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I have two samples $s_1$ and $s_2$ of count data. The sample size is > 1000 each. The distributions look similar to a Poisson distribution but the variance is much larger than the mean.

How do I test whether the mean of $s_1$ is larger than the mean of $s_2$?

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  • $\begingroup$ What are the counts? Are they all small numbers or do they vary over a wide range? $\endgroup$ – Peter Flom Oct 11 '13 at 20:13
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Give your large sample sizes, you could probably use a t-test on the means. If your sample sizes are equal, you are in pretty good shape whether you want to use a pooled estimate of the variance or unpooled (Welch's test). Do a one sided test, if you are sure that the population of s1 has a mean at least as large as the mean of the population of s2.

Note: If the variances are much larger than the means, your counts are not Poisson. But what matters here is the distribution of the sample averages, and that should be nearly normal, unless the data are super-skewed. In that case, you could do a non-parametric test like the Kruskal-Wallis.

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  • $\begingroup$ Since there are large sample sizes, z-test is an alternative option. $\endgroup$ – iliasfl Jan 14 '14 at 9:25
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I would suggest you fit a Poisson or loglinear regression model with just one dummy variable created for the two groups and then test the slope parameter, say, $H_a: \beta_1 >0$. Any test method (LRT, Wald, or score) under the maximum likelihood framework can be used. As for over-dispersion problem, you may consider other count models such as negative binomial or generalized Poisson models. This should essentially give you a two-sample test for count data.

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A t-test has already been proposed, a Poisson-like generalized linear regression has been proposed. With > 1000 sample size, how about bootstrapping the difference between both samples? It's easy, it's fast, it gives not only a point estimate but also a distribution an it gets rid of all assumptions of normality or poisson or negative-binomial and so on. Even if they are small counts, bootstrapping will do the job.

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