similar to: Confidence interval for the difference between proportions - method used in prop.test()

Dear Josh, Thanks for your help! Does your answer mean, that you agree the two methods should do the same, and what I was guessing, despite the small differences? What I prefer about ci.pd is, that the help clearly says which method is implemented, which is not the case for prop.test. But I do not know who has programmed the function. Best wishes Steffi Stefanie von Felten, PhD Statistician

Hi everyone, Is it possible to use tapply(x,y,mean) if not all groups of x by y are of the same length (for example if you have one missing observation)? I tried tapply(x,y,mean,na.omit=T) but it doesn't work! Steffi -- --------------------------------- Stefanie von Felten Doktorandin ETH Z?rich Institut f?r Pflanzenwissenschaften ETH Zentrum, LFW A 2 Telefon: 044 632 85 97 Telefax: 044

Hi, I would like to conduct a MANOVA. I know that there 's the manova() funciton and the summary.manova() function to get the appropriate summary of test statistics. I just don't manage to specify my model in the manova() call. How to specify a model with multiple responses and one explanatory factor? If I type:

Hi everyone, Can someone explain me how to calculate SAS type III sums of squares in R? Not that I would like to use them, I know they are problematic. I would like to know how to calculate them in order to demonstrate that strange things happen when you use them (for a course for example). I know you can use drop1(lm(), test="F") but for an lm(y~A+B+A:B), type III SSQs are only

Hi, I have done different discriminant function analysis of multivariat data. With the CV=True option I was not able to perform the predict() call. What do I have to do? Or is there no possibility at all? You also need the predicted values to produce a plot of the analysis, as far as I know. Here my code: pcor.lda2<-lda(pcor~habarea+hcom+isol+flowcov+herbh+inclin+windprot+shrubcov+baregr,

Greetings! I suspect that there is an error in the code for the function ci.pd() in the Epi package. This function is for computing confidence intervals for a difference of proportions between two independent groups of 0/1 responses, and implements the Newcombe ("Nc") method and the Agrasti-Caffo "AC" method. I think there is an error in the computation for the Newcombe

Dear R-help, I'm using prop.test() to compute a confidence interval for a proportion under R version 2.4.1, as in: prop.test(x = 340, n = 400)$conf [1] 0.8103309 0.8827749 I have two questions: 1) from the source code my understanding is that the confidence interval is computed according to Wilson, E.B. (1927) Probable inference, the law of succession, and statistical inference. J. Am.

Hi, I wish to install R on my computer but I do not know how to do this. I have a Windows ME. Please advise on how to install R. Thank you. Stefanie [[alternative HTML version deleted]]

Hi all, Presuming that my reply on r-help this morning was correct, attached is a patch file against the current svn trunk version of prop.test.Rd to add the references for the methods. Any corrections are welcome. Regards, Marc Schwartz

Preparing a paper for a medical journal. Using the prop.test() function in R (v2.4.0) to compare two groups' response to data like the following. A sample of 100 individuals from Population I, 18 with positive readings from a certain test, vs. A sample of 148 individuals from Population II, 61 with positive readings. Results look like this: R version 2.4.0 Patched (2006-11-25

Hello, I run into a problem: > ftable(table(Fire, Standard, StoAll), col.vars=c("Fire","Standard")) Error in table(Fire, Standard, StoAll) : object 'Fire' not found I do not understand that because when I read the table everything seems correct. > Stocking_all<-read.table("P://Benchmark//analysis//r//stocking_10//stock

Hi, I would like to prioritize VOIP traffic when we use the phone, but other times not do traffic shaping at all. Right now I have my openwrt router set up with htb to do shaping. In order to get it to work well I had to set my upload and download speeds much lower than my line speed. With these settings, I get good VOIP reception even while surfing the net and doing a long download. However,

I can't get the y axis to extend the full range that I need, which is -8 to 8 Here's my code. I tried using ylim, but it's still truncating at the extremes in my data. plot(x, y, type ="o", xlab="Panelist", ylab="T value",lwd=1.5,lty=1) xlim=range(1:14),ylim=range(-8:8), las=1) Any suggestions? Thanks! -- View this message in context:

Hi Group, I have a matrix, and I would like to replace numbers less than 0 by the smallest minimum number. Below is an small matrix, and the loop I used. I would like to get suggestions on the "R way" to do this. Thanks, Juliet # example data set mymat <- structure(c(-0.503183609420937, 0.179063475173256, 0.130473004669938, -1.80825226960127, -0.794910626384209, 1.03857280868547,

Dear R-devs, Below is a small example of what I am trying to achieve, that is trivial in R and I would like to learn how to do in C, for very large matrices: > (mymat <- matrix(c(1,0,0,2,2,1), nrow = 2)) [,1] [,2] [,3] [1,] 1 0 2 [2,] 0 2 1 And I would like to produce: [1] "a*C" "B*c" Which can be trivially done in R via something like: foo

Dear R-experts, I have found a splom-modification online which is given below. This works perfectly, but I would like to have a matrix of given correlation values to be used in the lower triangular part (lower.panel) of the splom-plot instead of calculated correlation values. Here is the matrix I would like to use (it can be any other convenient data structure):

Hi all, as I am trying to move slowly from just "working" to "good" code, I'd like to ask if there's a smarter way than using a for-loop in tasks like the example below. I need to obtain the extrema of the cumulated sum of a detrended time series. The following code is currently used, please have a look at the comments for my questions and remarks: system.time({ X

Dear all, I have about 30,000 matrix (512x512), with values from 1 to N. Each value on a matrix represent a habitat patch on my matrix (i.e. my landscape). Non-habitat are stored as ZERO. No I need to change each 1-to-N values for the same random number. Just supose my matrix is: mymat<-matrix(c(1,1,1,0,0,0,0,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 3,3,0,0,0,0,0,0,0,4,4,

Hi everybody! I have a matrix of class "myClass", for example: myMat <- matrix(rnorm(30), nrow = 6) attr(myMat, "class") <- "myClass" class(myMat) When I extract part of ''myMat'', the corresponding class ''myClass'' unfortunately disappear: myMat.p <- myMat[,1:2] class(myMat.p) Please for any advice / suggestions, how

I noticed something surprising (in R 2.2.1 on WinXP) According to the documentation, rownames and colnames are character vectors. Assigning a vector of class POSIXct or POSIXlt as rownames or colnames therefore is not strictly according to the rules. In some cases, R performs a reasonable typecast, but in some other cases where the same typecast also would be possible, it does not. Assigning a