Question 1197340
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


<img src = "http://theo.x10hosting.com/2022/102001.jpg">


second one is with sample size = 218


<img src = "http://theo.x10hosting.com/2022/102002.jpg">


your answer should be a sample size of at least 218 to get a margin of error less than 30.