search for: qchisq

Full_Name: David Clayton Version: 1.8.1 OS: Linux Submission from: (NULL) (131.111.126.242) qchisq behaves very strangely when ncp is passed as zero (forcing internal qnchisq to be called) when first argument is small. Eg > qchisq(1-1e-6, 1, ncp=0, lower.tail=TRUE) qchisq(1-1e-6, 1, ncp=0, lower.tail=TRUE) [1] 1024 while, if ncp is unspecified, > qchisq(1-1e-6, 1) qchisq(1-1e-6, 1) [1...

Full_Name: Drouilhet R?my Version: 1.8.1 OS: Linux Submission from: (NULL) (195.221.43.136) qchisq(1,10) works well but qchisq(1,10,ncp=0) does not work whereas ncp=0 is the default value of the function qchisq(1,10). (of course, 10 will be replaced by any integer value). Let us notice that this bug occurs only when applying probability one. (qchisq(seq(0,.9,.1),10,ncp=0) works very well).

Full_Name: Viktor Witkovsky Version: 2.9.2 OS: Windows XP Submission from: (NULL) (78.98.89.227) Hello, I have found strange behavior of the function qchisq (the non-central qchisq is based on inversion of pchisq, which is further based on pgamma). The function gives wrong results without any warning. For example: qchisq(1e-12,1,8.94^2,lower.tail=FALSE) gives 255.1840972465858 (notice that here the correct value should be 255.1841334848075), but qchis...

Hi, I want to compute the quantiles of Chi^2 distributions with different degrees of freedom like x<-cbind(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975, 0.99, 0.995) df<-rbind(1:100) m<-qchisq(x,df) and hoped to get back a length(df) times length(x) matrix with the quantiles. Since this does not work, I use x<-c(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975, 0.99, 0.995) df<-c(1:100) m<-qchisq(x,df[1]) for (i in 2:length(df)) { m<-rbind(m,qchisq(x,df[i])) } di...

...##### p26=function(x,alpha){ # dummy variable j=1 ci=matrix(ncol=2,nrow=3) while (j<4){ if (j==2) {x=x+c(-1,1)*0.5} ci[j,]= x+qnorm(1-alpha/2)^2/2+ c(-1,1)*qnorm(1-alpha/2)* sqrt(x+qnorm(1-alpha/2)^2/4) j=j+1 if (j==3) { # exact x=x-c(-1,1)*0.5 ci[j,]=c( qchisq(1-alpha/2,df=2*x)/2, qchisq(alpha/2,df=2*x+2)/2) j=j+1 } } rownames(ci)=c('without','with','exact') colnames(ci)=c('lower','upper') return(round(ci,2)) } ################################################### Most bits are fine. The problem part...

Hello all, Here's my query: Running R 1.6.2 on FreeBSD 5.0, and on WinXP, and I find that the following hangs the process: dchisq(alpha=0.01, df=1, ncp=295) but it does work for ncp < 294.92. Is this general? Best wishes to all, Andrew Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885 6226 University of Idaho E : andrewr at uidaho.edu PO

Hello, I want to use the quantile function so I read the doc but I don't understand with this > qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1)) [1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39 63160.27 63205.65 63250.33 63295.04 63340.48 63387.48 63437.03 63490.53 63550.14 63619.68 [18] 63707.24 63837.16 Can you help me please? __________________________________________...

Hi, How do we get the value of a chi square as we usually look up on the table on our text book? i.e. Chi-square(0.01, df=8), the text book table gives 20.090 > dchisq(0.01, df=8) [1] 1.036471e-08 > pchisq(0.01, df=8) [1] 2.593772e-11 > qchisq(0.01, df=8) [1] 1.646497 > nono of them give me 20.090 Thanks, cruz

Hello developers and users: My system fails (the computer freezes) when I use the ncp parameter, with the lower.tail=FALSE option in the qchisq function. qchisq(0.025,31,ncp=1,lower.tail=FALSE) Thank you very much for your help. Kenneth Cabrera Universidad Nacional de Colombia ICNE Sede Medellin krcabrer at perseus.unalmed.edu.co PS I am using: $platform "i386-pc-mingw32" $arch "x86" $os "Win32" $syst...

