Let's say a coin has a probability of P=2/3 of landing on heads on any independent flip. After flipping the coin N=10000 times, you observe heads on F=60% of the flips.

Suppose the width of the 95% confidence interval of the maximum likelihood estimator of P is w. What changes to P, N or F (keeping all other values constant) would approximately decrease the width of the 95% confidence interval by a factor of 10?


1 Answer 1


The traditional (Wald) 95% CI for success probability $p$ uses the MLE $\hat p = x/n,$ where $x$ is the number of successes in $n$ trials. It is an asymptotic CI intended for use with large $n$ where the normal approximation is accurate. It is of the form $$\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}}.$$

In judging the $n$ required for a given margin of error $E = 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}},$ it is customary to assume $\hat p = 1/2,$ which gives the maximum margin of error for a particular $n.$

The Agresti-Cooil CI uses $n^+ = n+4$ instead of $n$ and $p^+ = \frac{x+2}{n+4}$ instead of $\hat p.$ It comes closer to the intended 95% coverage of parameter $p$ for smaller values of $n.$

You can answer your questions about decreasing the width of the CI by a factor of 10, by deciding how to decrease $E$ by a factor of 10.


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