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Q1. A portfolio manager was analyzing the price-earnings ratio for this year's performance. His boss said thatthe average price-earnings ratio was 20 for the many stocks that his firm had traded, but the portfoliomanager felt that the figure was too high. He randomly selected a sample of 50 price-earnings ratios and found a mean of 18.17 and a standard deviation of 4.60. Assume that the population is normally distributed, and test at the 0.01 level of significance. Which of the following is the correct decision rule for the manager to use in this situation?

A. Because 2.81 is greater than 2.33, reject H0. At the 0.01 level, the sample data suggest that the average price-earnings ratiofor the stocks is less than 20.
B. If t > 2.68 or if t
C. Because -2.81 falls in the rejection region, reject H0. At the 0.01 level, the sample data suggest that the average priceearningsratio for the stocks is less than 20.
D. If z > 2.33, reject H0.

Q2. Determine which of the following four population size and sample size combinations would not requirethe use of the finite population correction factor in calculating the standard error.

A. N = 1500; n = 300
B. N = 150; n = 25
C. N = 15,000; n = 1,000
D. N = 2500; n = 75

Q3. In the statement of a null hypothesis, you would likely find which of the following terms?

A. >
B.
C. =
D. ?

Q4. A human resources manager wants to determine a confidence interval estimate for the mean test scorefor the next office-skills test to be given to a group of job applicants. In the past, the test scores have beennormally distributed with a mean of 74.2 and a standard deviation of 30.9. Determine a 95% confidenceinterval estimate if there are 30 applicants in the group.

A. 63.14 to 85.26
B. 64.92 to 83.48
C. 13.64 to 134.76
D. 68.72 to 79.68

Q5. Consider a null hypothesis stating that the population mean is equal to 52, with the research hypothesisthat the population mean is not equal to 52. Assume we have collected 38 sample data from which wecomputed a sample mean of 53.67 and a sample standard deviation of 3.84. Further assume the sample data appear approximately normal. What is the p-value you would report for this test?

A. 0.0037
B. 0.0041
C. 0.0074
D. 0.4963

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