Kenneth Cabrera <krcabrer@epm.net.co> writes: > Hello developers and users: > > My system fails (the computer freezes) when I use the ncp parameter, > with the lower.tail=FALSE option in the qchisq function. > > qchisq(0.025,31,ncp=1,lower.tail=FALSE) Yup, that's a bug. We have in pnchisq.c 48 for (ux = 1.0; pnchisq(ux, n, lambda, lower_tail, log_p) < p; ux *= 2); 49 for (lx = ux; pnchisq(lx, n, lambda, lower_tail, log_p) > p; lx *= 0.5); but if we look...

...ard <p.dalgaard@biostat.ku.dk> writes: PD> Kenneth Cabrera <krcabrer@epm.net.co> writes: >> Hello developers and users: >> >> My system fails (the computer freezes) when I use the ncp parameter, >> with the lower.tail=FALSE option in the qchisq function. >> >> qchisq(0.025,31,ncp=1,lower.tail=FALSE) PD> Yup, that's a bug. We have in pnchisq.c PD> 48 for (ux = 1.0; pnchisq(ux, n, lambda, lower_tail, log_p) < p; PD> ux *= 2); PD> 49 for (lx = ux; pnchisq(lx, n, lambda, lower_t...

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 instead of 0

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 2 2.773 3 4.108 Thanking you Prasanna Prasanna Balaprakash, Universit? Libre de Bruxelles, 50, Av. F. Roosevelt, CP 194/6, B-1050 Brussels, Belgium...

...sigma/mu # CV > .33 may give poor results, so warn the user if(k>.33) warning("Confidence interval may be very approximate.") method <- casefold(method) # In case we see "McKay" if(method=="mckay"){ # McKay method. See equation 15. t1 <- qchisq(1-alpha/2,v)/v t2 <- qchisq(alpha/2,v)/v u1 <- v*t1 u2 <- v*t2 lower <- k/sqrt(( u1/(v+1) -1)*k*k + u1/v) upper <- k/sqrt(( u2/(v+1) -1)*k*k + u2/v) } else if (method=="naive"){ # Naive method. See equation 17. t1 <- qchisq(1-alpha/2,v)/v...

...poulos" <dimitris.rizopoulos at med.kuleuven.ac.be> >To: "Prasanna Balaprakash" <pbalapra at ulb.ac.be> >Cc: <r-help at stat.math.ethz.ch> >Sent: Friday, January 21, 2005 1:05 PM >Subject: Re: [R] chi-Squared distribution > > >> if you check ?qchisq, you'll see that the second > >My mistake, the third argument > >> argument of the function denotes the non-centrality parameter! Try >> >> qchisq(0.75, 1:3) >> >> to get your answer. >> >> Best, >> Dimitris >> >> ---- >&gt...

...hod. I think there is an error in the computation for the Newcombe method. Specifically, where the code says # Put the data in the right form x1 <- aa; n1 <- aa+cc; p1 <- x1/n1 x2 <- bb; n2 <- bb+dd; p2 <- x2/n2 pd <- x1/n1 - x2/n2 z <- qnorm(1-alpha/2) zz <- qchisq(1-alpha/2,1) I believe zz (which is used in the Newcombe method, while z is used for Agresti-Caffo) should be zz <- qchisq(1-alpha,1) (though I think they could just as well have written xx <- z^2), since z^2 seems to be what they want for the Newcombe method, and qnorm(1-alpha/2)^2 = qc...

...urn(identify(X)) invisisble(X) } qqplot.mv.interactive <- function (data, xlim=NULL, ylim=NULL) { x <- as.matrix(data) # n x p numeric matrix center <- colMeans(x) # centroid n <- nrow(x); p <- ncol(x); cov <- cov(x); d <- mahalanobis(x,center,cov) # distances qqInteractive(qchisq(ppoints(n),df=p),d, # ppoints(n) makes a sequence from 0 to 1. with stepsize 1/n main="QQ Plot Assessing Multivariate Normality", # qchisq() makes a chi squared distribution function for the given probabilities in ppoints(n) and degress of freedom df ylab=...

...e question: Can I find somewhere in R the significance values for a Kolmogorov distribution (I know the degrees of freedom and I have already the maximum deviation). ks.test is not really doing what I want. All I need is the values, like one can get the values for a chi-squared distribution by 'qchisq(0.05, 375)'. tnx, Kurt.

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 one-billionth, and I am not s...

Hi, Attention chi-squared distribution, unlike F distribution, has only df1 as parameter, not df1 and df2. So correct into: outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1)) ^^^^^^^^^^^^^^^^^^^^ Regards, Vito you wrote: 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...