SOLUTION: SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is interested in estimating the average SAT score of first-y

Question 1202134: SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the 86% confidence interval to 15 points, how many students should the administrator sample? Make sure to give a whole number answer.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
mean is 1500
standard deviation is 300
confidence interval is .86
standard error = standard deviation / square root of sample size = s/sqrt(n)
s is the standard error
n is the sample size
they want to limit the margin of error to less than or equal 15 points.
z-score formula is z = (x-m)/s
x is the sample mean
m is the population mean
s is the standard error
(x-m) is the margin of error
with (x-m) equal to 15, the formula becomes:
z = 15 / s
at 86% confidence interval, the critical z-score is plus or minus 1.475791028.
the formula becomes:
1.475791028 = 15 / s
s = standard error = standard deviation / square root of sample size = 300/ sqrt(n).
the formula becomes:
1.475791028 = 15 / (300 / sqrt(n))
simplify to get:
1.475791028 = 15 / 300 * sqrt(n)
solve for sqrt(n) to get:
sqrt(n) = 1.475791028 * 300 / 15 = 29.51582055.
when sqrt(n) = 29.51582055, s = 300 / 29.51582055 = 10.16404065.
when s = 10.16404065, the z-score formula becomes 1.475791028 = (x-m) / 10.16404065
solve for (x-m) to get:
(x-m) = 1.475791028 * 10.16404065 = 15.
margin of error = 15 is confirmed to be accurate using the normal distribution calculator at https://www.hackmath.net/en/calculator/normal-distribution
the results are shown below.

the mean was input as 1500.
the standard error was input as 10.16404065.
the square root of n was equal to 29.51582055.
you can see that the margin of error was plus or minus 15 because the 86% confidence interval was from 1485 to 1515.
solve for n to get n = sqrt(n) squared = 871.183663.
round that to the next higher integers to get n = 872.
that would be the minimum sample size for the margin of error to be less than 15.

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