similar to: Noncentral t & F distributions

Hi everyone. I have a problem that I have been unable to determine either the best way to proceed and why the methods I'm trying to use sometimes fail. I'm using the pf() function in an optimization function to find a noncentrality parameter that leads to a specific value at a specified quantile. My goal is to have a general function that returns the noncentrality parameter that

Hello I could not find whether there is any R-function that implements the inverse of a noncentral Beta. Could someone out there tell me where I can find it? Or how to implement it? Many thanks Ed [[alternative HTML version deleted]]

I have a data with 236 observations. After plotting the histogram, I found that it looks like non-central t distribution. I would like to get MLE for mu and df. I found an example to find MLE for gamma distribution from "fitting distributions with R": library(stats4) ## loading package stats4 ll<-function(lambda,alfa) {n<-200 x<-x.gam

Hello all. It seems that the pf() function when used with noncentral parameters can behave badly at times. I've included some examples below, but what is happening is that with some combinations of df and ncp parameters, regardless of how large the quantile gets, the same probability value is returned. Upon first glance noncentral values greater than 200 may seem large, but they are in

Hi there: I'd thought these two versions of noncentral t-distribution are essentially the same: > qqplot(rt(1000,df=20,ncp=3),qt(runif(1000),df=20,ncp=3)) But, the scales of the x-axis and the y-axis are quite different according to the QQ-plot. Did I make any mistakes somewhere? Thanks, Long ---------------------------------

I sent a message a couple days ago about doing calculations for power of the ANOVA. Several people got back to me very quickly which I really appreciated. I'm working now on a similar problem, but instead of a balanced ANOVA, I have an unbalanced one. The first part of the question was: You assume that the within-population standard deviations all equal 9. You set the Type 1 error rate at á

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))

Hello Everyone, I've been told that the memory allocation in R and S-Plus is different, with R being better at doing simulations for whatever reason(s). I was wondering if anyone could enlighten me on the differences in the memory usage/allocation (preferably in layman's terms!). The reason I ask this is because of a simulation I am trying to run. In it, I need to use the pt( )

hello R-friends, I'm looking for a quantile function for the noncentral f-distribution in the area of equivalence hypotheses testing. Can somebody help me? Many thanks ----------------------------------------------------------------- Dipl. Inform. J. Hedderich Institut f?r Medizinische Informatik Phone : 0431 / 5973182 und Statistik im Klinikum an der CAU

Hi, Actually I am facing a similar problem. I would like to fit both an ordinary (symmetric) and a non-central t distribution to my (one-dimensional) data (quite some values.. > 1 mio.). For the symmetric one, fitdistr or funInfoFun (using fitdistr) from the qAnalyst package should do the job, and for the non-central one.. am I right to use gamlss(x ~ 1, family=GT()) ? Anyway, I am a little

Full_Name: Version: 1.9.0 OS: Windows XP Submission from: (NULL) (134.126.93.191) I believe the noncentrality parameter, lambda, in power.anova.test is incorrect. The noncentrality paramter is defined as lambda <- (groups - 1) * n * (between.var/within.var). The n should not be there. The function pf defines the noncentrality parameter as the sum of squares of the means. Therefore, the

Full_Name: Long Qu Version: 2.3.1 OS: Windows XP Submission from: (NULL) (64.113.93.235) The QQ-plot of two versions of simulating noncentral F-distributed random numbers has quite different scales: > qqplot(rf(1000,2,15,3),qf(runif(1000),2,15,3)) The rf() function reads: > rf function (n, df1, df2, ncp = 0) { if (ncp == 0) .Internal(rf(n, df1, df2)) else rchisq(n, df1,

Full_Name: Jerry W. Lewis Version: 2.7.0 OS: Windows XP Professional Submission from: (NULL) (198.180.131.16) If you are saying that there is no need to solve for the noncentrality parameter, please justify this amazing assertion. If you are saying that this need is already adequately addressed in R, then please enlighten me.

For those who are interested, I have made available a R function for noncentral hypergeometric distribution at http://www.geocities.com/jg_liao/software/Hypergeometric/hypergeometric_in_R.txt The paper that describes the algorithm will appear in The American Statistician. The function does not run on S-plus as the R's scoping rule is used. Here is how the function can be used: > n1

Hello! I have a problem with integrate() in my function nctspa(). Integrate produces an error message "evaluation of function gave a result of wrong length". I don't know what that means. Could anyone suggest me what is wrong with my function? These are the examples of function calls that work OK: nctspa(a=1:10,n=5) nctspa(a=1:10, n=5, mu=2, theta=3, renorm=0) This does not work:

Does anybody have any experience fitting data to the non-central chi-squared or chi-squared distribution? I am trying to fit some data to this distribution but there is error after error. audaces fortuna iuvat --------------------------------- [[alternative HTML version deleted]]

Hi, I would like to know if there is a program written in R to get the CDF (cumulative distribution function) of a bivariate non-central chi-square distribution. Hope someone will reply. Thank you, Rossita M Yunus yunus@usq.edu.au This email (including any attached files) is confidentia...{{dropped:19}}

Full_Name: Mike Meredith Version: 2.3.0 OS: WinXP SP2 Submission from: (NULL) (210.195.228.29) Using dt() with a non-centrality parameter and near-zero values for 'x' results in erratic output. Try this: tst <- c(1e-12, 1e-13, 1e-14, 1e-15, 1e-16, 1e-17, 0) dt(tst,16,1) I get: 0.2381019 0.2385462 0.2296557 0.1851817 0.6288373 3.8163916 (!!) 0.2382217 The 0.238 values are okay,

Hello All, Has anyone written a function for the distribution function of a *doubly* non-central F-distribution? I looked through the archives, but didn't find anything. Thanks! Wolfgang

The problem occurs when the sample odds ratio is Inf, such as in the following example. Given the fact that both upper bounds of the two 95% confidence intervals are Inf, I would have expected that the two lower bounds be equal, but they aren't. x <- matrix(c(9,4,0,2),2,2) x # [,1] [,2] #[1,] 9 0 #[2,] 4 2 rbind("two.sided.95CI"=fisher.test(x)$conf.int,