search for: satterthwaites

Dear R experts, I was wondering if there are any R functions that give the tail area of a sum of chisquare distributions of the type: a_1 X_1 + a_2 X_2 where a_1 and a_2 are constants and X_1 and X_2 are independent chi-square variables with different degrees of freedom. Thanks, Klaus -- "Feel free" - 5 GB Mailbox, 50 FreeSMS/Monat ...

I'm fairly new to R and am wondering if anybody knows of R code to calculate confidence intervals for parameters (fixed effects and variance components) from mixed effects models based on Sattherthwaite's method? I'm also interested in Satterthwaite-based confidence intervals for linear combinations (mostly sums) of various variance components. [[alternative HTML version

This is not a bug report but didn't see another way to ask a question. For the approximate t-test assuming unequal variances, the R docs cite Welch's method for the df of the approximating distribution. I have several methods books, and they all uses Satterthwaite's method. Why does R use Welch's method where can I learn about Welch's method? Sincerely, David Allen

>>>>> "TL" == Thomas Lumley <tlumley@u.washington.edu> >>>>> on Sun, 6 Oct 2002 09:19:27 -0700 (PDT) writes: TL> On Sat, 5 Oct 2002 roxburg@kih.net wrote: >> This is not a bug report but didn't see another way to >> ask a question. TL> Well, you could try the r-help or r-devel mailing lists >> For

Dear All: Significance Analysis of Microarrays(SAM) As we know sam do multiple t.test as following ## Default S3 method: t.test(x, y = NULL, alternative = c("two.sided", "less", "greater"),mu = 0, paired = FALSE, var.equal = FALSE,conf.level = 0.95, ...) var.equal: a logical variable indicating whether to treat the two variances as being equal. If 'TRUE'

Hi All, I'm taking an Experimental Design course this semester, and have spent many long hours trying to coax the professor's SAS examples into something that will work in R (I'd prefer that the things I learn not be tied to a license). It's been a long semester in that regard. One thing that has really frustrated me is that lme has an extremely counterintuitive way for

(This time with the r-help in the recipients...) Be careful when mixing lme4 and lmerTest together -- lmerTest extends and changes the behavior of various lme4 functions. >From the help page for lme4-anova (?lme4::anova.merMod) > ?anova?: returns the sequential decomposition of the contributions > of fixed-effects terms or, for multiple arguments, model >

Hi Ted, yes this was the problem. Thank you very much. best idaios On Wed, Jan 9, 2013 at 4:51 PM, Ted Harding <Ted.Harding@wlandres.net>wrote: > Ah! You have aqssigned a parameter "equal.var=TRUE", and "equal.var" > is not a listed paramater for t.test() -- see ?t.test : > > t.test(x, y = NULL, > alternative = c("two.sided",

Please reread my point #1: the tests of the (individual) coefficients in the model summary are not the same as the ANOVA tests. There is a certain correspondence between the two (i.e. between the coding of your categorical variables and the type of sum of squares; and for a model with a single predictor, F=t^2), but they are not the same in general. The t-test in the model coefficients is simply

Dear all, I'm performing a t-test on two normal distributions with identical mean & standard deviation, and repeating this tests a very large number of times to describe an representative p value distribution in a null case. As a part of this, the program bins these values in 10 evenly distributed bins between 0 and 1 and reports the number of observations in each bin. What I have noticed

I am running a multilevel growth curve model to examine predictors of social anhedonia (SA) trajectory through ages 12, 15 and 18. SA is a continuous numeric variable. The age variable (Index1) has been coded as 0 for age 12, 1 for age 15 and 2 for age 18. I am currently using a time varying predictor, stress (LSI), which was measured at ages 12, 15 and 18, to examine whether trajectory/variation

I am running a multilevel growth curve model to examine predictors of social anhedonia (SA) trajectory through ages 12, 15 and 18. SA is a continuous numeric variable. The age variable (Index1) has been coded as 0 for age 12, 1 for age 15 and 2 for age 18. I am currently using a time varying predictor, stress (LSI), which was measured at ages 12, 15 and 18, to examine whether trajectory/variation

Dear R users: I'm happy to announce the first CRAN release of the clubSandwich package: https://cran.r-project.org/web/packages/clubSandwich clubSandwich provides several variants of the cluster-robust variance estimator for ordinary and weighted least squares linear regression models, including the bias-reduced linearization estimator of Bell and McCaffrey (2002). The package includes

Dear R users: I'm happy to announce the first CRAN release of the clubSandwich package: https://cran.r-project.org/web/packages/clubSandwich clubSandwich provides several variants of the cluster-robust variance estimator for ordinary and weighted least squares linear regression models, including the bias-reduced linearization estimator of Bell and McCaffrey (2002). The package includes

Dear R users: Suppose I have two different response variables y1, y2 that I regress separately on the different explanatory variables, x1 and x2 respectively. I need to compare points on two regression lines. These are the x and y values for each lines. x1<-c(0.5,1.0,2.5,5.0,10.0) y1<-c(204,407,1195,27404313) x2<-c(2.5,5.0,10.0,25.0) y2<-c(440,713,1520,2634) Suppose we need to

Dear all Somebody may have asked this before but I could not find any answers in the web so let me ask a question on lme. When I have a fixed factor of, say, three levels (A, B, C), in which each level has different size (i.e. no. of observations; e.g. A>B>C). When I run an lme model, I get the same degree of freedom for all the contrast t-tests (e.g. AvsB or BvsC). I have tried this to

Dear list, I ran a mixed effect model using R 1.4.1 and SAS 8.0 on the SIMS data found in the SASmixed package and found that the degrees of freedoms for fixed effects are very different. From R, df = n - v -1 where n is total # of observations, v is the # of levels for the grouping factor. From SAS df = v -1. Am I wrong about this or can somebody explain which is correct and why? Thanks a

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

Hi R community! I've fitted three mixed-effects regression models to a thousand bootstrap samples (case-resampling regression) using the lme4 package in a custom-built for-loop. The only output I saved were the inferential statistics for my fixed and random effects. I did not save any output related to the performance to the machine learning algorithm used to fit the models (REML=FALSE).

Dear Srinidhi, You are trying to fit 1 random intercept and 2 random slopes per individual, while you have at most 3 observations per individual. You simply don't have enough data to fit the random slopes. Reduce the random part to (1|ID). Best regards, Thierry ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK