If you are testing H0: p = 0.6 vs H1: p != 0.6 with a sample of size 10 and you observe X=2, then Pr(X <= 2) + Pr(X >= 8) is not what you want. You can argue that you want Pr(X <= 2) + Pr(X >= 10). Both 2 and 10 are 4 away from the null.
binom.test(2, 10, 0.6, alternative="two.sided") # 0.01834 sum(dbinom(c(0:2, 10), 10, 0.6)) # 0.01834117 You don't want the same number of outcomes on each side unless p=0.5 is the null. Chris -----Original Message----- From: Robert Zimbardo [mailto:robertzimba...@gmail.com] Sent: Saturday, December 13, 2014 2:09 PM To: r-help@r-project.org Subject: [R] Two-tailed exact binomial test with binom.test and sum(dbinom(...)) Hi R experts, I have a few related questions that are actually a combination of an R and a hopefully not too trivial (?) statistics question, namely regarding the computation of an exact two-tailed binomial test. Let's assume the following scenario: - number of trials = 10 - p of success = 0.6 (a) Let's also assume we have an H1 that there are more than 6 successes and the number of successes we get is 8. In that case, we do sum(dbinom(8:10, 10, 0.6)) # 0.1672898 binom.test(8, 10, 0.6, alternative="greater") # 0.1673 (b) Now let's assume we have an H1 that there are fewer than 6 successes and the number of successes we get is 2. In that case, we do sum(dbinom(0:2, 10, 0.6)) # 0.01229455 binom.test(2, 10, 0.6, alternative="less") # 0.01229 So far no problem. My questions are now concerned with a two-tailed test: (1). My understanding would be that, if we have an H1 that says "the number of successes won't be 6", then we can add up the two probabilities from above: sum(dbinom(8:10, 10, 0.6)) + sum(dbinom(0:2, 10, 0.6)) # 0.1795843, or just sum(dbinom(c(0:2, 8:10), 10, 0.6)) # 0.1795843 However, that is not what binom.test(..., alternative="two.sided") does: binom.test(2, 10, 0.6, alternative="two.sided") # 0.01834, which is the method of small(er) p-values: sum(dbinom(0:10, 10, 0.6)[dbinom(0:10, 10, 0.6)<=dbinom(2, 10, 0.6)]) # 0.01834117 Thus, question 1) is, is there a reason binom.test is implemented the way it is rather than the other way? (2) I am struggling to understand two-tailed scenarios like this one: - number of trials = 235 - p of success = 1/6 - successes = 51 That is, cases where my logic of taking the successes+1 extreme cases on each tail don't work: adding the point probabilities of 51:235 is fine, but it of course makes no sense to add the point probabilities for 0:185 to that sum(dbinom(51:235, 235, 1/6)) # 0.02654425 sum(dbinom(0:185, 235, 1/6)) # 1 (!) So, while binom.test again does its small(er) p-value thing, ... binom.test(51, 235, 1/6, alternative="two.sided") # 0.04375 sum(dbinom(0:235, 235, 1/6)[dbinom(0:235, 235, 1/6)<=dbinom(51, 235, 1/6)]) # 0.04374797 ... I am wondering how my approach with adding the probabilities of the same number of events from each tail would be done here ...? (3) What is people's view on computing the two-tailed test like this, which leads to an ns result unlike binom.test? 2*sum(dbinom(51:235, 235, 1/6)) # 0.05308849 Any input would be much appreciated! R.Z. ********************************************************** Electronic Mail is not secure, may not be read every day, and should not be used for urgent or sensitive issues ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
