search for: aove

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

Hi, I've done a repeated ANOVA looking as follows: model<-aov(y~a*b*c+Error(d)) Now I want to get rid of some or all of the interactions. But neither AIC nor anova work. What is the appropriate function to test my model?

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

Greetings to all, This is my model: aov.fit<-aov(Y~A+B+C+D+E+A:C+A:E) In summary(aov.fit) all F values are comptuted by eg MS(A)/MS(Residuals). This is not correct (or what I want), except for F(B) and F(A:E). I suppose P values are not correct either. Is it possible with aov to define the way F computations will be done? I 'd like them to be like this: F(A)=MS(A)/MS(E),

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

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 ~

Hi, I am a method developer in drug discovery. I have developed a similarity searching method for small (drug-like) molecules. I want to compare the performance of my method to the performance of other methods. Similarity searching methods are commonly assessed by their ability to (re)discover a set of molecules that is avtive versus a given target, given one randomly selected query molecule.

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 <-

> tmp <- data.frame(y=rnorm(12), a=factor(rep(letters[1:4],3))) > tmp y a 1 -0.60866099 a 2 0.55500538 b 3 0.12231693 c 4 -0.24613790 d 5 -0.09253593 a 6 -1.54652581 b 7 0.17204210 c 8 -1.22778942 d 9 1.22151194 a 10 -0.43982577 b 11 -1.25444287 c 12 -0.97251060 d > tmp.aov <- aov(y ~ a, data=tmp) > summary(tmp.aov) Df Sum Sq Mean Sq F value

...53.2,56.0) npk <- data.frame(block=gl(6,4), N=factor(N), P=factor(P), K=factor(K), yield=yield) npk.aov <- aov(yield ~ block + N*P*K, npk) proj(npk.aov) ## as a test, not particularly sensible options(contrasts=c("contr.helmert", "contr.treatment")) npk.aovE <- aov(yield ~ N*P*K + Error(block), npk) proj(npk.aovE) ## ## Change the rownames and it fails ## rownames(npk) <- 2:25 npk.aovE <- aov(yield ~ N*P*K + Error(block), npk) proj(npk.aovE)

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:

This concerns whether p-values from lmer can be trusted. From simulations, it seems that lmer can produce very small, and probably spurious, p-values. I realize that lmer is not yet a finished product. Is it likely that the problem will be fixed in a future release of the lme4 package? Using simulated data for a quite standard mixed-model anova (a balanced two-way design; see code for the

Dear all, I'm trying to access the Sums of Squares resulting from a summary(aov(....)) so I can use them in a function. Is there an easy way to access these? The structure of a summary.aov object is something like this: > str(summary(fhc.aov)) List of 5 $ Error: PPNR :List of 1 ..$ :Classes anova and `data.frame': 1 obs. of 5 variables: .. ..$ Df : num 70 .. ..$

R developers, I have a first attempt to make an aov function. Eventually I want to build in Error() structure, but first I am trying to get this presentable for balanced data with only a single stratum, just using residual error. I am following R. M. Heiberger's Computation for the Analysis of Designed Experiments, Wiley (1989) I a using a wrapper (aov.bal) to call the

How do we specify a contrast interaction matrix for an ANOVA model? We have a two-factor, repeated measures design, with Cue Direction (2) x Brain Hemisphere(2) Each of these has 2 levels, 'left' and 'right', so it's a simple 2x2 design matrix. We have 8 subjects in each cell (a balanced design) and we want to specify the interaction contrast so that: CueLeft>CueRght

<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

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

I performed an aov() analysis and got the following results: Df Sum Sq Mean Sq F value Pr(>F) block 1 0.0040 0.0040 0.3282 0.5672 Residuals 269 3.2766 0.0122 Can anyone tell me how to extract the F value column and Pr(>F) column from the summary output of aov analysis? Many thanks in advance, Liqing Eco. Evol. Biol. UCIrvine

Hi everyone, I am fairly new to R, and I am aware that others have had this problem before, but I have failed to solve the problem from previous replies I found in the archives. As this is such a standard procedure in psychological science, there must be an elegant solution to this...I think. I would much appreciate a solution that even I could understand... ;-) Now, I want to calculate a