similar to: R: prop.test() and the simultaneous confidence interval for multiple proportions in R

Dear list members, I want to perform in R the analysis "simultaneous confidence interval for multiple proportions", as illustrated in the article of Agresti et al. (2008) "Simultaneous confidence intervals for comparing binomial parameter", Biometrics 64, 1270-1275. If I am not wrong the R function implementing the Agresti et al. method is prop.test(). I ask an help because I

Hello, Does anyone know which method from Newcombe (1998)* is implemented in prop.test for comparing two proportions? I would guess it is the method based on the Wilson score (for single proportion), with and without continuity correction for prop.test(..., correct=FALSE) and prop.test(..., correct=TRUE). These methods would correspond to no. 10 and 11 tested in Newcombe, respectively. Can

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

Dear all, Does someone know an R function implementing the method of Sison and Glaz (1995) (see full ref below) for computing Simultaneous confidence intervals for multinomial proportions? As alternative method, I think to boostrap the mean of each proportion and get in that way confidence interval of the mean. I observed 21 times a response that could be one out of 8 categories

Full_Name: Richard Johnston Version: 2.4.0 OS: OS X Submission from: (NULL) (69.169.0.241) The confidence interval calculation for prop.test appears incorrect when alternative="greater" . The upper limit is always set to 1.0000. The lower limit appears to be correct. > total=c(250,250) > success=c(55,31) >

The confidence interval calculation in prop.test appears to be incorrect when alternative="greater". The upper limit is always set to 1.000. Am I missing something? > total=c(250,250) > success=c(55,31) > prop.test(success,total,alternative="greater",correct=TRUE) 2-sample test for equality of proportions with continuity correction data: success out of

Hello! I want to calculate simultaneous confidence intervals for a nominal variable with three categories: "yes", "no", "partially" and I expect that far more than 5 samples fall into each category. I have read that Glaz & Sison's method is only appropriate for variables with 7 or more categories. Therefore, the Goodman method seems like a good idea. I have

Full_Name: Robert W. Baer, Ph.D. Version: 1.6.2 OS: Windows 2000 Submission from: (NULL) (198.209.172.106) Problem: prop.test() does not seem to produce appropriate confidence intervals for the case where the vector length of x and n is one. (I am not certain about higher vector lengths.) As an example, I include x=6 and n=42 which has a mean proportion of 0.115. When I calculate the 95% CI

Dear R-help, > prop.test(9, 137, p=0.066) > prop.test(9, 137, p=0.05) give two different 95% confidence intervals. I thought the confidence interval calculation should be independent of testing calculations (and thus the null hypothesis)? Splus 2000 has similar problems but give slightly different answer. Using R1.3.0 on windows. Mai Zhou

Hi, This code: n=40 x=17 phat=x/n SE=sqrt(phat*(1-phat)/n) zstar=qnorm(0.995) E=zstar*SE phat+c(-E,E) Gives this result: [1] 0.2236668 0.6263332 The TI Graphing calculator gives the same result. Whereas this test: prop.test(x,n,conf.level=0.99,correct=FALSE) Give this result: 0.2489036 0.6224374 I'm wondering why there is a difference. D. -- View this message in context:

This is partly an R and partly a general statistics question. I'm trying to get confidence intervals of proportions (sometimes for subgroups) estimated from complex survey data. Because a function like prop.test() does not exist for the "survey" package I tried the following: 1) Define a survey object (PSU of clustered sample, population weights); 2) Use svyglm() of the package

List, Would someone be so kind as to point me to a function that will calculate simultaneous confidence intervals of regression coefficients based upon their distribution as (under the assumption of normal errors, with \mathbf{X} as the design matrix): $\hat{\mathbf{\beta}} \sim N(\mathbf{\beta}, \sigma^2(\mathbf{X}^T\mathbf{X})^{-1})$. 'confint' calculates individual coefficients so

All, How come both of these are the same. Both say "1-sample proportions test without continuity correction." I would suspect one would say "without" and one would say "with." > prop.test(118,236,.5,correct=FALSE,conf.level=0.95) 1-sample proportions test without continuity correction data: 118 out of 236, null probability 0.5 X-squared = 0, df = 1,

Hi, I'm having no luck figuring out how to find Bonferroni simultaneous confidence intervals to obtain a family of estimates in R. Does anyone know how to do this? Thank you!

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 everyone, Suppose I have a count the occurrences of positive results, and the total number of occurrences: pos <- 14 total <- 15 testing that the proportion of positive occurrences is greater than 0.5 gives a p-value and confidence interval: prop.test( pos, total, p=0.5, alternative='greater') 1-sample proportions test with continuity correction data: 14 out of

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

I have just joined this list (and just started using R), so please excuse any etiquette breaches as I do not yet have a feel for how the list operates. I am in the process of teaching myself statistics using R as my utility as my ultimate goals cannot be satisfied by Excel or any of the plug-ins I could afford. I am currently looking at chap12 page 552 of Weiss's Introductory Statistics

I have the clinical study data. Year 0 Year 3 Retinol (nmol/L) N Mean +-sd Mean +-sd Vitamin A group 73 1.89+-0.36 2.06+-0.53 Trace group 57 1.83+-0.31 1.78+-0.30 where N is the number of male for the clinical study. I want to test if the mean serum retinol has increased over 3 years among subjects in the vitamin A group. > 1.89+0.36

Dear R users this may be a simple question - but i would appreciate any thoughts does anyone know how you would get one lower and one upper confidence interval for a set of data that consists of proportions. i.e. taking a usual confidence interval for normal data would result in the lower confidence interval being negative - which is not possible given the data (which is constrained between