On Mon, 6 Oct 2003, albinali wrote:
> I am trying to do a regression on a data set with a fixed intercept
> at some number (75) other than zero, the problem is that the package I
> am using can fix the intercept only at the origin.
I take it you have a theoretical model of the form
y = 75 + B*x
where B is the slope parameter to be estimated. (If you actually have
multiple predictors x1, x2, ..., these same comments apply, mutatis
mutandis).
You would do better (than you propose) by carrying out a standard linear
regression to estimate A and B in the model
y = A + B*x
(yielding estimates a and b in the result y^ = a + b*x )
and then ask whether a = 75. If "a" is close to 75, your estimate of B
(which I take it is what you really want) will be superior to the
estimate arising from the fixed-intercept procedure. If "a" is far from
75, you then need to worry about WHY that is.
The thing is, "a" is the value predicted for y when x=0, and "x=0" is
often a condition not met in the data set, perhaps not even possible to
encounter "in real life", and the "specialness" of that condition may
have unanticipated effects on the behavior of the function f
(as in "y = f(x)") in the neighborhood of x=0. E.g., that function may
be, unbeknownst to you or anyone else, quite reasonably linear in x for
values of x distant from zero, but containing a step function between
x=0 and x=<eps> for some small value <eps>; should this be the case,
ordinary linear regression will give you a good estimate of the slope in
the region where the function IS linear; and regression through a
forced intercept will misrepresent the slope _everywhere_.
< snip, details of forcing through a non-zero intercept >
> ... Is that valid?
Might be, might not be. See above, and references cited by another
respondent.
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Donald F. Burrill [EMAIL PROTECTED]
56 Sebbins Pond Drive, Bedford, NH 03110 (603) 626-0816
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