I've a small query and really confused as how to solve such problems.

$n$ people are question about internet connection 87 of them have high speed connection. We are given confidence interval(CI) of population proportion is $0.119 < p < 0.1771$.

  1. I found the mid-point of CI as 0.145. I need to understand how to find $n$
  2. It's given this interval is $\alpha \% $ confidence interval, how can we find $\alpha$

Any help is greatly appreciated.

Any resource that can help me to solve such problems will also be of great help.

Regards, Arif


$n$ is pretty simple. If you assume that $p = 0.145$, and $p = \frac{\Sigma x}{n}$, then $n=\frac{\Sigma x}{p}$ (left as an exercise ;).

While Zen gives a standard Wald confidence interval, Agresti & Coull have demonstrated that confidence intervals for proportions (given a nominal variable $x$) are better estimated using:

(1) $\tilde{p} \pm z_{\alpha/2}\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{\tilde{n}}},$

where $\tilde{n}=n+2z_{\alpha/2}$, and $\tilde{p} = \frac{\Sigma x + z_{\alpha/2}}{\tilde{n}}$.

Note that the Agresti-Coull confidence interval is an interval for $p$ and not for $\tilde{p}$ ($\tilde{p}$ is simpy instrumental in calculating the interval).

For example, suppose you have $n=50$ and $p=0.60$ (implying that $\Sigma x = np = 30$), and want to construct a 95%CI, then:

$\tilde{n} = 50+2\times1.96=53.96$,

$\tilde{p}=\frac{30+1.96}{53.96}=\frac{31.96}{53.96}=0.592$, and

the 95% Agresti-Coull CI for 0.60 is:
$0.592 \pm 1.96\sqrt{\frac{0.592\left(1-0.592\right)}{53.96}} = 0.592 \pm 0.131 = 0.461, 0.723$.


Agresti, A. and Coull, B. A. (1998). Approximate is better than “exact” for interval estimation of binomial proportions. The American Statistician, 52(2):119–126.

  • $\begingroup$ (+1) Very nice, Alexis. I believe this is a textbook question, and good old Wald is implicit on it. $\endgroup$ – Zen Jun 11 '14 at 4:27
  • $\begingroup$ (+1) Interesting paper, but confused me rather. If you have to assume a prior distribution on the unknown parameter to show that an approximate procedure for calculating a confidence interval has higher coverage on average than the exact procedure, why not go the whole hog & calculate a Bayesian credible interval? $\endgroup$ – Scortchi - Reinstate Monica Jun 11 '14 at 11:22
  • $\begingroup$ @Scortchi I think of it as a fancy continuity correction. :) $\endgroup$ – Alexis Jun 11 '14 at 13:02

If the estimate for the proportion is $\hat{\theta}$, then an approximate $(1-\alpha)\%$ confidence interval is $$ \left[\hat{\theta} - z_{\alpha/2} \times \sqrt{\frac{\hat{\theta}(1-\hat{\theta})}{n}},\; \hat{\theta} + z_{\alpha/2} \times \sqrt{\frac{\hat{\theta}(1-\hat{\theta})}{n}}\; \right] \, , \qquad (*) $$ in which $z_{\alpha/2}=\Phi^{-1}(1-(\alpha/2))$. As you already noticed, the middle point of the given interval is equal to $\hat{\theta}$. But $\hat{\theta}=87/n$, giving you the value of $n$. Now use $(*)$ to determine $z_{\alpha/2}$, and compute $\alpha$ in R (our friend, and fellow SE member, Prof. Dilip Sarwate would prefer a standard normal table ;-) doing $\alpha=2(1-\Phi(z_{\alpha/2}))$.

2 * (1 - pnorm(z))

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.