similar to: goodness-of-fit test

Try changing your method to "ML" and try again. I tried the run the first example from the documentation and it failed with the same error. Changing the estimation method to ML worked. @List: Can anyone else verify the error I got? I literally ran the following two lines interactively from the example for goodfit: dummy <- rnbinom(200, size = 1.5, prob = 0.8) gf <- goodfit(dummy,

I have troubles understanding how goodfit() function in the vcd package computes the Pearson coefficient. Can anybody provide more information on the computation? In particular, for HorseKicks data in vcd package, goodfit() yields > oo <- goodfit(HorseKicks,type="poisson",method="MinChisq") > summary(oo) Goodness-of-fit test for poisson distribution

Hello, Sample code: library("vcd") dummy <- rnbinom(200, size=1.5, prob=0.8) gf <- goodfit(dummy, type="nbinomial", method="MinChisq") plot(gf) I would like to: 1. add a lgened stating the bars show the actual counts and the red dots - the fit. 2. show the goodness-of-fit values calculated somewhere on an empty white space ob the plot. But... the legend

Dear All, I run the goodness of fit test using goodfit() in vcd package. The result is as follow: Goodness-of-fit test for poisson distribution X^2 df P(> X^2) Pearson 1.053348 2 0.5905661 Warning message: In summary.goodfit(gf) : Chi-squared approximation may be incorrect I want to save the the test statistics(X^2), df, and p-value. How can I save the result.

I'm looking for a package that has a start-of-the-art method of imputation of missing values in a data frame with both continuous and factor columns. I've found transcan() in 'Hmisc', which appears to be possibly suited to my needs, but I haven't been able to figure out how to get a new data frame with the imputed values replaced (I don't have Herrell's book). Any

Hi All, I have to say upfront that I am a complete neophyte when it comes to programming. Nevertheless I enjoy the challenge of using R because of its incredible statistical resources. My problem is this .........I am running a regression tree analysis using "rpart" and I need to run the calculation repeatedly (say n=50 times) to obtain a distribution of results from which I will pick

I have a data frame: > head(df) Time Temp Conc Repl Log10 1 0 -20 H 1 6.406547 2 2 -20 H 1 5.738683 3 7 -20 H 1 5.796394 4 14 -20 H 1 4.413691 5 0 4 H 1 6.406547 7 7 4 H 1 5.705433 > str(df) 'data.frame': 177 obs. of 5 variables: $ Time : Factor w/ 4 levels

Hello all! I am once again analyzing patient survival data with chronic liver disease. The severity of the liver disease is given by a number which is continuously variable. I have referred to this number as "meld"--model for end stage liver disease--which is the result of a mathematical calculation on underlying laboratory values. So, for example, I can generate a Kaplan-Meier plot

Hi Listers, Is it possible to produce an ordination plot in 2d, where bubbles represent the location of sites (this part is easy enough) and the size of the bubbles is proportional to the sites location in 3d space (I am stuck on this option). So sites that are very near the 2d plane of the xy axes would be larger while sites that are actually further away in 3 d space would be proportionally

How does one ideally handle and display multidimenstional contingency tables in R v. 2.6.2? E.g.: > prob1<- data.frame(victim=c(rep('white',4),rep('black',4)), + perp=c(rep('white',2),rep('black',2),rep('white',2),rep('black',2)), + death=rep(c('yes','no'),4), count=c(19,132,11,52,0,9,6,97)) > prob1 victim perp

Are there any packages in R that reproduce the package "resample" of S-Plus? The sample() function in R doesn't provide equivalent flexibility of bootstrap() and bootstrap2(). ================================================================ Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: ral at lcfltd.com Least Cost Formulations, Ltd. URL: http://lcfltd.com/ 824

I need to construct confidence intervals for the binomial variance. This is the usual estimate v = x*(n-x)/n or its unbiased counterpart v' = x*(n-x)/(n-1) where x = binomial number of successes observed in n Bernoulli trials from proportion p. The usual X^2 method for variance confidence intervals will not work, because of the strong non-normal character of the sampling

I'm using R 2.10.1 with the latest version of all packages (updated today). I'm confused as to why I'm getting a hard singularity in a simple set of experimental data: > blots ID Lot Age Conc 1 1 A 3 4.44 2 2 A 3 4.56 3 3 B 41 4.03 4 4 B 41 4.57 5 5 C 229 4.49 6 6 C 229 4.66 7 7 D 238 3.88 8 8 D 238 3.93 9 9 E 349 4.43 10 10 E 349

Hi, I have a dataset that has 2 groups of samples. For each sample, then response measured is the number of success (no.success) obatined with the number of trials (no.trials). So a porportion of success (prpop.success) can be computed as no.success/no.trials. Now the objective is to test if there is a statistical significant difference in the proportion of success between the 2 groups of

I have Vista Home with R-2.9.0, and installed and tried to test the package 'roxygen': > utils:::menuInstallPkgs() trying URL 'http://lib.stat.cmu.edu/R/CRAN/bin/windows/contrib/2.9/roxygen_0.1.zip' Content type 'application/zip' length 699474 bytes (683 Kb) opened URL downloaded 683 Kb package 'roxygen' successfully unpacked and MD5 sums checked The

Hello, I am a student and for project I need R. I have one problem regarding function goodfit(). Does goodfit() require frequency count of numbers or numbers themselves? For example suppose I have data say 150 readings.Do I need to use goodfit() directly on data or should I make suitable no of bins and then apply goodfit()? Aswad [[alternative HTML version deleted]]

I have R 2.6.2, and have tried downloading and installing the package "tseries". I get the same error when I use two different mirror sites: > utils:::menuInstallPkgs() trying URL 'http://cran.mirrors.hoobly.com/bin/windows/contrib/2.6/tseries_0.10-14.zip' Content type 'application/zip' length 400799 bytes (391 Kb) opened URL downloaded 391 Kb package

-----Original Message----- From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of r-help-request at r-project.org Sent: Monday, February 11, 2008 12:00 PM To: r-help at r-project.org Subject: R-help Digest, Vol 60, Issue 11 Send R-help mailing list submissions to r-help at r-project.org To subscribe or unsubscribe via the World Wide Web, visit

Hi everybody, somehow i dont get the shapiro wilk test for normality. i just can?t find what the H0 is . i tried : shapiro.test(rnorm(5000)) Shapiro-Wilk normality test data: rnorm(5000) W = 0.9997, p-value = 0.6205 If normality is the H0, the test says it?s probably not normal, doesn ?t it ? 5000 is the biggest n allowed by the test... are there any other test ? ( i know qqnorm

I'm trying to figure out how to carry out a Poisson regression fit to longitudinal data with a gamma distribution with unknown shape and scale parameters. I've tried the 'lmer4' package's glmer() function, which fits the Poisson regression using: library('lme4') fit5<- glmer(seizures ~ time + progabide + timeXprog + offset(lnPeriod) + (1|id), data=pdata,