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# Re: st: Standardized interaction terms - which p-values hold?

 From David Hoaglin To statalist@hsphsun2.harvard.edu Subject Re: st: Standardized interaction terms - which p-values hold? Date Wed, 16 Jan 2013 09:10:16 -0500

```As Nick pointed out, we can ignore the standard deviations (i.e., the
scaling).  When the model includes an interaction term, what matters
is the centering.

It may help to look at a model in which the predictors are two
quantitative variables and their interaction (product).  With no
centering (and omitting the error term) we can write the model as
Y = b0 + b1*X1 + b2*X2 + b3*(X1*X2).
If we center X1 at m1 and X2 at m2 (not necessarily their means), we
can write the model as
Y = c0 + c1*(X1 - m1) + c2*(X2 - m2) + c3*[(X1 - m1)*(X2 - m2)].
Multiplying out the predictors in the centered model, we see that c3 =
b3.  The other three coefficients differ between the two models:
b2 = c2 - m1*c3
b1 = c1 - m2*c3
b0 = c0 - m1*c1 - m2*c2 + m1*m2*c3.

The interpretations of the coefficients change accordingly.  For
example, b1 summarizes the change in Y per unit change in X1,
adjusting for simultaneous change in X2 and X1*X2; and c1 summarizes
the change in Y per unit change in X1 (which is the same as change in
X1 - m1), adjusting for simultaneous change in X2 and (X1 - m1)*(X2 -
m2).

In her initial message Elisabeth said that she needs to compare the
size of two interactions.  In practice, what does that mean?  Is it
simply the size of the two coefficients, or should one look at the
contribution of each interaction to the R^2 for the model?  Unless the
two interactions are orthogonal to each other, each of those
contributions would come from comparing the model that includes that
interaction against  the model that omits that interaction (but
includes the other interaction).  The same problem arises in trying to
say how much each predictor contributes to the regression.

David Hoaglin
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