similar to: p-value from survreg, library(survival)

dear r experts: It seems my message got spam filtered, another try: i would appreciate advice on how to get the p-value from the object 'sr' created with the function survreg() as given below. vlad sr<-survreg(s~groups, dist="gaussian") Coefficients: (Intercept) groups -0.02138485 0.03868351 Scale= 0.01789372 Loglik(model)= 31.1 Loglik(intercept only)= 25.4

The question was how to get the p-value from the fit below, as an S object sr<-survreg(s~groups, dist="gaussian") Coefficients: (Intercept) groups -0.02138485 0.03868351 Scale= 0.01789372 Loglik(model)= 31.1 Loglik(intercept only)= 25.4 Chisq= 11.39 on 1 degrees of freedom, p= 0.00074 n= 16 ---- In general, good places to start are > names(sr) >

Hi all. Belove is the example from the cluster-help page wtih the output. I simply cannot figure out how to relate the estimate and robust Std. Err to the p-value. I am aware this a marginal model applying the sandwich estimator using (here I guess) an emperical (unstructered/exchangeable?) ICC. Shouldent it be, at least to some extend, comparable to the robust z-test, for rx :

Hi, Is it normal to get intercept in the list of covariates in the output of survreg function with standard error, z, p.value etc? Does it mean that intercept was fitted with the covariates? Does Value column represent coefficients or some thing else? Regards, ------------------------------------------------- tmp = survreg(Surv(futime, fustat) ~ ecog.ps + rx, ovarian,

Hi, I would like to fit parametric survival models to time-to-event data that are left truncated. I have checked the help page for survreg and looked in the R-help archive, and it appears that the R function survreg from the survival library (version 2.16) should allow me to take account of left truncation. However, when I try the command

On Wed, 13 Nov 2002, Jan de Leeuw wrote: > > No problemo. And, in fact, I get the same results in > the R-1.6.0 Carbon version. I don't. Could there be a G3/G4 issue? -thomas > --- Jan > > On Wednesday, November 13, 2002, at 02:05 PM, tim@timcohn.com wrote: > > > Full_Name: Tim Cohn > > Version: 1.6.1 > > OS: Macintosh OS X > > Submission

Full_Name: Tim Cohn Version: 1.6.1 OS: Macintosh OS X Submission from: (NULL) (130.11.34.250) The Mac version of survreg does not handle left-censored data correctly (at least the results are not what I get doing it other ways, and they are not the same as I get running R 1.6.1 in Windows 98se; the Windows 98 results are correct). On the windows version of R 1.6.1. >

Thank you for looking into this so quickly. As you correctly surmise, I was using the Carbon version of R-1.6.1 on Mac OS 10.2.2 (Jaguar) when I got the "wrong" answers. One other observation: The right censoring seems to work fine. Thanks again, Tim On Thursday, November 14, 2002, at 11:09 AM, Jan de Leeuw wrote: > I take that back. I now get the "correct" result

---------------------------- Original Message ---------------------------- Subject: Different results from survreg with version 2.6.1 and 2.10.1 From: nathalcs at ulrik.uio.no Date: Fri, March 19, 2010 16:00 To: r-help at r-project.org -------------------------------------------------------------------------- Dear all I'm using survreg command in package survival.

Hi everybody. I'm trying to fit a weibull survival model with a spline basis for the predictor, using the survival library. I've noticed that it doesn't seem to be possible to use the aic method to choose the degrees of freedom for the spline basis in a parametric regression (although it's fine with the cox model, or if the degrees of freedom are specified directly by the user),

What objective function is maximized by survreg with the default Weibull model? I'm getting finite parameters in a case that has the likelihood maximzed at Infinite, so it can't be a simple maximum likelihood. Consider the following: ############################# > set.seed(3) > Stress <- rep(1:3, each=3) > ch.life <- exp(9-3*Stress) > simLife <- rexp(9,

Hi All, I have conducted the following survival analysis which appears to be OK (thanks BRipley for solving my earlier problem). > surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset, dist="weibull", scale = 1) > summary(surv.mod1) Call: survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset, dist = "weibull", scale = 1)

Full_Name: Andrew Manners Version: 2.5.1 OS: windows xp prof 2003 Submission from: (NULL) (130.102.0.177) To whom it may concern, I'm trying to get an ANOVA table within survreg but it always produces NA's in the p-value, regardless of the data set. The data set below comes from Tableman and Kim 2004. I had the same problem on a number of my own data sets. I searched the R site for

Dear R-users, I need to use the aftreg function in package 'eha' to estimate failure times for left truncated survival data. Apparently, survreg still cannot fit such models. Both functions should be fitting the accelerated failure time (Weibull) model. However, as G?ran Brostr?m points out in the help file for aftreg, the parameterisation is different giving rise to different

Dear list, I would like to simulate individual survival times from a model that has been fitted using the survreg procedure (library survival). Output shown below. My plan is to extract the shape and scale arguments for use with rweibull() since my error terms are assumed to be Weibull, but it does not make any sense. The mean survival time is easy to predict, but I would like to simulate

I was wondering if someone familiar with survival analysis can help me with the following. I would like to fit a Weibull curve, that may be dependent on a covariate, my dataframe "labdata" that has the fields "cov", "time", and "censor". Do I do the following? wieb<-survreg(Surv(labdata$time, labadata$censor)~labdata$cov,

Hello, I have a question regarding the use of an offset term with survreg(), in the Survival library. In particular, I am trying to figure out on what scale the offset term should be. Here's a simple example with no censoring and no coefficients: --------- y = rlnorm(1000, meanlog = 10, sdlog = 2) delta = rep(1, 1000) int = rep(1, 1000) survreg(Surv(y,delta)~offset(10*int), dist =

Hi, I am trying to fit the following model: sr.reg.s4.nore <- survreg(Surv(age_sym4,sym4), as.factor(lifedxm), data=bip.surv) Where age_sym4 is the age that a subject develops clinical thought problems; sym4 is whether they develop clinical thoughts problems (0 or 1); and lifedxm is mother's diagnosis: BIPOLAR, MAJOR DEPRESSION, or CONTROL. I am interested in whether or not

Dear R Listers- I am attempting to analyse the survival of seeds in cages (exclosures) that differ in their permeability to rainforest mammals. Because I did not observe the moment of seed disappearance, my data is interval censored. This limits my options for analysis (as I understand it) to survreg, in the survival package. Because I repeated the experiment in 8 sites, I have a random

Hello all, I want to extract the survival function estimate from a model fitted by survreg(). Using predict.survreg(..., type="quantile", p=seq(0,1,0.001)), gives the quantiles, which I managed to turn around into a survival function estimate (Prob{T > t} as function of t). Is there a more straightforward way of doing this? I have had difficulties using pweibull() with the