Interval Estimates - Confidence Interval for p (2024)

Generally, symmmetry is presumed when using a two-sided confidence interval so that \(E_1 = E_2 = E\) and therefore the interval looks like

\begin{equation*}P(\tilde{p} - E \lt p \lt \tilde{p} + E) = 1 - \alpha.\end{equation*}

In this case, E is known as the margin of error.

To determine E carefully, note that from the central limit theorem

\begin{equation*}\frac{Y-np}{\sqrt{np(1-p)}} = \frac{\tilde{p} - p}{\sqrt{p(1-p)/n}}\end{equation*}

is approximately standard normal for large n. Presuming that \(\tilde{p} \approx p\) and replacing the unknown p terms on the bottom with \(\tilde{p}\) gives

\begin{equation*}z = \frac{\tilde{p} - p}{\sqrt{\tilde{p}(1-\tilde{p})/n}}\end{equation*}

where z is a standard normal distribution variable. So, using the central limit theorem and the standard normal distribution, you can find the value \(z_{ \alpha/2}\) where

\begin{equation*}P( -z_{ \alpha/2} \lt z \lt z_{ \alpha/2}) = 1 - \alpha\end{equation*}

\begin{equation*}P( -z_{ \alpha/2} \lt \frac{\tilde{p} - p}{\sqrt{\tilde{p}(1-\tilde{p})/n}} \lt z_{ \alpha/2}) = 1 - \alpha\end{equation*}

or by rearranging the inside inequality

\begin{equation*}P( \tilde{p} - z_{ \alpha/2}\sqrt{\tilde{p}(1-\tilde{p})/n} \lt p \lt \tilde{p} + z_{ \alpha/2}\sqrt{\tilde{p}(1-\tilde{p})/n}) = 1 - \alpha.\end{equation*}

Setting \(E = z_{ \alpha/2}\sqrt{\tilde{p}(1-\tilde{p})/n}\) gives a way to determine a confidence interval centered on \(\tilde{p} = \frac{Y}{n}\) for p with "confidence level" \(1-\alpha\text{.}\)

To complete the interval, one needs a specific value for \(z_{ \alpha/2}\) using an inverse normal distribution calculator[STRUCT].[NUM]. Generally, one chooses confidence levels on the order of 90%, 95%, or 99% with 95% being the usual choice. Fortunately this value is easily computed using graphing calculators or other automatic methods although your ancient teacher might have been required to use tables. On a TI calculator, use

\begin{equation*}z_{\alpha /2} = \text{InvNorm}( 1 - \frac{\alpha}{2} )\end{equation*}

For 90% confidence level, you need to find a z-value so that

\begin{equation*}P( -z_{ \alpha/2} \lt z \lt z_{ \alpha/2}) = 0.9 = 1 - 0.1 .\end{equation*}

Using the symmetry of the normal distribution, this can be rewritten

\begin{equation*}F(z_{ \frac{0.1}{2}}) = P( z \lt z_{ \frac{0.1}{2}}) = 0.95 = 1 - \frac{0.1}{2} .\end{equation*}

Using the inverse of the standard normal distribution (on the TI calculator this is InvNorm(0.95)) gives \(z_{ 0.05} \approx 1.645\text{.}\)

Similarly, for a 95 % confidence level, find where

\begin{equation*}F(z_{ \frac{0.05}{2}}) = P( z \lt z_{ \frac{0.05}{2}}) = 0.975 = 1 - \frac{0.05}{2} .\end{equation*}

The calculators InvNorm(0.975) gives \(z_{ 0.025} \approx 1.960\text{.}\)

For a 99 % confidence level, find where

\begin{equation*}F(z_{ \frac{0.01}{2}}) = P( z \lt z_{ \frac{0.01}{2}}) = 0.995 = 1 - \frac{0.01}{2} .\end{equation*}

The calculators InvNorm(0.995) gives \(z_{ 0.005} \approx 2.576\text{.}\)

To illustrate this process, enjoy this computational cell.

Notice that when computing the confidence intervals above that we choose to just replace some of the p terms with \(\tilde{p}\) so that only one p term was left and could be isolated in the middle. There are other ways to deal with this. The easiest is to take the worst case scenario for the p terms in the denominator above. Indeed, the confidence interval is made wider (and therefore more likely to contain the actual p) if the square root term is as large as possible, using basic calculus it is easy to see that p(1-p) is maximized when p = 1/2. Therefore, a second alternative is to create your confidence interval using

\begin{equation*}z = \frac{\tilde{p} - p}{\frac{1}{2\sqrt{n}}}\end{equation*}

and therefore \(E = \frac{z_{ \alpha/2}}{2\sqrt{n}}\text{.}\) This method should be used only when trying to create the roughest and "safest" interval.

To relate the Wilson Score with the standard approach for creating a confidence interval for p seen above, note that

\begin{equation*}\big | p - \tilde{p} \big | = z_{\alpha /2} \sqrt{\frac{p(1-p)}{n}}\end{equation*}

can be simplified by squaring both sides to get

\begin{equation*}\big ( p - \tilde{p} \big )^2 = z_{\alpha /2}^2 \frac{p(1-p)}{n}.\end{equation*}

Replacing \(\tilde{p}\) with the relative frequency gives

\begin{equation*}\big ( p - \frac{Y}{n} \big )^2 = z_{\alpha /2}^2 \frac{p(1-p)}{n}\end{equation*}

or by simplifying

\begin{equation*}(n+z_{\alpha /2}^2 )p^2 - (2Y+z_{\alpha /2}^2) p + \frac{Y^2}{2} = 0.\end{equation*}

Solving for p using the quadratic formula and simplifying ultimately results in the described interval.