similar to: binom.test

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

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

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

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

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)

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've been using binconf (package Hmisc) at a range of alpha values and noticed that using the 'Wilson' method when alpha is larger (i.e. narrow CI), results in the upper value being smaller than the lower value. The 'exact' and 'asymptotic' methods give results in the realm I'd expect. But the help file suggests: "Following Agresti and Coull, the

Dear All I am trying to graph a proportion and CI95% by a factor with ooplot (any other better solution ?) It works well until I try to add the confidence interval. this is the error message and and a description of the data: > dat1 PointEst TT1 1 3.6 TT2 2 5.0 TT3 3 5.8 TT4 4 11.5 TT5 5 7.5 TT5 6 8.7 TT7 7 17.4 > dat2

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

Hello all, A friend just showed me how ks.test fails to work with pbinom for small "size". Example: x<-rbinom(10000,10,0.5) x2<-rbinom(10000,10,0.5) ks.test(x,pbinom,10,0.5) ks.test(x,pbinom,size = 10, prob= 0.5) ks.test(x,x2) The tests gives significant p values, while the x did come from binom with size = 10 prob = 0.5. What test should I use instead ? Thanks, Tal

Hello sir: As to the 2*2 table format for reporting results comparing a new test to true diagnosis,when I got the sensitivity and specificity,how can I calculate the exact 95% confidence intervals (based on the binomial distribution) for sensitivity and specificity via R? Thanks a lot! My best! ------------------------------ ******************************************* Xin Meng Capitalbio

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) --

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

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] >

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

I believe the binom.test procedure is producing one tailed p values rather than the two tailed value implied by the alternative hypothesis language. A textbook and SAS both show 2*9.94e-07 = 1.988e-06 as the two tailed value. As does the R summation syntax from R below. It looks to me like the alternative hypothesis language should be revised to something like " ... greater than or equal

I am trying to find a definition for the two-tailed exact binomial test but have been unsuccessful. Can you help? --------------------------------- [[alternative HTML version deleted]]

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

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