similar to: anova.lm fails with test="Cp"

Hello to the whole group. I am a newbie to R, but I got my way through and think it is a lot easier to handle than other software packages (far less clicks necessary). However, I have a problem with respect to the summary of regression results. The summary function gives sth like: Residuals: Min 1Q Median 3Q Max -0.46743 -0.09772 0.01810 0.11175 0.42252

Hello: I am trying to understand the method 'hatvalues(...)', which returns something similar to the diagonals of the plain vanilla hat matrix [X(X'X)^(-1)X'], but not quite.  A Fortran programmer I am not, but tracing through the code it looks like perhaps some sort of correction based on the notion of 'leave-one-out' variance is being applied. Whatever the

Hello dear R help members. I am trying to understand the anova.rq, and I am finding something which I can not explain (is it a bug?!): The example is for when we have 3 nested models. I run the anova once on the two models, and again on the three models. I expect that the p.value for the comparison of model 1 and model 2 would remain the same, whether or not I add a third model to be compared

I am trying to find methods for testing and visualizing calibration to Cox models with time-depended coefficients. I have read this nice article <http://journals.sagepub.com/doi/10.1177/0962280213497434>. In this paper, we can fit three models: fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) p <- log(predict(fit0, newdata = data1, type = "expected")) lp

First, as others have said please obey the mailing list rules and turn of First, as others have said please obey the mailing list rules and turn off html, not everyone uses an html email client. Here is your code, formatted and with line numbers added. I also fixed one error: "y" should be "status". 1. fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) 2. p

Dear List, After including cluster() option the coxreg (from eha package) produces results slightly different than that of coxph (from survival) in the following time-dependent treatment effect calculation (example is used just to make the point). Will appreciate any explaination / comment. cheers, Ehsan ############################ require(survival) require(eha) data(heart) # create weights

Dear R Helpers, I have noticed that when I use lmer to analyse data, the summary function gives different values for the AIC, BIC and log-likelihood compared with the anova function. Here is a sample program #make some data set.seed(1); datx=data.frame(array(runif(720),c(240,3),dimnames=list(NULL,c('x1','x2','y' )))) id=rep(1:120,2); datx=cbind(id,datx) #give x1 a

Hello, All: ????? The default "simulate" method for lm and glm seems to ignore the sampling variance of the parameter estimates;? see the trivial lm and glm examples below.? Both these examples estimate a mean with formula = x~1.? In both cases, the variance of the estimated mean is 1. ??? ??????? * In the lm example with x0 = c(-1, 1), var(x0) = 2, and

Dear r-helpers, Spencer Graves and Manual Morales proposed the following methods to simulate p-values in lme4: ************preliminary************ require(lme4) require(MASS) summary(glm(y ~ lbase*trt + lage + V4, family = poisson, data = epil), cor = FALSE) epil2 <- epil[epil$period == 1, ] epil2["period"] <- rep(0, 59); epil2["y"] <- epil2["base"]

On 2019-12-27 04:34, Duncan Murdoch wrote: > On 26/12/2019 11:14 p.m., Spencer Graves wrote: >> Hello, All: >> >> >> ? ????? The default "simulate" method for lm and glm seems to ignore the >> sampling variance of the parameter estimates;? see the trivial lm and >> glm examples below.? Both these examples estimate a mean with formula = >>

Folks: I'm having trouble doing a forward variable selection using step() First, I fit an initial model: fit0 <- glm ( est~1 , data=all, subset=c(n>=25) ) then I invoke step(): fit1 <- step( fit0 , scope=list(upper=est~ pcped + pchosp + pfarm ,lower=est~1)) I get the error message: Error in eval(expr, envir, enclos) : invalid 'envir' argument I looked at the

Hi everybody, I'm using the method described here to make a linear regression: http://www.apsnet.org/education/advancedplantpath/topics/Rmodules/Doc1/05_Nonlinear_regression.html > ## Input the data that include the variables time, plant ID, and severity > time <- c(seq(0,10),seq(0,10),seq(0,10)) > plant <- c(rep(1,11),rep(2,11),rep(3,11)) > > ## Severity

Dear R developers, I noticed a bug in the stats4 package, specifically in the confint method applied to ?mle? objects. In particular, when some ?fixed? parameters define the log likelihood, these parameters are stored within the mle object but they are not used by the ?confint" method, which retrieves their value from the global environment (whenever they still exist). Sample code: >

Dear all, I've replicated the cheese tasting example on p175 of GLM's by McCullagh and Nelder. This is a 4 treatment (rows) by 9 ordinal response (cols) table. Here's my simple code: #### cheese library(MASS) options(contrasts = c("contr.treatment", "contr.poly")) y = c(0,0, 1, 7, 8,8,19, 8,1, 6,9,12,11, 7,6, 1, 0,0, 1,1, 6, 8,23,7,

Hi , I want to apply the following code fo my data with 400 predictors. I was wondering if there ia an alternative way instead of typing 400 predictors for the following code. I really appreciate your help. fit0<-lm(Y~1, data= mydata) fit.final<- lm(Y~X1+X2+X3+.....+X400, data=mydata) ??? step(fit0, scope=list(lower=fit0, upper=fit.final), data=mydata, direction="forward")

Hi, I am just wondering if there is a test available for testing if a linear fit of an independent variable in a Cox regression is enough? Thanks for any suggestions. John Zhang ____________________________________________________________________________________ [[elided Yahoo spam]]

windows xp R 2.8.1 I am trying to plot the -log(log(survival)) to visually test the proportional hazards assumption of a Cox regression. The plot, which should give two lines (one for each treatment) gives only one line and a warning message. I would appreciate help getting two lines, and an explanation of the warning message. My problem may the that I have very few events in one of my strata,

Hello, I have spent a good deal of time searching for an answer to this but have come up empty-handed; I apologize if I missed something that is common knowledge. I am trying to figure out how to get an ANOVA table that shows the sum of squares. degrees of freedom, etc, for the full model versus the error (aka residuals). Here is an example of the kind of table I'd like to get:

R 2.8.1 Windows XP I am trying to plot the results of a coxph using plot(survfit()). The plot should, I believe, show two lines one for survival in each of two treatment (Drug) groups, however my plot shows only one line. What am I doing wrong? My code is reproduced below, my figure is attached to this EMail message. John > #Create simple survival object >

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)) >