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Just a note

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:
• F0.975 is the quantile below which 0.975 of that F-distribution lies.
N.B. If you are using published tables, look up F0.025 - in other words the quantile above which 0.025 of that F-distribution lies.
• CL and CU 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,
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