similar to: Scale-Inverse Chi square Distribution

I am running lrm() with a single factor. I then run anova() on the fitted model to obtain a p-value associated with having that factor in the model. I am noticing that the "Model L.R." in the lrm results is almost the same as the "Chi-Square" in the anova results, but not quite; the latter value is always slightly smaller. anova() calculates the p-value based on

Hi, This is more of a general question than a pure R one, but I hope that is OK. I want to combine one-tailed independent p-values using the weighted version of fisher's inverse chi-square method. The unweighted version is pretty straightforward to implement. If x is a vector with p-values, then I guess that this will do for the unweighted version: statistic <- -2*sum(log(x)) comb.p <-

Dear list, I had some confusion regarding what function too use in order too relate results from spec.pgram() too a chi-square distribution. The documentation indicates that the PSD estimate can be approximated by a chi-square distribution with 2 degrees of freedom, but I am having trouble figuring out how to do it in R, and figuring out what specifically that statement in the documentation

Dear R-help,   Chi Square Test for Goodness of Fit     Problem Faced :   I have got a discrete data as given below (R script)   No_of_Frauds <-c 1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,2,1,2,2,2,1,1,2,1,1,1,1,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,5,1,2,1,1,1,1,1,1,1,3,2,1,1,1,2,1,1,2,1,1,1,1,1,2,1,3,1,2,1,2,14,2,1,1,38,3,3,2,44,1,4,1,4,1,2,2,1,3)   I am trying to fit

Chi Square Test for Goodness of Fit   I have got a discrete data as given below (R script)   No_of_Frauds<-c(1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,2,1,2,2,2,1,1,2,1,1,1,1,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,5,1,2,1,1,1,1,1,1,1,3,2,1,1,1,2,1,1,2,1,1,1,1,1,2,1,3,1,2,1,2,14,2,1,1,38,3,3,2,44,1,4,1,4,1,2,2,1,3)   I am trying to fit Poisson distribution to this data using R.  

if i want to plot the chi-square distribution with a different degree of freedom how can i plot it in the graph?Sometimes i plot the histogram and cut it in a lot of piece.It's distribution like a chi-square.So i want to plot the chi-square with a different degree of freedom to compare it . -- View this message in context:

Dear list, For educational purposes I have been working with the script below. I have a observation: line 31 #CScal[i] = (amo^2) # IT IS WRONG, I KNOW, BUT IT MAKE R TO CRASHES! I'm thinking this is a possible bug in the R! And I have a couple of doubts: 1) line 51 #curve(dchisq(x, n-1), add = T, col = 'red' I think that it is correct, but the function no

Dear help list. I would like to plot the chi-square distribution for n=1 through 10, on one plot. How do I go about doing that? I know how to plot one chi-square (using curve() for example), but plotting 10 on one plot, that is too difficult for me. I greatly appreciate your help. -- fb isabellesup at fastmail.fm --

Hello, I've been attempting to fit the data below with an inverse gamma distribution. The reason for this is outside proprietary software (@Risk) kicked back a Pearson5 (inverse gamma) as the best fitting distribution with a Chi-Sqr goodness-of-fit roughly 40% better than with a log-normal fit. Looking up "Inverse gamma" on this forum led me the following post:

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

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

Dear All, Is there any function in R for computing "multivariate chi-square distribution"? How about "multivariate gamma distribution"? I appreciate any comment on this subject. Thank you, Amin Zollanvari PhD student Department of Electrical and Computer Engineering, Texas A&M University, College Station, TX

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

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

Sir, I have a problem here while applying chisquare test to the following Data ( below the subject of this mail) ...when I wanted to test the significance using three different free statistical packages, here R, EpiInfo and OpenEpi. *Only OpenEpi accepts the test based on Cochran's Recommendations. * R says " chi squared approximation may be incorrect." Does it mean the same as

Hi to all I found an formular of an ** ***p-Value Calculator for the Chi-Square test* *http://www.danielsoper.com/statcalc/calc11.aspx* *with the formula* *http://www.danielsoper.com/statkb/topic11.aspx* *what's the gamma function of this formula in r?* *df=5* *ch2=25.50878* *the following code does not give the result <0.001 for the values above * *p=

Hi. I have made 100 experiments of an M/M/1 queue, and for each one I have calculated both, mean and variance of the queue size. Now, a professor has told me that variance is usually chi-squared distributed. Is there a way in R that I can find the parameter that best fits a chi-square to the variance data? I know there's fitdistr()m but this function doesn't handle chi-square. I believe

Hi, I want to do a chi square test and I have two tab delimited text files with Expected and Observed values to compare. Each file contains only the values and are 48 rows by 116 columns. I have managed to do something with them, but I don't think it is right as I got a p value of 1. In this case I used the read.table() function to read the values from the files. But I don't know if

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

Dear all, I met a problem while doing SEM by sem package. I got a negative chi-square of null model. Because the theoretical value of chi-square cannot be negative, I checked the source code of sem.R in sem package and I found the Chi-square of null model was computed by the following expression: result$chisqNull <- (N - 1) * (sum(diag(S %*% diag(1/diag(S)))) + log(prod(diag(S)))) I think