My Favorite One-Liners: Part 65

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

I’ll use today’s one-liner just before I begin some enormous, complicated, and tedious calculation that’s going to take more than a few minutes to complete. To give a specific example of such a calculation: consider the derivation of the Agresti confidence interval for proportions. According to the central limit theorem, if $n$ is large enough, then

$Z = \displaystyle \frac{ \hat{p} - p}{ \displaystyle \sqrt{ \frac{p(1-p) }{n} } }$

is approximately normally distributed, where $p$ is the true population proportion and $\hat{p}$ is the sample proportion from a sample of size $n$ . By unwrapping this equation and solving for $p$ , we obtain the formula for the confidence interval for a proportion:

$z \displaystyle \sqrt{\frac{p(1-p)}{n} } = \hat{p} - p$

$\displaystyle \frac{z^2 p(1-p)}{n} = \left( \hat{p} - p \right)^2$

$z^2p - z^2 p^2 = n \hat{p}^2 - 2 n \hat{p} p + n p^2$

$0 = p^2 (z^2 + n) - p (2n \hat{p} + z^2) + n \hat{p}^2$

We now use the quadratic formula to solve for $p$ :

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm \sqrt{ \left(2n\hat{p} + z^2 \right)^2 - 4n\hat{p}^2 (z^2+n)}}{2(z^2+n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm \sqrt{4n^2 \hat{p}^2 + 4n \hat{p} z^2 + z^4 - 4n\hat{p}^2 z^2 - 4n^2 \hat{p}^2}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm \sqrt{4n (\hat{p}-\hat{p}^2) z^2 + z^4}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm \sqrt{4n \hat{p}(1-\hat{p}) z^2 + z^4}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm \sqrt{4n \hat{p} \hat{q} z^2 + z^4}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm z \sqrt{4n \hat{p} \hat{q} + z^2}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm z \sqrt{4n^2 \displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle 4n^2 \frac{z^2}{4n^2}}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm 2nz \sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle \frac{z^2}{4n^2}}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + 2n \displaystyle \frac{z^2}{2n} \pm 2nz \sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} +\displaystyle \frac{z^2}{4n^2}}}{2n \displaystyle \left(1 + \frac{z^2}{n} \right)}$

$p = \displaystyle \frac{\hat{p} + \displaystyle \frac{z^2}{2n} \pm z \sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle \frac{z^2}{4n^2}}}{\displaystyle 1 + \frac{z^2}{n} }$

From this we finally obtain the $100(1-\alpha)\%$ confidence interval

$\displaystyle \frac{\hat{p} + \displaystyle \frac{z_{\alpha/2}^2}{2n}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} } - z_{\alpha/2} \frac{\sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle \frac{z_{\alpha/2}^2}{4n^2}}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} } < p < \displaystyle \frac{\hat{p} + \displaystyle \frac{z_{\alpha/2}^2}{2n}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} } + z_{\alpha/2} \frac{\sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle \frac{z_{\alpha/2}^2}{4n^2}}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} }$ .

Whew.

So, before I start such an incredibly long calculation, I’ll warn my students that this is going to take some time and we need to prepare… and I’ll start doing jumping jacks, shadow boxing, and other “exercise” in preparation for doing all of this writing.

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My Favorite One-Liners: Part 65

Published by John Quintanilla

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