similar to: ANCOVA with random factor

I have an ANOVA model with 2 factors "Environment" and "Site", "Diameter" is the response variable. Site should be nested within Environment. Site is also a random factor while Environment is fixed. I can do this analysis using the "aov" function by using these commands: >model<-aov(Diam~Env+Error(Env%in%Site),data=environ) >summary(model)

Dear all, I would like to use the biological symbols for male and female as plotting symbols in a scatterplot (ideally filled and non-filled). R does not seem to have these symbols using pch= in plot() nor are they implemented via expression() or at least I did not find them. I found that the symbols are e.g. available in the wasysym and the marvosym package in LaTeX. I have coded two very rough

Dear all, I would like to use the biological symbols for male and female as plotting symbols in a scatterplot (ideally filled and non-filled). R does not seem to have these symbols using pch= in plot() nor are they implemented via expression() or at least I did not find them. I found that the symbols are e.g. available in the wasysym and the marvosym package in LaTeX. I have coded two very rough

Consider the following analysis of a class experiment done as a Latin Square: > spinner <- gl(4,4,16,label=c("Murray","Angela","Shasha","Stephen")) > order <- gl(4,1,16) > treat <- scan() 1: 1 2 4 3 5: 4 3 1 2 9: 3 4 2 1 13: 2 1 3 4 17: Read 16 items > coin <-

Dear all, I'm quite sure that this is a stupid question, but I'll ask anyway. I want to perform an ANCOVA with two continuous factors and three categorical factors. Plant population growth rate (GR) = dependent variable Seed reduction due to herbivory (SR) = continuous explanatory variable Herbivore species (HS, 2 levels) = categorical explanatory variable Population (Pop, 24 levels) =

Could you please help me on the following ANCOVA issue? This is a part of my dataset: sampling dist h 1 wi 200 0.8687212 2 wi 200 0.8812909 3 wi 200 0.8267464 4 wi 0 0.8554508 5 wi 0 0.9506721 6 wi 0 0.8112781 7 wi 400 0.8687212 8 wi 400 0.8414646 9 wi 400 0.7601675 10 wi 900 0.6577048 11 wi 900

The list of your interest is R-help not R-sig-mac stefano Il giorno 26/gen/06, alle ore 01:20, Sylvain Charlat ha scritto: > Hi, > > Is there any simple way to get histogram for different levels of > factor? > > Say you have the following data set: > > Island Sp.diam > Moorea 1.21 > Moorea 1.27 > Moorea 1.28 > Moorea 1.22 > Moorea 1.28 > Rurutu

Dear R user: I have a question about doing ANCOVA in S-plus or R. I know that many users use lm to do the regression and check the ANCOVA. But is there a way to get the traditional Table form of the ANCOVA test through S-plus (like what we would get from SPSS or SAS)? The problem I’m interested in is whether or not there is a treatment effect on some medical measurement. I will

Here are several ways to get there, but your original loop is fine once it is corrected: > for (i in 1:2) smean[i] <- mean(toy$diam[toy$group==i][1:nsel[i]]) > smean [1] 0.271489 1.117015 Using sapply() to hide the loop: > smean <- sapply(1:2, function(x) mean((toy$diam[toy$group==x])[1:nsel[x]])) > smean [1] 0.271489 1.117015 Or use head() > smean <- sapply(1:2,

Hi All. I am a new R user. Trying to do scatterplot. Not sure how to resolve this error message A<-subset (ErablesGatineau, station=="A") > B<-subset (ErablesGatineau, station=="B") > > plot(diam ~ biom) > abline(lm(diam ~ biom), col = "red") > > goodcases <- !(is.na(diam) | is.na(biom)) > lines(lowess(diam[goodcases] ~

Hi Pedro, This may not be much of an improvement, but it was a challenge. selvec<-as.vector(matrix(c(nsel,unlist(by(toy$diam,toy$group,length))-nsel), ncol=2,byrow=TRUE)) TFvec<-rep(c(TRUE,FALSE),length.out=length(selvec)) toynsel<-rep(TFvec,selvec) by(toy[toynsel,]$diam,toy[toynsel,]$group,mean) Jim On 4/3/16, Pedro Mardones <mardones.p at gmail.com> wrote: > Dear all; >

Dear all, A question related to the following has been asked on R-help before, but I could not find any answer to it. Input will be much appreciated. I got an unexpected sign of the "slope" parameter associated with a covariate (diam) using zeroinfl(). It led me to compare the estimates given by zeroinfl() and hurdle(): The (significant) negative estimate here is surprising, given

Hi! as my subject says I am struggling with the different of a two-way ANOVA and a (two-way) ANCOVA. I found the following examples from this webpage: http://www.statmethods.net/stats/anova.html # One Way Anova (Completely Randomized Design) fit <- aov(y ~ A, data=mydataframe) # Randomized Block Design (B is the blocking factor) fit <- aov(y ~ A + B, data=mydataframe) # Two Way

Dear all, I would like to know how to run an analysis of covariance in R. For example, I have a data frame ("data") consisting of two second-degree categorical variables ("diagnosis" and "gender"), one continous independent variable ("age") and one continous dependent variable ("response"). I ran a simple anova to see the effects of diagnosis

dear all, I want to perform a posthoc test for my ANCOVA: a1<-aov(seeds~treatment*length) With summary(glht(a1, linfct = mcp(treatment = "Tukey"))) R tells me: "covariate interactions found -- please choose appropriate contrast" How do I build these contrasts? Ideally, I would like to have the posthoc test for the ANCOVA including a block-effect

Hi there, The following data is obtained from a long-term experiments. > mydata <- read.table(textConnection(" + y year Trt + 9.37 1993 A + 8.21 1995 A + 8.11 1999 A + 7.22 2007 A + 7.81 2010 A + 10.85 1993 B + 12.83 1995 B + 13.21 1999 B + 13.70 2007 B + 15.15 2010 B + 5.69 1993 C + 5.76 1995 C + 6.39 1999

I am trying to run an ANCOVA with defined error terms. Thus I have to use AOV and not lm. my response variable is proportion of mice paw prints on track plates. These plates were placed on plots that had vegetation and fruit manipulated to two levels each (present or absent), and were sampled monthly for 14 months (repeated measures). The fully crossed factor design was also blocked. My sample

Hi I compouted a multiple linear regression with repeated measures on one explanatory variable: BOLD peak (blood oxygenation) as dependent variable, and as independent variables I have: -age.group (binaray:young(0)/old(1)) -and task-difficulty measured by means of the reaction-time 'rt'. For 'rt' I have repeated measurements, since each subject did 12 different tasks. -> so

Dear all; This must have a rather simple answer but haven't been able to figure it out: I have a data frame with say 2 groups (group 1 & 2). I want to select from group 1 say "n" rows and calculate the mean; then select "m" rows from group 2 and calculate the mean as well. So far I've been using a for loop for doing it but when it comes to a large data set is

Hello! I have measurements (length and volume) of fish collected in two years. I want to know if the the relationship between length and volume is the same for both years. The number of fish measured is different for each year. I don't know whether lm or aov is more appropriate to use. Here are the two output options: Call: lm(formula = Volume ~ Length * Year) Residuals: Min 1Q