[R] understanding the output

[Ronaldo Reis Jr.]

Em Quinta 20 Junho 2002 09:11, Peter Dalgaard BSA escreveu:
> "Ronaldo Reis Jr." <chrysopa at insecta.ufv.br> writes:
> >             Estimate Std. Error t value Pr(>|t|)
> > (Intercept)   4.8660     0.9808   4.961 0.000101 ***
> > gen2g         3.8504     1.3870   2.776 0.012459 *
> > ---
>
> ...
>
> > > tapply(peso,gen,mean)
> >
> >     1g     2g
> > 4.8660 8.7164
> >
> > The values is the same in estimate.
> > But the SE, when I try using tapply the SE is different os the SE of
> > estimates. look
> >
> > > tapply(peso,gen,sd)/sqrt(tapply(peso,gen,length))
> >
> >        1g        2g
> > 0.8665567 1.0829789
> >
> > Why this difference? What SE I can use with mean???
>
> (I assume you are only looking at the intercept, i.e. the mean of
> group "1g"). glm() assumes a constant SD and uses the pooled SD.
>
> Try for illustration:
>
> sds <- tapply(peso,gen,sd)
> lens <- tapply(peso,gen,len)
> dfs <- lens - 1
> sd.pool <- sqrt(sum(sds^2*dfs)/sum(dfs))
> sd.pool/sqrt(lens)
>
> It's a matter of taste which SE you'd put on a plot. Some people feel
> that you "shouldn't mix in the other groups" in a mainly descriptive
> plot, but as soon as any modelling is involved (e.g., age-adjusted
> means) you pretty much have to extract SE's from the model.

Its Ok,
I choice to use the model SE. So, my table of mean +- SE maybe:

trat	mean	SE
1g	4.8660	0.9808
2g	8.7164	1.3870

or

trat	mean	SE
1g	4.8660	0.9808
2g	8.7164	1.3870+0.9808

or 

trat	mean	SE
1g	4.8660	0.9808
2g	8.7164	0.9808

Or neither??

I'm a bit confused

And for poisson errors? The mean to use is antilog of estimate, and the 
associated error, is the glm calculated or need to be transformed?

Sorry by these newbie questions :))

Thanks
Ronaldo
-- 
[It is] best to confuse only one issue at a time.
		-- K&R
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