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Re: st: RE: Interpretation of quadratic terms


From   Rosie Chen <jiarongchen2002@yahoo.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: RE: Interpretation of quadratic terms
Date   Mon, 8 Mar 2010 13:26:30 -0800 (PST)

Nick, thank you for the guidance. The model I am estimating is a logistic regression. What I did to check the plot was to save the residual of the model, and then plotted the standardized residual against the predictor. I didn't really found a curve-linear relationship. 

Is there anything wrong with the way I plot the residual? If not, then why the inclusion of a quadratic term actually improves the model fitting when I made a model comparison using the -2log-likelihood? In addition, the nonsignificant predictor in the original form turned to be significant after using the quadratic term? Your further advice would be appreciated.

Rosie








----- Original Message ----
From: Nick Cox <n.j.cox@durham.ac.uk>
To: statalist@hsphsun2.harvard.edu
Sent: Mon, March 8, 2010 1:43:29 PM
Subject: st: RE: Interpretation of quadratic terms

I don't know what kind of guidance you need, but the first step is
surely to plot this curve and think about its substantive interpretation
within the entire range of the data. That should include bringing in
whatever science is behind this analysis. 

Nick 
n.j.cox@durham.ac.uk 

Rosie Chen
    I have a question regarding how to interpret quadratic terms in
regression, and would appreciate your help very much. 

    Because the non-linear nature of the relationship between X and Y; I
need to include quadratic terms in the model. To avoid multicollinearity
problem with the original variable and its quadratic term, I centered
the variable first (X) and then created the square term (Xsq). The model
with the quadratic term (Xsq) was proved to be significantly better.
Suppose the output is like the following (both coefficients are
significant), how to interpret the results? The two signs are opposite.
Could anyone provide some insight? Thank you very much in advance!
--Rosie

y= a + 1.3*X - 0.2*Xsq + e

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