similar to: Longitudinal study with three treatments

Dear All, I would like to find something ( references, code,..) to implement a comparison of three treatments in an observational study using the 'Propensity Score'. Any help is much appreciated. Thanks! Giovanni -- dr. Giovanni Parrinello Department of Biotecnologies Medical Statistics Unit University of Brescia Viale Europa, 11 25123 Brescia email: parrinel at med.unibs.it Phone:

Hello R user, I have a data set from a longitudinal study ( sample below) where subjects are followed over time. Second column (status) contains info about if subject is dead or still in the study and third column is time measured in the week. Here is what I need: if status is not dead or unknown take the last week, if status is dead or unknown I need to have corresponding week. Desired resulst:

Hi In R,how to do sample size calculation for three-way incomplete block crossover study where within subject residual standard deviation,treatment difference and power is given. Thanks in advance. Regards Jose [[alternative HTML version deleted]]

Dear R users, I was wondering if anyone knows how to obtain(subset) the first and/or the last record of a subject in a longitudinal setup. Normally in SAS one uses first.variable1 and last.variable1. So my question is that is there an R way of doing this. Regards, -- Luwis Diya, Phd student (Biostatistics), Biostatistical Center, School Of Public Health, Catholic University of Leuven, U.Z. St

Hi all, I asked this before the holiday, didn't get any response. So would like to resend the message, hope to get any fresh attention. Since this is not purely lme technical question, so I also cc-ed R general mailing list, hope to get some suggestions from there as well. I asked some questions on how to analyze longitudinal study with only 2 time points (baseline and a follow-up)

Dear list, I have been struggling with this for some time now, and for the last hour I have been struggling to make a working example for the list. I hope someone out there have some experience with plotting longitudinal data that they will share. My data is some patient data with three different time stamps. First the patients are identified at different times (first time stamp). Second, they

Hi! How to calculate the correct SE of mean in a nested or spliplot anova? Nested example: --------------------- m <- aov(Glycogen~Treatment+Error(Treatment/Rat/Liver)) > m Call: aov(formula = Glycogen ~ Treatment + Error(Treatment/Rat/Liver)) Grand Mean: 142.2222 Stratum 1: Treatment Terms: Treatment Sum of Squares 1557.556 Deg. of Freedom 2 Estimated

Hello, I am analyzing the results of a survey of the student body on the use of certain technologies. The student body was divided in strata according to two criteria, college and user class. The resulting 24 strata were sampled. Subjects filled out a survey with a number of questions, most of them of the yes-no variety. I created a data.frame with the results for the survey. One row for

Why do I see this error? > library(stats) > require(stats) [1] TRUE > > tHSD <- TukeyHSD(aov) Error in TukeyHSD(aov) : no applicable method for "TukeyHSD" In case it helps: > aov Call: aov(formula = roi ~ (Cue * Hemisphere) + Error(Subject/(Cue * Hemisphere)), data = roiDataframe) Grand Mean: 8.195069 Stratum 1: Subject Terms: Residuals Sum

I think there are two bugs in aov() that shows up when the right hand side of `formula' contains both `-1' and an Error() term, e.g., aov(y ~ a + b - 1 + Error(c), ...). Without `-1' or `Error()' there is no problem. I've included and example, and the source of aov() with suggested fixes below. The first bug (labeled BUG 1 below) creates an extra, empty stratum inside

I believe you are right, but can you please explain why anyone would want to fit this model? It differs only in the coding from aov(y ~ a + b + Error(c), data=test.df) and merely lumps together the top two strata. There is a much simpler fix: in the line if(intercept) nmstrata <- c("(Intercept)", nmstrata) remove the condition (and drop the empty stratum later if you

I WROTE THIS FUNCTION BELOW test <- function(x, ...) UseMethod('test', x) test.data.frame = function(x, model, which, error, ...) { av <- aov(formula(model), data = x) res <- test.aovlist(av, which = which, error = error) return(res) } test.aovlist <- function(x, which, error, ...) { mm <- model.tables(x, "means") tabs <- mm$tables[-1]

I am analyzing from a very simple experiment. I have measured plants of two different colours (yellow and purple) in 9 different populations. So, I have two different factors : a fixed effect (Colour with two levels) and a random one (Population with 9 levels). I first analyzed the data with the aov function LargS is the variable aov(formula = LargS ~ Col + Error(Col/Pop)) Terms:

Greetings, The aov() function may mislabel the random effects as in the example below: Has anybody else noticed this? Cordially, Giles Crane, MPH, ASA, NJPHA gilescrane at verizon.net > m2 Call: aov(formula = y ~ ap + pe + Error(ju), data = d) Grand Mean: 77.50667 Stratum 1: ju Terms: ap Sum of Squares 4322.538 Deg. of Freedom 12 13 out of 25 effects not

Dear R-help, I'm facing a problem with defining a repeated measures anova with weighted data. Here's the code to reproduce the problem: # generate some data seed=11 rtrep <- data.frame(rt=rnorm(100),ti=rep(1:5,20),subj=gl (20,5,100),we=runif(100)) # model with within factor for subjects/repeated measurements, no problem aov(rt~ti + Error(subj/ti),data=rtrep) #model with weights

Hello R listing, I did two-way anova on lm. Further question the investigator interested in is what two treatments are different? I am looking for a command which could do pairwise comparison for every treatment. Could anyone help me out? Thanks a bunch Kevin

Hi R-users I'd like to know if there is some packages or functions to do comparison of treatments after that manova pointed differences between them. Any suggestions are much appreciated. __________________________________________________________ Eng. Agr., M.Sc. Eduardo Dutra de Armas __________________________________________________________ Centro de Energia Nuclear na Agricultura

Hi Folks, I'm a bit puzzled by the following (example): N<-factor(sample(c(1,2,3),1000,replace=TRUE)) unique(N) # [1] 3 2 1 # Levels: 1 2 3 So far so good. Now: contrasts(N)<-contr.treatment(3, base=1, contrasts=FALSE) contrasts(N) # 1 2 # 1 1 0 # 2 0 1 # 3 0 0 whereas: contr.treatment(3, base=1, contrasts=FALSE) # 1 2 3 # 1 1 0 0 # 2 0 1 0 # 3 0 0 1 contr.treatment(3, base=1,

Hi there, I need some help to figure out what is the proper model in survival analysis for my data. Subjects were randomized to 3 treatments in trial 1, some of them experience the event during the trial; After period of time those subjects were randomized to 3 treatments again in trial 2, but different from what they got in 1st trial, some of them experience the event during the 2nd trial (I

I'm using the current version of xtable for 1.9.0 and I have an interesting error: Error in match.names(clabs, names(xi)) : names don't match previous names: F value, Pr(>F) In addition: Warning message: longer object length is not a multiple of shorter object length in: clabs == nmi This is produced in the following manner: >data.trans <-