similar to: Problem with code for bootstrapping chi square test with count data

Hi, I have fitted a model using a glm() approach and using a gls() approach (but without correcting for spatially autocorrelated errors). I have noticed that although these models are the same (as they should be), the AIC value differs between glm() and gls(). Can anyone tell me why they differ? Thanks, Geertje ~~~~ Geertje van der Heijden PhD student Tropical Ecology School of Geography

Hi all, I have probably a basic question, but I can't seem to find the answer in the literature or in the R-archives. I would like to do a robust ANCOVA (using either rlm or lmRob of the MASS and robust packages) - my response variable deviates slightly from normal and I have some "outliers". The data consist of 2 factor variables and 3-5 covariates (fdepending on the model).

Hi, I have collected data on trees from 5 forest plots located within the same landscape. Data within the plots are spatially autocorrelated (calculated using Moran's I). I would like to do a ANCOVA type of analysis combining these five plots, but the assumption that there is no autocorrelation in the residuals is obviously violated. Does anyone have any ideas how to incorporate these spatial

Hi, I am trying to fit a ANCOVA model using lmRob. The P-values of the variables in the model differ hugely between the summary() function and the anova() function (from >0.8 in the summary to <0.001in the anova for the same variable). I understand that with an ANCOVA the order in which the variables are added to the model matters and that this influences the P-value, but can this make such

Hi, I am trying to calculate Moran's I test for the residuals for a regression equation, but I have trouble converting my coordinates into nb format. I have used the dnearneigh() funtion now with an arbitrarily high upper distance to make it include all plots. However, when I do the lm.morantest() I get a Moran's I value which is the same as the expected value and a P-value of 1. I

Hi all, I need to estimate S.E. of a certain indicator. The function to compute the value of indicator contains two arguments. Can anybody tell me how to do it? Example: We have data: a <- c(1:10) b <- c(11:20) data <- data.frame(a, b) Function to compute value of the indicator: indicator <- function(X, Y) sum(X)/(sum(Y)*2) Next I need to do the

Puede que no me haya expresado certeramente. Es lo que pasa al tocar de oído. Tengo unos datos (df) de los que se alguien calculó unas distancias (dis.df) de las que se hicieron/hacen unos clusters (plot.clus.dis.df). Todo esto usando un programa on line al que no tengo acceso para ver cómo hace los cálculos. Yo tengo los datos, y el programa también me da la matriz de distancias. Quiero,

Hello all, I am doing a Likelihood Ratio (LR) test in my simulation and I have a vector LR values (each with 1 degree of freedom) at the end of my simulation. Can anybody tell me how I can write a 'R' code which gives me the p-value for each of those LR values. Thanks [[alternative HTML version deleted]]

hi all - is there a package or library that contains a function for partitioning the chi-square statistic of an I X J contingency table into its respective independent parts? i looked around for this, but i didn't find anything. perhaps there's another name for this sort of analysis? i know it as "g-squared". thanks, chris. [[alternative HTML version deleted]]

Dear all, sorry to bother you but I was looking for the "scale Inverse Chi square distribution" and I could not find it! If I remember well , when X~Scale-Inv-Chi-Square (a,b) then X~Inv-Gamma(a/2,a*b/2) but with the shape, scale and rate parameters I always get confused! Any suggestions? Thanks a lot for your prompt reply. Best regards, Giorgio Di Gessa

I am trying to compute the periodicity of a time series. I would like to know which function in R does it. Also, how do I plot a Enright / Chi-square periodogram using R ? ( Enright, J.T., 1965, Journal of Theoret. Biol. 8,426-468) Greatly appreciate your help. Thanks in advance, Sukhaswami Malladi *************************************************************************** The contents

Dear all, I have a problem that I could not solve neither with spss nor with R. Please, excuse me if it is atrivial question but I did not find any soultion. I have the followig practical problem: we a product that has A, B, C effects (we differentiated about 30) on health and we want to know whether these effects are associated with its physical properties

I am looking for an R package with which one can calculate an effect size for a set of contrasts given an omnibus chi-square test statistic (more than 1 degree of freedom). Is there such a package? Presumably, it would implement the procedure (or something like it) described by Roznow and Rosenthal 1996 Psychological Methods 1: 331-340. many thanks! Steven Orzack Fresh Pond Research Institute

Hi, I want to determine the confidence interval on the sum of two sigma's. Is there an easy way to do this in R? I guess I have to use some sort of chisquare convolution algorithm??? Thanx, Roy -- The information contained in this communication and any atta...{{dropped}}

Hi all. I am seeking for a statistical suggestion. My data set comprises 382 measures each having 169 variables. Each measure is the outcome of a nuclear magnetic resonance experiment, so each of the 169 points has the same unit. As I want to do some multivariate calibration using these data, I checked whether some multivariate outliers existed. I calculated Mahalanobis distances and did a

Hi there, How accurate is the aproximation R makes to the Chi-Square distribution? For example, if I run: > qchisq(1/1000000,6) [1] 0.03650857 how accurate is 0.0365 compared to the theoretical percentile? What kind of approximations have been made in the software's algorithm? It woudl be useful to know since I am working with tiny percentiles such as one one-millionth and one

Hi (again), Please find here a correction of the function I send previously --------------------------------------------------------- tableChisqMH=function(x) { n=sum(x) G2=(n-1)*(tablepearson(x)$estimate^2) dll=1 p.value=1-pchisq(G2,dll) out=list(estimate=G2,dll=dll,p.value=p.value,dim=dim(x),name="Mantel-Haenszel Chi-square") return(out) }

Hi All, I woul like to ask you a couple of questions on chisq.test. First, I have 40 flies, 14 males and 26 females and I want to test for an a priori hypothesis that the sex ratio is 1:1 sex<-c(14,26) pr<-c(1,1)/2 chisq.test(se, p=pr, correct=TRUE) Chi-squared test for given probabilities data: sex X-squared = 3.6, df = 1, p-value = 0.05778 If my calculations are correct, this is

Hi R People: Sorry about the off topic question. Does anyone know the reference for "-2 Log Lambda is approx dist. Chi square", please? It may be Bartlett, but I'm not sure.... thanks in advance! Sincerely, Laura Holt mailto: holtlaura at gmail.com

Please could you tell me how to make code to outpot chi-square distribution table? Please help