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second
2004 Jun 17
1
Using predict.lm()
...I continue to be puzzled by the
result. I had thought (perhaps erroneously) that lm() would return a model
object that would permit prediction. Indeed:
lm(old$y~old$f/(1+old$x)-1)
...results in:
Call:
lm(formula = old$y ~ old$f/(1 + old$x) - 1)
Coefficients:
old$fFIRST old$fSECOND old$fFIRST:old$x old$fSECOND:old$x
-0.08489 -0.05839 1.15351 0.72981
which clearly provides a model fit for each factor, and identifies the factor
from which each model coefficient was extracted, so lm() does provide the
capability to predict over the...