Question 1

A study was recently conducted to estimate the mean cholesterol for adult males over the age of 55 years. From a random sample of n = 10 men, the sample mean was found to be 242.6 and the sample standard deviation was 73.33. To find the 95 percent confidence interval estimate for the mean, the correct critical value to use is:

Accepted Answer

The correct critical value for a 95% confidence interval when the sample size is small (n < 30) and the population standard deviation is unknown is found using the t-distribution. The degrees of freedom (df) in this case is n - 1 = 10 - 1 = 9. Looking up the t-table for 9 degrees of freedom at a 95% confidence level gives a critical value of approximately 2.262.

Question 2

The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded:  Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

Accepted Answer

The answer of The Wisconsin Dairy Association is interested in...

Question 3

A study was recently conducted to estimate the mean cholesterol for adult males over the age of 55 years. The following random sample data were observed:  Given this information, what is the point estimate for the population mean?

Accepted Answer

The answer of A study was recently conducted to estimate...

Question 4

When σ is unknown, the margin of error is computed by using:

Accepted Answer

When σ (the population standard deviation) is unknown and the sample size is relatively small, the t-distribution is used instead of the normal distribution to compute the margin of error. This accounts for the added uncertainty of estimating the population standard deviation with the sample standard deviation.

Question 5

The following data represent a random sample of bank balances for a population of checking account customers at a large eastern bank. Based on these data, what is the critical value for a 95 percent confidence interval estimate for the true population mean?

Accepted Answer

The answer of The following data represent a random sample...

Question 6

An educational organization in California is interested in estimating the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day. A previous study showed that the population standard deviation was 21.5 minutes. The organization selected a random sample of n = 200 children between the ages of 6 and 18 and recorded the number of minutes of TV that each person watched on a particular day. The mean time was 191.3 minutes. If the leaders of the organization wish to develop an interval estimate with 98 percent confidence, what critical value should be used?

Accepted Answer

Since the population standard deviation is known and the sample size is large (n = 200), the appropriate distribution to use for constructing the confidence interval is the standard normal (Z) distribution. For a 98% confidence level, the critical value from the Z distribution is approximately 2.33.

Question 7

A hospital emergency room has collected a sample of n = 40 to estimate the mean number of visits per day. It has found the standard deviation is 32. Using a 90 percent confidence level, what is its margin of error?

Accepted Answer

The margin of error is calculated using the formula: E = Z * (σ/√n). For a 90% confidence level, the Z-score is approximately 1.645. Plugging in the values: E = 1.645 * (32/√40) ≈ 8.3 visits.

Question 8

Past experience indicates that the variance in the time it takes for a "fast lube" operation to actually complete the lube and oil change for customers is 9.00 minutes. The manager wishes to estimate the mean time with 99 percent confidence and a margin of error of ± 0.50 minutes. Given this, what must the sample size be?

Accepted Answer

To determine the sample size needed for estimating the mean with a specific confidence level and margin of error, the formula $n = \left(\frac{Z*\sigma}{E}\right)^2$ is used, where $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the standard deviation, and $E$ is the margin of error. Given a variance of 9, the standard deviation $\sigma$ is 3 (since $\sigma^2 = 9$). For a 99% confidence level, the Z-score is approximately 2.576. Plugging these values into the formula with $E = 0.5$ gives $n = \left(\frac{2.576*3}{0.5}\right)^2 \approx 239$.

Question 9

An educational organization in California is interested in estimating the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day. A previous study showed that the population standard deviation was 21.5 minutes. The organization selected a random sample of n = 200 children between the age of 6 and 18 and recorded the number of minutes of TV that each person watched on a particular day. The mean time was 191.3 minutes. If the leaders of the organization wish to develop an interval estimate with 98 percent confidence, what would be the upper and lower limits of the interval estimate?

Accepted Answer

To calculate the confidence interval, we use the formula for the margin of error: $ME = Z * (\sigma / \sqrt{n})$, where $Z$ is the Z-score corresponding to the confidence level (for 98% confidence, $Z \approx 2.33$), $\sigma$ is the population standard deviation (21.5 minutes), and $n$ is the sample size (200). Calculating the margin of error gives us $ME = 2.33 * (21.5 / \sqrt{200}) \approx 3.54$. Adding and subtracting this margin of error from the sample mean (191.3 minutes) gives us the confidence interval: $191.3 \pm 3.54$, which is approximately 187.76 minutes to 194.84 minutes.

Question 10

A traffic engineer plans to estimate the average number of cars that pass through an intersection each day. Based on previous studies the standard deviation is believed to be 52 cars. She wants to estimate the mean to within ±10 cars with 90 percent confidence. The needed sample size for n is:

Accepted Answer

To calculate the sample size needed to estimate the mean with a certain level of confidence and margin of error, the formula is $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, where $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the standard deviation, and $E$ is the margin of error. For a 90% confidence level, the Z-score is approximately 1.645. Given $\sigma = 52$ cars and $E = 10$ cars, substituting these values into the formula gives $n = \left(\frac{1.645 \cdot 52}{10}\right)^2 \approx 73.68$, which rounds up to 74 days.

