honeycfa

An airline was concerned about passengers arriving too late at the airport to allow for the additional security measures. The airline collected survey data from 1,000 passengers on their time from arrival at the airport to reaching the boarding gate. The sample mean was 1 hour and 20 minutes, with a sample standard deviation of 30 minutes. 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, thirty minutes.

B) One hour, fifty minutes.

C) Two hours, ten minutes.

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We can use standard distribution tables because the sample is so large.

From a table of area under a normally distributed curve, the z value corresponding to a 95%, one-tail test is: 1.65. (We use a one-tailed test because we are not concerned with passengers arriving too early, only arriving too late.)

Here, we do not divide by the standard error, because we are interested in a point estimate of making our flight.

The answer is one hour, twenty minutes + 1.65(30 minutes) = 2 hours, 10 minutes.

honeycfa

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.

honeycfa

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) $0.5100 to $1.4900.

C) $0.7713 to $1.2287.

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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.

honeycfa

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) $27,575 to $28,425.

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

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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.

honeycfa

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%).

honeycfa

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) 68.50 to 83.50 hours.

B) 65.75 to 86.25 hours.

C) 73.63 to 78.37 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.

honeycfa

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.)

honeycfa

From a sample of 41 monthly observations of the S& Mid-Cap index, the mean monthly return is 1% and the sample variance is 36. For which of the following intervals can one be closest to 95% confident that the population mean is contained in that interval?

A) 1.0% ± 1.9%.

B) 1.0% ± 6.0%.

C) 1.0% ± 1.6%.

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If the distribution of the population is nonnormal, but we don’t know the population variance, we can use the Student’s t-distribution to construct a confidence interval. The sample standard deviation is the square root of the variance, or 6%. Because there are 41 observations, the degrees of freedom are 40. From the Student’s t distribution, we can determine that the reliability factor for t_0.025, is 2.021. Then the 95% confidence interval is 1.0% ± 2.021(6 / √41) or 1.0% ± 1.9%.

honeycfa

A random sample of 25 Indiana farms had a mean number of cattle per farm of 27 with a sample standard deviation of five. Assuming the population is normally distributed, what would be the 95% confidence interval for the number of cattle per farm?

A) 25 to 29.

B) 23 to 31.

C) 22 to 32.

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The standard error of the sample mean = 5 / √25 = 1
Degrees of freedom = 25 ? 1 = 24
From the student’s T table, t_5/2 = 2.064
The confidence interval is: 27 ± 2.064(1) = 24.94 to 29.06 or 25 to 29.

honeycfa

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.