similar to: Chi-squared test

Hi, I try to calculate Kendall's W coefficient and I have a bizarre error. little.app.mat<-matrix(c(1,3,4,2,6,5,2,4,3,1,5,6,3,2,5,1,5,4),nrow=3,byrow=TRUE) print(kendall.w(little.app.mat[-1,])) >>> Kendall's W for ordinal data >>> W = 0.7753623Error in if (is.na(x$p.table)) { : argument is of length zero

Deall all, We need to perform a statistical analysis of a large database (40,000 entries with approximately 500 fields in each entry) currently handled in Oracle. The data contains categorical variables only. At the current stage we suggest classification and clustering analysis. We are planning to perform the analysis in R and would be very grateful for any

Hi, I am a new user of R. I have 2 series of nominal data (2 series of answers for the same question) and I want to calculate the correlation between these 2 series. I've tried to use the correlation function (corr, ...) but all are for numeric data... Does anyone know what function should I use for the nominal data? Thanks a lot, Bianca

Hi, how can I import really big plain text data files (several GB) from an ERP-System (SAP-Tables) to R? The Header of these files are always similar, for example: Tabelle: T009 Angezeigte Felder: 7 von 7 Feststehende F??hrungsspalten: 2 Listbreite 0250 ---------------------------------------------------------------------- |X|MANDT|PERIV|XKALE|XJABH|ANZBP|ANZSP|LTEXT

Hi there, > > I am using the function pgls in the CAPER package, everything seems to run > fine except for one of my variables. > When using this variable as continuous the anova works without problem, > but if using the same variable as factor (in fact, this is what it is), the > program returns this answer: > > Error in terms.formula(formula, data = data) : >

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

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

At my physics lab we have 30 servers with 1TB disk packs. I am in need of monitoring for disk failures. I have been reading about SMART and it seems it can help. However, I am not sure what to look for if a drive is about to fail. Any thoughts about this? Is anyone using this method to predetermine disk failures? TIA

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

You really do need ECC RAM, but for the naysayers: http://www.cs.toronto.edu/%7Ebianca/papers/sigmetrics09.pdf -- richard

Hello! I have several variables. Each of them has a different distribution. I was thinking to use a Generalized Linear Model, glm(), but I need to introduce the family. Do you know if R has any tests for matching data to any distribution ( I am aware of shapiro.test). All the best, -- BAS [[alternative HTML version deleted]]

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]]

Folks, I'm trying to get Asterisk to load my voicemail configuration from MySQL. I've followed the instructions at: http://www.voip-info.org/wiki-Asterisk+voicemail+database I restarted Asterisk, but no luck: the voicemail.conf does not get updated. I started with a sample voicemail.conf that I found on the Wiki. Or was it from Voicepulse? I can't remember. For initial testing, I

Dear Opus team, We are a PC, Tablets and Smartphones manufacturer in Brazil http://www.positivoinformatica.com.br/ . We were informed that the VP9 Video Codec will be added by Microsoft in the new versions of our contracts with them. We reseached this codec and sow that is developed by you and is royalty free. Would you please help me to check whether to sign any tipe of contract for the

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]]

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

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))