答案和详解如下： 6.The average U.S. dollar/Euro exchange rate from a sample of 36 monthly observations is $1.00/Euro. The population variance is 0.49. What is the 95 confidence interval for the mean U.S. dollar/Euro exchange rate? A) $0.5100 to $1.4900. B) $0.8075 to $1.1925. C) $0.8657 to $1.1343. D) $0.7713 to $1.2287. The correct answer was D) The population standard deviation is the square root of the variance (√0.49 = 0.7). Because we know the population standard deviation, we use the z-statistic. The z-statistic reliability factor for a 95% confidence interval is 1.960. The confidence interval is $1.00 ± 1.960($0.7/√36) or $1.00 ± $0.2287. 7.A 95 percent confidence interval for the mean number of monthly customer visits to a grocery store is 28,000 to 32,000 customers. Which of the following is an appropriate interpretation of this confidence interval? A) If we repeatedly sample the population and construct 95 percent confidence intervals, 95 percent of the resulting confidence intervals will include the population mean. B) There is a 95 percent chance that next month the grocery store will have between 28,000 and 32,000 customer visits. C) We are 95 percent confident that if a sample of monthly customer visits is taken, the sample mean will fall between 28,000 and 32,000. D) If we repeatedly sample the population and compute the mean each time, 95 percent of the resulting sample means will be between 28,000 and 32,000. The correct answer was A) There are two interpretations of this confidence interval: a probabilistic and a practical interpretation. Probabilistic interpretation: We can interpret this confidence interval to mean that if we sample the population of customers 100 times, we can expect that 95 (95%) of the resulting 100 confidence intervals will include the population mean. Practical interpretation: We can also interpret this confidence interval by saying that we are 95% confident that the population mean number of monthly customer visits is between 28,000 and 32,000. 8.An airline was concerned about passengers arriving too late at the airport to allow for the additional security measures. Based on a survey of 1,000 passengers, the mean time from arrival at the airport to reaching the boarding gate was 1 hour, 20 minutes, with a standard deviation of 30 minutes. If the airline wants to make sure at the 95 percent confidence level that passengers have sufficient time to catch their flight, how much time ahead of their flight should passengers be advised to arrive at the airport? A) One hour, fifty minutes. B) Two hours, ten minutes. C) Two hours, thirty minutes. D) Two hours, forty-five minutes. The correct answer was B) We can use standard distribution tables because the sample is so large. From a table of area under a normally distributed curve, the Z value corresponding to a 95 percent, one-tail test is: 1.65. (We use a one-tailed test because we are not concerned with passengers arriving too early, only arriving too late.) Here, we do not divide by the standard error, because we are interested in a point estimate of making our flight. The answer is One hour, twenty minutes + 1.65(30 minutes) = 2 hours,10 minutes. 9.A traffic engineer is trying to measure the effects of carpool-only lanes on the expressway. Based on a sample of 1,000 cars at rush hour, he finds that the mean number of occupants per car is 2.5, with a standard deviation of 0.4. Assuming that the population is normally distributed, what is the confidence interval at the 5 percent significance level for the number of occupants per car? A) 2.455 to 2.555. B) 2.288 to 2.712. C) 2.371 to 2.629. D) 2.475 to 2.525. The correct answer was D) The Z-score corresponding with a 5 percent significance level (95 percent confidence level) is 1.96. The confidence interval is equal to: 2.5 ± 1.96(0.4 / √1,000) = 2.475 to 2.525. (We can use Z-scores because the size of the sample is so large.) 10.Books Fast, Inc., prides itself on shipping customer orders quickly. Downs Shipping Service has promised that mean delivery time will be less than 72 hours. Books Fast sampled 27 of its customers and found a mean delivery time of 76 hours, with a sample standard deviation of 6 hours. Based on this sample and assuming a normal distribution of delivery times, what is the confidence interval at 5 percent significance? A) 68.50 to 83.50 hours. B) 65.75 to 86.25 hours. C) 70.34 to 81.66 hours. D) 73.63 to 78.37 hours. The correct answer was D) The confidence interval is equal to 76 + or – (2.056)(6 / √27) = 73.63 to 78.37 hours. Because the sample size is small, we use the t-distribution with (27 – 1) degrees of freedom. | 
发表于 2008-4-14 14:58
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