search for: aov1

Hello, I wrote the following function to compute multiple comparisons in a one way anova and randomized blocks anova. aov1 <- function(y,g,s=NULL,comp="mca",meth="Sidak") { # fun <- function(x) c(mean(x,na.rm=T),sd(x,na.rm=T),length(x[!is.na(x)])) # li <- length(unique(g)) cat(" Analysis of Variance with Multiple comparisons\n\n") cat(" Groups : ",li,"...

I've posted the following to R-help on May 15. It has reproducible R code for real data -- and a real (academic, i.e unpaid) consultion background. I'd be glad for some insight here, mainly not for myself. In the mean time, we've learned that it is to be expected for anova(*, "marginal") to be contrast dependent, but still are glad for advice if you have experience. Thank

The message for aov1 was "Estimated effects <may> be unbalanced". The effects are not unbalanced. The design is 'orthogonal'. The problem is that there are not enough degrees of freedom to estimate all those error terms. If you change the model to: aov1 <- aov(RT~fact1*fact2*fact...

Hello, I am running error rate analysis. It is my results below. When I compare aov1 and aov2, X square = 4.05, p = 0.044, which indicates that adding the factor "Congruity" improved the fitting of model. However, the following Z value is less than 1 and p value for Z is 1, which means that "Congruity" is not significant at all. Therefore, these two parts are no...

...ndement ~ Lignee * Pollinisateur, data = mca2) Lignee Pollinisateur Lignee:Pollinisateur 20 10 2 Of course, summary(aov(Rendement ~ Pollinisateur * Lignee, data = mca2)) gives wrong tests of random effects. But, summary(aov1 <- aov(Rendement ~ Error(Pollinisateur * Lignee), data = mca2)) gives no test at all, and I have to do it like this : tab1 <- matrix(unlist(summary(aov1)), nc=5, byrow=T)[,1:3] Femp <- c(tab1[1:3, 3]/tab1[c(3,3,4), 3]) names(Femp) <- c("Pollinisateur", "Lignee",...

...I put it in another way : I want to test random effects in this random effect model : Rendement ~ Pollinisateur (random) + Lignee (random) + Pollinisateur:Lignee (random) Of course : summary(aov(Rendement ~ Pollinisateur * Lignee, data = mca2)) gives wrong tests for random effects. But : summary(aov1 <- aov(Rendement ~ Error(Pollinisateur * Lignee), data = mca2)) gives no test at all, and I have to do it with mean squares lying in summary(aov1). With "lme4" package (I did'nt succeed in writing a working formula with lme from "nlme" package), I can "see" sta...

...clear enough in my querry. I have the following program: tst<- function() { x <- c(32.7,32.3,31.5,32.1,29.7,29.1,35.7,35.9,33.1, 36.0,34.2,31.2,31.8,28.0,29.2,38.2,37.8,31.9, 32.5,31.1,29.7) g <- rep(1:7,rep(3,7)) s <- rep(1:3,7) cat(" Only x and g \n") aov1(x,g) cat("\n\n Now x, g and s \n") aov1(x,g,s=s) } aov1 <- function(y,g,s=NULL,comp="mca",meth="Sidak") { # fun <- function(x) c(mean(x,na.rm=T),sd(x,na.rm=T),length(x[!is.na(x)])) # li <- length(unique(g)) cat(" Analysis of Variance with...

........ As you can notice each subject repeated the evaluation in 2 conditions (EXP1 and EXP2). What I am interested in is to know if there are significant differences between the evaluations of the males and the females. This is the command I used to perform the ANOVA with repeated measures: aov1 = aov(response ~ stimulus*sex + Error(subject/(stimulus*sex)), data=scrd) summary(aov1) I get the following error: > aov1 = aov(response ~ stimulus*sex + Error(subject/(stimulus*sex)), data=scrd) Warning message: In aov(response ~ stimulus * sex + Error(subject/(stimulus * sex)),...

Dear all, I need to understand a thing in the beheaviour of the two functions aov and anova in the following case involving an analysis of ANOVA with repeated measures: If I use the folowing command I don´t get any problem: >aov1 = aov(response ~ stimulus*condition + Error(subject/(stimulus*condition)), >data=scrd) > summary(aov1) Instead if I try to fit the same model for the regression I get an error: > fit1<- lm(response ~ stimulus*condition + Error(subject/(stimulus*condition)), >data=scrd) > Error...

...le over internet in R...is there among you someone so kind to give me an hint with a R example please? For example, the aov result of my analysis says that there is a statistical difference between stimuli (there are 7 different stimuli). ...I would like to know which stimuli are involved. > aov1 = aov(response ~ stimulus*condition + Error(subject/(stimulus*condition)), >data=scrd) > summary(aov1) Error: subject Df Sum Sq Mean Sq F value Pr(>F) Residuals 14 227.57 16.255 Error: subject:stimulus Df Sum Sq Mean Sq F value Pr(>F) stimulus...

