On Mon, 27 Jun 2005, Göran Broström wrote:
On Sun, Jun 26, 2005 at 11:28:04PM +0100, Prof Brian Ripley wrote:
Here gamma is the usual gamma function, see ?gamma. (I notice in the R
documentation of the Weibull distribution that "E(X) = b Gamma(1+1/a)",
which is an error; the G should be g (lowe
On Sun, Jun 26, 2005 at 11:28:04PM +0100, Prof Brian Ripley wrote:
> >Here gamma is the usual gamma function, see ?gamma. (I notice in the R
> >documentation of the Weibull distribution that "E(X) = b Gamma(1+1/a)",
> >which is an error; the G should be g (lowercase).)
>
> It is not an error. R's
> Here gamma is the usual gamma function, see ?gamma. (I notice in the R
> documentation of the Weibull distribution that "E(X) = b Gamma(1+1/a)",
> which is an error; the G should be g (lowercase).)
It is not an error. R's function gamma() is said to implement
\eqn{\Gamma}{Gamma}: `see ?gamma'!
On Fri, Jun 24, 2005 at 11:27:28AM -0400, [EMAIL PROTECTED] wrote:
> Hi,
> I was wondering if someone can help me
> interpret the results of running
> weibreg.
>
> I run the following and get the
> following R output.
> > weibreg(Surv(time, censor)~covar)
> fit$fail = 0
> Call:
> weibreg(formu
Hi,
I was wondering if someone can help me
interpret the results of running
weibreg.
I run the following and get the
following R output.
> weibreg(Surv(time, censor)~covar)
fit$fail = 0
Call:
weibreg(formula = Surv(time,
censor)~covar)
Covariate Mean Coef
Rel.Risk L-R p
