manchester.ac.uk> writes:
>
> On 07-Aug-10 09:29:41, Michael Bedward wrote:
> > Thanks for that clarification Peter - much appreciated.
> >
> > Is there an R function that you'd recommend for calculating
> > more valid CIs ?
> > Michael
>
> It depends on what you want to mean by "more valid"!
Michael Bedward wrote:
> On 7 August 2010 19:56, Martin Maechler wrote:
>
>> I'm coming late to the thread,
>> but it seems that nobody has yet given the advice which I would
>> very *strongly* suggest to anyone asking for confidence
>> intervals in GLMs:
>>
>> Use confint()
>
> confint was actu
On 07-Aug-10 09:29:41, Michael Bedward wrote:
> Thanks for that clarification Peter - much appreciated.
>
> Is there an R function that you'd recommend for calculating
> more valid CIs ?
> Michael
It depends on what you want to mean by "more valid"! If you have
a 95% CI for the linear predictor (
On 7 August 2010 19:56, Martin Maechler wrote:
> I'm coming late to the thread,
> but it seems that nobody has yet given the advice which I would
> very *strongly* suggest to anyone asking for confidence
> intervals in GLMs:
>
> Use confint()
confint was actually mentioned in the second post on
> "PD" == Peter Dalgaard
> on Sat, 07 Aug 2010 10:37:49 +0200 writes:
PD> Michael Bedward wrote:
>>> I was aware of this option. I was assuming it was not ok to do fit +/-
1.96
>>> se when you requested probabilities. If this is legitimate then all the
>>> better.
Thanks for that clarification Peter - much appreciated.
Is there an R function that you'd recommend for calculating more valid CIs ?
Michael
On 7 August 2010 18:37, Peter Dalgaard wrote:
>
> Probably, neither is optimal, although any transformed scale is
> asymptotically equivalent. E.g., neith
Michael Bedward wrote:
>> I was aware of this option. I was assuming it was not ok to do fit +/- 1.96
>> se when you requested probabilities. If this is legitimate then all the
>> better.
>
> I don't think it is. I understood that you should do the calculation
> in the scale of the linear predi
> Stefano
>
> -Messaggio originale-
> Da: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org]per conto di Troy S
> Inviato: Friday, August 06, 2010 6:31 PM
> A: Michael Bedward
> Cc: r-help@r-project.org
> Oggetto: Re: [R] Confidence Intervals for
> I was aware of this option. I was assuming it was not ok to do fit +/- 1.96
> se when you requested probabilities. If this is legitimate then all the
> better.
I don't think it is. I understood that you should do the calculation
in the scale of the linear predictor and then transform to
proba
rgument at the end of the
page.
Stefano
-Messaggio originale-
Da: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org]per conto di Troy S
Inviato: Friday, August 06, 2010 6:31 PM
A: Michael Bedward
Cc: r-help@r-project.org
Oggetto: Re: [R] Confidence Intervals for logistic reg
Michael,
Thanks for the reply. I believe Aline was sgiving me CI's on coefficients
as well.
So c(pred$fit + 1.96 * pred$se.fit, pred$fit - 1.96 *
pred$se.fit) gives me the CI on the logits if I understand correctly? Maybe
the help on predict.glm can be updated.
Thanks!
On 6 August 2010 01:46,
Sorry about earlier reply - didn't read your email properly (obviously :)
You're suggestion was right, so as well as method for Aline below,
another way of doing the same thing is:
pred <- predict(y.glm, newdata= something, se.fit=TRUE)
ci <- matrix( c(pred$fit + 1.96 * pred$se.fit, pred$fit - 1.
Dear Troy,
use this commend, your will get IC95% and OR.
logistic.model <- glm(formula =y~ x1+x2, family = binomial)
summary(logistic.model)
sum.coef<-summary(logistic.model)$coef
est<-exp(sum.coef[,1])
upper.ci<-exp(sum.coef[,1]+1.96*sum.coef[,2])
lower.ci<-exp(sum.coef[,1]-1.96*sum.coef[,2])
Dear UseRs,
I have fitted a logistic regression using glm and want a 95% confidence
interval on a response probability. Can I use
predict(model, newdata, se.fit=T)
Will fit +/- 1.96se give me a 95% of the logit? And then
exp(fit +/- 1.96se) / (exp(fit +/- 1.96se) +1) to get the probabilities?
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