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/ . =================================================================
