Reading 10: Sampling and Estimation-LOS e 习题精选 standard, selected, interpret, face, 2011

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Session 3: Quantitative Methods: Application
Reading 10: Sampling and Estimation

LOS e: Calculate and interpret the standard error of the sample mean.

 

 

From a population of 5,000 observations, a sample of n = 100 is selected. Calculate the standard error of the sample mean if the population standard deviation is 50.

A) 50.00.

B) 4.48.

C) 5.00.

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The standard error of the sample mean equals the standard deviation of the population divided by the square root of the sample size: 50 / 100^1/2 = 5.

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A traffic engineer is trying to measure the effects of carpool-only lanes on the expressway. Based on a sample of 100 cars at rush hour, he finds that the mean number of occupants per car is 2.5, and the sample standard deviation is 0.4. What is the standard error of the sample mean?

A) 0.04.

B) 1.00.

C) 5.68.

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The standard error of the sample mean when the standard deviation of the population is not known is estimated by the standard deviation of the sample divided by the square root of the sample size. In this case, 0.4 / √100 = 0.04.

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If the number of offspring for females of a certain mammalian species has a mean of 16.4 and a standard deviation of 3.2, what will be the standard error of the sample mean for a survey of 25 females of the species?

A) 0.64.

B) 1.28.

C) 3.20.

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The standard error of the sample mean when the standard deviation of the population is known is equal to the standard deviation of the population divided by the square root of the sample size. In this case, 3.2 / √25 = 0.64.

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A population has a mean of 20,000 and a standard deviation of 1,000. Samples of size n = 2,500 are taken from this population. What is the standard error of the sample mean?

A) 0.04.

B) 400.00.

C) 20.00.

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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: s_x = s / n^1/2 = 1000 / (2500)^1/2 = 1000 / 50 = 20.

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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) 2.0.

B) 0.4.

C) 0.8.

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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 4^1/2 = 2. Applying the formula: s_x = s / n^1/2 = 2 / (25)^1/2 = 2 / 5 = 0.4.

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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.

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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 / n^1/2, so n^1/2 = 5, so n = 25.

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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.

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The standard error of the sample mean equals the standard deviation of the population divided by the square root of the sample size: s_x = s / n^1/2 = 15 / 25^1/2 = 3.

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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) 2.00; 0.37 to 5.629.

B) 1.34; ?0.66 to 4.589.

C) 1.34; 0.37 to 5.629.

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The standard error of the sample is the standard deviation divided by the square root of n, the sample size. 6/20^1/2 = 1.34%.

The confidence interval = point estimate +/- (reliability factor × standard error)

confidence interval = 3 +/- (1.96 × 1.34) = 0.37 to 5.629

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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.

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The standard error of the mean = σ/√n = 0.16/√64 = 0.02.

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Frank Grinder is trying to introduce sampling into the quality control program of an old-line manufacturer. Grinder samples 38 items and finds that the standard deviation in size is 0.019 centimeters. What is the standard error of the sample mean?

A) 0.00308.

B) 0.00204.

C) 0.00615.

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If we do not know the standard deviation of the population (in this case we do not), then we estimate the standard error of the sample mean = the standard deviation of the sample / the square root of the sample size = 0.019 / √38 = 0.00308 centimeters.