Proportion Estimation

Proportion estimation is another common task for sample statistics.

We have sample proportion

${\hat {p}}={\frac {y}{n}}$

where $y$ is the number of subjects in the sample with a particular trait, and $n$ is the sample size.

We have

$\mu _{\hat {p}}=p,\sigma _{\hat {p}}={\sqrt {\frac {p(1-p)}{n}}}$

and standard error

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

Assumptions

We assume that

• A random sample was taken
• $y\geq 5$ and $n-y\geq 5$
  • rooted in normal approximation of binomial

Wilson-Adjusted CI for p

Correcting the sample proportion narrows the confidence interval. We do this with the Wilson-Adjusted estimate for $p$

${\widetilde {p}}={\frac {y+2}{n+4}}$

with standard error

$SE({\widetilde {p}})={\sqrt {\frac {{\widetilde {p}}(1-{\widetilde {p}})}{n+4}}}$

Remember that the confidence interval is ca

${\widetilde {p}}$ is slightly skewed towards $0.5$, but results in better CIs for $p$. I don't know why.

Confidence Interval

We use normal distribution since $p$ is bounded between 0 and 1, and we don't have extra error from extra parameters such as multiple sample mean.

Remember that the confidence interval is just mean plus-or-minus error margin, and the error margin is just the z score multiplied by standard error (since we are using normal distribution).

Notaby, it is possible to have a bound above 1 or below 0. This usually happens when the point estimate is close to 0 or 1. In this case, instead of listing the impossible bounds, we report that they have been cut off.