答案和详解如下： 11.The average return on small stocks over the period 1926-1997 was 17.7 percent, and the standard error of the sample was 33.9 percent. The 95 percent confidence interval for the return on small stocks in any given year is: A) –16.2% to 51.6%. B) 16.2% to 51.6%. C) –48.7% to 84.1%. D) 16.8% to 18.6%. The correct answer was C) A 95 percent confidence level is 1.96 standard deviations from the mean, so 0.177 ± 1.96(0.339) = (–48.7%, 84.1%). 12.A sample of 100 individual investors has a mean portfolio value of $28,000 with a standard deviation of $4,250. The 95 percent confidence interval for the population mean is closest to:
A) $27,575 to $28,425. B) $23,750 to $32,250. C) $27,159 to $28,842. D) $19,500 to $28,333. The correct answer was C) Confidence interval = mean = ± tc {S / √n} = 28,000 ± (1.98) 4250 / √100 or 27,159 to 28,842 If you use a Z-statistic because of the large sample size, you get 28,000 ± (1.96)4,250√100 = 27,167 to 28,833, which is closest to the correct answer. 13.A traffic engineer is trying to measure the effects of carpool-only lanes on the expressway. Based on a sample of 20 cars at rush hour, he finds that the mean number of occupants per car is 2.5, with a standard deviation of 0.4. If the population is normally distributed, what is the confidence interval at the 5 percent significance level for the number of occupants per car? A) 2.410 to 2.589. B) 2.387 to 2.613. C) 2.465 to 2.535. D) 2.313 to 2.687. The correct answer was D) The reliability factor corresponding with a 5 percent significance level (95 percent confidence level) for the Student’s t-distribution with (20 – 1) degrees of freedom is 2.093. The confidence interval is equal to: 2.5 ± 2.093(0.4 / √20) = 2.313 to 2.687. (We must use the Student’s t-distribution and reliability factors because of the small sample size.) 14.A local high school basketball team had 18 home games this season and averaged 58 points per game. If we assume that the number of points made in home games is normally distributed, which of the following is most likely the range of points for a confidence interval of 90 percent? A) 26 to 80. B) 24 to 78. C) 62 to 74. D) 34 to 82. The correct answer was D) This question has a bit of a trick! To answer this question, remember that the mean is at the midpoint of the confidence interval. The correct confidence interval will have a midpoint of 58. (34 + 82) / 2 = 58. 15.A random sample of 25 Indiana farms had a mean number of cattle per farm of 27 with a sample standard deviation of five. Assuming the population is normally distributed, what would be the 95 percent confidence interval for the number of cattle per farm? A) 22 to 32. B) 23 to 31. C) 24 to 30. D) 25 to 29. The correct answer was D) The standard error of the sample mean = 5 / √25 = 1
Degrees of freedom = 25 – 1 = 24
From the student’s T table, t5/2 = 2.064
The confidence interval is: 27±2.064(1) = 24.94 to 29.06 or 25 to 29 | 
发表于 2008-4-14 15:00
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