similar to: Mid-P value for a chi-squared test

Hi R-users, I would like to find the goodness of fit using Chi-suare test for my data below: xobs=observed data, xtwe=predicted data using tweedie, xgam=predicted data using gamma > xobs <- c(223,46,12,5,7,17) > xtwe <- c(217.33,39,14,18.33,6.67,14.67) > xgam <- c(224.67,37.33,12.33,15.33,5.33,15) > chisq.test(xobs, xtwe = xtwe, rescale.p = TRUE) Error in chisq.test(xobs,

I don't quite understand the difference between the two methods for performing a chi-squared test on contingency tables: summary(table()) and chisq.test() They may different results. E.g.: aa <- gl(2, 10) bb <- as.factor(c(1,2,2,2,1,2,1,2,2,2,1,2,2,2,1,1,1,2,1,1)) aa <- c(aa, aa) bb <- c(bb, bb) table(aa, bb) summary(table(aa, bb)) chisq.test(aa, bb) Could somebody give me

Dear UseRs, I'm running a chi-squared test where the expected matrix is the same as the observed, after rounding. R reports a X-squared of zero with a p value of one. I can justify this because any other result will deviate at least as much from the expected because what we observe is the expected, after rounding. But the formula for X-squared, sum (O-E)^2/E gives a positive value. What

Hello, I received the following warning when running chi-square; n Is there a way to catch the 'error' code of 'warning' after run chisq.test(x)? n What does this error mean? Thank you for your help. [[alternative HTML version deleted]]

Hello, I'm trying to calculate a chi-squared test to see if my data are different from the theoretical distribution or not: chisq.test(rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49, 60),c(80,80,80, 80, 80, 80, 80, 80, 80, 80, 80, 80))) Pearson's Chi-squared test data: rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49, 60), c(80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80,

Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 ..... 1 1.323

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 R-helpers, I need to generate standard variates normal to 'create' chi-squared variates. To make you more understand, (1)   a<-rnorm(3,0,1) *after do (1), I need to squared and summed the three values. My problem is, how am I going to continue the programming if I had to repeat the process for 15 times, which in the end I will get 15 values from the whole programme.Hope you can

Dear R users, I am a master student in Mathematics and I am writing my thesis in statistics. I need to use R and unfortunately I do not have any experience with a computer program. Could you please help me about chi squared goodness of fit test with R? In R-help website I saw a message about how to do that but I do not know how to cut the data into bins and calculate the expected numbers in each

R 2.12 windows 7 I am summary data that I would like to make into a 2x2 table representing counts positive vs. negative counts: 28/289 20/276 My table should look something like the following: group1 group2 Positive 28 20 Negative 289 276 How can a (1) create the 2x2 table (2) run a chi square test on the table? I have tried the following code, but I

I am getting "Chi-squared approximation may be incorrect in: chisq.test(x)" with the data bleow. Frequency distribution of number of male offspring in families of size 5. Number of Male Offspring N 0 518 1 2245 2 4621 3 4753 4 2476 5

Hi! I have some data that looks like this up down percentaje uew_21 20 14 58.82 uew_20_5 27 40 40.29 uew_20 8 13 38.09 uew_19_5 17 42 28.81 So I have 4 experimental conditions and I am counting number of animals in the up and down compartment and the calculating the percentage, I want to know which one of the conditions is different from each other. If the data wouldn't be percentage

Dear all, I have a question regarding performing test if the data fits chi-squared distribution. For example, using ks.test() I found in the examples how to fit it to gamma or weibull x<-rnorm(100) ks.test(x, "pweibull", shape=2,scale=1) for the gamma, pgamma can be used But I cannot find the value of this second parameter for the chi-squared distribution. Maybe someone

How i can perform a Pearson's Chi-squared Test in this data set: | Outcome -----------------+-----------+----------------------------------+ Treatment | Sex | None |Some | Marked | Total -----------------+------------+--------+--------+-------------+ Active | Female | 6 | 5 | 16 | 27

Dear R helpers: Thanks for the previous reply. I am using Friedman racing test. According the the book "Pratical Nonprametric Statistic" by WJ Conover, after computing the statistics, he suggested to use chi-squared or F distribution to accept or reject null hypothesis. After looking into the source code, I found that R uses chi-sqaured distribution as below: PVAL <-

Hi all, I know Chi-squared test can be done with the frequency data by R function "chisq.test()", but I am not sure if it can be applied to the percentage data ? The example of my data is as follow: ############################################# KSL MHL MWS CLGC LYGC independent (%) 96.22 92.18 68.54 93.80 85.74

Hi, I have written out the log-likelihood function to fit some data I have (called ONES20) to the non-central chi-squared distribution. >library(stats4) >ll<-function(lambda,k){x<-ONES20; 25573*0.5*lambda-25573*log(2)-sum(-x/2)-log((x/lambda)^(0.25*k-0.5))-log(besselI(sqrt(lambda*x),0.5*k-1,expon.scaled=FALSE))} > est<-mle(minuslog=ll,start=list(lambda=0.05,k=0.006))

Timothy, I believe you are mistaken. Fisher's exact test give the correct answer even in the face of small expected values for the cell counts. Pearson's Chi-square approximates Fisher's exact test and can give the wrong answer when expected cell counts are low. Chi-square was developed because it is computationally "simple". Fisher's exact test, particularly with tables

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]]

Full_Name: Jerry W. Lewis Version: 2.6.1 OS: Windows XP Professional Submission from: (NULL) (24.147.191.250) pchisq(0,0,ncp=lambda) returns 0 instead of exp(-lambda/2) pchisq(x,0,ncp=lambda) returns NaN instead of exp(-lambda/2)*(1 + SUM_{r=0}^infty ((lambda/2)^r / r!) pchisq(x, df + 2r)) qchisq(.7,0,ncp=1) returns 1.712252 instead of 0.701297103 qchisq(exp(-1/2),0,ncp=1) returns 1.238938