similar to: problem with intervals in mixed model

Hello. I am trying to understand the syntax of the nonlinear least squares function (nls) when the function definition is made outside of the call. Here is the context. 1. If I specify the following command, it works fine: > fit2<-nls( + A~Am*(1-exp(-alpha*(I-LCP))),data=dat1, + start=list(Am=3.6,alpha=0.01,LCP=20)) 2. Now, I want to be able to specify the function definition

Hello. I am fitting a two-level mixed model which assumes equality of variance in the lowest-level residuals across groups. The call is: fit3<-lme(CLnNAR~CLnRGR,data=meta.analysis, + na.action="na.omit",random=~1+CLnRGR|study.code) I want to test the assumption of equality of variances across groups at the lowest level. Can someone tell me how to do this? I know that one

Hi all, I was wondering if there is a function out there, or someone has written code for making confidence intervals around model averaged predictions (y~á+âx). The model average estimates are from the dRedging library? It seems a common thing but I can't seem to find one via the search engines Examples of the models are: fit1 <- glm(y~ dbh, family = binomial, data = data) fit2 <-

Dear all, Consider a completely randomized block design (let's use data(Oats) irrespoctive of the split-plot design it was arranged in). Look: library(nlme) fit <- lme(yield ~ nitro, Oats, random = ~1|Block, method="ML") fit2 <- lm(yield ~ nitro + Block, Oats) anova(fit, fit2) gives this: Model df AIC BIC logLik Test L.Ratio p-value fit 1 4 624.3245

R 2.3 Windows XP I am trying to understand lme. My aim is to run a random effects regression in which the intercept and jweek are random effects. I am comparing output from SAS PROC MIXED with output from R. The point estimates and the SEs are the same, however the DFs and the p values are different. I am clearly doing something wrong in my R code. I would appreciate any suggestions of how I can

I have a generalized linear model to solve. I used package "geepack". When I use the correlation structure "unstructured", I get a messeage that- R GUI front-end has stopped working. Why this happens? What is the solution? The r codes are as follows: a<-read.table("d:/bmt.txt",header=T")

Greetings, I am using rpart for classification with "class" method. The test data is the Indian diabetes data from package mlbench. I fitted a classification tree firstly using the original data, and then exchanged the order of Body mass and Plasma glucose which are the strongest/important variables in the growing phase. The second tree is a little different from the first one. The

Dear list, I am trying to re-analyse something. I do have two time series, one of which (ts.mar) might help explaining the other (ts.anr). In the original analysis, no-one seems to have cared about the data being time-series and they just did OLS. This yielded a strong positive correlation. I want to know if this correlation is still as strong when the autocorrelations are taken into account.

the data is : >table.8.3<-data.frame(expand.grid( marijuana=factor(c("Yes","No"),levels=c("No","Yes")), cigarette=factor(c("Yes","No"),levels=c("No","Yes")), alcohol=factor(c("Yes","No"),levels=c("No","Yes"))), count=c(911,538,44,456,3,43,2,279))

I stumbled across a mild glitch when trying to compare the result of gam() fitting with the result of lm() fitting. The following code demonstrates the problem: library(gam) x <- rep(1:10,10) set.seed(42) y <- rnorm(100) fit1 <- lm(y~x) fit2 <- gam(y~lo(x)) fit3 <- lm(y~factor(x)) print(anova(fit1,fit2)) # No worries. print(anova(fit1,fit3)) # Likewise. print(anova(fit2,fit3)) #

Greetings, I checked the Indian diabetes data again and get one tree for the data with reordered columns and another tree for the original data. I compared these two trees, the split points for these two trees are exactly the same but the fitted classes are not the same for some cases. And the misclassification errors are different too. I know how CART deal with ties --- even we are using the

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

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

Hi the list, I am experiencing several issues with profile.mle (and consequently with confint.mle) (stat4 version 2.9.2), and I have to spend a lot of time to find workarounds to what looks like interface bugs. I would be glad to get feedback from experienced users to know if I am really asking too much or if there is room for improvement. * Problem #1 with fixed parameters. I can't

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 <-

Hi guys I have data that contains the variances vt of the yields of 1, 2, 3, 4, 5,10, 20 year bonds. Assuming the Hull-White model for the yield of a t-year zero-coupon bond, I have to estimate the ? of the Hull-White model using nonlinear least squares and give a 95% con?dence interval for each parameter. Please can you guys tell how to find out ? using R. Any suggestion regarding what functions

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

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

i create inbound confi my confi is: [incoming] exten=> 18888246463,,1,Dial(SIP/8003,60,rT) exten=> 6463,1,Dial(SIP/8003,60,rT) exten=> 18888246463,,n,Wait(5) exten=> 18888246463,,n,Hangup but y calling and send this error in my CLI: [Apr 15 09:58:48] NOTICE[26985]: chan_sip.c:14383 handle_request_invite: Call from '101396_procall' to extension '8888246463' rejected

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,