Confidence Intervals

statistics

for a single mean

$\mu =\bar{x}\pm z_{\frac{\alpha }{2}}{\sigma }_{\bar{x}}$

Error bound

I’m not certain if this is identical to the “for a single mean” above, but calculating an error bound is done via $z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$

If we don’t have enough $n$ for the z-score.. we can use $t_{\frac{\alpha }{n}}\frac{s}{\sqrt{n}}$ instead.

proportion testing??

NOTE: Unless $n$ is extremely large, the math below doesn’t work great if $p$ is near 0 or 1. See Wilson’s adjustment for estimating $p$

If you sample the probability of a population, the distribution $\hat{p}$ is an unbiased estimator of the probability, $p$. The standard deviation of the sample is ${\sigma }_p=\sqrt{\frac{pq}{n}}$ where $q=1-p$. The sample is approximately normal for large samples, where large means both $n\hat{p}\ge 15$ and $n\hat{q}\ge 15$.

I’m not certain, but I believe these conditions are required:

1. Sample is randomly selected
2. np > 15 (or 10??) and nq > 15 (10?)
3. N >= 10n

To calculate a confidence interval, the formula is:

\begin{math} \hat{p} \pm z_{\frac{a}{2}}\sigma_{\hat{p}} = \hat{p} \pm z_{\frac{a}{2}}\sqrt{\frac{pq}{n}} \approx \hat{p} \pm z_{\frac{a}{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}} \end{math}

where $\hat{p}=\frac{x}{n}$ and $\hat{q}=1-\hat{p}$.

Values of $pq$ for different values of $p$.

p	pq	$\sqrt{pq}$
.5	.25	.5
.6 or .4	.24	.49
.7 or .3	.21	.46
.8 or .2	.16	.4
.9 or .1	.09	.3

Wilson’s adjustment for estimating $p$.

If $p$ is near 0 or 1, the sample size has to be “extremely large” to make the math above work. Alternatively, we can use this formula:

$\tilde{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\tilde{p}(1-\tilde{p})}{n+4}}$ where $\tilde{p}=\frac{x+2}{n+4}$