similar to: Chi-squared test (PR#9350)

Full_Name: Sahotra Sarkar Version: 2.2.0 OS: Windows XP Professional Submission from: (NULL) (146.6.130.180) The following two commands should give the same results for the eigenvectors but do not (there is a sign reversal for the first one): > summary(princomp(bumpus),loading = TRUE) Importance of components: Comp.1 Comp.2 Comp.3 Comp.4 Comp.5

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]]

Olly Betts writes: > On Wed, Jan 04, 2017 at 07:29:58AM +0100, Jean-Francois Dockes wrote: > > Olly Betts writes: > > > The ticket has a patch which attempts to handle the OR case (which seems > > > to be the part you actually care about) but this suffers from issues with > > > object lifetimes which get a bit involved in the details. Since there >

Folks -- I've read the FAQ at samba.org/samba/docs/FAQ, I've searched the mailing list archives for "unexpected network error", I've searched Google for "samba unexpected network error", and been unable to find anything. If there's another source of information that I should check, I'd greatly appreciate a pointer. I'm trying to copy files to Samba

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]]

Thanks. I assume there is no way to control this via. environment variables or configure settings? Obviously that would be great for something like this which affects tests and seems to be a known problem for older C standard libraries. Best, Kasper On Thu, May 4, 2017 at 9:12 AM, Tomas Kalibera <tomas.kalibera at gmail.com> wrote: > > As a quick fix, you can undefine HAVE_CTANH

Yes, the potential issue I see is that make check fails when I explicitly set TZ. However, I set it to be the same as what the system reports when I login. Details: The system (RHEL) I am working on has $ strings /etc/localtime | tail -n 1 EST5EDT,M3.2.0,M11.1.0 $ date +%Z EDT $ echo $TZ US/Eastern On Fri, Oct 2, 2020 at 9:48 AM Sebastian Meyer <seb.meyer at fau.de> wrote: > Thank

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

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

Full_Name: Alan Bain Version: 2.4.0 OS: XP Submission from: (NULL) (155.140.122.227) Code for pnchisq contains following if (tSml) { if (x> f+ theta+ 5*sqrt(2*(f+2*theta) ))){ return 1.; /* better than 0 --- but definately FIXME */ } } This needs to check which tail has been requested; it is only correct if the default lower_tail=1 has been requested; for upper tail should return 0

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

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

Hi, pchisq -> distribution function dchisq -> density function pval is the area under the curve, to calculte it you use distribution function which is the integral of density function. See: http://www.itl.nist.gov/div898/handbook/eda/section3/eda362.htm http://mathworld.wolfram.com/DistributionFunction.html f(x) density function F(x) distribution function =Pr(X<x)= integral(f(x))

Dear R helpers, >From the lrm( ) model used for binary logistic regression, we used the L.R. model value (or the G2 value, likelihood-ratio chi-squared statistic) to evaluate the goodness-of-fit of the models. The model with the lowest G2 value consequently, has the best performance and the highest accuracy. However our model includes rsc() functions to account for non-linearity. We

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, Can you tell me what R function to use to do a two-sample chi-squared test? I want to see if two distributions are significantly different from each other, and I don't specify the theoretical distribution of either. For example, I have the following fake count data: x <- sample(1:10,50,replace=TRUE) y <- sample(1:10,100,replace=TRUE) I saw chisq.test in the stats package, but

I've got a dataset which looks like this in the beginning: cbr dust smoking expo 1 0 0.20 1 5 2 0 0.25 1 4 3 0 0.25 1 8 4 0 0.25 1 4 5 0 0.25 1 4 (till no. 1240, anyway, a huge set) I have to analyse cbr and smoking, I know it works with chisq.test() for the whole set, but I only need cbr and smoking, and I

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

I would like to request access to bugzilla to file a bug report on make check for R-devel. Following changes to all.equal.POSIXt, make check now reports an error if the environment variable TZ is set to TZ="US/Eastern" (and likely other values). This can be addressed by using the argument check.tzone=FALSE in the call to all.equal.POSIXt in reg-tests-2.R. A patch has been posted by

Thanks, I'm going to shut up then. Despite having been reported multiple times, it was not at all clear to me that this was being worked on with the intention of addressing make check - that is one of the limitations of the communication system we use. Best, Kasper On Fri, Oct 30, 2020 at 11:38 AM Tomas Kalibera <tomas.kalibera at gmail.com> wrote: > Dear Kasper, > > if you