Question 1

Two confidence interval estimates were developed from the same sample of a population. The wider interval will be the one that has the higher confidence level.

Accepted Answer

A wider confidence interval corresponds to a higher confidence level because it indicates that there is a greater certainty that the interval contains the population parameter, thus requiring a broader range to achieve this higher level of confidence.

Question 2

If a decision maker desires a small margin of error and a high level of confidence, it is certain that the required sample size will be quite large.

Accepted Answer

The answer of If a decision maker desires a small...

Question 3

A grocery store manager is interested in estimating the mean weight of apples received in a shipment. If she wishes to have the estimate be within ±.05 pound with 90 percent confidence, the sample size should be 103 apples if she believes that the standard deviation is .08 pound.

Accepted Answer

The sample size required for estimating the mean weight of apples with a specific margin of error and confidence level is calculated using the formula for the sample size of a mean, which is $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, where $Z$ is the Z-score corresponding to the desired confidence level, $\sigma$ is the estimated standard deviation, and $E$ is the desired margin of error. Without calculating the exact number, the statement provides a specific sample size (103 apples) without showing the calculation or considering the Z-score for a 90% confidence level, which makes it impossible to verify the correctness of the statement without additional information or calculation.

Question 4

In estimating a population mean, increasing the confidence level will result in a higher margin of error for a given sample size.

Accepted Answer

Increasing the confidence level means you want to be more certain that the population mean falls within your confidence interval, which requires widening the interval, thus increasing the margin of error.

Question 5

An emergency room in a hospital wants to determine the sample size needed for estimating their mean number of visits per day. If they want a 99 percent confidence level the correct critical value to use is 2.33.

Accepted Answer

The critical value for a 99 percent confidence level is approximately 2.576, not 2.33. The value of 2.33 is closer to the critical value for a 98 percent confidence level.

Question 6

The main purpose of a pilot sample in an application involving an estimate for a population mean is to determine what the margin of error will likely be.

Accepted Answer

The main purpose of a pilot sample is to help estimate the variability in the population, which can then be used to determine the appropriate sample size needed for the main study to achieve a desired margin of error.

Question 7

An analyst for a financial investment firm recently went through the effort to determine the required sample size for estimating the mean number of transactions per year for the clients of his firm. The calculations, which were based on a 95 percent confidence level and a margin of error of ±3, gave a required sample size of 300. Given this information, the value used for the population standard deviation must have been about 26.5 transactions.

Accepted Answer

The sample size formula for estimating a mean with a given confidence level and margin of error is n = (Z*σ/E)^2. For a 95% confidence level, Z is approximately 1.96. Plugging in the values: 300 = (1.96*σ/3)^2, solving for σ gives approximately 26.5.

Question 8

If a pilot sample of n = 40 items has been used as a first step in determining a required sample size of n = 360, the decision maker can go ahead and use these 40 and take a sample of only 320 more items.

Accepted Answer

The initial pilot sample of 40 items can be included in the total required sample size of 360, so the decision maker only needs to collect an additional 320 items to reach the total required sample size.

Question 9

In a sample size determination situation, reducing the margin of error by half will double the required sample size.

Accepted Answer

Reducing the margin of error by half will actually quadruple the required sample size, as the sample size is inversely proportional to the square of the margin of error.

Question 10

In an effort to estimate the mean length of stay for motel guests at a major national motel chain, the decision makers asked for a 95 percent confidence, and a margin of error of ±0.25 days. Further, it was known that the population standard deviation is 0.50 days. Given this, the required sample size to estimate the mean length of stay is about 16 customers.

Accepted Answer

The formula for sample size is n = (Z^2 * s^2) / e^2, where Z is the z-score for the desired confidence level, s is the population standard deviation, and e is the margin of error. Plugging in the given values, we get n = (1.96^2 * 0.50^2) / 0.25^2 = 15.36. Rounding up to the nearest whole number, the required sample size is 16 customers.

Question 11

The State Transportation Department wishes to estimate the mean speed of vehicles on a certain stretch of highway. They wish to estimate the mean within ±0.75 mph and they wish to have a confidence level equal to 99 percent. Based on this information only, they can determine that the required sample size is 320 vehicles.

Accepted Answer

The information provided does not include the population standard deviation or an estimate of it, which is necessary to calculate the required sample size for estimating the mean speed of vehicles with a specified margin of error and confidence level.

