CL =	[1 + F0.025,v1,v2(q + 1/n)/p]−1

and F has v1 and v2 degrees of freedom: v1 = 2(nq + 1) and v2 = 2np.

CU = [1 +	q	]	−1
	(1/n + p)F0.975,v3,v4

and F has v3 and v4 degrees of freedom, v1 = 2(np+1) and v2 = 2nq

where:
• F_0.975 is the quantile below which 0.975 of that F-distribution lies.
  N.B. If you are using published tables, look up F_0.025 - in other words the quantile above which 0.025 of that F-distribution lies.
• C_L and C_U are the lower and upper 95% confidence limits
• p and q are the proportions in the sample with and without the character of interest,
• n is the sample size,