The standard deviation for the sampling distribution of the mean (standard error) is given by SD/sq of n, so if the variance = 100, the SD is (the square root of that) 10. 10/2 (the square-root of 4) gives the standard error of 5, as you've done. I agree that's right.
The mean of the sampling distribution equals mu, the population mean (in this case, 100), by the Central Limit Theorum. That is, if I sample repeatedly from any population and compute means for each of my same-sized samples, the average of those averages will be the population mean, provided I have enough samples. So, SEM = 5 (by computation in accord with definition) Mean = 100 (by the CLT) !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!STAT IS LIFE!!!!!!!!!!!!!!!!!!!!!!!!!!! On 9 Mar 2002 08:17:01 -0800, [EMAIL PROTECTED] (Green Bay Blows) wrote: >Can anyone straighten me out?? > >Suppose a random sample of "n" measurements is selected from a >population with mean u=100 and variance =100. For "n=4" give the mean >and standard deviation of the sampling distribution of the sample mean >(x-bar). > >I can figure out the standard deviation, sigma/sq root of n how does >one arrive at the "mean"?? > >For this problem, the standard deviation of the sampling distribution >would be 10/sq root of 4 or 10/2 = 5. Anyone agree?? . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
