similar to: R-alpha: class for chisquare tests; Thoughts on print & summary

I sort of asked this before, but perhaps not explicitly enough. In my ctest collection, there are several chisquare-based tests. For some of them, it may be useful to also return information on expected (and observed) counts. The question is, how should this be done. Of course, there is no problem adding the corresponding components to the list returned by the functions. However, as these are

Hello all you helpful people out there! I am stil R Beginner using R 2.5.1 on a Apple Power Book G4 with Mac OS X 10.4.10 . I have two tabels like this: MalTabChi X1 X4 X6 X8 X10 X14 X21 X24 X29 X38 X43 X50 X67 X76 X78 X80 X82 X84 Anth_cap 1 1 1 1 6 5 1 45 12 4 12 6 56 5 2 4 1 1 Anth_crin 12 1 1 2 76 5 1

In this years biostat teaching I will include Rcommander (it indeed simplifies syntax problems that makes students frequently miss the core statistical problems). But I could not find how to make a simple chisquare comparison between observed frequencies and expected frequencies (eg in genetics where you expect phenotypic frequencies corresponding to 3:1 in standard dominant/recessif

Hello! First of all I must appologize if this has been raised previously, but search provided by Robert King at the University of Newcastle seems to be down these days. Additionally let me know if such a question should be sent to R-help. I did a contribution to function hwe.hardy in package 'gap' during the weekend. That functions performs Hardy-Weinberg equilibrium test using MCMC. The

Hello, I've posted a question about this subject yesterday, but since there was no R code to comment, no one did. I'm trying to have the print method for class 'htest' print some extra information common in some test, like the time series linearity related tests. Many of them have an 'order' parameter, representing a lag or embedding dimension, and it would be a nice

Hi all, I just used power.t.test() in Version 1.9.0 beta (2004-03-31) today under FC1. The following is the output: > power.t.test(delta = .5, power = .95, type = "paired") $n [1] 53.94062 $delta [1] 0.5 $sd [1] 1 $sig.level [1] 0.05 $power [1] 0.95 $alternative [1] "two.sided" $note [1] "n is number of *pairs*, sd is std.dev. of *differences* within

Hello, everyone, I don't quite understand the following message: > TTT <- t.test(1:10, y=c(7:20)) > class(TTT) [1] "htest" > TTT@p.value Error: Trying to get slot "p.value" from an object whose class ("htest") is not defined > TTT$p.value [1] 1.855282e-05 Why the message says the class of TTT is not defined while class(TTT) gets

Hi everyone, I am quite new to R. my command is t.test(c(1:5,7:11), y=c(1:10),alternative = c("two.sided"), paired = TRUE) The output is Paired t-test data: c(1:5, 7:11) and c(1:10) t = 3, df = 9, p-value = 0.01496 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.1229738 0.8770262 sample estimates: mean of the differences

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}}

It's not a problem per se to put additional information into class htest objects (hey, it's S3 after all...) and there is a precedent in chisq.test which returns $observed and $expected. Getting such information printed by print.htest is more tricky, although it might be possible to (ab)use the $estimate slot. The further question is whether one would really want to do that (change the

I have a dataset called "raw-data" . I am trying to use the following code - col_name<-names(raw_data) for (i in 1:(length(names(raw_data))-2)) { tbl=table(raw_data$Pay.Late.Dummy, raw_data$col_name[i]) chisqtest<-chisq.test(tbl) } Say the 1st column of my raw_data is Column1. The idea is when i=1 then raw_data$col_name[i] will automatically become raw_data$Column1 ,

>>>>> peter dalgaard >>>>> on Fri, 22 Feb 2019 12:38:14 +0100 writes: > It's not a problem per se to put additional information > into class htest objects (hey, it's S3 after all...) and > there is a precedent in chisq.test which returns $observed > and $expected. It seems the consent is to simply return the SE but *not*

Hi I just discovered the answer thanks to a previous thread from Peter Daalgart The command is: sapply(test, "[[", "statistic") Cheers, Daniel ----- Original Message ----- To: r-help at r-project.org Subject: extracting values from a "by" function Date: Sun, 6 Jul 2008 05:59:04 +0800 Hello, I am trying to extract t and pvalues from a 1000 ttests using the

Hello, I am trying to extract t and pvalues from a 1000 ttests using the by-function but everythinhg I tried did not work. Unfortunately googling "by" is not very helpful. Any help will be very appreciated. Cheers, Danile Stall *creating a data set library(MASS) dataset <- mvrnorm(160, mu, Sigma) dataset <- as.data.frame(dataset) dataset$GROUP <- rep(1:10, each=16) data.uni

Hello- I've been trying to scour both * and linux user group archives for a solution to this particular problem, but I am just plan stuck. I got the latest zaptel sources from cvs, uncommented ztdummy.o in Makefile, ran make; make install then, did depmod -a. All is well up until now. When i do modprobe ztdummy, i get this error: dv@sidetrack:/usr/local/src/zaptel# depmod -a

Hi all, I'm getting contradictory results from bptest and ncvTest on a model calculated by GLS as: olslm = lm(log(rr)~log(aloi)*reg*inv, data) varlm = lm(I(residuals(olslm)^2)~log(aloi)*reg*inv, data) glslm = lm(log(rr)~log(aloi)*reg*inv, data, weights=1/fitted(varlm)) Testing both olslm and glslm with both ncvTest and bptest gives: > ncvTest(olslm) Non-constant Variance Score Test

Hi, I am trying to test for pairwise associations between genotypes ( Rows=individuals, Columns =genes, data are up to 4 genotypes per gene, some with 2,3 or 4) where each chisquare comparison is different depending on the genes tested. The test is the observed multilocus (across columns for each individual) genotypes vs the expectation, which is the product of the individual frequency for each

o <- (structure(list(sand.silt = structure(list(statistic = structure(185, .Names = "W"), parameter = NULL, p.value = 0.0478835773838087, null.value = structure(0, .Names = "location shift"), alternative = "two.sided", method = "Wilcoxon rank sum test with continuity correction", data.name = ".column by site"), .Names =

Hi, New to both R and SEM, so this may be a very simple question. I am trying to run a very simple path analysis using the sem package. There are 2 exogenous (FARSCH, LOCUS10) and 2 endogenous (T_ATTENT, RMTEST) observed variables in the model. The idea is that T_ATTENT mediates the effect of FARSCH and LOCUS10 on RMTEST. The RAM specification I used is FARSCH -> T_ATTENT, y1x1, NA

Hi John, Thanks for your reply. Of course I could write a package and of course I would find that trivial to do. The point is this is a main entry point to R for probably (at this point) hundreds of thousands of students. I?d like them to be able to get a basic quantity of interest from a t-test without four subsequent function calls. I also don?t really see the point about the object class,