Return to Answer

In this post, @whuber provides a similar approach: add $1/2$ to both $x$ and $y$ and add $1$ to both the $n$ and $m$. In this particular case, add $1/2$ to $a$ and $b$ and add $1$ to $(a+c)$ and $(b+d)$ for the positive LR and add $1/2$ to $c$ and $d$ and add $1$ to $(a+c)$ and $(b+d)$ for the negative LR.

In this post, @whuber provides a similar approach: add $1/2$ to both $x$ and $y$ and add $1$ to both the $n$ and $m$. In this particular case, add $1/2$ to $a$ and $b$ and add $1$ to $(a+c)$ and $(b+d)$ for the positive LR and add $1/2$ to $c$ and $d$ and add $1$ to $(a+c)$ and $(b+d)$ for the negative LR.

Confidence intervals by the normal approximation

Confidence intervals using a normal approximation

In this post, @whuber provides a similar approach: add $1/2$ to both $x$ and $y$ and add $1$ to both the $n$ and $m$. In this particular case, add $1/2$ to $a$ and $b$ and add $1$ to $(a+c)$ and $(b+d)$ for the positive LR and add $1/2$ to $c$ and $d$ and add $1$ to $(a+c)$ and $(b+d)$ for the negative LR.