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