= Solution
The <posterior mean> is a weighted average of the sample proportion and prior mean:
$$
\frac{9+\alpha}{90+\kappa}
=\frac{90}{207.75}(0.10)+\frac{117.75}{207.75}(0.05).
$$
Thus it is pulled from 10% toward the prior's 5%, giving about 7.17%. This is an <affine shrinkage estimator for a binomial proportion>. The prior carries substantial concentration relative to the 90 observations, so the posterior is also more precise under this model: its standard deviation is about $0.01785$, compared with the binomial plug-in standard error $\sqrt{0.1(0.9)/90}=0.03162$. The posterior 95% <credible interval> $[0.04076,0.11035]$ is narrower and shifted down compared with the simple Wald interval $[0.03802,0.16198]$ based on $9/90$ alone.
\b[The point estimate shrinks toward the historical mean, and the interval is narrower under the informative prior.] A <credible interval> and a frequentist <confidence interval> have different <probability> interpretations; the comparison does not make the prior's relevance automatic.
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