Statistics- A medical clinic in a small city in the state of Washington wants

QUESTION

1．A medical clinic in a small city in the state of Washington wants to estimate the mean serumcholesterol level (measured in mg of cholesterol per 100 mL of blood) of teenage males. Based on thefollowing summary data taken from a random sample of 10 teenage males from this community,compute a 90% confidence interval for the mean serum cholesterol level of teenage males in this city.Assume that the sample was taken from a normal distribution. Round your confidence limits to 2decimal places.Summary Data: n = 10, x¯=223.562, s = 24.920Select one:A. (207.15, 240.67)B. (209.12, 238.01)C. (210.12, 239.56)D. (213.34, 242.83)E. None of the above2．Pschological tests are often used to determine the hostility levels in people. High scores on the HLTpschological test corresponds to high hostility levels. Suppose that 26 college students, all with equallyhigh HLT test scores, were separated into two groups randomly. Group 1 (10 students) was treated forhigh hostility levels by using Method A; whereas Group 2 (16 students) was treated for high hostilitylevels by using Method B. Both groups were treated for a period of 5 months, at which time all 26students were again given the HLT test. Below is the summary data on the HLT test for the two groups.Group 1: sample mean is 79 and sample standard deviation is 7Group 2: sample mean is 84 and sample standard deviation is 8 Compute a 95% confidence level for thedifference in the means of the two treatment methods and select the best answer below. Assume thatthe independent random samples were taken from normal distributions and that the two populationvariances are equal.Select one:A. We are 95% confident that there is no statistically significant difference in the mean treatmentmethods for hostility.B. We are 95% confident that Method A is a better treatment for hostility than Method B.C. We are 95% confident that Method B is a better treatment for hostility than Method A.D. We are 95% confident that there is a statistically significant difference in the mean treatment methodsfor hostility.3．Questions on a statistics exam are considered good questions provided the questions discriminatebetween students who have studied for the exam and those who have not studied. Suppose that on aparticular statistics exam the students were separated into two groups, the group that studied and theother group that had not studied. Data was collected and a 95% confidence interval for the difference inthe proportion of those passing the exam from the group that studied and the proportion of thosepassing the exam from the group that had not studied. The confidence interval turned out to be(−0.005,0.125). Note that sample 1 is from the group that studied and sample 2 is from the group thatdid not study. Select the best answer.Select one:A. The data fails to show with 95% confidence that these questions discriminated between those whostudied and those who did not study.B. We are 95% confident that those who studied had a higher passing rate than those who did not study.C. We are 95% confident that those who studied had a lower passing rate than those who did not study.D. None of the these are correct.4．A random sample of size 100 was taken from a population. A 94% confidence interval to estimate themean of the population was computed based on the sample data. The confidence interval for the meanis: (107.62, 129.75). What is the z-value that was used in the computation. Round your z-value to 2decimal places.To answer the question input only the actual number. Do not include units. Do not give your answer insentence form — just include the numerical answer rounded to exactly 2 decimal places.Answer:5．The U.S. Food and Drug Administration (FDA) performed a study that compared the carbohydratecontent (as a percentage of weight) of several major brands of corn and potato chips. Fourteen brands ofcorn chips were sampled and 14 brands of potato chips were sampled. Data was collected and thefollowing 95% confidence interval was computed for the difference in means. Intrepret the 95%confidence interval for the difference in the mean carbohydrate content between the Corn Chip andPotato Chip groups.The C.I. is: (27.292,39.836)where μ1 = the average carbohydrate content as a percentage of weight in corn chips.and μ2 = the average carbohydrate content as a percentage of weight in potato chips.Select the best answer below.Select one:A. We are 95% confident that there is no statistically significant difference in the average carbohydratecontent as a percentage of weight between corn chips and potato chips.B. We are 95% confident that the average carbohydrate content as a percentage of weight in corn chipsis lower than the average carbohydrate content as a percentage of weight in potato chips.C. We are 95% confident that the average carbohydrate content as a percentage of weight in corn chipsis higher than the average carbohydrate content as a percentage of weight in potato chips.D. None of these are correct.6．The width of a confidence interval is defined to be the upper bound of the confidence interval minusthe lower bound of the confidence interval. For example, if a confidence interval is (−2.34,9.87), then thewidth of this confidence interval is 9.87−(−2.34)=12.21. Assume that we are estimating the mean of apopulation. Assuming that the confidence coefficient, the sample mean and sample standard deviationstay the same, if we decrease the the sample size of a confidence interval from 100 to 50, then the widthof that confidence interval will also decrease.Select one:TrueFalse

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