similar to: anova planned comparisons/contrasts

Dear R users, i am using the following code to produce barcharts with lattice: Compound<-c("Glutamine", "Arginine", "Glutamate", "Glycine", "Serine", "Glucose", "Fructose", "Raffinose", "Glycerol", "Galacglycerol", "Threitol", "Galactinol", "Galactitol")

Hi all, I would like to test the homoegeneity of variances between several linear model for some analysis of covariance. It seems that the Bartlett's test is a good test to use but I am having problem using with linear model and I cannot find any examples on the internet. There are some examples for comparisons of variances but not linear models. If I take the hellung data set, which is

Dear list, I am trying to use MuMIn to compare all possible mixed models using the dredge function on binomial data but I am getting an error message that I cannot decode. This error only occurs when I use glmer. When I use an lmer analysis on a different response variable every works great. Example using a simplified glmer model global model: mod<- glmer(cbind(st$X2.REP.LIVE,

Hi folks... check this out.. > GLU<-lme(gluc~rt*cd4+sex+age+rf+nadir+pharmac+factor(hcv)+factor(hbs)+ + haartd+hivdur+factor(arv), + random= ~rt|id, na.action=na.omit) > intervals(GLU)$fixed lower est. upper (Intercept) 67.3467070345 7.362307e+01 7.989944e+01 rt *0.0148050160* 6.249304e-02 1.101811e-01 cd4

Hello. I'm trying to find a way in R to calculate between and within subject variances and confidence intervals for some analytical method development data. I've found a reference to a method in Burdick, R. K. & Graybill, F. A. 1992, Confidence Intervals on variance components, CRC Press. This example is for Balanced Data confidence interval calculation from Pg 62. The data are

Hi, I have a dataset with a continuous response variable and, among other predictors, an ordinal variable. Here is what it could look like treatment <- factor(rep(c("AA", "AC", "AD","AE", "AB"), each = 10)) length <- c(75, 67, 70, 75, 65, 71, 67, 67, 76, 68, 57, 58, 60, 59, 62, 60, 60, 57, 59, 61, 58,

So I''m trying to do that hot new thing with AJAXy forms: http://idiet.toasterwaffles.com/foods/list Here''s the relevant code in list.rhtml (for the form portion) <tbody> <%= render_collection_of_partials "list_stripes", @foods %> </tbody> <tfoot> <tr id="addFood"><%= form_remote_tag( :html =>

Hi, I'm new. I tried to search out this answer but I suspect I was using the wrong terms, or simply not understanding some of the answers. Anyway here is my question: I want to have a 2x2 panel figure with 4 line graphs all in the same scale. Actually I have that. The thing I seem to be lacking is a way to Label each panel with a letter. I want it to look something like this:

Dears members of R list, I would like that a more experienced R user help me to complete this analysis: r = gl(3, 8, label = c('r1', 'r2', 'r3')) e = rep(gl(2, 4, label = c('e1', 'e2')), 3) y = c(26.2, 26.0, 25.0, 25.4, 24.8, 24.6, 26.7, 25.2, 25.7, 26.3, 25.1, 26.4, 19.6, 21.1, 19.0, 18.6, 22.8, 19.4, 18.8, 19.2, 19.8, 21.4, 22.8, 21.3) df =

Hi, I have a matrix: mymat <- matrix(runif(10*4), ncol=4) I wish to subtract the column means, down the colums, subtract the row means from the rows and add back the grand mean - all the means should be the means of mymat rather than of the new matrix. How can I do this? Any help much appreciated. Thanks Tom _________________________________________________________________

Hello, I have been looking for functions for calculating the within-groups variance and between-groups variance, for the case where you have several numerical variables describing samples from a number of groups. I didn't find such functions in R, so wrote my own versions myself (see below). I can calculate the within- and between-groups variance for the Sepal.length variable (iris[1]) in

Dear all; Thanks. Here are the dput results as Duncan suggested. Regards, Greg structure(list(Sub_Pathways = structure(c(3L, 3L, 3L, 3L, 3L), .Label = c("Acetylated_Peptides", "Advanced_Glycation_End-product", "Alanine_and_Aspartate", "Aminosugar", "Ascorbate_and_Aldarate", "Carnitine", "Ceramides", "Creatine",

Hi all, This is probably going to come off as unnecessary (and show my ignorance) but I am trying to understand the parameter estimates I am getting from R when doing an ANCOVA. Basically, I am accustomed to the estimate for the categorical variable being equivalent to the respective cell means minus the grand mean. I know is the case in JMP - all other estimates from these data match the

Works fine for me. What do you object to in the following? Calling the above df "d", > dm <- as.matrix(d) > dm Sub_Pathways BMI_beta SAT_beta VAT_beta 1 "Alanine_and_Aspartate" " 0.23820" "-0.02409" " 0.94180" 2 "Alanine_and_Aspartate" "-0.31300" "-1.97510" "-2.22040" 3

On 19/09/2017 9:47 AM, greg holly wrote: > Hi all; > > I have data at 734*22 dimensions with rows and columns names are > non-numeric.When I convert this data into matrix then all values show up > with quotes. Then when I use > x1= noquotes(x) to remove the quotes from the matrix then non-numeric row > names remain all other values in matrix disappear. > > Your help is

Hi all; I have data at 734*22 dimensions with rows and columns names are non-numeric.When I convert this data into matrix then all values show up with quotes. Then when I use x1= noquotes(x) to remove the quotes from the matrix then non-numeric row names remain all other values in matrix disappear. Your help is greatly appreciated. Greg [[alternative HTML version deleted]]

Dear all, Here is my code which am using to combine 5th column from different data sets. Here is the function to do my job genesymbol.append.file <-NULL gene.column <- NULL readGeneSymbol <- function(files,genesymbol.column=5){ for(i in fnames){ temp <- read.table(fnames,header=T,sep="\t",stringsAsFactors=F,quote="\"")

Thank you for reading. I am trying to get sphericity values, and I understood I need to use mlm, but how do I implement a nested within subject design in mlm? I already read the R newsletter, fox chapter appendix, EZanova, and whatever I could find online. My original ANOVA anova(aov(resp ~ sucrose*citral, random =~1 | subject, data = p12bl, subset = exps==1)) Or anova(aov(resp ~