similar to: Extract p value from aov

How can I get the outlier in this boxplot of "Score" to be represented by the corresponding value in "SubNo"? score=c(6,6,7,14,5,7,6,8) SubNo=1:8 mydata=data.frame(SubNo, score) boxplot(mydata$score) Thanks! Kevin [[alternative HTML version deleted]]

Dear Users, I am trying to understand the inner workings of a repeated measures linear model. Take for example a situation with 6 individuals sampled twice for two conditions (control and treated). set.seed(12) ctrl <- rnorm(n = 6, mean = 2) ttd <- rnorm(n = 6, mean = 10) dat <- data.frame(vals = c(ctrl, ttd), group = c(rep("ctrl", 6), rep("ttd",

Hi, I need to determinate HSD value from a matrix like that Thesis Days A1 Cx 0 66.07 Cx 0 60.24 Cx 0 42.86 Tw 0 66.07 Tw 0 60.24 Tw 0 42.86 Aa 0 66.07 Aa 0 60.24 Aa 0 42.86 Qe 0 66.07 Qe 0 60.24 Qe 0 42.86 Cx 56 310.86 Cx 56 223.17 Cx 56 186.77 Tw 56 149.42 Tw 56 127.75 Tw 56 138.59 Aa 56 130.24 Aa 56 214.83 Aa 56 137.95 Qe 56 186.64 Qe 56 189.09 Qe 56 187.87 this is my script

I'm trying to write a function to automate doing a variance analysis, part of which involves doing some further calculations. The method I've been using isn't very robust, if variable names change then it stops working. For this dummy data > dput(assayvar,"") structure(list(Run = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L), .rk.invalid.fields =

Dear all How do I access individual elements of a "summary.aov" object? > data(iris) > AnovaModel.1 <- aov(Sepal.Length ~ Species, data=iris) > tmp <- summary(AnovaModel.1) > tmp Df Sum Sq Mean Sq F value Pr(>F) Species 2 63.2 31.6 119 <2e-16 *** Residuals 147 39.0 0.3 --- Signif. codes: 0 '***' 0.001 '**' 0.01

Hi everyone, I'm fairly new to R (and this forum!), so first of all apologies if this question has already been answered (I've tried searching the forum but haven't come across anything which has solved the problem I'm having). I have a data frame containing an "independent variable" and a series of potential "dependent variables", and I'd like to find

I think there are two bugs in aov() that shows up when the right hand side of `formula' contains both `-1' and an Error() term, e.g., aov(y ~ a + b - 1 + Error(c), ...). Without `-1' or `Error()' there is no problem. I've included and example, and the source of aov() with suggested fixes below. The first bug (labeled BUG 1 below) creates an extra, empty stratum inside

I believe you are right, but can you please explain why anyone would want to fit this model? It differs only in the coding from aov(y ~ a + b + Error(c), data=test.df) and merely lumps together the top two strata. There is a much simpler fix: in the line if(intercept) nmstrata <- c("(Intercept)", nmstrata) remove the condition (and drop the empty stratum later if you

Dear R developers A question about the class 'listof', defined in package 'stats'. Other than its definition and use in the code for 'anova', we can't see that the class 'listof' is used for anything else (in recommended packages, or elsewhere). In the spatstat package we have been using a 'listof' to represent a list of spatial objects of the same

Hi, all I've written some R script to calculate the linear regression of a matrix. Here below is my script: >x<-matrix(scan("h:/data/xxx.dat",0),nrow=46,ncol=561,byrow=TRUE) >year <- NULL >year <- cbind(year,as.matrix(x[,1])) >lm.sol<-lm(x~year) >xtrend<-coef(lm.sol)[2,] # get the matrix of regression coefficient >t.test<- ?

Hello, I have a list of data with multiple elements, and each element in the list has multiple variables in it. Here's an example: ### Make the fake data dv <- c(1,3,4,2,2,3,2,5,6,3,4,4,3,5,6) subject <- factor(c("s1","s1","s1","s2","s2","s2","s3","s3","s3",

Dear r-users, Suppose I have three datasets: Dataset-1: Date x y Jan-1,2005 120 230 Jan-2,2005 123 -125 Jan-3,2005 -110 300 Jan-4,2005 114 -21 Jan-7,2005 11 299 Mar-5,2005 200 311 Dataset-2: Date x y Jan-2,2005 123 -125 Jan-3,2005 -110 300 Jan-4,2005 114 -21

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 .. ..$

Buenas noches Quisiera saber como hago para sacar el valor p de esta lista. He intentado como b$Pr(>F) pero no lo he logrado str(b) List of 1 $ :Classes ‘anova’ and ''data.frame'': 2 obs. of 5 variables: ..$ Df : num [1:2] 1 18 ..$ Sum Sq : num [1:2] 0.000155 0.00058 ..$ Mean Sq: num [1:2] 1.55e-04 3.22e-05 ..$ F value: num [1:2] 4.82 NA ..$ Pr(>F) : num

hello all, Quick question, how do I get the p value out of the anova? Thanks, Paul > pb<-aov(as.numeric(diff[5,16:33]) ~ grF) > summary(pb) Df Sum Sq Mean Sq F value Pr(>F) grF 3 2.7860e+10 9.2867e+09 4.2236 0.02534 * Residuals 14 3.0783e+10 2.1988e+09 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 '

Hi, this is my script > anova <- aov(data ~ Ts*Te*t + Error(R/Ts*Te*t)) > results <- summary(anova$Within) this is results Df Sum Sq Mean Sq F value Pr(>F) Ts 2 1232.2 616.11 53.606 3.965e-10 *** Ts:Te 4 4889.5 1222.37 106.356 4.075e-16 *** Ts:t 4 6472.1 1618.01 140.780 < 2.2e-16 *** Ts:Te:t 8 4181.0 522.63 45.473 1.088e-13 ***

<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 list, I am trying to write a package for simulating meioses in R. We defined a class 'haplotype' which contains the basic units of our simulation: setClass("haplotype",representation(snp = "numeric",qtl = "list", hID = "numeric",phID0 = "numeric",phID1 = "numeric"),

I want to know how accurate are the p-values when you do linear regression in R? I was looking at the variable x3 and the t=10.843 and the corresponding p-value=2e-16 which is the same p-value for the intercept where the t-value for the intercept is 48.402. I tried to calculate the p-value in R and I got 0 x<-2*(1-pt(10.843,2838)) > x [1] 0 > G<-lm(y~x1+x2+x3+x4+x5) >

Let's say I have a within-subject experiment with 2 observables, obs1 and ob2 and 2 independent factors, fac1 and fac2. I can do summary( aov( obs1~fac1*fac2 + Error(Subject/(fac1*fac2)) ) ) summary( aov( obs2~fac1*fac2 + Error(Subject/(fac1*fac2)) ) ) to test the 2 observables separately. > summary( fit<-manova( cbind(obs1,obs2)~fac1*fac2 + Error(Subject/(fac1*fac2)) ) ) gives