similar to: Problems with aov function

Hello, I want to do this model: Y_ml=Mu+M+L(M)+Error M is a fixed factor with two levels, 1 and 2 L is a random factor and nested in M, with 9 levels (9 places inside each level of M) amarc is the dependent variable (animal abundance) datos is the matrix I used the next orders in R, but there is a problem with the Df calculation: > m [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

I'm working with a orthogonal and nested model (mixed). I have four factors, A,B,C,D; A and B are fixed and orthogonal C is nested in AB interaction and finally, D is nested in C. I would like to model the following Y_ijklm=Mu+A_i+B_j+AB_ij+C_k(ij)+D_l(k(ij))+Error_m(...) I used the next command >summary(aov(abund~A*B + C % in % A:B + D % in % C % in % A:B ,datos)) Is it the correct

Dear all I was looking at the aov documentation page and came across the following which seems like a contradiction to me: " This provides a wrapper to |lm| for fitting linear models to balanced or unbalanced experimental designs." (I presume 'This' refers to aov) and "|aov| is designed for balanced designs, and the results can be hard to interpret without

I am using R 1.1 with Redhat 6.2 and RW 1.001 with Win98 (the upkey doesn't work on my IBM either as has been previously reported by others). The function aov doesn't return either the residuals or the residual degrees of freedom for split-plot designs. If you use the following code from Baron and Li's "Notes on the use of R for psycology experiments and questionnaires"

I've recently run into the problem of using aov with nested factors, and wanting to get the type II and III sums of squares. Normally Anova from the car package would do fine, but it doesn't like having an Error included, so my.aov <-aov(Response ~ Treatment + Error(Treatment:Replicate)) Anova(my.aov, type="II") yields Error in Anova(nested.anova) : no applicable method

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

I notice something curious about how aov() treats a numeric factor: "score" is a dependent variable and "group" is a factor in a one-way ANOVA. But "group" contains numeric codes and is not a factor (checked with is.factor). An ANOVA done using: > aov(score~factor(group), data=mydata) gives the right answers. But > aov(score~group, data=mydata) also

Hi, I'm a new R user, coming from SPSS, and without a particularly strong stats background. I've got a data set that I'd like to do a mixed-design ANOVA with. No missing values. Here's the summary: summary(learnDat.ae) Type Subject idio struct TrainErrs cond 0:20 11 : 3 idio :28 ae :58 Min. : 0.00 idioae :28 2:19 12 : 3

Dear R Users, I want to write a function that applies to the dataframe and variables that were used in a previous call to lm or aov. In order to do this, I need to write a function that applies to the output of lm or aov, and yields the names of the dataframe and variables that were used in the lm or aov analysis. For example, suppose that I give the command: aov.out <- aov( Rt ~

Dear listmembers, I would like to define a class with a slot that takes either an object of class aov or NULL. I have been reading "S4 Classes in 15 pages more or less" and "Lecture: S4 classes and methods" #First I tried with list and NULL setClass(listOrNULL") setIs("list", "listOrNULL") setIs("NULL", "listOrNULL") #doesn't

Hi, I have a question of aov. e.g. aov.ex = aov(x~y) summary(aov.ex) The aov summary will print to the screen. How can I extract the aov result, in particular the values of Pr(>F) and F value into a vector so that I can use them for other use? Thanks. -- Waverley @ Palo Alto

Dear all, Type III SS time again. This case trying to reproduce some SPSS (type III) data in R for a repeated measures anova with a betwSS factor included. As I understand this list etc, if I want type III then I can do library(car) Anova(lm.obj, type="III") But for the repeated measures anova, I need to include an Error-term in the aov() call (Psychology-guide from Jonathan Baron)

Hi, I have been having some trouble using aov to do an anova, probably because I'm not understanding how to use this function correctly. For some reason it always tells me that "Estimated effects may be unbalanced", though I'm not sure what this means. Is the formula I am using written incorrectly? Below is the code I am using along with the data: > my.data response

Hello, I have a simple question about using the aov function syntax (ie. * + or :) for the interaction of 2 factors. I have read the help files, and researched other sites, and have included my source files. My goal is to measure the signifigance of the interaction between population and condition (aka. population:condition). I can't seem to figure it out. 1. In the first example the

Dear Madame, Dear Sir, I am able to obtain the coefficients from a 'summary' of 'lm', but NOT from a 'summary' of 'aov'. The following example shows my steps. ## Initialize rm(list = ls()) # remove (almost) everything in the working environment utils::data(npk, package="MASS") # get data model <- yield ~ block + N*P*K ## Using lm npk.lm <-

Hi everyone, I'm having a little trouble with working out what formula is better to use for a repeated measures two way anova. I have two factors, L (five levels) and T (two levels). L and T are both crossed factors (all participants do all combinations). So, I do: summary(aov(dat~L*T+Error(participant/(L*T)),data=dat4)) But get different results with:

Dear R-Community! The example "oats" in MASS (2nd edition, 10.3, p.309) is calculated for aov and lme without interaction term and the results are the same. But I have problems to reproduce the example aov with interaction in MASS (10.2, p.301) with lme. Here the script: library(MASS) library(nlme) options(contrasts = c("contr.treatment", "contr.poly")) # aov: Y ~

I have 2 questions on ANOVA with 1 between subjects factor and 2 within factors. 1. I am confused on how to do the analysis with aov because I have seen two examples on the web with different solutions. a) Jon Baron (http://www.psych.upenn.edu/~baron/rpsych/rpsych.html) does 6.8.5 Example 5: Stevens pp. 468 - 474 (one between, two within) between: gp within: drug, dose aov(effect ~ gp * drug *

<ripley at stats.ox.ac.uk> wrote: > Hint 2: in the absence of balance, ...., and lme can do that Would it be possible to make aov like wrappers to the various special lm variants allowing for a uniform syntax for anova? aov(resp~f1*f2+Error(S/(f1*f2))) ## uses lm aov.lme(resp~f1*f2+Error(S/(f1*f2))) ## uses lme aov.rlm(resp~f1*f2+Error(S/(f1*f2))) ## uses rlm ... I'd do it, but I

Dear useRs, I've just encountered a serious problem involving the F-test being carried out in aov() and anova(). In the provided example, aov() is not making the correct F-test for an hypothesis involving the expected mean square (EMS) of a factor divided by the EMS of another factor (i.e., instead of the error EMS). Here is the example: Expected Mean Square