Interpreting ordinal regression result and calculating individual percentage contributions of independent variables I ran an ordinal logistic regression in R using the polr function on a survey analysis dataset. The responses of the dependent variable range from Poor to Excellent.
The responses to the independent variables range from 1 to 5 (1 being Poor and 5 being Excellent). I obtained the following result:

I want to measure the individual percentage contribution of my independents (R1, R2,...,R17) to the dependent variable. 
Is there a way to do this.
Thanks for any help.
 A: Remember that you used maximum likelihood estimation which optimizes the likelihood function.  Use this to your advantage - it is the gold standard statistical measure, with only penalized likelihood or Bayesian analysis (which also uses the likelihood) being better.  Compute the proportion of the total likelihood ratio $\chi^2$ statistic for the model that is due to each variable in the model.  You can do this by running a series of anova() commands.  To get this easily using the more approximate Wald $\chi^2$ statistics instead of likelihood ratio $\chi^2$ statistics, use something like:
require(rms)
f <- orm(y ~ x1 + x2 + ....)  # default = proportional odds model
a <- anova(f)
a   # print table
plot(a)   # plot results on chi-square minus d.f. scale
plot(a, what='proportion chisq')   # plot what was described above

A: There is no general way to do this because the contribution of each variable will be different at different levels of the other variables. This is true in any logistic regression.  But you can look at the contribution at particular levels of the other variables. In SAS there is a slice statement that does this. There is probably something similar in polr but you'd have to look at the documentation. 
