1215

The average U.S. dollar/Euro exchange rate from a sample of 36 monthly observations is $1.00/Euro. The population variance is 0.49. What is the 95% confidence interval for the mean U.S. dollar/Euro exchange rate?

A) $0.8075 to $1.1925.

B) $0.7713 to $1.2287.

C) $0.5100 to $1.4900.

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The population standard deviation is the square root of the variance (√0.49 = 0.7). Because we know the population standard deviation, we use the z-statistic. The z-statistic reliability factor for a 95% confidence interval is 1.960. The confidence interval is $1.00 ± 1.960($0.7 / √36) or $1.00 ± $0.2287.

1215

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).

1215

A local high school basketball team had 18 home games this season and averaged 58 points per game. If we assume that the number of points made in home games is normally distributed, which of the following is most likely the range of points for a confidence interval of 90%?

A) 34 to 82.

B) 24 to 78.

C) 26 to 80.

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This question has a bit of a trick. To answer this question, remember that the mean is at the midpoint of the confidence interval. The correct confidence interval will have a midpoint of 58. (34 + 82) / 2 = 58.

1215

A traffic engineer is trying to measure the effects of carpool-only lanes on the expressway. Based on a sample of 20 cars at rush hour, he finds that the mean number of occupants per car is 2.5, with a standard deviation of 0.4. If the population is normally distributed, what is the confidence interval at the 5% significance level for the number of occupants per car?

A) 2.313 to 2.687.

B) 2.387 to 2.613.

C) 2.410 to 2.589.

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The reliability factor corresponding with a 5% significance level (95% confidence level) for the Student’s t-distribution with (20 ? 1) degrees of freedom is 2.093. The confidence interval is equal to: 2.5 ± 2.093(0.4 / √20) = 2.313 to 2.687. (We must use the Student’s t-distribution and reliability factors because of the small sample size.)

1215

A sample of 100 individual investors has a mean portfolio value of $28,000 with a standard deviation of $4,250. The 95% confidence interval for the population mean is closest to:

A) $19,500 to $28,333.

B) $27,159 to $28,842.

C) $27,575 to $28,425.

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Confidence interval = mean ± t_c{S / √n}

= 28,000 ± (1.98) (4,250 / √100) or 27,159 to 28,842

If you use a z-statistic because of the large sample size, you get 28,000 ± (1.96) (4,250 / √100) = 27,167 to 28,833, which is closest to the correct answer.

1215

The average return on small stocks over the period 1926-1997 was 17.7%, and the standard error of the sample was 33.9%. The 95% confidence interval for the return on small stocks in any given year is:

A) 16.8% to 18.6%.

B) –16.2% to 51.6%.

C) –48.7% to 84.1%.

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A 95% confidence level is 1.96 standard deviations from the mean, so 0.177 ± 1.96(0.339) = (–48.7%, 84.1%).

1215

Books Fast, Inc., prides itself on shipping customer orders quickly. Downs Shipping Service has promised that mean delivery time will be less than 72 hours. Books Fast sampled 27 of its customers and found a mean delivery time of 76 hours, with a sample standard deviation of 6 hours. Based on this sample and assuming a normal distribution of delivery times, what is the confidence interval at 5% significance?

A) 73.63 to 78.37 hours.

B) 68.50 to 83.50 hours.

C) 65.75 to 86.25 hours.

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The confidence interval is equal to 76 + or ? (2.056)(6 / √27) = 73.63 to 78.37 hours.
Because the sample size is small, we use the t-distribution with (27 ? 1) degrees of freedom.

1215

A traffic engineer is trying to measure the effects of carpool-only lanes on the expressway. Based on a sample of 1,000 cars at rush hour, he finds that the mean number of occupants per car is 2.5, with a standard deviation of 0.4. Assuming that the population is normally distributed, what is the confidence interval at the 5% significance level for the number of occupants per car?

A) 2.475 to 2.525.

B) 2.455 to 2.555.

C) 2.288 to 2.712.

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The Z-score corresponding with a 5% significance level (95% confidence level) is 1.96. The confidence interval is equal to: 2.5 ± 1.96(0.4 / √1,000) = 2.475 to 2.525. (We can use Z-scores because the size of the sample is so large.)

1215

The average salary for a sample of 61 CFA charterholders with 10 years experience is $200,000, and the sample standard deviation is $80,000. Assume the population is normally distributed. Which of the following is a 99% confidence interval for the population mean salary of CFA charterholders with 10 years of experience?

A) $172,514 to $227,486.

B) $160,000 to $240,000.

C) $172,754 to $227,246.

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If the distribution of the population is normal, but we don’t know the population variance, we can use the Student’s t-distribution to construct a confidence interval. Because there are 61 observations, the degrees of freedom are 60. From the student’s t table, we can determine that the reliability factor for t_α/2, or t_0.005, is 2.660. Then the 99% confidence interval is $200,000 ± 2.660($80,000 / √61) or $200,000 ± 2.660 × $10,243, or $200,000 ± $27,246.

1215

From a sample of 41 orders for an on-line bookseller, the average order size is $75, and the sample standard deviation is $18. Assume the distribution of orders is normal. For which interval can one be exactly 90% confident that the population mean is contained in that interval?

A) $71.29 to 78.71.

B) $70.27 to $79.73.

C) $74.24 to $75.76.

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If the distribution of the population is normal, but we don’t know the population variance, we can use the Student’s t-distribution to construct a confidence interval. Because there are 41 observations, the degrees of freedom are 40. From the student’s t table, we can determine that the reliability factor for t_α/2, or t_0.05, is 1.684. Then the 90% confidence interval is $75.00 ± 1.684($18.00 / √41), or $75.00 ± 1.684 × $2.81 or $75.00 ± $4.73