similar to: test of hazard ratios

Hi, I am computing the Nelson-Aalen (NA) estimate of baseline cumulative hazard in two different ways using the "survival" package. I am expecting that they should be identical. However, they are not. Their difference is a monotonically increasing with time. This difference is probably not large to make any impact in the application, but is annoyingly non-trivial for me to just

Dear R help forum, I am using the function 'coxph' to obtain hazard ratios for the comparison of a standard treatment to new treatments. This is easily obtained by fitting the relevant model and then calling exp(coef(fit1)) say. I now want to obtain the hazard ratio for the comparison of two non-standard treatments. >From a statistical point of view, this can be achieved by dividing

Dear R-helpers, I try to perform glm's with negative binomial distributed data. So I use the MASS library and the commands: model_1 = glm.nb(response ~ y1 + y2 + ...+ yi, data = data.frame) and predict(model_1, newdata = data.frame) So far, I think everything should be ok. But when I want to perform a glm with a subset of the data, I run into an error message as soon as I want to predict

Hi, I wish to analyze with R the results of a perception experiment in which subjects had to recognize each stimulus among three choices (this was a forced-choice design). The experiment runs under two different conditions and the data is like the following: N1 : count of trials in condition 1 p11, p12, p13: proportions of choices 1, 2, and 3 in condition 1 N2 : count of trials in

Dear R users, I am using lme and nlme to account for spatially correlated errors as random effects. My basic question is about being able to correct F, p, R2 and parameters of models that do not take into account the nature of such errors using gls, glm or nlm and replace them for new F, p, R2 and parameters using lme and nlme as random effects. I am studying distribution patterns of 50 tree

Hi, I want to do a global likelihood ratio test for the proportional odds logistic regression model and am unsure how to go about it. I am using the polr() function in library(MASS). 1. Is the p-value from the likelihood ratio test obtained by anova(fit1,fit2), where fit1 is the polr model with only the intercept and fit2 is the full polr model (refer to example below)? So in the case of the

System: R 2.3.1 on Windows XP machine. I am building a logistic regression model for a sample of 100 cases in dataframe "d", in which there are 3 binary covariates: x1, x2 and x3. ---------------- > summary(d) y x1 x2 x3 0:54 0:50 0:64 0:78 1:46 1:50 1:36 1:22 > fit <- glm(y ~ x1 + x2 + x3, data=d, family=binomial(link=logit)) >

Greetings, I am new to R and have been comparing CPH survival analysis hazard ratios between R and SAS PhReg. The binary covariates' HRs are the same, however the continuous variables, for example age, have quite different HRs although in the same direction. SAS PhReg produces HRs which are the change in risk for every one increment change in the independent variable. How do I

Hello, I am running R 2.6.1 on windows xp I am trying to fit a cox proportional hazard model with a shared Gaussian frailty term using coxme My model is specified as: nofit1<-coxme(Surv(Age,cen1new)~ Sex+bo2+bo3,random=~1|isl,data=mydat) With x1-x3 being dummy variables, and isl being the community level variable with 4 levels. Does anyone know if there is a way to get the standard error

Hi, I am new to time-dependent Cox model to estimate time dependent hazard ratios. Let me use aml dataset from survival package: > aml3<-survSplit(aml2,cut=c(5,10,20),end="time",start="start", event="status",episode="i") If I want to esimate hazard ratio for each of the time intervals 0-5, 5-10, 10-20 and >=20, would the following calculate

Dear R users, Could somebody please help me to find a way of comparing nonlinear, non-nested models in R, where the number of parameters is not necessarily different? Here is a sample (growth rates, y, as a function of internal substrate concentration, x): x <- c(0.52, 1.21, 1.45, 1.64, 1.89, 2.14, 2.47, 3.20, 4.47, 5.31, 6.48) y <- c(0.00, 0.35, 0.41, 0.49, 0.58, 0.61, 0.71, 0.83, 0.98,

Hello, Can anyone suggest a simple way to generate a Kaplan-Meier plot with 2 survfit objects, just like this one:? https://drive.google.com/file/d/1fEcpdIdE2xYtA6LBQN9ck3JkL6-goabX/view?usp=sharing Suppose I have 2 survfit objects: fit1 is for the curve on the left (survtime has been truncated to the cutoff line: year 5), fit2 is for the curve on the right (minimum survival time is at the

Hi, I am trying to do logistic regression for data of 104 patients, which have one outcome (yes or no) and 15 variables (9 categorical factors [yes or no] and 6 continuous variables). Number of yes outcome is 25. Twenty-five events and 15 variables mean events per variable is much less than 10. Therefore, I tried to analyze the data with penalized regression method. I would like please some of the

Thanks for the reproducable example. I can confirm that it fails on my machine using survival 2-37.5, the next soon-to-be-released version, The issue is with NextMethod, and my assumption that the called routine inherited everything from the parent, including the environment chain. A simple test this AM showed me that the assumption is false. It might have been true for Splus. Working this

Dear R-Helpers, I am a novice in survival analysis. I have the following code: for (i in 3:12) print(coxph(Surv(time, status)~a[,i], data=a)) I used it to fit the Cox Proportional Hazard models separately for every available parameter (columns 3:12) in my data set - with intention to compare the Hazard Ratios. However, some of my variables are in range 0.1 to 1.6, others in range 5000 to

Hello all, I wish to extract the terms from an rpart object. Specifically, I would like to be able to know what is the response variable (so I could do some manipulation on it). But in general, such a method for rpart will also need to handle a "." case (see fit2) Here are two simple examples: fit1 <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) fit1$call fit2 <-

You also need to reply-all so the mailing list stays in the loop. -- Sent from my phone. Please excuse my brevity. On December 19, 2017 4:00:29 PM PST, Timothy Axberg <axbergtimothy at gmail.com> wrote: >Sorry about that. Here is the code typed directly on the email. > >qe = (Qmax * Kl * ce) / (1 + Kl * ce) > >##The data >ce <- c(15.17, 42.15, 69.12, 237.7, 419.77)

Hello R users, I've used the following help two compare two regression line slopes. Wanted to test if they differ significantly: Hi, I've made a research about how to compare two regression line slopes (of y versus x for 2 groups, "group" being a factor ) using R. I knew the method based on the following statement : t = (b1 - b2) / sb1,b2 where b1 and b2 are the two slope

I try to send this message To Gordon Smyth at smyth at vehi,edu.au but it bounced back, so here it is to r-help I am trying to use limma, just downloaded it from CRAN. I use R 2.0.1 on Win XP see the following: > library(RODBC) > chan1 <- odbcConnectExcel("D:/Data/mgc/Chips/Chips4.xls") > dd <- sqlFetch(chan1,"Raw") # all data 12000 > # > nzw <-

I have the following results for an ANOVA comparing two nested models. I wasn't sure how I am supposed to report this result in the area of psychology. Specifically, am I supposed to report the DF's or just the F ratio? I could manually calculate the degrees of freedoms, but there must be a reason why R does not give this information, i.e. those are not conventionally used in the