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 750 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 135 miles per month of ?
Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole
number that satisfies the requirements).
(If necessary, consult a list of formulas.)

Ask by Chadwick Schmidt. in the United States

Mar 27,2025

Question

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 750 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 135 miles per month of ?
Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole
number that satisfies the requirements).
(If necessary, consult a list of formulas.)

Ask by Chadwick Schmidt. in the United States

Mar 27,2025

UpStudy AI · Accepted Answer

We start with the margin of error formula for estimating the mean:
$$
E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}
$$
where:
- $$ E = 135 $$ (the desired margin of error),
- $$ \sigma = 750 $$ (the standard deviation), and
- $$ z_{\alpha/2} $$ for a 95% confidence level is approximately 1.96.

Rearranging the formula to solve for $$ n $$:
$$
\sqrt{n} = z_{\alpha/2} \cdot \frac{\sigma}{E}
$$
$$
n = \left(z_{\alpha/2} \cdot \frac{\sigma}{E}\right)^2 
$$

Substitute the known values:
$$
n = \left(1.96 \cdot \frac{750}{135}\right)^2
$$

Compute the intermediate quotient:
$$
\frac{750}{135} \approx 5.556
$$

Multiply by $$ 1.96 $$:
$$
1.96 \times 5.556 \approx 10.889
$$

Square the result:
$$
n \approx (10.889)^2 \approx 118.665
$$

Since we need the minimum whole number that satisfies the requirements, we round up:
$$
n = 119
$$

Thus, the minimum sample size needed is $$ \boxed{119} $$.
The minimum sample size needed is 119.

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