On 14 May 2004 11:14:44 -0700, [EMAIL PROTECTED] (mac55) wrote: > I have a questionnaire with about 20 questions that will result in > ordered categorical variables. I also have about 10 background > questions related to demographics and background. MY outcome is > dichotomous. > > How can I build a logistic regression model with so many ordinal > categorical variables. The amount of dummy variables required (about > 3-4 per question) would make the model unstable. - true, if you don't have (perhaps) thousands of cases. > > My question is what is the best way to proceed with this data. Can > they be entered as continuous variables so long as there is a linear > relationship with the log odds? Or should I just build a model with
I think you mean "monotonic relationship" -- and the answer is "probably yes." If it is a linear relation already, the answer is "Certainly, Yes, Without a doubt, Why are you asking?" > the background questions and adjust the 20 variables with those > factors. Of course this will miss a fair amount of confounding and > interaction. > > I know a lot of this is vague without the clinical > context.....sorry...... If this is a "clinical context," I guess that I have expectations that you are describing a questionnaire with related items, and you probably can say more about individual items: How reliable? related to how many other items? how serious in the literature? Is it feasible to make one or more composite scores, after assuming scale-steps are equal intervals? Ten background questions to control for is also a large number. A composite-score strategy might help with those, too, if they are (something like) known risk-factors. -- Rich Ulrich, [EMAIL PROTECTED] http://www.pitt.edu/~wpilib/index.html . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
