# Averaging binomial confidence intervals

I have a general statistical question. Three schools A, B, and C have 0.23, 0.56, and 0.8 fraction of female teachers. The upper and lower confidence intervals (CI) for the three schools can be tabulated as shown below:

school  female_instructors  upper_CI  lower_CI
A       0.33                0.53      0.13
B       0.55                0.65      0.45
C       0.80                0.90      0.70


If I were asked to report the average fraction of female instructors in the three schools combined I would have informed: (0.33+0.55+0.8)/3 or 0.56. How would I define the CI for the average fraction?? Average given CIs as well i.e. upper: 0.69 and lower:0.43?? Is this is the best way to do such calculations?

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Two points

First point: You could also perform a weighted average of the proportion of female instructors at the three schools (e.g. if A has ten times as many total instructors as B and C combined, one might argue that it should count more heavily). For example, $(n_{A}\times 0.23 + n_{B}\times 0.56 + n_{C} \times 0.8)/(n_{A} + n_{B} + n_{C})$. If the total number of instructors at each institution is the same size, then the weighted average reduces to the simple arithmetic mean as in your example.

Second point: If you recall that for Bernoulli distribution, the variance is determined by $p$ (the proportion), as in $\sigma^{2}_{p}=p(1-p)$, so that $\sigma_{p}=\sqrt{p(1-p)}$, and $\sigma_{\hat{p}}=\sqrt{p(1-p)/n}$ (where $n = n_{A} + n_{B} + n_{C}$), one could readily generate a CI (say $\hat{p} \pm z_{\alpha/2}\sigma_{\hat{p}}$) under the assumption that A, B, and C are all drawn from the same population.

Bonus point: Agresti and Coull have shown that the nominal coverage of the CI I just indicated performs suboptimally for small $n$. They provide an alternative which gives better nominal CI coverage for $\hat{p}$ thus:

$$\tilde{p} \pm z_{\alpha/2}\sqrt{\tilde{p}(1-\tilde{p})\tilde{n}}$$

where $\tilde{n} = n+2z_{\alpha/2}$, and $\tilde{p} = \frac{\left(\Sigma x\right) +z_{\alpha/2}}{\tilde{n}}$. Understand that the use of $\tilde{p}$ is purely instrumental, and the Agresti-Coull confidence interval is for $\hat{p}$. As sample size gets big, the nominal coverage of the standard Wald-type CI and that of the Agresti-Coull CI converge. More details about these and other binomial proportion confidence intervals on Wikipedia.

References

Agresti, A. and Coull, B. A. (1998). Approximate is better than “exact” for interval estimation of binomial proportions. The American Statistician, 52(2):119–126.

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+1 I'd only add that the Wikipedia page on binomial proportion CI's has a fairly extensive discussion of various approximations for the small-$n$ case, including Agresti-Coull. Some people use the simple "add 2 successes and 2 failures" ... which I think is what Agresti-Coull gives with $z=2$. – Glen_b Jul 6 '14 at 22:15
@Glen_b excellent suggestion, will add the link. The 2 successes and 2 failures of the 95% Agresti-Coull CI, obtains because $2 \approx 1.96$. – Alexis Jul 6 '14 at 22:46
+1. We can reverse-engineer the data even without knowing which CI method was used, because (1) the UCL and LCL are symmetrically placed around the proportions and therefore (2) for a sample size $n$ the UCL should exceed the proportion $p$ by approximately $Z_{0.975}\sqrt{p(1-p)}/\sqrt{n}$. Solving for $n$ (and rounding) gives $n\approx 21, 95, 61$, respectively. Therefore the fraction of females is approximately $0.33(21)+0.55(95)+0.80(61),$ or about $108$ out of $177$. We can erect a 95% CI around these data. Using the same crude Normal approximation gives $0.61$, 95% CI = $(0.54,0.68)$. – whuber Jul 6 '14 at 23:10
@whuber, I may not be understanding you, but no, at least for the Agresti-Coull CI, the interval for $\hat{p}$ is symmetric about $\tilde{p}$ (not $\hat{p}$). – Alexis Jul 6 '14 at 23:36