similar to: Possible bug in summary of residuals with lm and weights

Hi all: Can someone advice me on how to hold the residuals Mean sq value on a string so it can be used in other calculations. I was trying something like this: Msquare<-dfr$Mean sq but fails..Thanks dfr <- read.table(textConnection("percentQ Efficiency 1.565 0.0125 1.94 0.0213 0.876 0.003736 1.027 0.006 1.536 0.0148 1.536 0.0162 2.607 0.02 1.456 0.0157 2.16 0.0103

Hello all! I have a problem to extract values greater that for example 1820. I try this code: x[x[,1]>1820,]->x1 Please help me! Thank you! The data structure is: structure(c(2.576, 1.728, 3.434, 2.187, 1.928, 1.886, 1.2425, 1.23, 1.075, 1.1785, 1.186, 1.165, 1.732, 1.517, 1.4095, 1.074, 1.618, 1.677, 1.845, 1.594, 1.6655, 1.1605, 1.425, 1.099, 1.007, 1.1795, 1.3855, 1.4065, 1.138, 1.514,

I am wondering how to interpret the parameter estimates that lm() reports in this sort of situation: y = round(rnorm(n=24,mean=5,sd=2),2) A = gl(3,2,24,labels=c("one","two","three")) B = gl(4,6,24,labels=c("i","ii","iii","iv")) # Make both observations for A=1, B=4 missing y[19] = NA y[20] = NA data.frame(y,A,B) nonadd = lm(y ~

Dear R fans, I would like to create a scatterplot showing the relationship between 4 continuous variables. I thought of using the package "scatterplot 3d" to have a 3-dimensional plot and then using the size of the symbols to represent the 4th variable. Does anybody know how to do this? I already tried to create this graph using the colour of the symbols, but I was unable to generate

Dear all, I have the following dataset: each row corresponds to count of forest floor small mammal captured in a plot and vegetation characteristics measured at that plot > sotr plot cnt herbc herbht 1 1A1 0 37.08 53.54 2 1A3 1 36.27 26.67 3 1A5 0 32.50 30.62 4 1A7 0 56.54 45.63 5 1B2 0 41.66 38.13 6 1B4 0 32.08 37.79 7 1B6 0 33.71 30.62

dear all i have this dataset: x,y, datavalue > dati[,c(1,2,5)] [,1] [, 2] [,3] [1,] 2.386 3.077 1.740 [2,] 2.544 1.972 1.335 [3,] 2.807 3.347 1.610 [4,] 4.308 1.933 2.150 [5,] 4.383 1.081 1.565 [6,] 3.244 4.519 1.145 [7,] 3.925 3.785 0.894 [8,] 2.116 3.498 0.525 [9,] 1.842 0.989 0.240 [10,] 1.709 1.843 0.625 [11,] 3.800 4.578 3.873 [12,] 2.699 1.199 1.425

Hi all: I have always used SPSS to estimate weekly covariance based on a linear regression model but have to hard code the model Std. Error and the Mean-Square and then execute one week a the time. I was wondering if someone could give me an idea on how to estimate weekly(WK) covariance using the summary and anova of "dfr"(lineal model below). I have to do this for 52

Hello, I’ve fitted a Garch(2,1) model with function 'garchFit()' from the package 'fGarch': > m1 <- garchFit(formula = ~garch(2,1),data = X,trace = F) * See 'summary(m1)' OUTPUT BELOW * PROBLEM: My alpha1 term is not significant and I would like to make a NEW model, say m2, that does not contain alpha1, but I am not sure how to specify this with the garchFit()

Hello everybody, I have a problem with the model.table of anova. I have data (datos2) with 4 cluster (V7), calculated with daisy and hclust. str(datos2) 'data.frame': 56 obs. of 7 variables: $ Estacion: Factor w/ 56 levels "Abradelo","AltoDoRodicio",..: 1 2 3 4 5 6 7 8 9 10 ... $ Invierno: num 36 53.9 37.1 63.6 12.5 ... $ X : int 643449 616292 562796

Dear R-help This is a follow-up to my previous post here: http://groups.google.com/group/r-help-archive/browse_thread/thread/d9b6f87ce06a9fb7/e9be30a4688f239c?lnk=gst&q=dobomode#e9be30a4688f239c I am working on developing an open-source automated system for running batch-regressions on very large datasets. In my previous post, I posed the question of obtaining VIF's from the output of

Hi, I am trying to find m breakpoints of a linear regression model. I used the segmented package. It works fine for small number of predicators and breakpoints.(3 r.v. 3 points). However, my model has 14 variables it even would not work even for just one breakpoints!. The error message is always estimated breakpoints are out of range. Since my problem is time related problem. So I