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I am attempting to compare two diagnostic odds ratios (DORs). I would like to know of a statistical test which will allow me to do this. Please help! Thank you!

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    $\begingroup$ what's the raw information you have?... i.e. how did you come to two DORs? $\endgroup$ – John Aug 8 '10 at 1:37
  • $\begingroup$ If you have the raw information available that you used to calculate the odds ratios a statistical test will be possible. Alternatively there might be some reasonable simulation approaches if you know how many observations went into each odds ratio. P.S. For those of us who only have a passing familiarity with DORs would you please provide the formula used for calculating one? $\endgroup$ – russellpierce Aug 8 '10 at 18:17
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Assuming the odds ratios are independent, you can proceed as you would in general with any estimate, only you have to look at the log odds.

Take the difference of the log odds, $\delta$. The standard error of $\delta$ is $\sqrt{SE_{1}^2 + SE_{2}^2}$. Then you can obtain a p-value for the ratio $z = \delta/SE(\delta)$ from the standard normal.

UPDATE

The standard error of $\log OR$ is the square root of the sum of the reciprocals of the frequencies:

$SE(\log OR) = \sqrt{ {1 \over n_1} + {1 \over n_2} + {1 \over n_3} + {1 \over n_4} }$

In your case, each $n_i$ correspond to TP, FP, TN, FN.

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  • $\begingroup$ Raw data is from a 2 x 2 diagnostic contingency table. DOR equation = (TPTN)/(FPFN) $\endgroup$ – Jay Aug 10 '10 at 11:10
  • $\begingroup$ Where TP is Test Positive and TN is Test Negative and FP is False Positive and FN is False Negative? $\endgroup$ – russellpierce Aug 17 '10 at 7:20
  • $\begingroup$ How would the question asker obtain SE^2_1 and SE^2_2? $\endgroup$ – russellpierce Aug 17 '10 at 7:52
  • $\begingroup$ @Jay: didn't see your comment sooner, but the above is applicable if the DOR are independent. @drknexus: updated the answer for how to obtain SE. $\endgroup$ – ars Aug 17 '10 at 18:59
  • $\begingroup$ @ars: I have asked a follow-up question to this post (stats.stackexchange.com/questions/350637/…) in case you are willing/available to comment further. $\endgroup$ – ksroogl Jun 10 '18 at 2:30
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If you have the 95% CIs for all of your odds ratios, you can do a quick check for statistical difference by looking for overlap in the odds ratios. If 95% CI's for two OR overlap, they aren't significantly different.

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    $\begingroup$ I believe they can overlap a little (maybe ~25%) & still be significant at the 5% level. Remember that the 95% CI you see is for the individual OR, but the test of 2 ORs is about the difference between them. However, if they don't overlap at all, then they are definitely significantly different, & if the 95% CI's overlap the other OR point estimate, they definitely don't. $\endgroup$ – gung - Reinstate Monica Sep 25 '14 at 15:11
  • $\begingroup$ @gung Would it not be possible that one of the 95% CI's overlaps the other ORs point estimate, but that they are still significantly different (at the same alpha = 0.05 level) if the two estimators are negatively correlated? $\endgroup$ – baruuum Sep 12 '19 at 7:52
  • $\begingroup$ @baruuum, you should probably ask that as a new question, or even search the site for existing answers first. At any rate, I don't think it's possible for them to be significantly different if 1 CI overlaps the other point. $\endgroup$ – gung - Reinstate Monica Sep 12 '19 at 11:38

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