similar to: repost: problems with lm for nested fixed-factor Anova (ANOVA I)

Dear R users, I want to run nested fixed-factor Anova in R on different experiments. In this toy example I have 3 levels of the main factor x1 and 7 levels of the nested factor z1 x1 and continuous response variable y1. x1 [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 1 2 2 2 2 2 2 2 2 2 2 2 2 [38] 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 [75] 3 3 3

Hi, there I'm trying to get the value of the Mean Square from the ANOVA model summary that comes from specifying the error term, and am wondering if one can actually do this ( I know it's possible when using anova(lm) objects and the like, but I'm having a tough time with it under this framework). There does appear to be some indexing in the output of this type, but perhaps not

Dear R users, I have trouble obtaining the same results for nested Anova with two fixed factors when using lm and aov functions. The formulas are: > e1=aov(y~x/z) > e2=lm(y~x/z) summary(e1) Df Sum Sq Mean Sq F value Pr(>F) x 47 260.0 5.5 18.0088 < 2.2e-16 *** x:z 195 169.6 0.9 2.8318 < 2.2e-16 *** Residuals 14425

Hello all R-users _question 1_ I need to make a statistical model and respective ANOVA table but I get distinct results for the T-test (in summary(lm.object) function) and the F-test (in anova(lm.object) ) shouldn't this two approach give me the same result, i.e to indicate the same significants terms in both tests??????? obs. The system has two restrictions: 1) sum( x_i ) = 1 2) sum(

Dear All. I have the following table that I want to analyze using multinom function freq segments sample 4271 Seg1 tumour 4311 Seg2 tumour 3515 Seg1 normal 3561 Seg2 normal I want to compare model with both factors to the one where only sample is present. model1=multinom(freq~segments+sample,data=table) model2=multinom(freq~ sample,data=table)

Dear All, I would be very appreciative of your help with the following 1). I am running multivariate multiple regression through the manova() function (kindly suggested by Professor Venables) and getting two different answers for test=c("Wilks","Roy","Pillai") and tests=c("Wilks","Roy",'"Pillai") as shown below. In the

Dear All, I am trying to perform MANOVA. I have table with 504 columns(species) and 36 rows) with two grouping (season and location) Zx <- Z[c(4:504)] Zxm <- as.matrix(Z) m<- manova(Zxm~Season*location, data=Z) when I do summary.aov, I get respond for each species but summary.manova summary.manova(m) :" residuals have rank" 24<501. What can it be the reason for this error

hello I make a subroutine that give-me a (mathematical) function in string format. I would like transform this string into function ( R function ). thanks for any tips. cleber #e.g. fun_String = "-100*x1 + 0*x2 + 100*x3" fun <- function(x1,x2,x3){ return( ############ evaluation( fun_String ) ############ ) True String mathematical function :-( :-( > nomes [1]

Hi list, I am unsuccessfully trying to produce a serious of trellis barcharts from within a for-loop. The barcharts work outside the loop. What am I missing? Example attached. Thanks Herry #XXXXXXXXXXXXXXXXXXXXXX trellis.device(bg="white") trellis.par.get("fontsize")->fontsize fontsize$default<-16 trellis.par.set("fontsize",fontsize)

Hello again, after obtaining the keytab file I tried to use kinit keytab.file followed by the spn $ samba-tool spn list z1 z1 User CN=z1,CN=Users,DC=home,DC=lan has the following servicePrincipalName: zookeeper/ap42.home.lan $ samba-tool domain exportkeytab z1.ktab --principal=z1 $ samba-tool domain exportkeytab z1.ktab --principal=zookeeper/ap42.home.lan $ kinit -V -k -t z1.ktab

Hi Team, I am trying to very simple plot with command plot. Question : I am trying to plot (x,y) based on the value of Column e. If column e value is greater than 0 then plot(x,y) otherwise do not plot it. Data : structure(list(x = c(1, 1, 1, 2, 2, 2, 3, 3, 3), y = c(1, 2, 3, 1, 2, 3, 1, 2, 3), e = c(0, -1, -2, 1, 0, -1, 2, 1, 0)), row.names = c(NA, -9L), .Names = c("x",

