Re: [R] p values in coxph()
1) The p values in the printout are a Wald test. The Wald, score, and likelihood ratio tests are asymptotically equivalent, but may differ somewhat in finite samples. (The Wald and score are both Taylor series approximations to the LR). If you want to do an LR test, fit the two models and use the anova command. But beware if your second variable has missing values: the two fits have to be on the same sample. 2) Yes, coxph(Surv(time, status) ~1) is a valid Cox model. Not a particularly interesting one -- it's the LR for the overall fit of the baseline hazard which is equivalent to a Kaplan Meier when there are no covariates. Terry T. ---begin inclusion -- I'm interested in building a Cox PH model for survival modeling, using 2 covariates (x1 and x2). x1 represents a 'baseline' covariate, whereas x2 represents a 'new' covariate, and my goal is to figure out where x2 adds significant predictive information over x1. Ideally, I could get a p-value for doing this. Originally, I thought of doing some kind of likelihood ratio test (LRT), where i measure the (partial) likelihood of the model with just x1, then with x1 and x2, then it becomes a LRT with 1 degree of freedom. But when i use the summary() function for coxph(), i get the following output (shown at the bottom). I have two questions: 1) What exactly are the p-values in the Pr(|z|) representing? I understand that the coefficients have standard errors, etc., but i'm not sure how the p-value there is calculated. 2) At the bottom, where it shows the results of an LRT with 2df, i don't quite understand what model the ratio is being tested against. If the current model has two variables (x1 and x2), and those are the extra degrees of freedom, then the baseline should then have 0 variables, but that's not really a Cox model? thanks for any help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] p values in coxph()
Hi, I'm interested in building a Cox PH model for survival modeling, using 2 covariates (x1 and x2). x1 represents a 'baseline' covariate, whereas x2 represents a 'new' covariate, and my goal is to figure out where x2 adds significant predictive information over x1. Ideally, I could get a p-value for doing this. Originally, I thought of doing some kind of likelihood ratio test (LRT), where i measure the (partial) likelihood of the model with just x1, then with x1 and x2, then it becomes a LRT with 1 degree of freedom. But when i use the summary() function for coxph(), i get the following output (shown at the bottom). I have two questions: 1) What exactly are the p-values in the Pr(|z|) representing? I understand that the coefficients have standard errors, etc., but i'm not sure how the p-value there is calculated. 2) At the bottom, where it shows the results of an LRT with 2df, i don't quite understand what model the ratio is being tested against. If the current model has two variables (x1 and x2), and those are the extra degrees of freedom, then the baseline should then have 0 variables, but that's not really a Cox model? thanks for any help. Brian summary(coxph(Surv(myTime,Event)~x1+x2)) Call: coxph(formula = Surv(myTime, Event) ~ x1 + x2) n= 211 coef exp(coef) se(coef) z Pr(|z|) x1 0.03594 1.03660 0.17738 0.203 0.83942 x2 0.53829 1.71308 0.17775 3.028 0.00246 ** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 exp(coef) exp(-coef) lower .95 upper .95 x1 1.037 0.96470.7322 1.468 x2 1.713 0.58371.2091 2.427 Rsquare= 0.111 (max possible= 0.975 ) Likelihood ratio test= 21.95 on 2 df, p=1.714e-05 Wald test= 20.29 on 2 df, p=3.924e-05 Score (logrank) test = 22.46 on 2 df, p=1.328e-05 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] p-values from coxph?
On Oct 15, 2010, at 9:21 AM, ?hagen Patrik wrote: Dear List, I each iteration of a simulation study, I would like to save the p- value generated by coxph. I fail to see how to adress the p-value. Do I have to calculate it myself from the Wald Test statistic? No. Look at help(coxph.object). This list the components of a coxph object and explains what they are. You will find that fit - coxph(. fit$wald.test contains the Wald test statistic. I prefer the likelihood ratio test myself 2*diff(fit$loglik), with fit$df degrees of freedom. Hunting with str(...) is a good strategy, but even better is the documentation (when it exists). Terry T. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] p-values from coxph?
Dear List, I each iteration of a simulation study, I would like to save the p-value generated by coxph. I fail to see how to adress the p-value. Do I have to calculate it myself from the Wald Test statistic? Cheers, Paddy __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] p-values from coxph?
On Oct 15, 2010, at 9:21 AM, Öhagen Patrik wrote: Dear List, I each iteration of a simulation study, I would like to save the p- value generated by coxph. I fail to see how to adress the p-value. Do I have to calculate it myself from the Wald Test statistic? No. And the most important reason is that would not give you the same value as is print()-ed by coxph(). If you ask for the the str(print(coxph(...)) you get NULL (after the side-effect of prinitng. The print function only produces side- effects. On the other hand you can use the summary function and it gives you a richer set of output. Using the first example on the help page for coxph: str(summary(coxph(Surv(time, status) ~ x + strata(sex), test1))) List of 12 $ call: language coxph(formula = Surv(time, status) ~ x + strata(sex), data = test1) $ fail: NULL $ na.action : NULL $ n : int 7 $ loglik : num [1:2] -3.87 -3.33 $ coefficients: num [1, 1:5] 0.802 2.231 0.822 0.976 0.329 ..- attr(*, dimnames)=List of 2 .. ..$ : chr x .. ..$ : chr [1:5] coef exp(coef) se(coef) z ... $ conf.int: num [1, 1:4] 2.231 0.448 0.445 11.18 ..- attr(*, dimnames)=List of 2 .. ..$ : chr x .. ..$ : chr [1:4] exp(coef) exp(-coef) lower .95 upper .95 $ logtest : Named num [1:3] 1.087 1 0.297 ..- attr(*, names)= chr [1:3] test df pvalue $ sctest : Named num [1:3] 1.051 1 0.305 ..- attr(*, names)= chr [1:3] test df pvalue $ rsq : Named num [1:2] 0.144 0.669 ..- attr(*, names)= chr [1:2] rsq maxrsq $ waldtest: Named num [1:3] 0.95 1 0.329 ..- attr(*, names)= chr [1:3] test df pvalue $ used.robust : logi FALSE So the fifth element of coefficients leaf of the list structure has the same p-value as that print()-ed. Try: summary(fit)$coefficients[5] [1] 0.3292583 (It does seem to me that the name for that leaf of the fit object is not particularly in accord with what I would have considered coefficients., but I am really in no solid position to criticize Terry Therneau to whom we all owe a great deal of gratitude.) -- David. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
