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.

lestimate a 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 90% confident that its estimate is within 100 miles per month of ju?

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).



To determine the minimum sample size required for the consumer group to estimate the mean monthly mileage with 90% confidence, within 100 miles of the true mean \( \mu \), we can use the formula for the sample size of a mean estimate:

\[
n = \left( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right)^2
\]

Where:
- \( Z_{\alpha/2} \) is the critical value corresponding to the desired confidence level.
- \( \sigma = 750 \) miles (the standard deviation).
- \( E = 100 \) miles (the margin of error).

### Step 1: Determine the critical value \( Z_{\alpha/2} \)
For a 90% confidence level, the critical value \( Z_{\alpha/2} \) corresponds to the 95th percentile of the standard normal distribution, since \( \alpha = 1 - 0.90 = 0.10 \) and \( \alpha/2 = 0.05 \).

The critical value \( Z_{\alpha/2} \) for 90% confidence is approximately 1.645.

### Step 2: Plug values into the formula

\[
n = \left( \frac{1.645 \times 750}{100} \right)^2
\]

\[
n = \left( \frac{1233.75}{100} \right)^2
\]

\[
n = (12.3375)^2
\]

\[
n \approx 152.21
\]

### Step 3: Round up to the nearest whole number
Since the sample size must be a whole number, and it should be the minimum whole number that satisfies the requirements, round up:

\[
n = 153
\]

### Final Answer:
The minimum sample size required is **153**.

Would you like more details or have any questions about these calculations? Here are some related questions:

1. How would the sample size change if the desired confidence level was 95% instead of 90%?
2. What happens to the required sample size if the standard deviation is larger?
3. Can you explain why we round up the sample size rather than rounding to the nearest whole number?
4. How does increasing the margin of error \( E \) affect the sample size?
5. What assumptions are made when using this formula for sample size calculation?

**Tip:** Always round up the sample size to ensure the desired confidence level and margin of error are met.

Learn how to calculate the minimum sample size required to estimate the mean monthly mileage of U.S. rental cars with 90% confidence and within a 100-mile margin of error. Detailed step-by-step solution using statistical formulas and the Central Limit Theorem.

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