similar to: lme vs. SAS proc mixed. Point estimates and SEs are the same, DFs are different

Dear All, I am trying to use the function ses from the forecast package. >From its help I have : Usage: ses(y, h = 10, level = c(80, 95), fan = FALSE, initial = c("optimal", "simple"), alpha = NULL, lambda = NULL, biasadj = FALSE, x = y, ...) My query is that if I do not mention the initial value will its default value be "optimal". A MWE would be

The first one, i.e. "optimal"; check help for match.arg() for the idiom. -pd > On 2 Oct 2017, at 11:48 , Ashim Kapoor <ashimkapoor at gmail.com> wrote: > > Dear All, > > I am trying to use the function ses from the forecast package. > > From its help I have : > > Usage: > > ses(y, h = 10, level = c(80, 95), fan = FALSE, initial =

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 <-

If I do fit<-lm(y~x) Is it possible to access the SE of the slope? (Analogous to getting the slope like this: fit$coef[2]) If not, it would be handy. (I want SE of 1/slope, and an approx way is fit$se[2]/(fit$coef[2]^2)) Thanks for any help. Bill Simpson -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

So I'm curious how SAS JBOD arrays and linux MDraid as implemented in CentOS6, and SES (SCSI/SAS Enclosure Services) backplane controllers 'get along' and how much configuration is needed to get the warning lights to work properly. scenario: whitebox server with a SAS backplane or two, daisy chained on a SAS HBA (like an LSI Logic 2008), and disks organized as several raid5/6

Hello R-list, I'm attempting to migrate from Stata to R for my complex survey work. It has been straight-forward so far except for the following problem: I have some code below, but first I'll describe the problem. When I compute predicted logits from a logistic regression, the standard errors of the predicted logits are way off (but the predicted logits are fine). Furthermore, the

Can someone direct me to an R function that properly computes standard errors of data obtained from a complex survery design, i.e. perform alnalyses similiar to those that can be performed with SUDAAN, particularly for a non-parametric one-way ANOVA, e.g. signed rank test? Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC, University of

Dear friends I am analysing the leaf expansion of a grass species and am interested in the speed of expansion. I produced exponential models for each of the treatments and got the equation for leaf size in function of time. I want to compare the coeficients that gives the initial inclination of the exponential curve, but as they are result of the previous modelling, I have them as means and SE. Is

Hi i'm student and my final project is related to Voip. I have Asterisk almost fully configured. The next step is to accept login of users, that data is in Universitys database which uses ActiveDirectory and also Ldap and Kerberos. It's possible? I don't want authentications in sip.conf, but in other remote database. The problem is i don't have ideas how to start with. I would

predict.lm(..., type="terms") gives wrong standard errors. Below, I have provided what I believe are the necessary fixes. However, there are subtleties, and the code needs careful checking. Some of the looping is surely not necessary, but it is surely best to begin with the minimum necessary changes. My tests, including checks against S-PLUS, have extended to fitting spline curves. I

>From e980153 Tue Apr 25 14:42:27 2000 To: r-help@stat.math.ethz.ch Subject: Wrong SEs in predict.lm(..., type="terms") For what it is worth, I am using RW-1.0.0 under Windows 98. I submitted this earlier to r-help. There is one change below to my proposed corrected code: predict.lm(..., type="terms") gives wrong standard errors. Below, I have provided what I believe are

I'm trying to replicate a time-series cross-sectional analysis (countries over years) with SEs clustered by country. ?The original analysis was done in Stata 10 with: xtreg [DV] [IVs] fe cluster(country). Using plm() in R (cran.r-project.org/web/packages/plm/index.html), I've replicated the coefficients. I sought to estimate country-clustered SEs with vcovHC(), and tried a variety of

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")

Dear R Helpers This is the R code (which I have slightly changed) I got in VGAM package for estimating the parameters of FRECHET. _________________________________________________________________ y = rfrechet(n <- 100, shape=exp(exp(0))) # (A) fit3 = vglm(y ~ 1, frechet3(ilocation=0), trace=TRUE, maxit=155) # (B) coef(fit3, matrix=TRUE) # (C)

Simple-mindedly I tried getting MLE and SE for one-parameter model in the same way as for multi-param models. out<-nlm(fn,p=c(2),hessian=T) But sqrt(diag(solve(out$hessian))) gives the answer 1. The Hessian has only one entry, not really a matrix. diag(x) gives 1 if x is just a single number. Is this what I should be doing to get SE for MLE? sqrt(solve(out$hessian)) Thanks very much for

Simple-mindedly I tried getting MLE and SE for one-parameter model in the same way as for multi-param models. out<-nlm(fn,p=c(2),hessian=T) But sqrt(diag(solve(out$hessian))) gives the answer 1. The Hessian has only one entry, not really a matrix. diag(x) gives 1 if x is just a single number. Is this what I should be doing to get SE for MLE? sqrt(solve(out$hessian)) Thanks very much for

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 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 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,

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))