A note on Wilson coverage intervals for proportions estimated from complex samples
Section 3. The logistic transformation

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The complex-sampling Wilson coverage interval turns out to be very similar to this two-sided coverage interval derived using a logistic transformation (see Brown et al., 2001):

$$f^{-1}\left\{f\left(p\right)-z_{1-\alpha /2}\sqrt{\text{var}\left[f\left(p\right)\right]}\right\}\le P\le f^{-1}\left\{f\left(p\right)+z_{1-\alpha /2}\sqrt{\text{var}\left[f\left(p\right)\right]}\right\},(3.1)$$

where $f\left(p\right)=\log \left(p\right)-\log \left(1-p\right),$ and $\mathrm{var}\left[f\left(p\right)\right]={\left[f^{\prime }\left(p\right)\right]}^2\mathrm{var}\left(p\right)={\left[1/p+1/\left(1-p\right)\right]}^{2}p\left(1-p\right)/n*=1/\left[n*p\left(1-p\right)\right].$ The original rationale for this interval appears to be that it has this desirable property: it cannot contain values less than 0 or greater than 1, which would be nonsensical for a proportion.

The left-hand side of equation (3.1) can be rewritten as $g\left(x-h\right),$  where

$$g\left(y\right)=f^{-1}\left(y\right)={\left[1+\exp \left(-y\right)\right]}^{-1}\text{},x=f\left(p\right)=\log \left(\frac{p}{1-p}\right),$$

and

$$h=\frac{z_{1-\alpha /2}}{\sqrt{n*p\left(1-p\right)}}.$$

The first and second derivatives of $g\left(y\right)$ are $g^{\prime }\left(y\right)=g\left(y\right)\left[1-g\left(y\right)\right],$ and $g^{\prime }{}^{\prime }\left(y\right)=g\left(y\right)\left[1-g\left(y\right)\right]\left[1-2g\left(y\right)\right].$ Invoking the mean value theorem, there is an $h*$ between 0 and $h$ such that

$$\begin{array}{l}\begin{array}{ll}g\left(x-h\right)\hfill & =g\left(x\right)-g^{\prime }\left(x\right)h+\frac{1}{2}{g}^{\prime }{}^{\prime }\left(x-h*\right)h^2\hfill \\ \hfill & =p-p\left(1-p\right)\frac{z_{1-\alpha /2}}{\sqrt{n*p\left(1-p\right)}}\hfill \\ \hfill & +\frac{1}{2}{\left[1+\left(\frac{1-p}{p}\right)e^{h*}\right]}^{-1}\left\{1-{\left[1+\left(\frac{1-p}{p}\right)e^{h*}\right]}^{-1}\right\}\left\{1-2{\left[1+\left(\frac{1-p}{p}\right)e^{h*}\right]}^{-1}\right\}\frac{z_{1-\alpha /2}^2}{n*p\left(1-p\right)}\hfill \\ \hfill & =p-p\left(1-p\right)\frac{z_{1-\alpha /2}}{\sqrt{n*p\left(1-p\right)}}\hfill \\ \hfill & +\frac{1}{2}\frac{p}{1+\left(1-p\right)\left(e^{-h*}-1\right)}\frac{\left(1-p\right)-\left(1-p\right)\left(e^{h*}-1\right)}{1+\left(1-p\right)\left(e^{h*}-1\right)}\frac{\left(1-2p\right)-\left(1-p\right)\left(e^{h*}-1\right)}{1+\left(1-p\right)\left(e^{h*}-1\right)}\frac{z_{1-\alpha /2}^2}{n*p\left(1-p\right)},\hfill \end{array}\\ \text{}\text{}\text{}\end{array}$$

using

$${\left[1+\left(\frac{1-p}{p}\right)e^{h*}\right]}^{-1}=\frac{p}{1+\left(1-p\right)\left(e^{h*}-1\right)}.$$

An analogous derivation can be made for the right-hand side of equation (3.1).

Consequently,

$$\begin{array}{l}p+\frac{1-2p}{n*}\frac{z_{1-\alpha /2}^2}{2}-z_{1-\alpha /2}{\left(\frac{p\left(1-p\right)}{n*}\right)}^{1/2}\text{}+o_P\left(\frac{1}{n*}\right)\le P\hfill \\ \le p+\frac{1-2p}{n*}\frac{z_{1-\alpha /2}^2}{2}+z_{1-\alpha /2}{\left(\frac{p\left(1-p\right)}{n*}\right)}^{1/2}\text{}+o_P\left(\frac{1}{n*}\right).\hfill \end{array}$$

After invoking the asymptotic equality in equation (2.3) and dropping $o_P\left(1/n*\right)$ terms, the last set of inequalities is equivalent to Wilson interval in equation (2.2) so long as $n*$ is sufficiently large and $P\left(1-P\right)>0,$ the latter meaning that the true proportion is neither 0 or 1.

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