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3、The approximate 99% confidence interval for the population mean based on a sample of 60 returns with a mean of 7% and a sample standard deviation of 25% is closest to:

A) 1.584% to 14.584%.

B) -1.584% to 15.584%.

C) 0.546% to 13.454%.

D) 1.546% to 13.454%.

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The correct answer is B

The standard error for the mean = s / (n)0.5 = 25% / (60)0.5 = 3.227%. The critical value from the t-table should be based on 60 ? 1 = 59 df. Since the standard tables do not provide the critical value for 59 df the closest available value is for 60 df. This leaves us with an approximate confidence interval. Based on 99% confidence and df = 60, the critical t-value is 2.660. Therefore the 99% confidence interval is approximately: 7% ± 2.660(3.227) or 7% ± 8.584% or -1.584% to 15.584%.

If you use a z-statistic, the confidence interval is 7% ± 2.58(3.227) = -1.326% to 15.326%, which is closest to the correct choice.

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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% 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).

D) 0.8 + 1.96(1.85).

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The correct answer is A

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 (zα/2) factor times the known standard deviation of the population divided by the square root of the sample size: x ± zα/2 × ( σ / n1/2) = 60 ± (1.96) × (16 / 751/2) = 60 ± (1.96) × (16 / 8.6603) = 60 ± (1.96) × (1.85).

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5、What is the appropriate test statistic for constructing confidence intervals for the population mean of a normal distribution when the population variance is unknown?

A) The t-statistic at α/2 with n degrees of freedom.

B) The z-statistic with n – 1 degrees of freedom.

C) The z-statistic at α with n degrees of freedom.

D) The t-statistic at α/2 with n – 1 degrees of freedom.

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The correct answer is D

Use the t-statistic at α/2 and n – 1 degrees of freedom when the population variance is unknown, regardless of sample size.

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6、What is the appropriate test statistic for constructing confidence intervals for the population mean of a nonnormal distribution when the population variance is unknown and the sample size is large (n ≥ 30)?

A) The z-statistic at α with n degrees of freedom.

B) The t-statistic at α with 29 degrees of freedom.

C) The z-statistic or the t-statistic.

D) The t-statistic at α/2 with n degrees of freedom.

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The correct answer is C

When the sample size is large, and the central limit theorem can be relied upon to assure a sampling distribution that is normal, either the t-statistic or the z-statistic is acceptable for constructing confidence intervals for the population mean. However, the t-statistic will provide a more conservative range (wider) at a given level of significance.

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7、The 95 percent confidence interval of the sample mean of the price earnings ratio for all traded stocks is 19 to 44. There are over 5,000 traded stocks and the sample size of this test is 100. Given that the expected value of the price earnings ratio is 31.5, the standard error of the ratio is closest to:

A) 1.96.

B) 2.58.

C) 12.50.

D) 6.38.

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The correct answer is

The confidence interval is 31.5 ± 1.96x, where x is the standard error. If we take the upper bound, we know that 31.5 +/– 1.96x = 44, or 1.96x = 12.5. Hence, x = 6.38.

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