Dear all,
Thank you for your remarks.
The data under analysis were multiply-imputed using Mice.
To compare the nested models, I used the following R codes by van Buuren:
pool.compare (Model2, Model1, method = c("wald"), data = NULL)
As far as I know the Wald statistic tests the null hypothesis that
Yes-- there's no paradox; the adjusted R^2 and deviance are looking
at/testing different things.
Also you don't say *what* deviance you are looking at, but
your interpretation of the deviance is probably wrong.
A significant test for
anova(model2, model1)
says that x3 & x4 add significantly to p
This list is about R programming. Statistics questions, which this is, are
generally off topic here. Try posting on a statistics list like
stats.stackexchange.com instead.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it.
Hello,
I am currently analysed two nested models using the same sample. Both the
simpler model (Model 1 ~ x1 + x2) and the more complex model (Model 2 ~ x1 + x2
+ x3 + x4) yield the same adjusted R-square. Yet the p-value associated with
the deviance statistic is highly significant (p=0.0047),
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