Question 12

The national sales manager for a textbook publishing company wishes to estimate the mean number of books sold per college. She wishes to have her estimate be within ± 30 copies and wants a 95 percent confidence interval estimate. If a pilot sample of 30 schools gave a sample standard deviation equal to 60 books, the required total sample size is less than the pilot sample already taken.

Accepted Answer

The sample size is already large enough to achieve the desired precision.

Question 13

The manager in charge of concessions at an NFL football stadium is interested in estimating the mean dollars that are spent per person attending the games. A pilot sample of n = 50 people has revealed a sample mean and standard deviation of $12.35 and $2.35 respectively. He wishes to estimate the population mean within ± $0.20 of the true mean and wishes to have a confidence level of 95 percent. Given this, he needs to sample an additional 481 people.

Accepted Answer

To find the required sample size, the formula for the sample size in estimating a population mean is used: n = (Z * σ / E)^2, where Z is the Z-score for the desired confidence level (1.96 for 95%), σ is the standard deviation, and E is the margin of error. Plugging in the values: n = (1.96 * 2.35 / 0.20)^2 ≈ 531. Since 50 people have already been sampled, an additional 481 people are needed.

Question 14

When finding sample size, cutting the margin of error in half requires that the sample size be four times larger.

Accepted Answer

The sample size needed for a given margin of error is inversely proportional to the square of the margin of error. Therefore, to cut the margin of error in half, the sample size must be increased by a factor of four.

Question 15

A bank manager wishes to estimate the mean waiting time spent by customers at his bank. He knows from previous experience that the standard deviation is about 4.0 minutes. If he desires a 90 percent confidence interval estimate and wishes to have a margin of error of 1 minute, the required sample size will be approximately 143.

Accepted Answer

The required sample size for estimating a mean with a specified margin of error (E), given the population standard deviation (σ) and desired confidence level (Z), is calculated using the formula $n = \left(\frac{Z\sigma}{E}\right)^2$. For a 90% confidence level, the Z value is approximately 1.645. Plugging in the values: $n = \left(\frac{1.645 \times 4}{1}\right)^2 = \left(6.58\right)^2 = 43.2964$, which rounds to approximately 43, not 143.

Question 16

When estimating sample size, a 90 percent confidence level will result in a smaller sample size than a 95 percent confidence level.

Accepted Answer

A higher confidence level (e.g., 95%) requires a larger sample size to ensure the same level of precision in estimates, compared to a lower confidence level (e.g., 90%), because it aims to cover a wider range of the population parameter.

Question 17

In determining the required sample size in an application involving an estimate for the population mean, if the population standard deviation is known, there is no compelling reason to select a pilot sample.

Accepted Answer

When the population standard deviation is known, it can be directly used in the formula to calculate the required sample size for estimating the population mean, eliminating the need for a pilot sample to estimate variability.

Question 18

After taking a speed-reading course, students are supposed to be able to read faster than they could before taking the course. A pilot sample of n = 25 students showed a mean increase of 300 words per minute with a standard deviation equal to 60 words per minute. To estimate the population mean with 95 percent confidence and a margin of error of ±10 minutes, the required sample size is approximately 139 students.

Accepted Answer

The sample size is calculated using the formula n = (Z^2 * s^2) / E^2, where Z is the z-score for the desired confidence level, s is the sample standard deviation, and E is the margin of error. Plugging in the given values, we get n = (1.96^2 * 60^2) / 10^2 = 138.24. Rounding up to the nearest whole number, the required sample size is approximately 139 students.

Question 19

When a decision maker determines the required sample size for estimating a population mean, a change in the confidence level will result in a change in the required sample size, provided that the margin of error is also modified accordingly.

Accepted Answer

The answer of When a decision maker determines the required...

Question 20

A university computer lab manager wishes to estimate the mean time that students stay in the lab per visit. She believes that the population standard deviation would be no larger than 10 minutes. Further, she wishes to have a confidence level of 90 percent and a margin of error of ±2.00 minutes. Given this, the sample size that she uses is approximately 60 students.

Accepted Answer

To calculate the sample size, the formula n = (Z*σ/E)^2 is used, where Z is the Z-score for the confidence level, σ is the standard deviation, and E is the margin of error. For a 90% confidence level, Z is approximately 1.645. Plugging in the values (1.645*10/2)^2 gives a sample size of approximately 68, not 60.

Quiz 8: Estimating Single Population Parameters