# How to calculate confidence interval of an MLE estimate for a Bernouli random variable

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?

• How can P, which is unknown, influence the estimation of the confidence interval? – stefgehrig Feb 17 at 16:30
• That was exactly my thought. Thanks for validating! – Shahzeb Naveed Feb 18 at 6:35

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.