One last thing. At the outset of this discussion I provided the results of a validation procedure on a model (see below). As discussed previously, the model overall seems to fair well, with the exception of the slope. With that in mind, is there a way to correct the coefficients of the model to account for the corrected slope so that future predictions on a new data set are more accurate? Or is that not recommended at all?
index.orig training test optimism index.corrected n Dxy 0.9932 0.9940 0.9905 0.0035 0.9897 363 R2 0.9291 0.9364 0.9163 0.0202 0.9089 363 Intercept 0.0000 0.0000 0.0233 -0.0233 0.0233 363 Slope 1.0000 1.0000 0.7836 0.2164 0.7836 363 Emax 0.0000 0.0000 0.0582 0.0582 0.0582 363 D 0.9118 0.9190 0.8915 0.0275 0.8844 363 U -0.0110 -0.0110 0.0124 -0.0234 0.0124 363 Q 0.9228 0.9299 0.8791 0.0508 0.8720 363 B 0.0205 0.0172 0.0239 -0.0067 0.0272 363 -- View this message in context: http://r.789695.n4.nabble.com/interpreting-bootstrap-corrected-slope-rms-package-tp3928314p3933121.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.