Reading 10: Sampling and Estimation LOSj习题精选 2010, following, interpret, normal

honeycfa

LOS j: Calculate and interpret a confidence interval for a population mean, given a normal distribution with (1) a known population variance, (2) an unknown population variance, and (3) an unknown variance and a large sample size.

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 normal, the population variance is known, and the sample size is less than 30.

B) The distribution is nonnormal, the population variance is unknown, and the sample size is at least 30.

C) The distribution is nonnormal, the population variance is known, and the sample size is at least 30.

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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.

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honeycfa

Based on Student's t-distribution, 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) -0.794% to 8.794%.

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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% confidence and df = 40, the critical t-value is 2.021. Therefore the 95% confidence interval is approximately: 4% ± 2.021(2.372) or 4% ± 4.794% or -0.794% to 8.794%.

honeycfa

A sample of 25 junior financial analysts gives a mean salary (in thousands) of 60. Assume the population variance is known to be 100. A 90% confidence interval for the mean starting salary of junior financial analysts is most accurately constructed as:

A) 60 + 1.645(2).

B) 60 + 1.645(10).

C) 60 + 1.645(4).

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Because we can compute the population standard deviation, we use the z-statistic. A 90% confidence level is constructed by taking the population mean and adding and subtracting the product of the z-statistic reliability (z_á/2) factor times the known standard deviation of the population divided by the square root of the sample size (note that the population variance is given and its positive square root is the standard deviation of the population): x ± z_á/2 * ( σ / n^1/2) = 60 +/- 1.645 × (100^1/2 / 25^1/2) = 60 +/- 1.645 × (10 / 5) = 60 +/- 1.645 × 2.

honeycfa

Based on Student's t-distribution, 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) -0.794% to 8.794%.

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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% confidence and df = 40, the critical t-value is 2.021. Therefore the 95% confidence interval is approximately: 4% ± 2.021(2.372) or 4% ± 4.794% or -0.794% to 8.794%.

honeycfa

Construct a 90% confidence interval for the mean starting salaries of the CFA charterholders if a sample of 100 recent CFA charterholders gives a mean of 50. Assume that the population variance is 900. All measurements are in $1,000.

A) 50 ± 1.645(3).

B) 50 ± 1.645(30).

C) 50 ± 1.645(900).

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Because we can compute the population standard deviation, we use the z-statistic. A 90% confidence level is constructed by taking the population mean and adding and subtracting the product of the z-statistic reliability (z_α/2) factor times the known standard deviation of the population divided by the square root of the sample size (note that the population variance is given and its positive square root is the standard deviation of the population): x ± z_α/2 × ( σ / n^1/2) = 50 ± 1.645 × (900^1/2 / 100^1/2) = 50 ± 1.645 × (30 / 10) = 50 ± 1.645 × (3). This is interpreted to mean that we are 90% confident that the above interval contains the true mean starting salaries of CFA charterholders.

honeycfa

The average return on the Russell 2000 index for 121 monthly observations was 1.5%. The population standard deviation is assumed to be 8.0%. What is a 99% confidence interval for the monthly return on the Russell 2000 index?

A) -0.4% to 3.4%.

B) 0.1% to 2.9%.

C) -6.5% to 9.5%.

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Because we know the population standard deviation, we use the z-statistic. The z-statistic reliability factor for a 99% confidence interval is 2.575. The confidence interval is 1.5% ± 2.575[(8.0%)/√121] or 1.5% ± 1.9%.

honeycfa

A sample size of 25 is selected from a normal population. This sample has a mean of 15 and the population variance is 4.

Using this information, construct a 95% confidence interval for the population mean, m.

A) 15 ± 1.96(2).

B) 15 ± 1.96(0.4).

C) 15 ± 1.96(0.8).

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Because we can compute the population standard deviation, we use the z-statistic.  A 95% confidence level is constructed by taking the population mean and adding and subtracting the product of the z-statistic reliability (z_α/2) factor times the known standard deviation of the population divided by the square root of the sample size (note that the population variance is given and its positive square root is the standard deviation of the population): x ± z_α/2 × ( σ / n^1/2) = 15 ± 1.96 × (4^1/2 / 25^1/2) = 15 ± 1.96 × (0.4).

honeycfa

A nursery sells trees of different types and heights. Suppose that 75 pine trees are sold for planting at City Hall. These 75 trees average 60 inches in height with a standard deviation of 16 inches.

Using this information, construct a 95% confidence interval for the mean height of all trees in the nursery.

A) 60 + 1.96(1.85).

B) 60 + 1.96(16).

C) 0.8 + 1.96(16).

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Because we know the population standard deviation, we use the z-statistic. A 95% confidence level is constructed by taking the population mean and adding and subtracting the product of the z-statistic reliability (z_α/2) factor times the known standard deviation of the population divided by the square root of the sample size: x ± z_α/2 × ( σ / n^1/2) = 60 ± (1.96) × (16 / 75^1/2) = 60 ± (1.96) × (16 / 8.6603) = 60 ± (1.96) × (1.85).

honeycfa

What is the 95% confidence interval for a population mean with a known mean of 96, a known variance of 9, and with a number of observations of 400?

A) 95.613 to 96.387.

B) 95.118 to 96.882.

C) 95.706 to 96.294.

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Because we can compute the population standard deviation, we use the z-statistic. A 95% confidence level is constructed by taking the population mean and adding and subtracting the product of the z-statistic reliability (z_α/2) factor times the known standard deviation of the population divided by the square root of the sample size (note that the population variance is given and its positive square root is the standard deviation of the population): x ± z_α/2 × ( σ / n^1/2) = 96 ± 1.96 × (9^1/2 / 400^1/2) = 96 ± 1.96 × (0.15) = 96 ± 0.294 = 95.706 to 96.294.