Question 1187600: You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately
σ=21.9
. You would like to be 95% confident that your estimate is within 1.2 of the true population mean. How large of a sample size is required?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population standard deviation = 21.9
confidence interval = 95%.
margin of error = plus or minus 1.2.
at 95% confidence interval, your critical z-score will be plus or minus 1.96.
the z-score formula is:
z = (x - m) / s
z is the z-score
x is the sample mean
m is the population or assumed mean
s is the standard error.
the standard error is equal to the standard deviation divided by the square root of the sample size.
replace s with that to get:
z = (x - m) / (sd / sqrt(ss))
sd is the standard deviation
ss is the sample size
you are given that the standard deviation is equal to 21.9, so replace that in the standard error formula to get:
z = (x - m) / (21.9 / sqrt(ss))
(x - m) needs to be plus or minus 1.2 because that's your margin of error.
replace (x - m) with 1.2 to get:
z = 1.2 / (21.9 / sqrt(ss))
since 1.2 is positive, replace z with 1.96 to get:
1.96 = 1.2 / (21.9 / sqrt(ss))
this is equivalent to:
1.96 = 1.2 / 21.9 * sqrt(ss) because division by a fraction is equivalent to multiplication by the reciprocal of that fraction.
you have:
1.96 = 1.2 / 21.9 * sqrt(ss)
multiply both sides of the equation by 21.9 and divide both sides of the equation by 1.2 to get:
1.96 * 21.9 / 1.2 = sqrt(ss)
solve for sqrt(ss) to get:
sqrt(ss) = 1.96 * 21.9 / 1.2 = 35.77
solve for ss to get:
ss = 35.77^2 = 1279.4929.
that's the minimum sample size you need to make sure the margin of error is less than or equal to 1.2.
confirm by replacing sqrt(ss) with 35.77 in the standard equation to get:
standard error = 21.9 / 35.77 = .612244898.
your critical z-score formula is, once again, z = (x - m) / s
with z = 1.96 and with s = .612244898, you get:
1.96 = (x - m) / .612244898
solve for (x - m) to get:
(x - m) = 1.96 * .612244898 = 1.2
that's your margin of error.
on the low side, the z-score formula would become:
-1.96 = -(x - m) / .612244898.
solve for -(x - m) to get:
-(x - m) = -1.96 * .612244898 = 1.2.
your margin of error is also 1.2 on the low side.
this margin of error will work with any mean as long as the standard deviation is always 21.9 and the sample size is 1279.4929.
those two value will always make the standard error equal to .612244898.
for example, assume the mean is 100.
1.96 = (x - 100) / .612244898.
solve for (x - 100) to get:
1.96 * .612244898 = (x - 100).
add 100 to both sides and solve for x to get:
x = 1.96 * .612244898 + 100 = 1.2 + 100 = 101.2
the margin of error is 1.2, regardless of what the mean is, as long as the standard error is equal to .612244898.
the standard error is equal to that as long as the standard deviation is 21.9 and the sample size is 35.77^2 = 1279.4929.
your solution is that the sample size needs to be greater than or equal to 1279.4919 so that the margin of error will be less than or equal to 1.2.
let me know if you have any questions.
theo
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