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I'm analyzing a 2×2 contingency table, and am required to a report a 95% CI on the Odds Ratio.

Is there any point in prefacing this with a Chi Square Test of Independence? I understand that if the Odds Ratio does not contain 1 then it is guaranteed that the Chi Square test would have been statistically significant, and if it does contain 1 it's guaranteed the Chi Square test would not have been statistically significant. Is the Chi Square Test providing any extra information of use, or is it totally redundant?

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There are several ways of calculating a statistic from a two by two table which asymptotically follows the chi-squared distribution (Pearson, Yates, deviance, ...). There are also several ways of estimating a confidence interval for the odds ratio. Since they do not all agree whoever told you about the guarantee that a test and a ci would always agree should explain what s/he meant. Better to quote both and if necessary explain why they point to different conclusions.

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You should do the 95% CI on the odds ratio. The equations can be found here: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2938757/

Generally you do not do the Chi-square test because the cases are not independent of each other.

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