>>>>> Sorkin, John <jsor...@som.umaryland.edu>
>>>>> on Sun, 22 Oct 2017 22:56:16 +0000 writes:
> David,
> Thank you for responding to my post.
> Please consider the following output (typeregional is a factor having two
levels, "regional" vs. "general"):
> Call:
> glm(formula = events ~ type, family = poisson(link = log), data = data,
> offset = log(SS))
> Deviance Residuals:
> Min 1Q Median 3Q Max
> -43.606 -17.295 -4.651 4.204 38.421
> Coefficients:
> Estimate Std. Error z value Pr(>|z|)
> (Intercept) -2.52830 0.01085 -233.13 <2e-16 ***
> typeregional 0.33788 0.01641 20.59 <2e-16 ***
> Let's forget for a moment that the model is a Poisson regression and
pretend that the output is from a simple linear regression, e.g. from lm.
> To get the estimate for "general" one simply needs to use the value of
the intercept i.e. -2.5830. Similarly to get the 95% CI of general one simply
needs to compute -2.52830-(1.96*0.01085) and -2.52830+(1.96*0.01085).
I'm pretty sure you can just use (the base R) functions
dummy.coef()
or
model.tables()
possibly with SE=TRUE to get coefficients for all levels of a factor..
I'd like to have tried to show this here, but for that we'd have
wanted to see a "MRE" or "ReprEx" (minimal reproducible example) ..
> To get the estimate for "regional" one needs to compute intercept +
typeregional, i.e. -2.52830 + 0.33788. To get the 95% CI is somewhat more
difficult as one needs to use results from the variance-covariance matix,
specifically the variance of intercept, the variance of "regional", and the
covariance of (intercept,"regional") which involves:
> var = var(intercept) + var(regional) +2*(covar(intercept,regional)),
> and then get the SE of the variance
> SE=sqrt(var)
> 95% CI = intercept + regional - 1.95*SE and intercept + regional +
1.95*SE.
> I was hoping that a contrast statement could be written that would give
me the point estimate and SE for "general" and its SE and another contrast
statement could be written that would give me the point estimate and SE for
"general" and it SE without my having to work directly with the
variance-covariance matrix. I tried doing this using the fit.contrast
statements (from the gmodels package):
> fit.contrast(model,type,c(1,0),showall=TRUE)
> fit.contrast(model,type,c(0,1),showall=TRUE)
> and received the error message,
> Error in `[[<-`(`*tmp*`, varname, value = c(0, 1)) :
> no such index at level 1
> Perhaps fit.contrast is not the way to accomplish my goal. Perhaps my
goal can be accomplished without a contrast statement, but I don't know how.
My guess is that "standard R" aka "base R" would be
sufficient to get what you'd want, notably if you'd consider
using se.contrast() additionally.
Martin
> Thank you,
> John
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
