similar to: proportions - finite population correction

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 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

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

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

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

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

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:

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 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:

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

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)

Dear r-helpers! I have trouble reading data from an Oracle data base using RODBC Version 1.0-3, R Version 1.7.1, Windows XP, Oracle8 ODBC Driver Version 8.1.6.4.0: > library(RODBC) > channel <- odbcConnect(dsn="PAV32", case="oracle", believeNRows=FALSE) > # ok, this was succesful > x <- sqlTables(channel) > x[37, ] TABLE_CAT TABLE_SCHEM TABLE_NAME

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

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, 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

Dear R-users! I would like to search for "association rules" (and compute support and confidence, for example) in a data set, and wonder if this can be done with R (version 1.5.0 for Windows)? Particulary, I wonder if a) anyone has done something like that with R or b)maybe someone has written an R-interface to some existing algorithm/software (e.g. to something like 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)

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

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] >

Dear R users! I am using R Version 1.7.1, Windows XP, package "foreign" (Version: 0.6-1), SPSS 11.5.1. There is one thing I noticed with "read.spss", and I'd like to ask if this is considered to be a feature, or possibly a bug: When reading character columns, character strings seem to get filled with blanks at the end. Simple example: In SPSS, create a file with one