Hi, R users, I'm trying to transform a matrix A into B (see below). Anyone knows how to do it in R? Thanks. Matrix A (zone to zone travel time) zone z1 z2 z3 z1 0 2.9 4.3 z2 2.9 0 2.5 z3 4.3 2.5 0 B: from to time z1 z1 0 z1 z2 2.9 z1 z3 4.3 z2 z1 2.9 z2 z2 0 z2 z3 2.5 z3 z1 4.3 z3 z2 2.5 z3 z3 0 The real matrix I have is much larger, with more than 2000 zones. But I think it should

hello, this is my script: #1) read in data: daten<-read.table('K:/Analysen/STRUCTURE/input_STRUCTURE_tab_excl_5_282_559.txt', header=TRUE, sep="\t") daten<-as.matrix(daten) #2) create empty matrix: indxind<-matrix(nrow=617, ncol=617) indxind[1:20,1:19] #3) compare cells to each other, score: for (s in 3:34) { #walks though the matrix colum by colum, starting at

Hi,   I use fCopulae package to draw different graphs of univariate and bivariate skew t.  But the plots titles overlap.  I tried using cex.main, font.main to adjust the size but they still overlaps.  Here is my code: par(mfrow = c(3, 1)) mu = 0 Omega = 1 alpha1 = 0 alpha2 = 1.5 alpha3 = 2 alpha4 = 0.5 Z1 = matrix(dmvst(x, 1, mu, Omega, alpha1, df = Inf), length(x)) Z2 = matrix(dmvst(x, 1, mu,

Dear R-group members, I use: platform i386-pc-mingw32 arch x86 os Win32 system x86, Win32 status major 1 minor 4.1 year 2002 month 01 day 30 language R I try to fit a 2 compartment model. The compartments are open, connected to each other and

Dear Rusers, I would like some comments about the following results (under R-2.2.0) > m = matrix(1:6 , 2 , 3) > m [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 > z1 = m[(m[,1]==2),] > z1 [1] 2 4 6 > is.matrix(z1) [1] FALSE > z2 = m[(m[,1]==0),] > z2 [,1] [,2] [,3] > is.matrix(z2) [1] TRUE Considered together, I'm a bit surprised about

Hi R-users,   I have this code below and I understand the error message but do not know how to correct it.  My question is how do I get rid of “with absolute error < 7.5e-06” attach to value of cdf so that I can carry out the calculation.   integrand <- function(z) { alp  <- 2.0165   rho  <- 0.868   # simplified expressions   a      <- alp-0.5   c1     <-

Hello, I would like to perform a multivariate regression analysis to model the relationship between m responses Y1, ... Ym and a single set of predictor variables X1, ..., Xr. Each response is assumed to follow its own regression model, and the error terms in each model can be correlated. Based on my readings of the R help archives and R documentation, the function lm() should be able to

Is there an easy way to compare complex numbers? Here is a small example: > (z1=polyroot(c(1,-.4,-.45))) [1] 1.111111-0i -2.000000+0i > (z2=polyroot(c(1,1,.25))) [1] -2+0i -2+0i > x=0 > if(any(identical(z1,z2))) x=99 > x [1] 0 # real and imaginary parts: > Re(z1); Im(z1) [1] 1.111111 -2.000000 [1] -8.4968e-21 8.4968e-21 > Re(z2); Im(z2) [1] -2

Hi, I was wondering if there is a way, or function in R to find confounders. For istance, > a = sample( c(1:3), size=10,replace=T) > X1 = factor( c('A','B','C')[a] ) > X2 = factor( c('Aa','Bb','Cc')[a] ) > Xmat = data.frame(X1,X2,rnorm(10),rnorm(10)) > dimnames(Xmat)[[2]] = c('z1','z2','z3','y') Now,