Calibration, weighting and post-stratification in audience measurement
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A note on observing zero successes
Say that you have a sample of size $n=1000$ and you observed $S_n=100$ successes. Traditionally you would use $\hat p=\frac{S_n}{n}=\frac{100}{1000}=0.1$ as a point estimate of the population proportion $p$. From a frequentist perspective you would probably also report a confidence interval: $$p_-=\hat p - z_\alpha\sqrt{\frac{\hat p(1-\hat p)}{n}}=0.1-1.96\sqrt{\frac{0.1 \times 0.9}{1000}}=0.08140581,$$ and $$p_+=\hat p + z_\alpha\sqrt{\frac{\hat p(1-\hat p)}{n}}=0.1-1.96\sqrt{\frac{0.1 \times 0.9}{1000}}=0.1185942,$$ using $z_\alpha=1.96$ for a 95% confidence interval (Assuming that the sample fraction is small, i.e. the universe size $N$ is large relative to $n$. Also, I will not go into how such a confidence interval needs to be interpreted.). So far, so good. Now say you have observed zero successes, i.e. $S_n=0$, and you want to apply the procedure above. To start with, you can't because it violates the non-zero sample proportion assumption. There are some alterna...