答案和详解如下： Q6. A sample of size n = 25 is selected from a normal population. This sample has a mean of 15 and a sample variance of 4. What is the standard error of the sample mean? A) 0.4. B) 2.0. C) 0.8. Correct answer is A) The standard error of the sample mean is estimated by dividing the standard deviation of the sample by the square root of the sample size. The standard deviation of the sample is calculated by taking the positive square root of the sample variance 41/2 = 2. Applying the formula: sx = s / n1/2 = 2 / (25)1/2 = 2 / 5 = 0.4. Q7. Melissa Cyprus, CFA, is conducting an analysis of inventory management practices in the retail industry. She assumes the population cross-sectional standard deviation of inventory turnover ratios is 20. How large a random sample should she gather in order to ensure a standard error of the sample mean of 4? A) 25. B) 20. C) 80. Correct answer is A) Given the population standard deviation and the standard error of the sample mean, you can solve for the sample size. Because the standard error of the sample mean equals the standard deviation of the population divided by the square root of the sample size, 4 = 20 / n1/2, so n1/2 = 5, so n = 25. Q8. From a population with a known standard deviation of 15, a sample of 25 observations is taken. Calculate the standard error of the sample mean.
A) 0.60. B) 3.00. C) 1.67. Correct answer is B) The standard error of the sample mean equals the standard deviation of the population divided by the square root of the sample size: sx = s / n1/2 = 15 / 251/2 = 3.
Q9. The mean return of Bartlett Co. is 3% and the standard deviation is 6% based on 20 monthly returns. What is the respective standard error of the sample and the confidence interval of a two tailed z-test with a 5% level of significance? A) 1.34; 0.37 to 5.629. B) 2.00; 0.37 to 5.629. C) 1.34; −0.66 to 4.589. Correct answer is A) The standard error of the sample is the standard deviation divided by the square root of n, the sample size. 6/201/2 = 1.34%.The confidence interval = point estimate +/- (reliability factor × standard error) confidence interval = 3 +/- (1.96 × 1.34) = 0.37 to 5.629 Q10. Joseph Lu calculated the average return on equity for a sample of 64 companies. The sample average is 0.14 and the sample standard deviation is 0.16. The standard error of the mean is closest to: A) 0.1600. B) 0.0025. C) 0.0200. Correct answer is C) The standard error of the mean = σ/√n = 0.16/√64 = 0.02. | 
发表于 2009-1-9 10:58
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