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Question 1177422: A survey reports it results by starting that standard error of the mean to be is 20.
The population standard deviation is 500.
1:How large is the sample used in this surveying?
2: What is the probability that the sample mean will be within 25 of the population mean?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! standard error is equal to population standard deviation divided by square root of sample size.
the formula is s = psd / sqrt(ss).
s is the standard error.
psd is population standard deviation.
ss is sample size.
when s = 20 and psd = 500, the formula becomes:
20 = 500 / sqrt(ss)
solve for sqrt(ss) to get:
sqrt(ss) = 500 / 20 = 25
solve for ss to get:
ss = 25^2 = 625
that's the sample size.
the formula for z-score is:
z = (x - m) / s
z is the z-score
x is the raw score
m is the raw mean
s is the standard error.
when (x - m) = 25, and s = 20, the formula becomes:
z = 25 / 20 = 1.25
since the normal distribution curve is symmetric bout the mean, then the confidence interval is z-score of -1.25 to 1.25.
the probability is the area under the normal distribution curve between those 2 z-scores.
that probability is .7887003221.
the probability that the sample mean will be within 25 of the population mean is .7887003221.
this works regardless of the mean.
for example:
assume the mean is 1500.
the z-score formula becomes plus or minus 1.25 = (x - 1500) / 20
solve for the raw score to get:
on the high side, x = 20 * 1.25 + 1500
on the low side, x = 20 * -1.25 + 1500
since 20 * 1.25 is always equal to 25, then you get:
x = 1500 - 25 or x = 1500 + 25.
the sample mean will always be within 25 of the population mean if the z-score is plus or minus 1.25 and the standard error is 20, regardless of what the mean is.
standard error = psd / sqrt(ss)
with a sample size of 625 and a psd of 500, this becomes:
standard error = 500 / sqrt(625) = = 500 / 25 = 20
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