Calculate Pass / Fail probabilty

pakalasa

I have an infinite number of independent events, each has 2 possible outcomes, Pass /Fail. The probability of each result is the same. How do I calulate this probability based on sample size.

i.e., what sample size do I need to take to get a certain confidence level.

Fremen

This page should help:
http://www.math.uah.edu/stat/interval/Bernoulli.xhtml

pakalasa

Sorry Fremen, I cant open that link.!
Could you put up the equations directly on screen.
Thanks

MathsManiac

I think you need to be much clearer about what you're asking.

Exactly what probability are you trying to find? The probability of r passes in n independently selected trials, given a known pass rate p?

Or, (a different question), are you looking for a confidence interval for the population parameter p given an observed pass rate in a given sample?

Or, (a different question), do you want to know how big a sample is needed to create a confidence interval of a given size for the population parameter p?

pakalasa

The last one,
Cheers


i.e -

MathsManiac wrote: »

Or, (a different question), do you want to know how big a sample is needed to create a confidence interval of a given size for the population parameter p?

MathsManiac

The width of the confidence interval is actually a function of p, so you can't determine n precisely in advance. However, you can get a value that will guarantee that the resulting confidence interval will be no more than a given size. The "rule of thumb" is: divide 1 by the square of the half-length of the desired 95% confidence interval.

e.g., If you want the 95% confidence interval to be your observed value plus or minus 5% (=0.05), then you need n=1/(.05)^2 = 400; if you want plus or minus 3%, you need n=1/(.03)^2 = 1111, etc.

pakalasa

Cheers, 'MathsManiac.
That's the one.

Fremen

Got a link to a derivation of that result MM?

MathsManiac

Quicker to do it than find it!

The SE for a proportion is sqrt(p(1-p)/n). This varies with p, but is maximimised for p=1/2, giving sqrt(1/(4n)). Since, taking 1.96 to be as near as bedamned to 2, the 95% C.I. is +/- twice the SE, which is thus seen to be bounded above by 1/sqrt(n).

As I understand it, this is pretty much the reason why opinion polls with just over a thousand voters are usually quoted as having a "margin of error" of 3%.