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, pl, 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 90% confident that its estimate is within 100 miles per month of u? 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.) 0 X 5 ?

Answer 1

Standard deviation = $\small \sigma$ = 850

Confidence level = 90%

z score = z = 1.645       (from z dcore table)

Margin of error = 100

Let n be the sample size.

$\small z\times \frac{\sigma}{\sqrt n}=Margin \ of \ error$

$\small or, \ 1.645\times \frac{850}{\sqrt n}=100$

$\small or, \ n=[1.645\times \frac{850}{100}]^2=195.510\approx 196$

Therefore, required sample size is 196

z score table :

Z-score Confidence Level Area between O and z-score Area in one tail (alpha/2) 50% 0.2500 0.2500 0.674 80% 0.4000 0.1000 1.282 90% 0.4500 0.0500 1.645 95% 0.4750 0.0250 1.960 98% 0.4900 0.0100 2.326 99% 0.4950 0.0050 2.576