Dick Startz <[EMAIL PROTECTED]> wrote in news:[EMAIL PROTECTED]:
> The suggestion below is the right idea, but I think is off in a > detail. Doing least squares we take the partial derivatives of the sum > of (y-b1-b2*x)^2 with respect to b1 and b2. I think the derivative > w.r.t x is given below by mistake. Yep, I realized that a little while after posting. Using the derivative with respect to b1 (2*(b1+b2*x-y)) works, though, since the subexpression that needs to be zero is the same.' . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
