 0.400 0.13 2 15 1.900 0.200 0.052 Difference = mu (1) - mu (2) Estimate for difference: 0.400 90% CI for difference: (0.156, 0.644) T-Test of difference = 0 (vs not =): T-Value = 2.93 P-Value = 0.013 DF = 23 A local claims that the waiting time for its customers to be served is the lowest in the area. A competitor's bank checks the waiting times at both banks. The sample statistics are listed below. Test the local bank's claim. Use  EMBED Equation.3  Local Bank Competitor Bank n1 = 15 n2 = 16  EMBED Equation.3  = 5.3 minutes  EMBED Equation.3  = 5.6 minutes S1 = 1.1 minutes S2 = 1.0 minutes Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 15 5.30 1.10 0.28 2 16 5.60 1.00 0.25 Difference = mu (1) - mu (2) Estimate for difference: -0.300 95% upper bound for difference: 0.344 T-Test of difference = 0 (vs <): T-Value = -0.79 P-Value = 0.217 DF = 29 Suppose that simple random samples of college freshman are selected from two universities - 15 students from school A and 20 students from school B. On a standardized test, the sample from school A has an average score of 1000 with a standard deviation of 100. The sample from school B has an average score of 950 with a standard deviation of 90. What is the 95%  HYPERLINK "http://stattrek.com/Help/Glossary.aspx?Target=Confidence_interval" confidence interval for the difference in test scores at the two schools, assuming that test scores came from normal distributions in both schools? Two-Sample T-Test and CI Difference = mu (1) - mu (2) Estimate for difference: 50.0 95% CI for difference: (-15.6, 115.6) T-Test of difference = 0 (vs not =): T-Value = 1.55 P-Value = 0.130 DF = 33 Both use Pooled StDev = 94.3719 Answer: 95% confidence interval is -15.6 to 115.6. That is, we are 95% confident that the true difference in population means is within: -15.6 to 115.6. The local baseball team conducts a study to find the amount spent on refreshments at the ball park. Over the course of the season they gather simple random samples of 50 men and 100 women. For men, the average expenditure was $20, with a standard deviation of $3. For women, it was $15, with a standard deviation of $2. What is the 99%  HYPERLINK "http://stattrek.com/Help/Glossary.aspx?Target=Confidence_interval" confidence interval for the spending difference between men and women? Assume that the two populations are independent and normally distributed. Two-Sample Z-Test and CI Difference = mu (1) - mu (2) Estimate for difference: 5.000 99% CI for difference: (3.759, 6.241) Z-Test of difference = 0 (vs not =): Z -Value = 10.66 P-Value = 0.000 Answer: The 99% confidence interval is $3.76 to $6.24. That is, we are 99% confident that men outspend women at the ballpark by at least $3.76 and at most $6.24. SAT prep courses are often very expensive, but are they worth it? The data below is from nine different randomly selected students who took the SAT twice (once before taking Kaplans test prep course and one after taking the Kaplan test prep course). At the 5% level of significance, test the claim that the after scores are better: Student123456789Before Score480510530540550560600620660After Score460500530520580580560640690Difference2010020-30-2040-20-30**The standard deviation of the differences is 25.2212 A salesman for a shoe company claimed that runners would record quicker times, on the average, with the company's brand of sneaker. A track coach decided to test the claim. The coach selected eight runners. Each runner ran two 100-yard dashes on different days. In one 100-yard dash, the runners wore the sneakers supplied by the school; in the other, they wore the sneakers supplied by the salesman. Each runner was randomly assigned the sneakers to wear for the first run. Their times, measured in seconds, were as follows: Runners12345678Company Sneaker 10.812.310.712.010.611.512.111.2School Sneaker11.412.510.811.710.911.812.211.7 Note. For the differences,  EMBED Equation.3  = -.225 and  EMBED Equation.3  = .276. Assume the population of differences is approximately normal. Paired T-Test and CI N Mean StDev SE Mean Diff. 8 -0.2250 0.2760 0.0976 95% CI for mean difference: (-0.4557, 0.0057) T-Test of mean difference = 0 (vs < 0): T-Value = -2.31; P-Value = 0.027 Twenty-four males age 25-29 were selected from the Framingham Heart Study. Twelve were smokers and 12 were nonsmokers. The subjects were paired, with one being a smoker and the other a nonsmoker. Otherwise, each pair was similar with regard to age and physical characteristics. Systolic blood pressure readings were as follows: People123456789101112Smokers 