SOLUTION: SAT scores are distributed with a mean of 1,500 and a standard deviation of 307. You are interested in estimating the average SAT score of first year students at your college. If y
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Question 1197340: SAT scores are distributed with a mean of 1,500 and a standard deviation of 307. You are interested in estimating the average SAT score of first year students at your college. If you would like to limit the margin of error of your confidence interval to 30 points with 85 percent confidence, how many students should you sample? (Round up to a whole number of students.) Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! mean is 1500.
standard deviation is 307.
margin of error = plus or minus 30 from the mean.
standard error = standard deviation / square root of sample size = 307/sqrt(n), where n = sample size.
z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard error.
with margin of error = 30, then (x - m) = 30
at 85% two tailed confidence interval, the critical z-score = plus or minus 1.439531471
z = (x - m) / s becomes:
plus or minus 1.439531471 = 30 / (307 / sqrt(n))
multiply both sides of this equation by 307 / sqrt(n)) to get:
plus or minus 1.4395315471 * 307 / sqrt(n) = 30
multiply both sides of this equaton by sqrt(n) and divide both side of this equation by 30 to get:
plus or minus 1.4395315471 * 307 / 30 = sqrt(n)
solve for sqrt(n) to get:
sqrt(n) = 14.73120539
s = 307 / 14.73120539 = 20.84011403
that's the standard error that should result in a margin of error or plus or minus 30.
the high side z-score formula becomes:
1.4395315471 = (x - 1500) / 20.84011403
solve for x to get:
x = 1.4395315471 * 20.84011403 + 1500 = 1530
the low side z-score formula becomes:
1.4395315471 * 20.84011403 + 1500 = 1470
your margin of error is plus or minus 30.
the sample size that allowed this to happen was calculate to be 14.73120539 squared = 217.0084121.
since the sample size has to be an integer, then use sample size of 218 (nearest higher integer).
this will result in a margin of error that will be slightly less than 30.
i used an online calculator to show you how this looks.
here are the results.
first one is with sample size = 217.0084121
second one is with sample size = 218
your answer should be a sample size of at least 218 to get a margin of error less than 30.
