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A consumer advocacy group is doing a large study on car rental practices

Math

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 ji using the mean of the sample. OO 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 95% confident that its estimate is within 150 miles per month of u? EO 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. ) X S ?

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sample size, n = 97

 

In the problem, we are asked to compute for the sample size needed in order to be 95% Confident
that the estimate is within 150 miles per month of M. In order to calculate the sample size, we
should follow the formula: N n = 2 12

 

where : Z Z value which corresponds to the desired level of confidence o . standard deviation E : margin of error . Margin of error ( E ) shows how many points or the amount by which the sample results may differ from the actual population. In the problem, our margin of error is equal to 150 miles. Now , we have to find the 2 value corresponding to 95% confidence. We can use the normal distribution table to get this value. For a 99% confidence, & is: . ( 1 - d ) x 10% = 95%, therefore, d = 0.05 . We need to determine the corresponding value for Zay = 20.05, = 2 0 -025 12 . Using the table, the Z-score corresponding to P ( 2 > ZA/2 ) = 0.025 or P ( 2 < Zx ) = 0.975 From the table, the z-score is 1'96. Now, we can compute for the sample size. 2 n = Zay * 5 E 1. 96 x 750 150 = 96.04 x 97 Therefore, our sample size is 97 .

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