Sounds like you want a 95% binomial confidence interval:
binom.test(N, P)
will compute this for you, and you can get the bounds directly with
binom.test(N, P)$conf.int
Actually, binom.test computes a two-sided confidence interval, which
corresponds roughly to 2.5 and 97.5 percentages in your approach. It doesn't
give you the 50% point either, but I don't think that's a meaningful quantity
with a two-sided test.
Hope this helps,
Stefan
On 9 Oct 2013, at 15:53, Benjamin Ward (ENV) <b.w...@uea.ac.uk> wrote:
> I got given some code that uses the R function pbionom:
>
> p <- mut * t
> sumprobs <- pbinom( N, B, p ) * 1000
>
> Which gives the output of a probability as a percentage like 5, 50, 95.
>
> What the code currently does is find me the values of t I need, by using the
> above two code lines in a loop, each iteration it increaces t by one and runs
> the two lines. When sumprobs equals 5, it records the value t, then again
> when sumprobs is equal to 50, and again when sumprobs is equal to 95 - giving
> me three t values. This is not an efficient way of doing this if t is large.
> Is it possible to rearrange pbinom so it gives me the small p (made of mut*t)
> as the result of plugging in the sumprobs instead, and is there an R function
> that already does this?
>
> Since pbinom is the binomial probability equation I suppose the question is -
> in more mathematical terminology - can I change this code so that instead of
> calculating the Probability of N successes given the number of trials and the
> probability of a single success, can I instead calculate the probability of a
> single success using the probability of N successes and number of trials, and
> the number of successes? Can R do this for me. So instead I plug in 5, 50,
> and 95, and then get the small p out?
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