similar to: Odds ratio from Logistic model in R

Hi all. A simple question. Is there a function to compute the Odds Ratio and its confidence intervall, from a logistic model (glm(.......,family=binomial....). I've written my own, but certainly someone did a better job. Thank you in advance, Stefano *********************************************** Stefano Calza Istituto di Statistica Medica e Biometria Universit? degli Studi di Milano Via

hi there tried to use subset with drop=TRUE, but all the 'old' levels are preserved, i.e. when calling e.g. ftable a lot of zeros are displayed >x<-subset(LREG, (kir=='AA' | kir=='BB') & (type=='t1' | otype=='t2'), drop=TRUE, select=c(event, kir, type)) > ftable(x) i explicit have to call factor like

Windows XP R 2.8.1 Colleagues, I am trying to run a function testone() and if the function completes without error do one set of instructions, and if the function generates either a warning or an error run another set of instructions. I have read try, and tryCatch help screens at least 20 times, have tried to experiment with code, but I can't understand how to accomplish my desired task.

I am having trouble getting a loop to work for the following problem. Any help would be much appreciated. Thanks. I need to find the slope and intercept from the linear regression of Drug Level on Day by Participant. There are a total of 37 Participants. I need to store the Participant, Label, Slope, and Intercept in a new data frame. This data is ordered by Participant number 37 total

Hi, Found no measure of association or correlation for 2x2 contingency tables in fullrefman.pdf or google. Can someone point to a package that implements such calculations? Thanx. -- Alexandre Santos Aguiar - consultoria para pesquisa em sa??de - R Botucatu, 591 cj 81 tel 11-9320-2046 fax 11-5549-8760 www.spsconsultoria.com

Full_Name: Marketa Kylouskova Version: 1.2.1 OS: Windows Submission from: (NULL) (147.231.7.2) Hi, I believe there is a bug in the extension of wilcox.test, namely in determination of confidence intervals. I concentrated on the two-sample test only. I examined and tested its performance on this working data set: x<-rnorm(10,3,1) # So I have two samples, I will test whether

Hi All, I've hit a problem using nls. I think it may be a restriction in the applicability of nls and I may have found a fix, but I've been wrong before. This example is simplified to the essentials. My real application is much more complicated. Take a function of matrix 'x' with additional arguments: matrix 'aMat' whose values are _not_ to be determined by nls vector

Hello, I have a question regarding the candisc package. My data are: species three five 1 2.95 6.63 1 2.53 7.79 1 3.57 5.65 1 3.16 5.47 2 2.58 4.46 2 2.16 6.22 2 3.27 3.52 I put these in a table and then a linear model >newdata <- lm(cbind(three, five) ~ species, data=rawdata) and then do a candisc on them >candata<-candisc(newdata)

Hi eveybody. I've seen posted before the same question, but no answer. Is there anybody who has implemented the Hosmer-Lemeshow test in R? TIA, Stefano -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

Dear all, I have a four variable: Stuy.Name, OR, 95%LCI and 95%UCI and I would like to create a meta analysis plot. I can't use meta.MH function in metaplot because I do not have n.trt, n.ctrl, col.trt, col.ctrl are not available! Is there an alternative way to do it? Many thanks in advance, Cheba [[alternative HTML version deleted]]

Full_Name: Wei Xu Version: 1.5.1 OS: WindowsME Submission from: (NULL) (63.215.238.92) The P-value for a two.sided test is not consistent with the confidence interval. For example, P-value=0.1372, but the 95% CI doesn't include the H0 value(1). > x [1] 0.80 0.83 1.89 1.04 1.45 1.38 1.91 1.64 0.73 1.46 > y [1] 1.15 0.88 0.90 0.74 1.21 >

Hello, I'm having some issues getting a ggplot figure to show up in the knitr output, when placed in a loop. Specifically, I have a loop inside a knitr chunk : ```{r fitting, warning=FALSE, fig.width=10, fig.height=10, fig.keep='high'} for (t in 1:T) { # do a regression of tgt.vals ~ predictors and compute coeffs and fitted values (fit.vals / fit.adj.vals) plot(

Muchísimas gracias Carlos, de verdad que te agradezco la ayuda, pero no es lo que voy buscando. Quiero colocar en el eje de abscisas la secuencia temporal de los meses, es decir, agosto septiembre, octubre, etc? pero no las fechas de las toma de datos, sino que aparezca la marca de un mes, y la siguiente marca sea la del siguiente mes, etc?, y además que las muestras estén separadas de acuerdo con

Hello everybody, I'm writting some functions for experimental designs. The one I'm working on is similar to "fac.design" on Splus. The problem I have is with the form of the output : When the argument fraction is submitted, Splus gives something like : A B C 1 - - - 2 + + - 3 + - + 4 - + + Fraction : ~A:B:C The first part of this output is the

Muchas gracias Carlos, previo a mi correo, entre las pruebas que hice estaba una parecida a la que apuntas de la siguiente manera: attach (Libro1) plot (xbar~as.Date(fechas,"%d/%m/%y"), ylim=c(400,660), xaxt="n", type="b", pch=19,cex=1) xlabels<-strptime(fecha,format="%d/%m/%Y") axis.Date (1,at=xlabels,format="%b-%y")

Hola de nuevo, acabo de encontrar la solución. He creado una variable ficticia con los días 1 de cada mes en la secuencia temporal que quería y después he actuado de la siguiente manera attach(Libro1) plot (xbar~as.Date(fecha,"%d/%m/%y"), type="b", pch=19,cex=2,xaxt="n") xlabels<-strptime(ofeje, format = "%d/%m/%y") axis.Date(1,

Hi Folks, I'm using the 'norm' package (based on Shafer's NORM) on some data. In outline, (X,Y) are bivariate normal, var(X)=0.29, var(Y)=24.4, cov(X,Y)=-0.277, there are some 900 cases, and some 170 values of Y have been set "missing" (NA). The puzzle is that, repeating the multiple imputation starting from the same random seed, I get different answers from the repeats

Am I correct in assuming that the output below essentially translates to "Males have a mean time that is significantly lower than Females"? Is this the correct way to interpret the fact that the coefficient is negative? Assume the variale sex is treated as a factor with Female =0 and Male=1. survmodel<-survreg(survobj~sex,data=data1, dist="weibull")

I'm a new user, so my apologies for what is likely a dumb question... I am having a hard time getting a table of regression results when using Koenker's code for quantile regression with fixed effects (http://www.econ.uiuc.edu/~roger/research/panel/rq.fit.panel.R). I use the example data parameters that Koenker provides (see below). m <- 3 n <- 10 s <- rep(1:n,rep(m,n)) x <-

Hi Thanks for your response. However I seem to be doing something wrong regarding the try block resulting in yet another error described below. I have a function that takes in a file name and does the fit for the data in that file. Hence based on your input, I tried try ( (fit = lm(y~x, data = data_fitting)), silent = T); I left the subsequent lines of my code unchanged. coeffs =