...ix subjects, a random effect. h1 <- c(7.8,8.7, 11.1,12.0, 11.7,10.0, # Each S does each task twice. 8.0,9.2, 11.3,10.6, 9.8,11.9, 4.0,6.9, 9.8,10.1, 11.7,12.6, 10.3,9.4, 11.4,10.5, 7.9,8.1, 9.3,10.6, 13.0,11.7, 8.3,7.9, 9.5,9.8, 12.2,12.3, 8.6,10.5) aov1 <- aov(h1~task*subj) anova(aov1) # See note below. lme1 <- lme(h1~task,random=~1|subj) lme2 <- lme(h1~task,random=~1|subj/task) anova(lme1,lme2) # for interaction anova(lme2) # for effect of task ------- The anova gives very close to the correct F values for subj (1.68, according to Hays...

...> >> Dear ExpeRts, >> >> Suppose I have a typical psychological experiment that is a >> within-subjects design with multiple crossed variables and a >> continuous response variable. Subjects are considered a random >> effect. So I could model >> > aov1 <- aov(resp~fact1*fact2+Error(subj/(fact1*fact2)) >> >> However, this only holds for orthogonal designs with equal numbers of >> observation and no missing values. These assumptions are easily >> violated so I seek refuge in fitting a mixed-effects model with the >&...

Dear ExpeRts, Suppose I have a typical psychological experiment that is a within-subjects design with multiple crossed variables and a continuous response variable. Subjects are considered a random effect. So I could model > aov1 <- aov(resp~fact1*fact2+Error(subj/(fact1*fact2)) However, this only holds for orthogonal designs with equal numbers of observation and no missing values. These assumptions are easily violated so I seek refuge in fitting a mixed-effects model with the nlme library. > lme1 <- lme(resp~f...

...I need an help because I am really not able to find over internet a good example in R to analyze an unbalanced table with Anova with repeated measures. For unbalanced table I mean that the questions are not answered all by the same number of subjects. For a balanced case I would use the command aov1 = aov(response ~ stimulus*condition + Error(subject/(stimulus*condition)), data=scrd) Does the same code still work for unbalanced design? Can anyone provide a R example of code in order to get the same analysis? Thanks in advance for any suggestion. Best regards [[alternative HTML...

I'm creating an ANOVA model. Is there a way to pull in consecutive columns of variables for the test? Example: aov1<-aov(y~x1+x2+REG[,2:num], data=REG) I'm not looking for interaction effects, I just want to create a model for the first few columns of variables (exact number and names will vary) and a few other predetermined variables (in the example I named them x1 and x2). Thanks in advance. - Jacy...

...variable "dist" with the two sets of helmert contrasts of the variable "train". I have tried so many things I think are wrong I won't reproduce them all here. Here is a representative example: My code: contrasts(dist) = c(-3, -1, 1, 3) contrasts(train) = contr.helmert(3) aov1 = aov(fatigue~dist*train) summary(aov1, intercept = T, split=list(train=1:2), expand.split = T) My output: ? ? ? ? ? ? ? Df Sum Sq Mean Sq ?F value ? ?Pr(>F) (Intercept) ? ? ? 1 890.03 ?890.03 801.0250 < 2.2e-16 *** dist ? ? ? ? ? ? ?3 ?85.19 ? 28.40 ?25.5583 1.199e-07 *** train ? ? ? ? ?...

Dear Colleagues, John Mandel ( Chain block designs with two-way elimination of heterogeneity. Biometrics 10, 251-272 ,1954). extended the class of chain block designs (Youden & Conner (1953) to elimination of both row and column (blocks) effects. These experimental designs can be useful in engineering and other fields. I am having difficulty obtaining his adjusted treatment means in his

...9.82, 10.07, 8.34, 9.94, 11.38, 9.19, 8.35, 10.17, 9.04)), .Names = c("subject", "block", "treatment", "session", "blood", "score"), row.names = c(NA, -30L), class = "data.frame") # some first try which however throws a warning aov1 <- aov(score~blood*treatment*session*block+Error(subject/(treatment*session*block)), data=data) summary(aov1) [[alternative HTML version deleted]]

Dear list, i am a biologist who needs to do some ttest between disease and non disease, sex, genotype and the serum levels of proteins on a large number of individuals. i have been using excel for a long time but it is very tedious and time consuming. i am posting the data below and ask your help in generating a code to get this analysis done in R. thanks gender disease genotype data M N CC

Dear R People: I have an AOV model that I get confidence intervals from via > confint(chick1.aov1) 2.5 % 97.5 % trtA 1.472085 1.607915 trtB 1.512085 1.647915 trtC 1.328751 1.464582 > I am using R2HTML to produce HTML output. However, the HTML code itself just has rounded values, i.e., 1.5 and 1.6. Has anyone run across this, please? Any suggestions would be much appreciated. Sin...