similar to: a statistic question,a bit off-topic,but important

Hi everyone, Here's how I solved a problem for my stats class. I'm pretty sure I understand what's going on, but I wonder if there's a more direct way to solve it. Problem summary: A recent poll indicated that Candidate A is leading B with 55% of the vote. How many voters need to be surveyed to ensure a margin of error of +/- 2.5% with 99% confidence. Here's what I did:

Steve, 1) go to /etc/asterisk 2) open modules.conf for editing using vi 3) add this line: noload=pbx_wilcalu.so 4) Save the file 5) Restart asterisk Lightup the candles, open the Cabernet Savignon ( or whatever your prefernce) and call your girlfriend. ;) Seshu -----Original Message----- From: asterisk-users-bounces@lists.digium.com

Full_Name: Uffe H?gsbro Thygesen Version: 2.2.0 OS: linux Submission from: (NULL) (130.226.135.250) Hello all. pbinom(q=0,size=0,prob=0.5) returns the value NaN. I had expected the result 1. In fact any value for q seems to give an NaN. Note that dbinom(x=0,size=0,prob=0.5) returns the value 1. Cheers, Uffe

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, I trying to determine the best way to compute the power for a one-sample one-sided binomial test. Specifically I need to sample a population of individuals and ask whether a sample rate of 0% is compatable with a minimum threshold of 3% and how many samples are needed. I have made use of power.prop.test but I am not sure if a) that is the correct (or best) function to use and b) if the

Hi all, I am wondering whether there is a small bug in the binom.test function of the ctest library (I'm using R 1.6.0 on windows 2000, but Splus 2000 seems to have the same behaviour). Or perhaps I've misunderstood something. the command binom.test(11,100,p=0.1) and binom.test(9,100,p=0.1) give different p-values (see below). As 9 and 11 are equidistant from 10, the mean of the

Full_Name: Petr Savicky Version: 2.4.0 OS: Fedora Core release 2 Submission from: (NULL) (62.24.91.47) the error is > binom.test(0.56*10000,10000) Error in binom.test(0.56 * 10000, 10000) : 'x' must be nonnegative and integer while > binom.test(5600,10000) yields correct result. The same error occurrs for > binom.test(0.57*10000,10000)

Hello everybody. Does anyone else find the last test in the following sequence odd? Can anyone else reproduce it or is it just me? > binom.test(100,200,0.13)$p.value [1] 2.357325e-36 > binom.test(100,200,0.013)$p.value [1] 6.146546e-131 > binom.test(100,200,0.0013)$p.value [1] 1.973702e-230 > binom.test(100,200,0.00013)$p.value [1] 0.9743334 (R 1.5.1, Linux RedHat 7.1) --

I wanted to know if there is a way to perform a two sample binomial test in R. I know you can use the proportion test i.e.: prop.test(c(19,5),c(53,39),p=NULL,alternative="two.sided"). But I was looking to use the exact binomial test, binom.test, however when I have tried replacing prop.test with binom.test I get an error. Is there any way to do this? -- View this message in context:

Hello all, I am trying to understand the different results I am getting from the following 3 commands: chisq.test(c(62,50), p = c(0.512,1-0.512), correct = F) # p-value = 0.3788 binom.test(x=62,n=112, p= 0.512) # p-value = 0.3961 2*(1-pbinom(62,112, .512)) # p-value = 0.329 Well, the binom.test was supposed to be "exact" and give the same results as the pbinom, while the chisq.test

R-experts: A quick question, please. >From a lab exp, I got 12 positives out of 50. To get 90% CI for this , I think binom.test might be the one to be used. Is there a better way or function to calculate this? > binom.test(x=12, n=50, p=12/50, conf.level = 0.90) Exact binomial test data: 12 and 50 number of successes = 12, number of trials = 50, p-value = 1 alternative

Hi, I have a seemingly simple proportional test. ?here is the question I am trying to answer: ? There is a test running each day in the lab, the test comes out as either positive or negative. So at the end of each month, we can calculate a positive rate in that month as the proportion of positive test results. The data look like: ? Month??? ??# positive?????? # total tests??? positive rate

Hi, I noticed this problem on my home desktop running FC4 and again on my laptop running FC5. Both have previously compiled and passed make check-all on 2.3.1 svn revisions from 10 days ago or so. On both these machines, make check-all is consistently failing (4 out of 4 attempts on the FC 4 desktop and 3 out of 3 on the FC 5 laptop) in the p-r-random-tests tests. This is with both default

R. 1.1.0 The example below is self explanatory. ## 1 ## # works fine > binom.test((50*.64),50,.5,alt='g') ... Exact binomial test ... ## 2 ## # WHAT ! ? > binom.test((50*.65),50,.5,alt='g') Error in binom.test((50 * 0.65), 50, 0.5, alt = "g") : x must be an

Hi Folks, The recent correspondence about "strange fisher.test result", and especially Peter Dalgaard's reply on Tue 03 April 2007 (which I want to investigate further) led me to take a close look at the code for binom.test(). I now have a query! The code for the two-sided case computes the p-value as follows: if (p == 0) (x == 0) else if (p == 1) (x == n)

Does anyone know of another version of the Mcnemar test that provides: 1. Odds Ratios 2. 95% Confidence intervals of the Odds Ratios 3. Sample probability 4. 95% Confidence intervals of the sample probability Obviously the Odds Ratios and Sample probabilities are easy to calculate from the contingency table, but I would appreciate any help on how to calculate the confidence

Hi, I was hoping to get some advice regarding the testing of interactions, when one factor is modelled as a random effect... I have a model with binomial error structure where the response variable is the proportion of time spent at the main sett (animals were tracked for 28 consecutive days in each season, and were recorded either at the main sett or an outlier sett, so the response variable is

Hi, With the following Sweave minimal file: ---<--------------------cut here---------------start------------------->--- \documentclass{article} \usepackage{Sweave} \begin{document} <<binom-sim>>= thetas <- seq(0, 1, by=0.001) prior <- rep(1, length(thetas)) / length(thetas) lik <- dbinom(1, 1, thetas) lik.p <- prior * lik post <- lik.p / sum(lik.p)

When I perform a two-tailed sign test with the following simple syntax, binom.test(59,100) R returns a P-value (0.088) but nothing else. As I want the result for a one-tailed test I take P/2 = 0.044). However, the journal to which I've submitted my results requests the test statistics, not just the P-values. How can I make R return the test statistics? Best regards, Johan Stenberg, Umea

No, since I'd like to test null: p <= p0 alternative: p > p0. and my understanding is that binom.test tests null: p = p0 (can only be a "simple" null hypothesis according to help(binom.test)) alternative: p > p0 (or p < p0 or p != p0). Thanks, Mirko. > -----Urspr?ngliche Nachricht----- > Von: Douglas Bates [mailto:bates at stat.wisc.edu] >