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| It turns out that ln(OR) has a variance of (1/a) + (1/b) + (1/c) + (1/d) | It turns out that {{{ Variance {ln(OR)} = (1/a) + (1/b) + (1/c) + (1/d) }}} |
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| (ln OR)* ln(OR) ) / variance ( ln(OR) ) | __ln (OR)^2__ variance ( ln(OR) ) |
Suppose we have a 2 x 2 table of frequencies
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The Odds Ratio is defined as ad/bc
It turns out that
Variance {ln(OR)} = (1/a) + (1/b) + (1/c) + (1/d)so it follows
ln (OR)^2 variance ( ln(OR) )
is chi-square on 1 degree of freedom.
If you have a zero cell then adding one half to all the frequencies enables an estimate of the odds ratio to be made.
Reference:
Everitt BS (1996) Making Sense of Statistics in Psychology A Second Level Course. OUP:Oxford.