similar to: Incorrect p value for binom.test?

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)

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

R 0630 for windows > library(ctest) > binom.test(7,10,p=0.3, alternative="two.sided") returns a p-value of =< 2.2e-016 and a warning In Splus 3.4 > binom.test(7,10,p=0.3, alternative="two.sided") returns a p-value of 0.0106 I think it is the max(v[v<=(1+eps)*PVAL]) causing the problem... max() of an empty vector....... Mai Z

running R 1.4.1 on MAC and 1.2.2 on Linux When I use run binom.test with small N the results are a little perplexing to me >binom.test(9,20,p=0.5) gives the below plus other stuff 95 percent confidence interval: 0.2305779 0.6847219 Now: >pbiom(9,20,0.6847219) [1] 0.02499998 # i.e., lower 2.5% of distribution >pbinom(9,20,0.2305779) [1] 0.9923132 >pbinom(8,20,0.2305779)

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

With this line: > binom.test(x=12, n=50, p=12/50, conf.level = 0.90) I get this output: > Exact binomial test > > data: 12 and 50 > number of successes = 12, number of trials = 50, p-value = 1 > alternative hypothesis: true probability of success is not equal to 0.24 > 90 percent confidence interval: > 0.1447182 0.3596557 > sample estimates: > probability

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] >

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) --

Hi there, as part of a 2 x 2 contingency table analysis I would like to estimate conditional probabilities (success rates) in a Bernoulli experiment. In particular I want to test a null hypothesis p <= p0 versus the alternative hypothesis p > p0. As far as I understand the subject, there are UMPU tests for these types of hypotheses. Now I know about R's "binom.test" but the

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 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

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)

Dear R crew: I am sorry this question has been posted before, but I can't seem to solve this problem yet. I have a simple dataset consisting of two variables: cestode intensity and chick size (defined as CAPI). Intensity is a count and clearly overdispersed, with way too many zeroes. I'm interested in looking at the association between these two variables, i.e. how well does chick

Just a general question concerning the woolf test (package vcd), when we have stratified data (2x2 tables) and when the p.value of the woolf-test is below 0.05 then we assume that there is a heterogeneity and a common odds ratio cannot be computed? Does this mean that we have to try to add more stratification variables (stratify more) to make the woolf-test p.value insignificant? Also in the

Dr. Flom, I was searching the web for any examples of one-inflated negative binomial regression, and ran across your post. Fittingly, I am working on the analysis of data from the NIDA Cooperative Agreement where I had the pleasure of working with Sherry Deren and other folks at NDRI. NBR does a poor job of modeling number of sex partners. (I am using Stata.) Did you have any luck modeling a

Hi, If any one has time I need some help understanding the P-values given in the lmer output. Using AIC for model selection I find my minimal model is FOLLOW~MOVERSTATUS+DISTANCE however it appears DISTANCE is not significant at 95% confidence, see output quoted below. However, removing DISTANCE gives a higher AIC=433.5, therefore I will keep it in, but am confused as to what is adds to the

Hello I am interested in Poisson or (ideally) Negative Binomial regression with an inflated number of 1 responses I have seen JK Lindsey's fmr function in the gnlm library, which fits zero inflated Poisson (ZIP) or zero inflated negative binomial regression, but the help file states that for ' Poisson or related distributions the mixture involves the zero category'. I had thought

Dear R gurus I am running a GLMM that looks at whether chimpanzees spend time in shade more than sun (response variable 'y': used cbind() on counts in the sun and shade) based on the time of day (Time) and the availability of shade (Tertile). I've included some random factors too which are the chimpanzee in question (Individual) and where they are in a given area (Zone). There are

Dear Sir/Ma, I Adelabu.A.A, one of the R-users from Nigeria. When am running a coxph command the below error was generated, and have try some idea but not going through. kindly please assist: > cox1 <- coxph(Surv(tmonth,status) ~ sex + age + marital + sumassure, X) Warning message: In fitter(X, Y, strats, offset, init, control, weights = weights, : Ran out of iterations and did not

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