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?

  • $\begingroup$ How can P, which is unknown, influence the estimation of the confidence interval? $\endgroup$ – stefgehrig Feb 17 at 16:30
  • $\begingroup$ That was exactly my thought. Thanks for validating! $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.