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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

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$n$ 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:
$0.592 \pm 1.96\sqrt{\frac{0.592\left(1-0.592\right)}{53.96}} = 0.592 \pm 0.131 = 0.461, 0.723$.


Reference

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.

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  • $\begingroup$ (+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
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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))
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