Hello,
Can you post your computations? Assuming a two-sided test, mine are
x <- 6
n <- 26
p <- 0.1
cmb <- sapply(x:n, \(i) choose(n, i))
sum(cmb * p^(x:n) * (1 - p)^(n - (x:n)))
#[1] 0.03985931
binom.test(x=6, n=26, p=0.1)$p.value
#[1] 0.03985931
The result are equal to one another.
Às 19:00 de 03/04/2022, Sigbert Klinke escreveu:
Hi,
for the specific example binom.test(x=6, n=26, p=0.1) I get as p-value
0.03986. The default approach to decide whether I can reject the null or
or not is to compare the p-value with the given significance level.
Using a significance level of 0.05 this will lead to reject the null
hypothesis.
However, computing things by hand it turned out that the critical values
are 0 and 6. Since the test statistic is also 6 I can not reject the
null hypothesis.
I found the discussion under
https://stat.ethz.ch/pipermail/r-help/2009-February/380341.html and I
understand that a p-value is not well defined if we have a asymmmetric
(discrete) distribution under the null.
At least I would have expected some hint in the documentation for
binom.test.
Sigbert
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