# Hypothesis testing with odds ratios

I was wondering about hypothesis testing with odds ratios. I know that in this situation the null hypothesis is $OR = 1$. However, what standard deviation and statistic should be used? For some reason, I can not find a thorough description of the procedure in that case. Maybe it is because people use Fisher's exact test.

Reading Wikipedia, there is a section for statistical inference, but they are using the logarithm of the odds ratio instead.

• You're asking about estimating odds ratios in 2x2 contingency tables? (They're estimated in logistic regression as well.) Dec 15, 2013 at 0:41
• Well, I expected a test statistic with a given distribution in which the null hypothesis is $OR=1$, instead of Fisher's exact test. Obviously, you can use a contingency table to obtain the odds ratio, but that's not necessary. Dec 15, 2013 at 0:50

First the null hypothesis can be anything you like; an odds ratio of one is saying that the odds are equal across two groups, but other nulls may be of interest. In a two-by-two contingency table the sample odds ratio $\hat\theta=\frac{n_{11}n_{22}}{n_{12}n_{21}}$, where $n_{ij}$ is the frequency in the $i$th row & $j$th column, can be used as an estimate of the population odds ratio $\theta$. The logarithm of the sample odds ratio converges more quickly to a Gaussian distribution, with a standard error estimate of $$\hat\sigma_{\log\hat\theta}=\sqrt{\frac{1}{n_{11}}+\frac{1}{n_{12}}+\frac{1}{n_{21}}+\frac{1}{n_{22}}}$$; you can use this to obtain (approximate) confidence intervals for the population odds ratio by exponentiating the (approximate) confidence interval bounds for its logarithm. For example 95% bounds can be calculated with $$\exp(\log\hat\theta\pm1.96\hat\sigma_{\log\hat\theta})$$. Hypothesis tests can be formed using the fact that $\frac{\log\hat\theta-\log\theta_0}{\sigma_{\log\hat\theta}}$ has a standard normal distribution under the null hypothesis $\theta=\theta_0$. For a null hypothesis of $\theta=1$, that's $\log\theta=0$.
• 1+. Thanks. That's pretty much what I read in Wikipedia. However, is there something like $t = \frac{\hat{\theta} - 1}{\sigma}$ as with means and population proportions? Dec 15, 2013 at 1:16