similar to: Confidence Interval for SMR

The concrete problem is that I am refereeing a paper where a confidence interval is presented for the risk ratio and I do not find it credible. I show below my attempts to do this in R. The example is slightly changed from the authors'. I can obtain a confidence interval for the odds ratio from fisher.test of course === fisher.test example === > outcome <- matrix(c(500, 0, 500, 8),

Dear list, I need to convert this data.frame > names(codesM) [1] "key" "AMR.pa1.M" "AMR.pa2.M" "AMR.pa3.M" "AMR.pa4.M" [6] "AMR.pa5.M" "AMR.pa6.M" "AMR.pa7.M" "AMR.pa8.M" "AMR.pa9.M" [11] "AMR.pa10.M" "AMR.ta1.M" "AMR.ta2.M" "AMR.ta3.M"

Dear all, I would like to ask one question related to statistics, for specifically on defining dummy variables. As of now, I have come across 3 different kind of dummy variables (assuming I am working with Seasonal dummy, and number of season is 4): > dummy1 <- diag(4) > for(i in 1:3) dummy1 <- rbind(dummy1, diag(4)) > dummy1 <- dummy1[,-4] > > dummy2 <- dummy1 >

Hi I am trying to carry out some maximum likelihood estimation and I'm not making much headway, and I'm hoping that someone will be able to point me in the right direction. I am modelling mortality statistics. One way to do this is to model the mortality rate (or, more accurately, log of the mortality rate, log_m) as (say) a constant plus a proportion of age, plus time, so: r_1 <-

At work after updating to R 2.5 I get an error using epitab from package epitools, when at home (R 2.4) I get no error. Could someone help me? Thanks Pietro Bulian Servizio di Onco-Ematologia Clinico-Sperimentale I.R.C.C.S. Centro di Riferimento Oncologico Via Franco Gallini 2 33081 Aviano (PN) - Italy phone: +39 0434 659 412 fax: +39 0434 659 409 e-mail: pbulian at cro.it (at work)

Hello, Is there a better way to use paste such as: a = paste(colnames(list.indep)[1],colnames(list.indep)[2],colnames(list.indep)[3],colnames(list.indep)[4],colnames(list.indep)[5],sep="+") > a [1] "aa+dummy1+dummy2+bb+cc" I tried a = paste(colnames(list.indep)[1:5],sep="+") > a [1] "aa" "dummy1" "dummy2"

I want to do kernel smoothing on the standardised mortality rates at specified location(x-y coordinates) and be able to produce contour map. I've tried to use the package called fields in R (function smooth.2d) but the problem is I get negative results after smoothing and I do not understand why? Your contributions will be highly appreciated or if there is any package that can do this I

I haven't looked inside to see what is causing this, but there's a big discontinuity in qgamma: curve(qgamma(x, shape=19), from=1e-10, to=2e-10) This appears in both R-patched and R-devel. Duncan Murdoch

Dear list members, is it possible to set some options only inside a function so that the original options are restored once the function is finished or aborted due to an error? Until now I do something like: dummy=function() { old.options=options(error=dummy1()) .... options(old.options) } This works for most cases but when the function terminates because of an error and its last

Hello ALL! Can anyone tell me how to specify qqmath() function with distribution qgamma? In help it is bit vague how to pass shape and scale parameters for gamma. Thank you in advance. Best, PM

hello ! i try to use a network technology on one single host, which wasn`t designed for that. to give a short overview of what i`m talking about: AoE is just like a "networked blockdevice" (just like nbd/enbd) - but without tcp/ip. AoE kernel driver is the "client end" (see this like an iSCSI initiator) - and an etherblade storage appliance is the "server end"

Hi, there I have a loop that is producing data, but is also generating an error message that I can't understand. Here's the loop and the error message: > bio<-matrix(NA, ncol=9, nrow=366) > W<-NULL > M<- length(Day) #number of days iterated > > for (i in 1:M) + { + + + if (Day[i]==1) W[i] <- Wo else W[i] <- (W[i-1]+(Gr[i]/Ef)) + + + C<-

Doese anyone know a R function to find the median of a gamma distribution? [[alternative HTML version deleted]]

I am relatively new to R and object oriented programming. I have relied on SAS for most of my data analysis. I teach an introductory undergraduate forecasting course using the Diebold text and I am considering using R in addition to SAS and Eviews in the course. I work primarily with univariate or multivariate time series data. I am having a great deal of difficulty understanding and working with

Dear all I have a work-flow issue with lm(). When I use > lm(y1~x1, anscombe) Call: lm(formula = y1 ~ x1, data = anscombe) Coefficients: (Intercept) x1 3.0001 0.5001 I get as expected the formula, "y1 ~ x1", in the print()ed results or summary(). However, if I pass through a formula object > (form <- formula(y1~x1)) y1 ~ x1 > lm(form, anscombe) Call:

Dear all, I have a problem with importing an excel file into R. I can open the file easily (either saving it as a CSV or using RODBC). But the original file is using merged cell in its first column, which gives the name of the observation. (I am dealing with repeated measurements for the same observation) So when I open the dataframe in R it looks like this Col1 Col2 Col3 name1 val1 val2

can you recommend a good manual for R that starts with a data set and gives demonstrations on what can be done using R? I downloadedR Langauage definition and An introduction to R but haven't found them overly useful. I'd really like to be able to follow some tutorials using a dataset or many datasets. The datasets I have available on R are Data sets in package 'datasets':

Full_Name: Version: 2.7.1 (2008-06-23) OS: windows vista Submission from: (NULL) (216.82.144.137) Hello, I have been working with various probability distributions in R, and it seems the gamma distribution is inaccurate for some inputs. For example, qgamma(1e-100, 5e-101, lower.tail=FALSE) gives: 1.0. However, it seems this is incorrect; I think the correct answer should be

Hello R-users I was puzzled by some strange results of an analysis and I found out what is for me a strange behaviour (I won't dare to say a bug) in both plot and points (and I suspect lines and other kind of lower level plots). If you try the following code: > data<-data.frame(dummy1=c(1:10),dummy2=c(1:10)) > plot(data$dummy1,data$dummy3) A plot of the values of dummy1 against

Hi there, First off, thanks to everyone who has helped me so far. Sorry to keep pestering you all. I'm including my code here, and I will comment down it where it is that I am having problems figuring out how to write this damn thing. > temper <- scan("temp2.dat", na.strings = ".", list(Day=0, Temp=0)) Read 366 records > > Day <- temper$Day ;