In article <[EMAIL PROTECTED]>, David C. Howell <[EMAIL PROTECTED]> writes > This morning I sent out the message below. I still have basically > the same question, but when I went back and looked at my resampling > program, I discovered an error. When I corrected that, the > distribution is normal. But I still don't know what is wrong with > my argument that it shouldn't be normal. > (trimmed) > > That would seem to settle the question--b is normally distributed. > > BUT, suppose that we standardized our variables. This is a linear > transformation for both X and Y, so would not change the > correlation nor the shape of distributions. With standardized > variables the slope is equal to the standardized slope (beta). And > it is easy to show that beta, with only one predictor, is equal to > the correlation coefficient. So when we take standardized > variables, which are just linear transformations of unstandardized > variables, r, b, and beta will all be numerically equal. > > Now, we all know that the sampling distribution for r is skewed > when r* (the parameter) is not 0. If we let r* = .60, the skew is > quite noticeable. But if r and beta are equal, then the sampling > distribution of beta will be equal to the sampling distribution of > r, and therefore it will also be skewed. And if the sampling > distribution of beta is skewed, so should be the sampling > distribution of b, because it is simply a linear transformation on > beta. > > So now I have shown that the sampling distribution is skewed, > though I began by quoting experts I respect saying that it is > normal. > > Finally, I used Resampling Stats to empirically generate the > sampling distribution. It is very definitely skewed. > > So where did I go wrong? The sampling distribution of b cannot be > both normal and skewed, at the same time. > In order to standardise the variables, you need to look at the data to work out the constant and scaling factor, which depend on the data. Y'i=cYi+d is linear but Y'i=Yi/stddev(Y)-mean(Y) is not. Linear functions don't give you grief if you present them with Yi=0 for all i.
After you have standardised the data points, they are no longer even independent of each other - since you fix both the mean and variance, if you show me all but two of the data points, I believe that I should be able to work out what the other two are - well almost. I think this boils down to Y1^2+Y2^2=G Y1+Y2=H which is the intersection of a straight line and a circle, so you have 0, 1, or 2 possible solutions. -- A. G. McDowell . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
