Binomial probability and binomail test

I have made a survey for n people where k people are of one sign - 12 hip. (so p = 1/12)

The Binomial probability is:

P= nCk * p^k * (1-p)^(n-k)

My question is: how can I evaluate this result with the true one (k/n)?

The z-score test is:

z = (k/n-p)/sqr(p*(1-p)/n)

But, in this formula, is p=1/12 or p=P?

Another question is the values and limits of confidence. I have:

If z < 1.64 --> the observed score is not significantly different from chance level

If z >= 1.64 --> the observed score is significantly different from chance at level p < .05

If z >= 1.96 --> the observed score is significantly different from chance at level p < .025

If z >= 2.33 --> the observed score is significantly different from chance at level p < .01

If z >= 2.57 --> the observed score is significantly different from chance at level p < .005

Can you give me, please, these values with more decimals? I know, for instance, that 2.33 is in fact 2.3263

Kind regards,

Kepler

Re: Binomial probability and binomail test

Hey kepler.

What exactly are you trying to do? Are you trying to test whether a proportion is 1/12 or are you trying to test whether something has evidence to follow a distribution?

For the first one, use the estimator of a proportion (you can use a Normal approximation if the sample size is big enough) and for the second one, use a chi-square goodness of fit (Pearsons chi-square) test.

Re: Binomial probability and binomail test

Hi Chiro,

I just want to know which value to use in the z-score test: 1/12 or P.

Kepler

Re: Binomial probability and binomail test

What are you actually testing? Are you doing a hypothesis test that p = 1/12 or are you testing something else?

Re: Binomial probability and binomail test

Hi,

The chances of one person be a Libra, for example, is 1/12; but the chance of k of n being a Libra is given by P as given in my first post. I want to evaluate this. So I want to make the Binomial Test. My doubt is wether I use 1/12 or P.

Regards, and thanks for being patient...

Kepler

Re: Binomial probability and binomail test

If you assume that the probability is 1/1 2 for all members (and that all people are independent) then just use p=1/12 and calculate P(X = k) for your answer.

If however you need to estimate the proportion from your data, then you will need to use the mean and standard error of the mean to get a point and interval estimation of the true parameter under some level of confidence.

Both things are very different: in the first you assume that you know the proportion and the in the second you estimate it from the data.

Re: Binomial probability and binomail test

If you are asking how can you test whether or not p = 1/12, you simply can generate a confidence interval for your result, and see whether or not what you obtained does indeed contain your hypothesized value of p. In the case of what you are asking p=1/12. k/n is whatever you got from your survey results.

If you are intent on using the z-score test (equally fine), it's simply a matter of calculating the z-score and comparing that to known values of "significance".