1) A medical clinic in a small city in the state of Washington wants to estimate the mean serum cholesterol level (measured in mg of cholesterol per 100 mL of blood) of teenage males. Based on the following summary data taken from a random sample of 36 teenage males from this community, compute a 95% confidence interval for the mean serum cholesterol level of teenage males in this city. Round your confidence limits to 2 decimal places. Summary Data: n = 36, a = 230 and s = 12 O (207.15, 240.67) O (209.12, 238.01) O (210.12, 239.56) O (226.08, 233.92) O None of the above 2) Psychological tests are often used to determine the hostility levels in people. High scores on the HLT psychological test corresponds to high hostility levels. Suppose that 26 college students, all with equally high HLT test scores, were randomly selected and separated into two groups randomly. Group 1 (10 students) was treated for high hostility levels by using Method A; whereas Group 2 (16 students) was treated for high hostility levels by using Method B. Both groups were treated for a period of 5 months, at which time all 26 students were again given the HLT test. The differences computed are based on Group 1 scores minus Group 2 scores The 95% confidence interval for the difference in the mean scores of the two treatment methods is (-11.34, 1.34). Based on this interval, select the best answer below. We are 95% confident that there is no statistically significant difference in the mean scores for the two treatment methods. O We are 95% confident that the mean score for Group 1 using Method A is larger than the mean score for Group 2 using Method B. We are 95% confident that the mean score for Group 1 using Method A is smaller than the mean score for Group 2 using Method B. O None of the above 3) Questions on a statistics exam are considered good questions provided the questions discriminate between students who have studied for the exam and those who have not studied. Suppose that on a particular statistics exam the students were separated into two groups, Group 1 that studied and Group 2 that had not studied. Data was collected and a 95% confidence interval for the difference in the proportion of those passing the exam from Group 1 that studied and the proportion of those passing the exam from Group 2 that had not studied. The confidence interval turned out to be (0.005, 0.125). Select the best answer. Let p1 = the proportion of Group 1 students that passed the exam Let p2 = the proportion of Group 2 students that passed the exam The data fails to show with 95% confidence that these questions discriminated between those who studied and those who did not study. We are 95% confident that those who studied had a higher passing rate than those who did not study. O We are 95% confident that those who studied had a lower passing rate than those who did not study. O None of the above 4) A random sample of size 100 was taken from a population. A 94% confidence interval to estimate the mean of the population was computed based on the sample data. The confidence interval for the mean is (107.62,129.75). What is the z-value that was used to compute this C.I. Round your z-value to 2 decimal places. To answer the question input only the actual number. Do not include units. Do not give your answer in sentence form -- just include the numerical answer rounded to exactly 2 decimal places. 5) The U.S. Food and Drug Administration (FDA) performed a study that compared the carbohydrate content (as a percentage of weight) of several major brands of corn and potato chips. Fourteen brands of corn chips were sampled and 14 brands of potato chips were sampled. Data was collected and the following 95% confidence interval was computed for the difference in means. Interpret the 95% confidence interval for the difference in the mean carbohydrate content between the Corn Chip and Potato Chip groups. The C.I. is: (27.292, 39.836) M1 = the average carbohydrate content as a percentage of weight in corn chips. (2 = the average carbohydrate content as a percentage of weight in potato chips. O We are 95% confident that there is no statistically significant difference in the average carbohydrate content as a percentage of weight between corn chips and potato chips. O We are 95% confident that the average carbohydrate content as a percentage of weight in corn chips is lower than the average carbohydrate content as a percentage of weight in potato chips. O We are 95% confident that the average carbohydrate content as a percentage of weight in corn chips is higher than the average carbohydrate content as a percentage of weight in potato chips. O None of the above 6) The width of a confidence interval is defined to be the upper bound of the confidence interval minus the lower bound of the confidence interval. For example, if a confidence interval is (-2.34, 9.87), then the width of this confidence interval is 9.87 - (-2.34) = 12.21. Assume that we are estimating the mean of a population. Assuming that the confidence coefficient, the sample mean and sample standard deviation stay the same, if we decrease the sample size of a confidence interval from 100 to 50, then the width of that confidence interval will also decrease. Select an option True False