similar to: Repeates Measures MANOVA for Time*Treatment Interactions

Dear R Masters, I'm an anesthesiology resident trying to make his way through basic statistics. Recently I have been confronted with longitudinal data in a treatment vs. control analysis. My dataframe is in the form of: subj | group | baseline | time | outcome (long) or subj | group | baseline | time1 |...| time6 | (wide) The measured variable is a continuous one. The null hypothesis in

Hi, I wish to compute multivariate test statistics for a within-subjects repeated measures design with anova.mlm. This works great if I only have two factors, but I don't know how to compute interactions with more than two factors. I suspect, I have to create a new "grouping" factor and then test with this factor to get these interactions (as it is hinted in R News 2007/2), but

Hi, I'm encountering a very odd problem with calls to anova.mlm() from within a function. Consider the following code (data.n is a matrix of numeric values): mlmfit <- lm(data.n ~ 1) mlmfit0 <- lm(data.n ~ 0) print(mlmfit) anova(mlmfit,mlmfit0,test="Spherical") If I run it just like this from the console, it works just fine. If, however, I call it from within a function,

Hi, yet another anova.mlm problem - it doesn't seem to end. This time, I have a setup with a few within-subject factors and a between-subject factor (SGROUP). Consider the most simple case with only one within-factor (apo): > mlmfit0 <- lm(data.n ~ 0 + SGROUP) > mlmfit1 <- lm(data.n ~ 1 + SGROUP) > anova(mlmfit1,mlmfit0,test="Spherical",M=~hemi,X=~1) Analysis of

Dear All,   #I have the following code   Dose<-1000 Tinf <-0.5 INTERVAL <-8 TIME8 <-matrix(c((0*INTERVAL):(1*INTERVAL))) TIME7 <-matrix(c((0*INTERVAL):(2*INTERVAL))) TIME6 <-matrix(c((0*INTERVAL):(3*INTERVAL))) TIME5 <-matrix(c((0*INTERVAL):(4*INTERVAL))) TIME4 <-matrix(c((0*INTERVAL):(5*INTERVAL))) TIME3 <-matrix(c((0*INTERVAL):(6*INTERVAL))) TIME2

Dear R-list, the following code had been running well over the last months: exam <- matrix(rnorm(100,0,1), 10, 10) gg <- factor(c(rep("A", 5), rep("B", 5))) mlmfit <- lm(exam ~ 1); mlmfitG <- lm(exam ~ gg) result <- anova(mlmfitG, mlmfit, X=~0, M=~1) Until, all of a sudden the following error occured: Fehler in

I am trying to plot where data points from a give subject are connected by a line. Each subject is represented by a single row of data. Each subject can have up to five observations. The first five columns of mydata give the time of observation, columns 6-10 give the values at each time point. Some subjects have all data, some are missing values. The code I wrote to draw the plot is listed below.

I have 100+ .csv files which have the basic format: > test X Substance1 Substance2 Substance3 Substance4 Substance5 1 Time1 10 0 0 0 0 2 Time2 9 5 0 0 0 3 Time3 8 10 1 0 0 4 Time4 7 20 2 1 0 5 Time5

Greetings listeRs - Given a data frame such as times time1 time2 time3 time4 1 70.408543 48.92378 7.399605 95.93050 2 17.231940 27.48530 82.962916 10.20619 3 20.279220 10.33575 66.209290 30.71846 4 NA 53.31993 12.398237 35.65782 5 9.295965 NA 48.929201 NA 6 63.966518 42.16304 1.777342 NA one can use "transform" to

On Sat, 12 Aug 2006, takahashi kohske wrote: > Dear list members: > > I'd like to one-way repeated measured anova by using mlm. > I'm using R-2.3.1 and my code is: > > dat<-matrix( c(9,7,8,8,12,11,8,13, 6,5,6,3,6,7,10,9, > 10,13,8,13,12,14,14,16, 9,11,13,14,16,12,15,14), > ncol=4, dimname=list(s=1:8, c=1:4)) >

Sorry for bothering you - but I have a little problem. I`m completely new in R and trying to use lmer. That works, but I m not sure if my code is right, so I would just like to get an opinion from the experts. I have a Growthrate of plants (RGR Fixed Factors: Fertilizer, Status (alien or native plants) and their interaction Random Factors: Age is a covariable; than I have plantfamily and

Dear list, I´m calculating time differences between series of time stamps and I noticed something odd: If I do this... > time1=strptime("2009 05 31 22 57 00",format="%Y %m %d %H %M") > time2=strptime("2009 05 31 23 07 00",format="%Y %m %d %H %M") > > diff(c(time1,time2),units="mins") Time difference of 10 mins .. I get the correct

Dear R Community, Below is a problem with a simple ggplot2 graph. The code returns the error message below. Error: stat_count() must not be used with a y aesthetic. My code is below and the data is attached as a ?text? file. # Graph of the probabilities library(digest) library(DT) datatable(probability) str(probability) probability$Fertilizer <- as.factor(probability$Fertilizer)

Hello R-help, I'm studying an example in the R book.? The data file is available from the link below.http://www.bio.ic.ac.uk/research/mjcraw/therbook/data/fertilizer.txt Could you explain Why the results from lme() and lmer() are different in the following case? In other examples, I can get the same results using the two functions, but not here...? Thank you.Miya library(lme4)library(nlme)#

Hi James, If you want to specify the y-values, you need to use stat="identity" as below: ggplot(probability, aes(x=Fertilizer, y=prob)) + geom_bar(stat="identity", aes(fill=Treatment)) best, huzefa On Tue, Apr 12, 2016 at 1:02 PM, James Henson <jfhenson1 at gmail.com> wrote: > Dear R Community, > > Below is a problem with a simple ggplot2 graph. The code

Dear list-members I am new to R and a statistics beginner. I really like the ease with which I can extract and manipulate data in R, and would like to use it primarily. I've been learning by checking analyses that have already been run in SAS. In an experiment with Y being a response variable, and group a 2-level between-subject factor, and time a 5-level within-subject factor. 2

Hi all, I analysed my data with lme and after that I spent a lot of time for mean separation of treatments (post hoc). But still I couldn’t make through it. This is my data set and R scripts I tried. replication fertilizer variety plot height 1 level1 var1 1504 52 1 level1 var3 1506 59 1 level1 var4 1509 54 1 level1 var2 1510 48 2 level1 var1 2604 47 2 level1 var4 2606 51 2 level1 var3

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

Hi! I try to explain the efffect of (1) forest where i took samples's soils (* Lugar*: categorical variable with three levels), (2) nitrogen addition treatments (*Tra*: categorical variable with two levels) on total carbon concentration's soil samples (*C: *continue* *variable) during four months of sampling (*Time:* categorical and ordered variable with four levels). I fitted the

Hi, I am having difficulty understanding some lme output -- I haven't found too many examples to help explain to me how to interpret the coefficients and would appreciate any help. I am fitting a model: fit <- lme(y ~ pre + group + time + group:time, random=~1|subject, na.action=na.omit, data=mydata) ...for a dataset where there are two groups being followed over time. pre is