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

An airline is concerned about passengers arriving too late at the airport to allow for the additional security measures. The airline collects survey data from 1,000 passengers on their time from arrival at the airport to reaching the boarding gate. The sample mean is 1 hour and 20 minutes, with a sample standard deviation of 30 minutes, and the distribution of arrival times is approximately normal. Based on this sample, how long prior to a flight should a passenger arrive at the airport to have a 95% probability of making it to the gate on time?

A) Two hours, ten minutes.

B) Two hours, thirty minutes.

C) One hour, fifty minutes.

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For a normal distribution, 95% of the probability is less than +1.65 standard deviations from the mean. We use the right-hand tail of the distribution because we are not concerned with passengers arriving too early, only arriving too late.

To have a 95% probability of arriving on time, the passenger should allow one hour and twenty minutes + 1.65 × 30 minutes
= 80 minutes + 49.5 minutes
= 129.5 minutes
= 2 hours, 9.5 minutes.

Note that we use the standard deviation, not the standard error, because we are interested in a single observation: one passenger arriving on time.

1215

A 95% confidence interval for the mean number of monthly customer visits to a grocery store is 28,000 to 32,000 customers. Which of the following is an appropriate interpretation of this confidence interval?

A) If we repeatedly sample the population and construct 95% confidence intervals, 95% of the resulting confidence intervals will include the population mean.

B) There is a 95% chance that next month the grocery store will have between 28,000 and 32,000 customer visits.

C) We are 95% confident that if a sample of monthly customer visits is taken, the sample mean will fall between 28,000 and 32,000.

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There are two interpretations of this confidence interval: a probabilistic and a practical interpretation. Probabilistic interpretation: We can interpret this confidence interval to mean that if we sample the population of customers 100 times, we can expect that 95 (95%) of the resulting 100 confidence intervals will include the population mean. Practical interpretation: We can also interpret this confidence interval by saying that we are 95% confident that the population mean number of monthly customer visits is between 28,000 and 32,000.

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 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(16).

B) 0.8 + 1.96(16).

C) 60 + 1.96(1.85).

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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).

1215

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.

1215

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(30).

B) 50 ± 1.645(3).

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.

1215

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) 0.1% to 2.9%.

B) -0.4% to 3.4%.

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%.

1215

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(10).

B) 60 + 1.645(4).

C) 60 + 1.645(2).

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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.