Question 632:

Asked on February 19, 2009

Tags: Sample Size , Margin of Error

1

Answer:

No answer provided yet.

For this question we need to estimate the sample size needed to have a margin of error to within +/- .25 of the average miles per gallon.  We basically work backwards from a confidence interval calculation to arrive at our sample size.

 

First, the Margin of Error is composed of the Standard Error of the Mean times a critical value from the normal distribution. Since we're interested in a 90% confidence interval we need to find the point under the normal curve which accounts for 90% of the area. We can look this value up using a z-table or the percentile to z-score calculator. For a 2-sided area we get the critical value of 1.645. So we know SEM*1.645 = .25.

 

Lets isolate the SEM to get SEM = .151976

 

Next, the SEM is composed of the standard deviation divided by the square root of the sample size. We do know the standard deviation is 2.0, but we don't know the sample size because we are solving for it, so it becomes our unknown.

 

2.0 / sqrt(n)= .151976

 

Square both sides

 

4/n = .023097

4 = .023097n

4/.023097 = n

n = 173.1856

 

Rounding up to the next whole run we get 174. So the manufacturer would need 174 sample runs to have a margin of error of +/- .25mpg around the mean.