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LOS j: Calculate and interpret a confidence interval for a population mean, given a normal distribution with 1) a known population variance, 2) an unknown population variance, or 3) an unknown variance and a large sample size.

In which one of the following cases is the t-statistic the appropriate one to use in the construction of a confidence interval for the population mean?

A)	The distribution is normal, the population variance is known, and the sample size is less than 30.

B)	The distribution is nonnormal, the population variance is known, and the sample size is at least 30.

C)	The distribution is nonnormal, the population variance is unknown, and the sample size is at least 30.

The t-distribution is the theoretically correct distribution to use when constructing a confidence interval for the mean when the distribution is nonnormal and the population variance is unknown but the sample size is at least 30.

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?

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.

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?

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

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?

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%.

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?

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.

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%?

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.

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?

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.)

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:

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.

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:

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?

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.

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?

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.)

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.

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.

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.

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.

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?

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.

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.

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).

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.

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).

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?

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.

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.

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.

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?

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%.

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:

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.

Based on Student's t-distribution, the 95% confidence interval for the population mean based on a sample of 40 interest rates with a sample mean of 4% and a sample standard deviation of 15% is closest to:

The standard error for the mean = s/(n)^0.5 = 15%/(40)^0.5 = 2.372%. The critical value from the t-table should be based on 40 – 1 = 39 df. Since the standard tables do not provide the critical value for 39 df the closest available value is for 40 df. This leaves us with an approximate confidence interval. Based on 95% confidence and df = 40, the critical t-value is 2.021. Therefore the 95% confidence interval is approximately: 4% ± 2.021(2.372) or 4% ± 4.794% or -0.794% to 8.794%.

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:

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.