SOLUTION: A researcher reports survey results by stating that the standard error of the mean is 20. The population standard deviation is 500. a) How large was the sample used in this surv

Question 1024705: A researcher reports survey results by stating that the standard error of the mean is 20. The population standard deviation is 500.
a) How large was the sample used in this survey?
b) What is the probability that the point estimate was within �25 of the population mean?
Answers:
a) 625
b) 0.7888
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
sd = 500
se = 20
n = sample size

the formula for standard error of the mean is:

se = sd / sqrt(n)

se = 20
sd = 500

formula becomes 20 = 500 / sqrt(n).

solve for sqrt(n) to get sqrt(n) = 500 / 20 = 25.

solve for n to get n = 25^2 = 625.

the formula for z-score is z = (x-m) / se.

(x-m) is equal to plus or minus 25.
se is equal to 20.

the formula for z-score becomes z = plus or minus 25/20.

this results in z-score = plus or minus 1.25.

use a z-score calculator to find that the z-score between -1.25 and 1.25 has a probability of occurrence of .788...

these solutions agree with the answers provided to you.

you could assume the mean is any number.
if the actual score is plus or minus 25 from that, you will get the same result.
this assumes the standard error is 20.

for example:

assume the mean is 5000.
assume the actual score is 4975 or 5025.

the z-score formula becomes z = (5025 - 5000) / 20 = 25/20 = 1.25.

the z-score formula also becomes z = (4975 - 5000) / 20 = -25/20 = -1.25.

the z-wscore table between a z-score of -1.25 and 1.25 will tell you that the probbility of this occuring is .788....

here's a picture of just what i told you.

$$$

this solution depends on the standard error of the mean being 20.

as an additional note, the standard error of the mean is equal to the standard deviation of the distribution of sample means from multiple samples of the same sample size.

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