similar to: Hypothsis simulation

I am considering testing: Ho: Odds Ratio =1 H1: Odds Ratio <>1 How can I generate data from the null distribution for a specific configuration of a Fisher exact test? Jim [[alternative HTML version deleted]]

I have two questions: (1) How do you 'create' an 2 x 2 table in R using say an Odd ratio of 3 or even 0.5 (2) If I have several 2 x 2 tables, how can I 'implement' dependence in the tables with say 25 of the Tables having an odds ratio of 1 and 75 of the tables having an odds ratio of 4? Jim [[alternative HTML version deleted]]

Dear all, I have something about function outer() that I can't understand. Just see the following example. The two NaNs are due to 0/0, but I can't figure out the cause of the last two errors. I wonder if some one can explain this for me. ___________________________________________________________________ > sx=rbinom(10,1,0.5);ot=rbinom(10,1,0.5);ag <- rbinom(10,100,0.3);ho <-

Hi R help list I'm simulating logistic regression data with a specified odds ratio (beta) and have a problem/unexpected behaviour that occurs. The datasets includes a lognormal exposure and diseased and healthy subjects. Here is my loop: ors <- vector() for(i in 1:200){ # First, I create a vector with a lognormally distributed exposure: n <- 10000 # number of study subjects

Hi, I noted a discrepancy between R and openepi when I ran a fisher test with the same matrix. In R: > a=matrix(c(1,2,6,17), nrow=2) > a [,1] [,2] [1,] 1 6 [2,] 2 17 > fisher.test(a, conf.int=T) Fisher's Exact Test for Count Data data: a p-value = 1 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.02061498

Hi, I have calculated chi-square goodness of fit test,Sample coming from Poisson distribution. please copy this script in R & run the script The R script is as follows ########################## start ######################################### No_of_Frauds<- c(4,1,6,9,9,10,2,4,8,2,3,0,1,2,3,1,3,4,5,4,4,4,9,5,4,3,11,8,12,3,10,0,7) lambda<- mean(No_of_Frauds) # Chi-Squared

I want to calculate chi square test of goodness of fit to test, Sample coming from Poisson distribution. please copy this script in R & run the script The R script is as follows ########################## start ######################################### No_of_Frouds<- c(4,1,6,9,9,10,2,4,8,2,3,0,1,2,3,1,3,4,5,4,4,4,9,5,4,3,11,8,12,3,10,0,7) N <- length(No_of_Frouds) # Estimation of

Hi all, I have the following data: rep1_treat rep2_treat rep1_control rep2_control 2 3 4 5 100 20 98 54 0 1 2 3 23 32 27

Hi. How can i simulate a binary data set from a logistic regression model?I need to manipulate parameters and so obtain my set of data. I want to show the improve in analyzing binary data with GLM(binomial) model instead of classical ANOVA or NON-MODELS procedures(relative risk-odds ratio-Pearson test of godness of fit...) Can you say me what is the right function to use? Do you know any

This code worked fine for me, then did some cleaning up of formatting using ESS (Emacs) and now I get this error, no idea what is causing it, all the brackets/parentheses seem to be balanced. What have I done wrong? Thanks Jim p0.trial01 <- 0.25 TruOR01 <- 0.80 num.patients.01 <- 50 num.trials.01 <- 5 LOR01.het.in <- 0.00 num.sims <- 1 simLOR01 <-

I'm trying to simulate trend data over a five year period. I want different trend profiles...the simplest being a linear trend. I've been using the following code: patBdta1 <- NULL for(i in 1:100) patBdta1 <- rbind(patBdta1,c(yr1= mean(rbinom(50,1,.50)), yr2 =mean(rbinom(50,1,.51)), yr3 =mean(rbinom(50,1,.52)),

I am hereby forwarding the data & method use to calculate the Kolmogorov-Smirnov goodness of fit test made manually by me in R launguage which deffers with the actual inbuilt formula as shown below. Further I have plot the graph in R. In that graph how to add trendline (i.e. straight line passing through maximum points in plot) to a Plot. R script is as follows please run this script to see

Hello, I have a simple (?) simulation problem. I'm doing a simulation with logistic model and I want to reapet it 600 times. The simulation looks like this: z <- 0 x <- 0 y <- 0 aps <- 0 tiss <- 0 for (i in 1:500){ z[i] <- rbinom(1, 1, .6) x[i] <- rbinom(1, 1, .95) y[i] <- z[i]*x[i] if (y[i]==1) aps[i] <- rnorm(1,mean=13.4, sd=7.09) else aps[i] <-

Hi, I have problem with simulating. This is my task... Suppose that there are N persons some of whom are sick with influenza. The following assumptions are made: * when a sick person meets a healthy one, the chance is á that the latter will be infected * all encounters are between two persons Write a function which simulates this model for various values of N (say, 10 000) and á (say, between

Hi Guy's I was wondering if someone could point me in the right direction. dbinom(10,1,0.25) I am using dbinom(10,1,0.25) to calculate the probabilty of 10 judges choosing a certain brand x times. I was wondering how I would go about simulating 1000 trials of each x value ? regards Brendan -- View this message in context:

Dear forum users, It's 3:35am and I am swamped with statistics homework lol I'm terrible with R and this time I have no idea what the prof wants. Here is the question: Consider the (two-??sample) Wilcoxon rank statistic T = ?rank(Xi). For n1=106 and n2=192, determine by simulation the ?=.05 critical point for testing H0: ?=0, H1:?<0. We can do this as follows: For m=10000 (no wimpy

In an early section on Action View, showing code being put directly into the rhtml file: <% require ''date'' DAY_NAMES = %w{ Sunday Monday Tuesday Wednesday Thursday Friday Saturday } today = Date.today %> <h1>Hello, Dave</h1> <p> It''s <%= DAY_NAMES[today.wday] %>. Tomorrow is <%= DAY_NAMES[(today + 1).wday %>. </p> I get this

Is anyone aware of a fast way of doing fisher's exact test for a series of 2 x 2 tables in R? The fisher.test is really slow if n1=1000 and n2 = 1000. -- Thanks, Jim. [[alternative HTML version deleted]]

hello, I have a problem in how to generate data in a simulation study. I have a logistic model to evaluate p by 3 covariates. I need to generate 4 variables: the binary outcome Y and 3 covariates: gender (binary) and aps and tiss (continuous variables). I have the logistic model which is the expected model: log(p(y=1)/(1-p(y=1))=-1.659-0.05*sex+0.063*aps+0.04*tiss0) I generate the outcome y like

Hola, Deseo simular un modelo logit con interacción, estimar sus coeficientes y comprobar si son o no parecidos al modelo teórico. Con este ejemplo obtengo que los coeficientes estimados no se asemejan mucho a los originales. ¿Se le ocurre a alguien cuál es el motivo de esta discrepancia? ¿y cómo solucionarlo? Muchas gracias Emilio logisticsimulation <- function(n){ dat <-