答案和详解如下： 16.From a sample of 41 monthly observations of the S& Mid-Cap index, the mean monthly return is 1 percent and the sample variance is 36. For which of the following intervals can one be closest to 95 percent confident that the population mean is contained in that interval? A) 1.0 percent ± 1.6 percent. B) 1.0 percent ± 1.9 percent. C) 1.0 percent ± 6.0 percent. D) 1.0 percent ± 0.9 percent. The correct answer was B) If the distribution of the population is nonnormal, but we don’t know the population variance, we can use the Student’s t-distribution to construct a confidence interval. The sample standard deviation is the square root of the variance, or 6%. Because there are 41 observations, the degrees of freedom are 40. From the Student’s t distribution, we can determine that the reliability factor for t0.025, is 2.021. Then the 95% confidence interval is 1.0% ± 2.021(6/√41) or 1.0% ± 1.9%. 17.From a sample of 41 orders for an on-line bookseller, the average order size is $75, and the sample standard deviation is $18. Assume the distribution of orders is normal. For which interval can one be exactly 90 percent confident that the population mean is contained in that interval? A) $74.24 to $75.76. B) $70.27 to $79.73. C) $71.29 to 78.71. D) $62.00 to $88.00. The correct answer was B) If the distribution of the population is normal, but we don’t know the population variance, we can use the Student’s t-distribution to construct a confidence interval. Because there are 41 observations, the degrees of freedom are 40. From the student’s t table, we can determine that the reliability factor for tα/2, or t0.05, is 1.684. Then the 90% confidence interval is $75.00 ± 1.684($18.00/√41),or $75.00 ± 1.684 x $2.81 or $75.00 ± $4.73
18.The average salary for a sample of 61 CFA charterholders with 10 years experience is $200,000, and the sample standard deviation is $80,000. Assume the population is normally distributed. Which of the following is a 99 percent confidence interval for the population mean salary of CFA charterholders with 10 years of experience? A) $172,754 to $227,246. B) $160,000 to $240,000. C) $197,811 to $202,189. D) $172,514 to $227,486. The correct answer was A) If the distribution of the population is normal, but we don’t know the population variance, we can use the Student’s t-distribution to construct a confidence interval. Because there are 61 observations, the degrees of freedom are 60. From the student’s t table, we can determine that the reliability factor for tα/2, or t0.005, is 2.660. Then the 99% confidence interval is $200,000±2.660($80,000/√61) or $200,000±2.660 x $10,243, or $200,000±$27,246. 19.In which one of the following cases is the t-statistic the appropriate one to use in the construction of a confidence interval for the population mean? A) The distribution is nonnormal, the population variance is unknown, and the sample size is at least 30. B) The distribution is nonnormal, the population variance is known, and the sample size is at least 30. C) The distribution is normal, the population variance is known, and the sample size is at least 30. D) The distribution is normal, the population variance is known, and the sample size is less than 30. The correct answer was A) The t-distribution is the theoretically correct distribution to use when constructing a confidence interval for the mean when the distribution is nonnormal and the population variance is unknown but the sample size is at least 30. 20.The approximate 99 percent confidence interval for the population mean based on a sample of 60 returns with a mean of 7% and a sample standard deviation of 25% is closest to:
A) 1.584% to 14.584%. B) 0.546% to 13.454%. C) -1.584% to 15.584%. D) 1.546% to 13.454%. The correct answer was C) The standard error for the mean = s/(n)0.5 = 25%/(60)0.5 = 3.227%. The critical value from the t-table should be based on 60 – 1 = 59 df. Since the standard tables do not provide the critical value for 59 df the closest available value is for 60 df. This leaves us with an approximate confidence interval. Based on 99 percent confidence and df = 60, the critical t-value is 2.660. Therefore the 99 percent confidence interval is approximately: 7% ± 2.660(3.227) or 7% ± 8.584% or -1.584% to 15.584%. If you use a Z-statistic, the confidence interval is 7% ± 2.58(3.227) = -1.326% to 15.326%, which is closest to the correct choice. 21.The 95% confidence interval for the population mean based on a sample of 40 interest rates with a sample mean of 4% and a sample standard deviation of 15% is closest to:
A) -0.851% to 8.851%. B) 1.261% to 6.739%. C) 1.851% to 9.851%. D) -0.794% to 8.794%. The correct answer was D) The standard error for the mean = s/(n)0.5 = 15%/(40)0.5 = 2.372%. The critical value from the t-table should be based on 40 – 1 = 39 df. Since the standard tables do not provide the critical value for 39 df the closest available value is for 40 df. This leaves us with an approximate confidence interval. Based on 95 percent confidence and df = 40, the critical t-value is 2.021. Therefore the 95 percent confidence interval is approximately: 4% ± 2.021(2.372) or 4% ± 4.794% or -0.794% to 8.794%. | 
发表于 2008-4-14 15:03
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