Question 11

The Internal Revenue Service (IRS) is interested in estimating the mean amount of money spent on outside tax service by income tax filers filing as single on their individual form. To do this, they have selected a random sample of n = 16 people from this population and surveyed them to determine the sample mean and sample standard deviation. The following information was observed:  Given this information, what is the 95 percent confidence interval for the mean dollars spent on outside tax assistance by taxpayers who file as single?

Accepted Answer

The answer of The Internal Revenue Service (IRS) is interested...

Question 12

The purpose of a pilot sample is:

Accepted Answer

The answer of The purpose of a pilot sample is:
...

Question 13

A large Midwestern university is interested in estimating the mean time that students spend at the student recreation center per week. A previous study indicated that the standard deviation in time is about 40 minutes per week. If the officials wish to estimate the mean time within ± 10 minutes with a 90 percent confidence, what should the sample size be?

Accepted Answer

To estimate the mean time within ±10 minutes with 90% confidence and a standard deviation of 40 minutes, the sample size can be calculated using the formula for the sample size of a mean: $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, where $Z$ is the Z-score corresponding to the confidence level (for 90% confidence, $Z = 1.645$), $\sigma$ is the standard deviation (40 minutes), and $E$ is the margin of error (10 minutes). Plugging in the values: $n = \left(\frac{1.645 \cdot 40}{10}\right)^2 = \left(6.58\right)^2 \approx 43.3$, which rounds up to 44. The total number of students at the university is not needed to calculate the sample size for estimating a mean with a specified confidence and margin of error.

Question 14

In order to reduce the cost of a study, a marketing manager wants to reduce the sample size for a study of customer response to a recent advertising campaign. What can she do that would lead to a reduction in sample size?

Accepted Answer

All of the options provided would lead to a reduction in sample size. Allowing a higher margin of error, reducing the level of confidence, and reducing the variation in the population would each decrease the required sample size for the study.

Question 15

A study has indicated that the sample size necessary to estimate the average electricity use by residential customers of a large western utility company is 900 customers. Assuming that the margin of error associated with the estimate will be ±30 watts and the confidence level is stated to be 90 percent, what was the value for the population standard deviation?

Accepted Answer

The sample size formula for estimating a mean with a specific margin of error at a given confidence level is $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, where $n$ is the sample size, $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $E$ is the margin of error. Given $n = 900$, $E = 30$ watts, and a 90% confidence level (which corresponds to a Z-score of approximately 1.645 for a two-tailed test), we can rearrange the formula to solve for $\sigma$: $\sigma = \frac{E \cdot \sqrt{n}}{Z}$. Plugging in the given values, we find that $\sigma$ is approximately 547.1 watts.

Question 16

An educational organization in California is interested in estimating the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day. A previous study showed that the population standard deviation was 21.5 minutes. The organization selected a random sample of n = 200 children between the age of 6 and 18 and recorded the number of minutes of TV that each person watched on a particular day. The mean time was 191.3 minutes. If the leaders of the organization wish to develop an interval estimate with 95 percent confidence, what will the margin of error be?

Accepted Answer

The margin of error (ME) for a 95% confidence interval can be calculated using the formula ME = z * (σ/√n), where z is the z-score corresponding to the desired confidence level (1.96 for 95%), σ is the population standard deviation (21.5 minutes), and n is the sample size (200). Plugging in the values: ME = 1.96 * (21.5/√200) ≈ 2.98 minutes.

Question 17

The Hilbert Drug Store owner plans to survey a random sample of his customers with the objective of estimating the mean dollars spent on pharmaceutical products during the past three months. He has assumed that the population standard deviation is known to be $15.50. Given this information, what would be the required sample size to estimate the population mean with 95 percent confidence and a margin of error of ±$2.00?

Accepted Answer

To calculate the required sample size for estimating the population mean with a specific confidence level and margin of error when the population standard deviation is known, the formula is: $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, where $n$ is the sample size, $Z$ is the Z-score corresponding to the confidence level (for 95% confidence, $Z = 1.96$), $\sigma$ is the population standard deviation ($15.50), and $E$ is the margin of error ($2.00). Plugging in the values: $n = \left(\frac{1.96 \cdot 15.50}{2}\right)^2 \approx 231$.

Question 18

When small samples are used to estimate a population mean, in cases where the population standard deviation is unknown:

Accepted Answer

When small samples are used to estimate a population mean and the population standard deviation is unknown, the t-distribution is used instead of the normal distribution to account for the additional uncertainty introduced by estimating the population standard deviation with the sample standard deviation. This approach helps to maintain the desired confidence level despite the small sample size.

Question 19

The following data represent a random sample of bank balances for a population of checking account customers at a large eastern bank. Based on these data, what is the 95 percent confidence interval estimate for the true population mean?

Accepted Answer

The answer of The following data represent a random sample...

Question 20

A major tire manufacturer wishes to estimate the mean tread life in miles for one of its tires. It wishes to develop a confidence interval estimate that would have a maximum sampling error of 500 miles with 90 percent confidence. A pilot sample of n = 50 tires showed a sample standard deviation equal to 4,000 miles. Based on this information, the required sample size is:

Accepted Answer

The answer of A major tire manufacturer wishes to estimate...

Quiz 8: Estimating Single Population Parameters

A study was recently conducted to estimate the mean cholesterol for adult males over the age of 55 years. The following random sample data were observed: Given this information, what is the point estimate for the population mean?