similar to: P-values in gls

Dear R-users, I have noticed small discrepencies in the reported estimate of the variance of the frailty by the print method for survreg() and the 'theta' component included in the object fit: # Examples in R-2.6.2 for Windows library(survival) # version 2.34-1 (2008-03-31) # discrepancy fit1 <- survreg(Surv(time, status) ~ rx + frailty(litter), rats) fit1 fit1$history[[1]]$theta

Hi, I am working with R version 2.1.0, and I seem to have run into what looks like a bug. I get the same error message when I run R on Windows as well as when I run it on Linux. When I call anova to do a LR test from inside a function, I get an error. The same call works outside of a function. It appears to not find the right environment when called from inside a function. I have provided

I'm writing a paper aimed at motivating the use of GEE within the field of economics. However, after computing using the geeglm function, I noticed there's one intercept in the summary output. I assume this means the function is pooling the data. That means my code is not what I want. I want a "fixed effects" model, meaning I want the intercept to vary by cluster. Here's

Hi, how to estimate a the following model in R: y(t)=beta0+beta1*x1(t)+beta2*x2(t)+...+beta5*x5(t)+beta6*y(t-1)+beta7*y(t-2)+beta8*y(t-3) 1) using "lm" : dates &lt;- as.Date(data.df[,1]) selection&lt;-which(dates&gt;=as.Date("1986-1-1") &amp; dates&lt;=as.Date("2007-12-31")) dep &lt;- ts(data.df[selection,c("dep")]) indep.ret1

Hello, Does anyone know whether the gls function in the nlme library uses the Newton-Raphson or EM algorithm to find the restricted log-likelihood or maximum log-likelihood estimates? Brendan Klick bklick@jhsph.edu [[alternative HTML version deleted]]

Hi All, I have a n x m matrix. The n rows are individuals, the m columns are variables. The matrix is in itself a collection of 1s (if a variable is observed for an individual), and 0s (is there is no observation). Something like: [,1] [,2] [,3] [,4] [,5] [,6] [1,] 1 0 1 1 0 0 [2,] 1 0 1 1 0 0 [3,] 1 0 1 1 0 0 [4,] 0 1 0

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

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

If first fit my data column V1 to column V2 using normal "lm" fitting, call it "fit1", then I used "acf(fit1$residuals, type='cov', 40) " function to obtain the autocovariance of the residuals, and then constructed a autocovariance matrix, I chose it to be 40x40. Call this autocovariance matrix B, I then use the following "lm.gls" function to

Dear list, I am getting weired with fitting data with a 1/x-polynomial. Suggest I have the following data: x <- c(1,2,3,4,5,6,7) y <- c(100,20,4,2,1,.3,.1) I may fit this with a linear model fit1 = lm(y ~ I(x)) Getting plot out of this model I applied library(polynom) pol1 = polynomial(fit1$coefficients) f1 = as.function(pol1) plot(x,y) lines(x, f1(x), col = 2) Clearly, this model

Hello R users & R friends, I just want to ask you if density() can produce a confidence interval, indicating how "certain" the density() line follows the true frequency distribution based on the sample you feed into density(). I've heard of loess.predict(loess(y ~ x), se=TRUE) which gives you a SE estimate of the smoothed scatterplot - but density() kernel smoothing is not the

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

Hello Instead of testing against 0 i would like to test regression slopes against -1. Any idea if there's an R script (package?) available. Thanks for any hint. Cheers Lukas ??? Lukas Indermaur, PhD student eawag / Swiss Federal Institute of Aquatic Science and Technology ECO - Department of Aquatic Ecology ?berlandstrasse 133 CH-8600 D?bendorf Switzerland Phone: +41 (0) 71 220

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