Reading 10: Sampling and Estimation - LOS j ～ Q1-5

答案和详解如下：

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

A)   0.1 percent to 2.9 percent.

B)   -0.4 percent to 3.4 percent.

C)   -6.5 percent to 9.5 percent.

D)   7.65 percent to 8.35 percent.

The correct answer was B)

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%

2.Construct a 90 percent 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(30).

B)   50 + 1.645(300).

C)   50 + 1.645(3).

D)   50 + 1.645(900).

The correct answer was C)

Because we can compute the population standard deviation, we use the z-statistic. A 90 percent 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 percent confident that the above interval contains the true mean starting salaries of CFA charterholders.

3.What is the 95 percent 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)   95.706 to 96.294.

C)   95.118 to 96.882.

D)   95.235 to 96.374.

The correct answer was B)

Because we can compute the population standard deviation, we use the z-statistic. A 95 percent 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.

4.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 percent confidence interval for the mean height of all trees in the nursery:

A)   60 + 1.96(16).

B)   0.8 + 1.96(16).

C)   60 + 1.96(1.85).

D)   0.8 + 1.96(1.85).

The correct answer was C)

Because we know the population standard deviation, we use the z-statistic. A 95 percent 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).

5.A sample of size 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 percent confidence interval for the population mean, m.

A)   15 + 1.96(0.4).

B)   15 + 1.96(2).

C)   15 + 1.96(0.8).

D)   15 + 1.96(4).

The correct answer was A)   

Because we can compute the population standard deviation, we use the z-statistic.  A 95 percent 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).