[EMAIL PROTECTED] (Green Bay Blows) wrote in message news:<[EMAIL PROTECTED]>... > 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??
i hope all will agree!!! the sampling distribution of xbar based on repeated samples of size 4 from the population you mention will have a mean of 100 and a standard deviation (the standard error of the mean) of 5. what will be its shape? well, it depends. if x is normal, so will xbar be normal. if x isn't normal, it gets trickier. . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
