similar to: F-statistic in lme

Why are the results not reliable? ________________________________ From: ESCHEN Rene [mailto:rene.eschen@unifr.ch] Sent: Wednesday, August 23, 2006 3:48 AM To: Spencer Graves; r-help@stat.math.ethz.ch Cc: Doran, Harold Subject: RE: [R] Random structure of nested design in lme The output of the suggested lmer model looks very similar to the output of aov, also when I ran the model

Dear R: I am trying to fit a doubly multivariate LME (DM) where I have two response variables measured on two occasions per person. Specifically, reading and math scores measured at the beginning and ending of a school year. The response variables have a correlation of r = .85. The response variables in the data matrix are stacked in a vector with a dummy code flagging each outcome and with

r-help at stat.math.ethz.ch on Wednesday, October 26, 2005 at 6:00 AM -0500 wrote: Ronaldo, Try Harold's suggestion. The df still won't agree, because lmer (at least in its current version) just puts an upper bound on the df. But that should be OK, because all those t tests are approximations anyways, and you can get better confidence intervals (credible intervals, whatever) by using the

Dear list In HLM, one can specify a covariate at one of the "levels". For example, if the data structure are repeated observations nested within students nested within schools, school size might be a covariate that is used at level 3, but not at the other levels. In HLM this is rather easy to do. However, how can one specify a covariate in R for only one of the levels? I have a

If you are indeed using lme and not lmer then the needed function is VarCorr(). However, 2 recommendations. First, this is a busy list and better emails subject headers get better attention. Second, I would recommend using lmer as it is much faster. However, VarCorr seems to be incompatible with lmer and I do not know of another function to work with lmer. Hence, a better email subject header

Good day R-Users, I have a problem accessing some values in the output from the summary of an lme fit. The structure of my data is as shown below (I have attached a copy of the full data). id trials endp Z.sas ST 1 1 -1 -1 42.42884 1 1 1 -1 48.12007 2 1 -1 -1 43.42878 2 1 1 -1

This question comes up periodically, probably enough to give it a proper thread and maybe point to this thread for reference (similar to the 'conservative anova' thread not too long ago). Moving from lme syntax, which is the function found in the nlme package, to lmer syntax (found in lme4) is not too difficult. It is probably useful to first explain what the differences are between the

Dear R users I have run an regression and want to extract the p value of the F statistics, but I can find a way to do that. x<-summary(lm(log(RV2)~log(IV.m),data=b)) Call: lm(formula = log(RV2) ~ log(IV.m), data = b[[11]]) Residuals: Min 1Q Median 3Q Max -0.26511 -0.09718 -0.01326 0.11095 0.29777 Coefficients: Estimate Std. Error t value Pr(>|t|)

Dear List: I am trying to figure out how to incorporate measurement error in an longitudinal educational data set using lme to create a "true score" model. As a by-product of the procedures used to scale educational tests, one can obtain a person-specific measurement error associated with each score, or a conditional standard error. For example, a score of 200 would have measurement

Paul: I may have found the issue (which is similar to your conclusion). I checked using egsingle in the mlmRev package as these individuals are strictly nested in this case: library(mlmRev) library(nlme) fm1 <- lme(math ~ year, random=~1|schoolid/childid, egsingle) fm2 <- lmer(math ~ year +(1|schoolid:childid) + (1|schoolid), egsingle) Checking the summary of both models, the output is

Actually, I re-read the post and think it needs clarification. We may both be right. If the question is "I am building a model and want to know if I should retain this random effect?" (or something like that) then the LRT should be used to compare the fitted model against another model. This would be accomplished via anova(). In other multilevel programs, the variance components are

Dear List: I am having difficulties with the fitted values at different levels of a multilevel model. My data set is a series of student test scores over time with a total of 7,280 observations, 1,720 students nested witin 60 schools. The data set is not balanced. The model was fit using eg.model.1<-lme(math~year, random=~year|schoolid/childid, data=single). When I call the random

[Please use the subject line!] In the help page for summary.lm, the "Value" section says that the returned object has a component called "fstatistic", which has the F-statistic and the associated numerator and denominator degrees of freedom. You can get the p-value by something like: fstat <- summary(speciallinearmodel)$fstatistic pval <- pf(fstat[1], fstat[2],

FYI, in R devel (to become 3.5.0), there's isFALSE() which will cut some corners compared to identical(): > microbenchmark::microbenchmark(identical(FALSE, FALSE), isFALSE(FALSE)) Unit: nanoseconds expr min lq mean median uq max neval identical(FALSE, FALSE) 984 1138 1694.13 1218.0 1337.5 13584 100 isFALSE(FALSE) 713 761 1133.53 809.5 871.5

Quite possibly, and I?ll look into that. Aside from the work I was doing, however, I wonder if there is a way such that sapply could avoid the overhead of having to call the identical function to determine the conditional path. From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:14 PM To: Doran, Harold <HDoran at air.org> Cc: Martin Morgan <martin.morgan

Greetings, 1) Is there a nice way of extracting the variance estimates from an lme fit? They don't seem to be part of the lme object. 2) In a series of simulations, I am finding that with ML fitting one of my random effect variances is sometimes being estimated as essentially zero with massive CI instead of the finite value it should have, whilst using REML I get the expected value. I guess

Dear R-users, I did do a thorough search and read many articles and forum threads on the lme and lmer methods and their pitfalls and problems. I, being not a good statistician but a mere "user", came to the conclusion that the most correct form of reporting statistics for a mixed linear model would be to report the parameter estimates and SEs, and, if the sample size is considerably

You?re right, it sure does. My suggestion causes it to fail when simplify = ?array? From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:11 PM To: Doran, Harold <HDoran at air.org> Cc: r-help at r-project.org Subject: Re: [R] Possible Improvement to sapply Wouldn't that change how simplify='array' is handled? > str(sapply(1:3,

Using ranef() (package nlme, version 3.1-75) with an 'lme' object I can obtain random effects for intercept and slope of a certain level (say: 1) - this corresponds to (say level 1) "residuals" in MLWin. Maybe I'm mistaken here, but the results are identical. However, if I try to get the standardized random effects adding the paramter "standard=T" to the

Could your code use vapply instead of sapply? vapply forces you to declare the type and dimensions of FUN's output and stops if any call to FUN does not match the declaration. It can use much less memory and time than sapply because it fills in the output array as it goes instead of calling lapply() and seeing how it could be simplified. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue,