similar to: Breslow Day Test

Hello all, I have a very simple set of data and I would like to analyze them with probit analysis. The data are: X Event Trial 100 8 8 75 8 8 50 6 8 25 4 8 10 2 8 0 0 8 I want to estimate the value of X that will give a 95% hit rate (Event/Trial) and the corresponding 95% CI. Anyone can offer some help? Thanks!! -

Hi, Thanks in advance for the help. I have a few vectors and I need to get all the combinations of the vector values. Normally I would use a few for-loops to do that. e.g x1_c(1,3,5) x2_c(2,5,6) for (i in x1) { for (j in x2) { ... }} My problem is that I don't know how many vectors there are ahead of time, so I don't know how many for-loops I need. Any

Hi all, I want to know how to perform the test Breslow-Day test for homogeneity of odds ratios (OR) stratified for svytable. This test is obtained with the following code: epi.2by2 (dat = daty, method = "case.control" conf.level = 0.95, units = 100, homogeneity = "breslow.day", verbose = TRUE) where "daty" is the object type table svytable consider it, but

Breslow-Day test A statistical test for the homogeneity of odds ratios. Homogeneity In <javascript:void(0);> systematic reviews homogeneity refers to the degree to which the results of studies included in a review are similar. "Clinical homogeneity" means that, in studies included in a review, the participants, interventions and outcome measures are similar or comparable.

Dear all, I have been trying to find a function to calculate the Breslow-Day test for homogeneity of the odds ratio in R. I know the test can be preformed in SAS but i was wondering if anyone could help me to perform this in r. In addition i have the fullrefman file to search for functions in the basic R packages, does anyone have any suggestions of an efficient way of searching for

Dear R Users, I'm looking for a package that allows to test hypothesis about a homogeneity of odds ratio in k 2x2 tables. I know that Breslow-Day is suitable but does anybody could me point out a package? I found diffR, but as far as I see this package is for IRT theory. Best, Robert

Hi all, The survivals functions can be tested by the Log-rank test and others, for example the Gehan-Breslow. The graham breslow work with the alpha values. But I don't know how is the Gehan-Breslow test with R. Somebody know a type function?.. or other suggestions? Any help will be really appreciated Jos? Bustos Marine Biologist Master Apllied Stat Program University of Concepci?n

Dear R-users, I am checking the proportional hazard assumption of a cox model for a given covariate, let say Z1, after adjusting for other relavent covariates in the model. To this end, I fitted cox model stratified on the discrete values of Z1 and try to get beslow estimator for the baseline cumulative hazard function (H(t)) in each stratum. As far as i know, if the proportionality assumption

Hi all, I am trying to run the k-means cluster analysis using the function kmeans in the package cluster. The data are: x = c(-0.26, -0.23, -0.05, -0.20, 0.30, -0.84, -0.10, -0.12, 0.10, -0.31, -0.19, 0.18, -0.26, -0.23, -0.37, -0.23) I've got two different solutions when I ran this function over a few times: kmeans(x, centers=2) The first solution gives the following: $cluster [1]

Dear list, I have a simple question. If I need to run linear regression on a subset of the dataset, will the following two commands give the same output: lm(y ~ as.factor(x1) + as.factor(x2), data, subset = z==Z[1]) and lm(y ~ as.factor(x1) + as.factor(x2), data[data$z==Z[1],] ) I have got different results running the two commands. Am I missing something here, or it there a bug on the

Hi all, Suppose I have a vector Fruits <- c('Apple','Orange','Pear','Banana','Mango') I want to print the statement "The fruits in the basket are: Apple, Orange, Pear, Banana, Mango" If I do: print(paste('The fruits in the basket are:',Fruits)) The output is: [1] "The fruits in the basket are: Apple" "The fruits in

How does R compute the p-value in the Wilcoxon Rank Sum Test? If I have the test statistics, can I get the p-value using the function pwilcox? Thanks -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- 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 R helpers, I have a 3D scatter plot that I have generated from scatterplot3d (which looks great- thanks!) and I can see that the points in my graph fall in a plane. Following the example 5 from 3D scatter plot (below) I have fitted a regression plane. Now what I would like to do is a rotation so that my new co-ordinate system is about the fitted plane (by finding the normal to the plane

  Hi there,   I am beginner in R and I have some basic question. Suppose I run a common procedure such as a t test or cox model like below:   out<-coxph( Surv(tstart,tstop, death1) ~ x1+x1:log(tstop+1) , test1,method=c("breslow"))    Which yields the following result:   Call: coxph(formula = Surv(tstart, tstop, death1) ~ x1 + x1:log(tstop +     1), data = test1, method =

Dear R users, I'm interested in measuring the stability of a heirarchical clustering, of the overall clustering and finding sub clusters (from cutting the heirarchical clustering at different levels) which demonstrate stability. I saw some postings on the R help from a while back about bootstrapping for clustering (using sample and generating a consesus tree with a web based tool CONSENSE)

Dear list, I want to create a POSIX time vector as follows: day <- as.character("110809") time.t <- 1:3600 t.min <- time.t %/% 60 t.sec <- time.t-t.min*60 DATE <- as.POSIXct(strptime(paste(day,t.min,t.sec),"%y%m%d %M%S")) Tail(DATE) The problem is that the last element (3600) returns a NA and I don't understand why. 600, 1200, 2400

Dear r-helpers, Prior to analysis of variance, I ran the Boxcox function (MASS library) to find the best power transformation of my data. However, reading the Boxcox help file, I cannot figure out if this function (through its associated log-likelihood function) corrects for * normality only * or if it also induces * homogeneity of variances *. I found in Biometry (Sokal and Rohlf, p. 419)

We occasionally utilize the coxph function in the survival library to fit multinomial logit models. (The breslow method produces the same likelihood function as the multinomial logit). We then utilize the predict function to create summary results for various combinations of covariates. For example:

I am confused when trying the function survfit. my question is: what does the survival curve given by plot.survfit mean? is it the survival curve with different covariates at different points? or just the baseline survival curve? for example, I run the following code and get the survival curve #### library(survival) fit<-coxph(Surv(futime,fustat)~resid.ds+rx+ecog.ps,data=ovarian)

Dear R Help list, My question is regarding extracting the standard error or Z-score from a cph or coxph call. My Cox model is: - modz=cph(Surv(TSURV,STATUS)~RAGE+DAGE+REG_WTIME_M+CLD_ISCH+POLY_VS, data=kidneyT,method="breslow", x=T, y=T) I've used names(modz) but can't see anything that will let me extract the Z scores for each coefficient or the standard errors in the same