A consumer advocacy group is doing a large study on car rental practices. Among other things, the consumer group would like to estimate the mean monthly mileage, , of cars rented in the U. S. over the past year. The consumer group plans to choose a random sample of monthly U. S. rental car mileages and then estimate using the mean of the sample. Using the value miles per month as the standard deviation of monthly U. S. rental car mileages from the past year, what is the minimum sample size needed in order for the consumer group to be confident that its estimate is within miles per month of

The minimum sample size needed 64 monthly U.S. rental car mileages.

Step-by-step explanation:

Note: This question is not complete as all the important data are omitted. The complete question is therefore provided before answering the question as follows:

A consumer advocacy group is doing a large study on car rental practices. Among other things, the consumer group would like to estimate the mean monthly mileage, u, of cars rented in the U.S. over the past year. The consumer group plans to choose a random sample of monthly U.S. rental car mileages and then estimate u using the mean of the sample. Using the value 850 miles per month as the standard deviation of monthly U.S. rental car mileages from the past year, what is the minimum sample size needed in order for the consumer group to be 95% confident that its estimate is within 175 miles per month of u?

The explanation of the answer is now given as follows:

The minimum sample size needed can be calculated using the following sample size formula:

n = ((Z * S) / E)^2 ………………………… (1)

Where:

n = sample size or minimum sample size = ?

Z = Confidence interval at 95% = 1.645

S = Standard deviation = 850

E = Accepted magnitude of error = 175

Substituting all the relevant values into equation (1), we have:

n = ((1.645 * 850) / 175)^2 = (1,398.25 / 175)^2 = 7.99^2 = 63.8401, or 64.

Therefore, the minimum sample size needed 64 monthly U.S. rental car mileages.