Question

10# 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, μ, 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 μ using the mean of the sample.Using the value 700 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 135 miles per month of μ?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).

Answer 1

ANSWER:

104

STEP-BY-STEP EXPLANATION:

Given:

Standard deviation (σ) = 700

Confidence level = 95%

Mean error (Eμ) = 135

We have for a confidence level of 95%:

$\begin{gathered} \alpha=100\%-95\%=5\%=0.05 \\ \\ \alpha\text{/2}=\frac{0.05}{2}=0.025 \\ \\ \text{ The corresponding Z value for \alpha/2 = 0.025 is as follows:} \\ \\ Z_{\alpha\text{/2}}=1.96 \end{gathered}$

Now, we calculate the minimum value of the sample size as follows:

$\begin{gathered} n=\left(\frac{Z_{\alpha\text{/2}}\cdot\sigma}{E}\right)^2 \\ \\ \text{ We replacing:} \\ \\ \:n=\left(\frac{1.96\cdot 700}{135}\right)^2 \\ \\ \:n=103.2858\cong104 \end{gathered}$

The minimum sample size needed is 104