Question 1205228: You work in the leasing office of GE capital and wish to predict the average mileage driven of your leased cars within a margin of error of 100,000 miles. Suppose your average customer drives 10,000 miles a year, standard deviation is 750 miles. How big of a sample size do you need to generate 80% confidence interval with a mean of 10,000 miles a year with a margin of error of 100 miles?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i think your margin of error is 100 miles and not 100,000 miles.
i'll assume margin of error of 100 miles.
the mean is 10,000 miles a year.
the standard deviation is 750 miles a year.
you want to predict with a confidence interval of 80%.
that means your two tailed alpha is 20% which means 10% on each end of the confidence interval.
10% alpha on each end required a critical z-score of plus or minus 1.28155 on each end of the confidence interval.
since you are looking for the mean of a sample of specified size, you need to use the standard error and not the standard deviation.
standard error = standard deviation / sqrt(sample size).
when n = sample size and s = standard error, your formula for standard error is s = 750 / sqrt(n)
your z-score formula is z = (x-m)/s
z is the critical z-score
x is the critical raw score
m is the sample mean
s is the standard error.
the margin of error is equal to (x-m).
solve for (x-m) to get:
(x-m) = z * s
since s = 750 / sqrt(n), that formula becomes:
(x-m) = z * 750 / sqrt(n)
since you want (x-m) to be equal to 100, that formula becomes:
100 = z * 750 / sqrt(n)
solve for sqrt(n) to get:
sqrt(n) = z * 750 / 100
on the high side of your confidence interval, the critical z-score is equal to 1.281551567.
the formula becomes:
sqrt(n) = 1.281551567 * 750 / 100 = 9.611636749.
since s = 750 / sqrt(n), that makes s = 78.03041454.
with that value of s, you should get plus or minus 100 as your margin of error.
on the high side of the confidence interval, the z-score formula becomes:
1.281551567 = (x-m)/78.03041454.
solve for (x-m) to get (x-m) = 100
on the low side of the confidence interval, the z-score formula becomes:
-1.281551567 = (x-m)/78.03041454.
solve for (x-m) to get (x-m) = -100
since the mean is given at 10,000 miles a year, then your confidence interval should be between 9,900 and 10,100 miles a year.
this is valid as long as the standard error is equal to 78.03041454.
here's what it looks like on a z-score graph.
you can see that the confidence interval is .8 = 80%, shown below the bottom of the graph.
since the square root of the sample size is equal to 9.611636749, then the sample size will be equal to that squared = 92.383561.
since the sample size needs to be an integer, then the minimum sample size will be equal to 93 if you round up, and equal to 92 if you round down.
if you round up, the margin of error will be slightly less than 100 at 80% confidence interval.
if you round down, the margin of error will be slightly more than 100 at 80% confidence interval.
this is to be expected.
when you round up, your standard error becomes 750 / sqrt(93) = 77.7713771 and your margin of error becomes 1.281551567 * 77.7713771 = 99.66803016.
when you round down, your standard eror becomes 750 / sqrt(92) = 78.19290527 and your margin of error becomes 1.281551567 * 78.19290527 = 100.2082402.
these can be seen on the following graphs.
rounding up to sample size of 93:
rounding down to sample size of 92:
because of the requirement that the sample size be an integer, the best you can do is get exactly 80% confidence interval with slightly different margin of error or exactly 100 margin of error with slightly different confidence interval, but not exactly 80% confidence interval and exactly 100 margin of error at the same time.
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