similar to: chi-square test

To whom it may concern: The help page for ?legend refers to a `title.cex` parameter, which suggests that the function has such a parameter. As far as I can tell, though, it doesn't; here's an example: > plot(1,1) > legend("topright",pch=1, legend="something", title="my legend", title.cex=2) Error in legend("topright", pch = 1, legend =

All, I have a data file with 56 entries that looks like this: City State JanTemp Lat Long Mobile, AL 44 31.2 88.5 Montgomery, AL 38 32.9 86.8 Phoenix, AZ 35 33.6 112.5 Little Rock, AR 31 35.4 92.8 Los Angeles, CA 47 34.3 118.7 San Francisco, CA 42 38.4 123.0 I would like to "read" this data into a dataframe. Is it possible to do without editing the datafile? D.

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]]

Suppose we have a data file containing: "Smith, John", 120, 90 "Thomson, Peter", 110, 85 there are 3 variables in it. If we use x <- read.table("tmp.txt", sep=",") to read the data to a data.frame, the result will be 4 columns. Splus 4.0 have no problem with this kind of data. Li Dongfeng

Suppose we have a data file containing: "Smith, John", 120, 90 "Thomson, Peter", 110, 85 there are 3 variables in it. If we use x <- read.table("tmp.txt", sep=",") to read the data to a data.frame, the result will be 4 columns. Splus 4.0 have no problem with this kind of data. Li Dongfeng

Dear R-help mailing list,   Kindly help me out with this problem:   I have a dataset that is in the format below, ID  time  Y  Age 1  0     195  23.1 1  2    204  23.3 1  4   202    23.5 2  0  170    22.0 2  3   234   22.2 3  0  208   24.4 3  2  194  24 .7 3  3   204  24.9   I wish to remove all the measurements at time point 0 and convert them to a baseline variable as follows;    ID  time  Y 

Dear all, I would like to use predict.lm() with an existing lm object but with new arbitrary coefficients. I modify 'fit$coef' (see example below) "by hand" but the actual model in 'fit' used for prediction does not seem to be altered (although fit$coef is!). Can anyone please help me do this properly? Thanks in advance, J?r?mie > dat <-

Hi, Please can someone tell me if this is correct for significance tests on count data; 1: If overdispersed, use quasipoisson, drop1(model, test="Chi"), 2: If not overdispersed, use (poisson or quasipoisson), drop1(model, test="Chi"). Thanks for your time, Martin. Martin Hoyle, School of Life and Environmental Sciences, University of Nottingham, University Park, Nottingham,

Hello all, I am doing a Likelihood Ratio (LR) test in my simulation and I have a vector LR values (each with 1 degree of freedom) at the end of my simulation. Can anybody tell me how I can write a 'R' code which gives me the p-value for each of those LR values. Thanks [[alternative HTML version deleted]]

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, sorry to bother you but I was looking for the "scale Inverse Chi square distribution" and I could not find it! If I remember well , when X~Scale-Inv-Chi-Square (a,b) then X~Inv-Gamma(a/2,a*b/2) but with the shape, scale and rate parameters I always get confused! Any suggestions? Thanks a lot for your prompt reply. Best regards, Giorgio Di Gessa

I am trying to compute the periodicity of a time series. I would like to know which function in R does it. Also, how do I plot a Enright / Chi-square periodogram using R ? ( Enright, J.T., 1965, Journal of Theoret. Biol. 8,426-468) Greatly appreciate your help. Thanks in advance, Sukhaswami Malladi *************************************************************************** The contents

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

Dear all, I have a problem that I could not solve neither with spss nor with R. Please, excuse me if it is atrivial question but I did not find any soultion. I have the followig practical problem: we a product that has A, B, C effects (we differentiated about 30) on health and we want to know whether these effects are associated with its physical properties

I am looking for an R package with which one can calculate an effect size for a set of contrasts given an omnibus chi-square test statistic (more than 1 degree of freedom). Is there such a package? Presumably, it would implement the procedure (or something like it) described by Roznow and Rosenthal 1996 Psychological Methods 1: 331-340. many thanks! Steven Orzack Fresh Pond Research Institute

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 want to determine the confidence interval on the sum of two sigma's. Is there an easy way to do this in R? I guess I have to use some sort of chisquare convolution algorithm??? Thanx, Roy -- The information contained in this communication and any atta...{{dropped}}

Hi all. I am seeking for a statistical suggestion. My data set comprises 382 measures each having 169 variables. Each measure is the outcome of a nuclear magnetic resonance experiment, so each of the 169 points has the same unit. As I want to do some multivariate calibration using these data, I checked whether some multivariate outliers existed. I calculated Mahalanobis distances and did a

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

Hi (again), Please find here a correction of the function I send previously --------------------------------------------------------- tableChisqMH=function(x) { n=sum(x) G2=(n-1)*(tablepearson(x)$estimate^2) dll=1 p.value=1-pchisq(G2,dll) out=list(estimate=G2,dll=dll,p.value=p.value,dim=dim(x),name="Mantel-Haenszel Chi-square") return(out) }