similar to: how to find "p" in binomial(n,p)

Hi, Sorry for bothering your guys again. I am trying to use "par" to put 6 figures into 3 rows and 2 columns. My code is: op <- par(mfrow = c(1, 2)) xyplot(x1+x2+x3 ~ y, data=resulta, type="l", pch=LETTERS[1:3], lty=c(1:3), main="For k=3, theata1=1") xyplot(x1+x2+x3 ~ y, data=resultb, type="l", pch=LETTERS[1:3], lty=c(1:3), main="For k=3,

Thank you for your guys reply for my previous question. But I got one more question about the boxplot. With the code in the R-help: boxplot(len ~ dose, data = ToothGrowth, boxwex = 0.25, at = 1:3 - 0.2, subset = supp == "VC", col = "yellow", main = "Guinea Pigs' Tooth Growth", xlab = "Vitamin C dose mg", ylab =

Hi, sorry for bothering your guys. I will trying to make some nice graph using boxplot. when I check the help file of boxplot, there is a sample code as: boxplot(len ~ dose, data = ToothGrowth, add = TRUE, boxwex = 0.25, at = 1:3 + 0.2, subset = supp == "OJ", col = "orange") legend(2, 9, c("Ascorbic acid", "Orange juice"),

Full_Name: Chanseok Park Version: R 2.2.1 OS: RedHat EL4 Submission from: (NULL) (130.127.112.89) pbinom(any negative value, size, prob) should be zero. But I got the following results. I mean, if a negative value is close to zero, then pbinom() calculate pbinom(0, size, prob). dbinom() also behaves similarly. > pbinom( -2.220446e-22, 3,.1) [1] 0.729 > pbinom( -2.220446e-8, 3,.1)

Full_Name: Wolfgang Resch Version: R 2.8.1 GUI 1.27 OS: OS X 10.4.11 Submission from: (NULL) (137.187.89.14) Strange behavior of qbinom: > qbinom(0.01, 5016279, 1e-07) [1] 0 > qbinom(0.01, 5016279, 2e-07) [1] 16 > qbinom(0.01, 5016279, 3e-07) [1] 16 > qbinom(0.01, 5016279, 4e-07) [1] 16 > qbinom(0.01, 5016279, 5e-07) [1] 0

Hi, Dear R-users: Sorry for bothering your guys again. I think I should rewrite my question. I know how to extend whisker by using range. The question is that I will set the range=1.5, and at the same time, I only want to show the extreme outlier, like 0.01% and 99.99% percentile, so what should I do? Thank you very much! Catherine -- View this message in context:

Hi, Dear R-users, I have a problem when I drawing a boxplot. I want to extend the whisker to the 5% and the 95% quantiles and only show the most extreme outlier, like 0.01% and 99.99% percentiles. What should I do? I saw something on boxplot.stat, but even I define the parameter in boxplot.stat, what I should do next? how to connect it to boxplot? Thank you very much! Catherine -- View

n_10;p_0.5;jjx_0:n;qbinom(pbinom(jjx,n,p),n,p) # This one works as expected n_100;p_0.5;jjx_0:n;qbinom(pbinom(jjx,n,p),n,p) # This one causes severe problems I cannot interrupt using ESC and I finally have to resort to the Windows Task manager to kill the R session. A friend of mine told me that he faced similar problems under Unix. --please do not edit the information below-- Version:

Dear R users, I?m trying to optimize simultaneously two binomials inequalities (used in acceptance sampling) which are nonlinear solution, so there is no simple direct solution. Please, let me explain shortly the the problem and the question as following. The objective is to obtain the smallest value of 'n' (sample size) satisfying both inequalities: (1-alpha) <= pbinom(c, n, p1)

Hi, sorry for bothering your guys again. I want to simulate 100 AR(1) data with cor(x_t, x_t-1)=rho=0.3. The mean of the first 70 data (x_1 to x_70) is 0 and the mean of the last 30 data (x_71 to x_100) is 2. Can I do it in the following way? x <- arima.sim(list=(ar=0.3), 100) mean <- c(rep(0, 70), rep(2, 30)) xnew <- x+mean If the above code to simulate 100 AR(1) data is right, what

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

Dear all Just for fun, I have just downloaded the paper mentioned below and checked it with R-1.6.1. Everything is ok with exception of Table 2b, where I get always 1 instead of 0.5: > pbinom(1e15,2e15,0.5) [1] 1 Which value should be correct? Best regards Christian Stratowa ============================================== Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead

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

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

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

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

On Feb 10, 2005, at 7:38 PM, George W. Gilchrist wrote: > Today I was running a graduate level stats lab using R and we > encountered a > major problem while using the current build of the Cocoa GUI: > >> From the GUI: >> system.time(pbinom(80, 1e5, 806/1e6)) > [1] 14.37 4.94 30.29 0.00 0.00 >> > >> From the command line on the same machine: >>

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)

I would like to solve the equation is is the sum from k = i to N of choose(N,k) * MR ^ k * (1 - MR) ^ (N - k) - 0.50 = 0 I want to solve for MR. This seems like a non-linear equation to me. But I am having a hard time writing the function that implements the above. I could use 'for(...) as a brute force appoarch but I would like a more "elegant" solution. The variables '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)