[R] Two-tailed exact binomial test with binom.test and sum(dbinom(...))

[Andrews, Chris]

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:robertzimbardo at gmail.com] 
Sent: Saturday, December 13, 2014 2:09 PM
To: r-help at 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.


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