Hi R-Help!
My question: I have lifetime/failure data of machines with different
stress levels and i think an weibull/extreme value distribution would
fit this data. So I did:
model1 - survreg(Surv(lfailure)~stress,data=steel,dist=extreme)
(where lfailure=log(failure))
Now I would like to
Hi R-Help!
My question: I have lifetime/failure data of machines with different
stress levels and i think an weibull/extreme value distribution would
fit this data. So I did:
model1 - survreg(Surv(lfailure)~stress,data=steel,dist=extreme)
(where lfailure=log(failure))
Now I would like to
Hi R-Help!
My question: I have lifetime/failure data of machines with different
stress levels and i think an weibull/extreme value distribution would
fit this data. So I did:
model1 - survreg(Surv(lfailure)~stress,data=steel,dist=extreme)
(where lfailure=log(failure))
Now I would
On Sun, Jan 23, 2005 at 03:10:18PM -0500, [EMAIL PROTECTED] wrote:
Hi R-Help!
My question: I have lifetime/failure data of machines with different
stress levels and i think an weibull/extreme value distribution would
fit this data. So I did:
model1 -
I am still trying to find a common intercept but a different slopes for each
group within my lifetime data. By stratifying the variable stress (the groups)
I get different scale parameters which is not my goal.
So I did this:
survreg(Surv(lfailure)~as.factor(stress),data=steel,dist=extreme)