122146120114124126118128130134116130Nonsmokers114134114116138110112116132126108116 List the differences A - B and verify that  EMBED Equation.3  = 6 and  EMBED Equation.3  = 8.40. Use a 5% level of significance to determine whether the data indicate a difference in mean systolic blood pressure levels for the populations from which the two groups were selected. You may assume that the population of differences is approximately normal. Paired T-Test and CI N Mean StDev SE Mean Diff. 12 6.00 8.40 2.42 95% CI for mean difference: (0.66, 11.34) T-Test of mean difference = 0 (vs not = 0): T-Value = 2.47 P-Value = 0.031 A local school district is concerned about the number of school days missed by its teachers due to illness. A random sample of 10 teachers is selected. The number of days absent in one year is listed below. An incentive program is offered in an attempt to decrease the number of days absent. The number of days absent in one year after the incentive program is listed below. Test the claim that the incentive program cuts down on the number of days missed by teachers. Use  EMBED Equation.3  = 0.05. Assume that the distribution is normally distributed TeacherABCDEFGHIJDays absent before incentive3872942075Days absent after incentive1770820155 A weight-lifting coach claims that weight-lifters can increase their strength by taking a certain supplement. To test the theory, the coach randomly selects 9 athletes and gives them a strength test using a bench press. The results are listed below. Thirty days later, after regular training using the supplement, they are tested again. The new results are listed below. Test the claim that the supplement is effective in increasing the athletes' strength. Use ( = 0.05. Assume that the distribution is normally distributed. AthleteABCDEFGHIBefore215240188212275260225200185After225245188210282275230195190 Confidence Interval (Two populations) 1. Is there a difference in the total scores for womens and mens basketball games? A random sample of n1 = 55 womens games had a mean winning score of  EMBED Equation.3  = 78. Another random sample of n2 = 60 mens games had a mean winning score of  EMBED Equation.3  = 90. Historical data suggests  EMBED Equation.3  = 10 and  EMBED Equation.3  = 16. Find a 95% confidence interval for the population difference  EMBED Equation.3 . 95% CI for difference: (-16.89, -7.11) 2. Two growth hormones are being considered. A random sample of 10 rats were given the first hormone and their average weight gain was  EMBED Equation.3  = 2.3 pounds with standard deviation  EMBED Equation.3  = 0.4 pound. For the second hormone, a random sample of 15 rats showed their average weight gain to be  EMBED Equation.3 = 1.9 pounds with standard deviation  EMBED Equation.3  = 0.2 pound. Assume the weight gains follow a normal distribution. Find a 90% confidence interval for the difference in average weight gains for the two growth hormones. Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 10 2.300 0.400 0.13 2 15 1.900 0.200 0.052 Difference = mu (1) - mu (2) Estimate for difference: 0.400 90% CI for difference: (0.194, 0.606) T-Test of difference = 0 (vs not =): T-Value = 3.32 P-Value = 0.003 DF = 23 Both use Pooled StDev = 0.2949 3. Stone Tires has developed a new tread which they claim reduces stopping distance on wet pavement. A random sample of 56 test drives with cars using tires with tread type I (old design) showed that the average stopping distance on wet pavement was  EMBED Equation.3  = 183 feet. A random sample of 61 test drives conducted under similar conditions, but with cars using tires with tread type II (new tread) showed that the average stopping distance was  EMBED Equation.3  = 152 feet. Historical data suggests  EMBED Equation.3 = 49 feet and  EMBED Equation.3  = 53 feet. Find a 90% confidence interval for the population mean difference  EMBED Equation.3 of stopping distances for the two types of tire tread. a) 15.5 to 46.5 b) 13.5 to 44.5 c) 14.5 to 43.5 d) 16.5 to 45. 5 e) 12.5 to 43.5 4. At a large office supply store, the daily sales of two similar brand-name laser printers are being compared. A random sample of 16 days showed that Brand I had mean daily sales  EMBED Equation.3  = $2464 with standard deviation  EMBED Equation.3  = $529. A random sample of 19 days showed that Brand II had mean daily sales  EMBED Equation.3  = $2285 with sample standard deviation  EMBED Equation.3 = $440. Assume sales follow an approximately normal distribution. Find a 90% confidence interval for the population mean difference in sales  EMBED Equation.3 . 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