You have a mean of 500 and a standard deviation of 100, suppose that
many samples of size n are taken from a large population and the mean
is computed for each sample. What is the mean & standard deviation of
the distribution of sample means for n=100 and for n=400.

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Request for Question Clarification by elmarto-ga on 06 Apr 2005 07:43 PDT

Hi jeffstud,
Although the answer the commenter provided is correct, do you still
need the mathematical proof for that answer?

Regards,
elmarto

the mean of the sample mean will always be the mean of the distribution 500

the variance of the sample mean will always be

V(X)/n

or 100/100 = 1 (n=100)

100/400 = 1/4  (n=400)

That would be nice if possible, thanks,

Jeff.

For the sample mean

E(Xbar) = E (1/n * (X1+...+Xn)) = 1/n E (X1+...+Xn) = 1/n * E(X1)+...+ 1/n * E(Xn)
        = 1/n * mu +...+ 1/n * mu = 1/n * mu (1+...+1) (n times) = 1/n * n * mu

so the mean of the sample mean is always the population mean
regardless of the distribution

V(Xbar) = V (1/n * (X1+...+Xn)) = 1/n^2 V (X1+...+Xn) = 1/n^2 V(X1)+...+1/n^2V(X2)
          by independence of the sample 
        = 1/n^2 sigma^2 +...+ 1/n^2 sigma^2 = 1/n^2 *sigma^2 (1+...+1) (n times)
        = 1/n^2 * n * sigma ^2 = sigma^2/n

Thus the variance of the sample mean is always the population variance divided by n

Any thing else you need?