similar to: pwr.prop.test and continuity correction

Hi, I am trying to do a power calculation for a difference in proportions test where I want to estimate the sample size required. I know (well estimate) that group one (n1) is 10% of the population and group 2 (n2) is 90% of the population. I know the effect size (h). pwr.2p2n.test only allows one variable to be left null whereas I would like both n1 and n2 to be determined where I know there

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

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

Hi All, I have a table based on ordial data and i want to compare proportions and i've seen in the pwr package i can use power.prop.test however i want to find out what the sig. value is based on n1,n2,p1,p2 and this package doesn't contain this.. Does anyone know of a package that does or is it a case of writting a function specifically for this? Many thanks in advance [[alternative

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,

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

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

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) >

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 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.

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

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

Hi all, I'm having trouble getting access to the pwr. This is on Ubuntu Lucid Lynx, 64 bit. I'm installing pwr via packages.install('pwr'), and loading it via library(pwr), both of which appear successful. Strangely, I never get access to the pwr object in R. I tried installing it to /usr/local/lib/R/site-library and ~/R/x86_64-pc-linux-gnu-library. I also tried installing it

On 12/22/2014 05:11 PM, Vince Hsu wrote: > The platform device does not use the common nouveau_pwr_init() to initialize > the PWR, but it does need the .prob() be assigned to avoid NULL pointer > dereference in graph/nve4.c. s/prob/pgob/ :-( Will fix in next version. > > Signed-off-by: Vince Hsu <vinceh at nvidia.com> > --- > > v2: this patch is added since v2. (v1

Dear all, For a population of about 200 I have a continuous variable and an ordinal variable. The question I would like to ask is whether the continuously increases (or decreases) as the rank of the ordinal variable increases. I was thinking that a Spearmen's rank correlation or or a chi squared trend might be appropriate. I don't have any experience dealing with ordinal variables so I

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

Hi everyone. I need to do a sample size calculation for a one sample based on a CI interval of +- 5% of the proportion, we expect to have a rate of 0.07 based on similar studies. I have looked at the pwr-package as well as the power.prop.test function, but I can't see how to make either do what I need. Could anyone point me to a function, or explain how to do this will the functions in pwr?

Hello, I would like to reshape my data for presentation purposes from something like this: > test <- data.frame(a=c("A","A","A","A","B","B","B"),b=c(1,1,2,2,1,1,1),c=1:7) > test a b c 1 A 1 1 2 A 1 2 3 A 2 3 4 A 2 4 5 B 1 5 6 B 1 6 7 B 1 7 to something like this: a b c 1 A 1 1,2 3 A 2 3,4 5 B 1 5,6,7 This seems

Hi, I am trying to produce a plot with an xlabel that reads (x >= 5%) with the >= turned into the correct epression. I can do this up to the percentage sign by specifiing xlab=expression(x>=5). Whatever I do to include the % sign as well doesn't seem to work. xlab=bquote(x>=5.("%")) almost works but includes brackets. Anyonw know how to solve this one Dan PS I am

Hello, I would like to scrape some html tables from a web page and convert them to a data.frame so I can perform further analysis. Could anyone tell me the best way to do this? Would it be more appropriate to use an external tool first? Thanks Dan -- ************************************************************** Daniel Brewer, Ph.D. Institute of Cancer Research Molecular Carcinogenesis