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I have two data sets, which are assumed to be log-normally distributed. How can I test whether their means are statistically different from each other?

I guess I cannot use the 2-sample t-test, because data is not normal. Are there any easy solutions for the issue?

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    $\begingroup$ You should still be able to use 2 sample t test since this only really relies on fact of CLT which only criteria is look at sum of random variables that are iid. But if you know they are log normal then you may be able to come up with a better test that relies on less approximation. Also note that for $X\sim N(\mu,\sigma^2)$ that $E(e^X)=e^{\mu+\frac{1}{2}\sigma^2}$ where $e^X$ is lognormal $\endgroup$
    – Kamster
    Dec 29, 2014 at 23:18
  • $\begingroup$ but like others have said that if you assume variances are the same as Glen_b said you can just check on logged data whether means are significantly different $\endgroup$
    – Kamster
    Dec 29, 2014 at 23:21

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If you can assume equality of the $\sigma$ parameters (the population standard deviation of the logs), then a test for equality of $\mu$ (the population mean of the logs) will be the same as a test for equality of the mean of the lognormal.

That is, under that assumption, you can take logs and do a two sample equal-variance t-test (a Welch test wouldn't work for that though).

Alternatively, under the same assumption of equality of $\sigma$ parameters, a Wilcoxon-Mann-Whitney test will also be a test of equality of means (against a location shift in the logs or equivalently a scale shift in the original variables). You don't have to take logs to do this test - it works the same either way.

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Take the log of your data, and run standard ANOVA.

If you have the data sets $x_i$ and $y_i$ both from lognormal distribution, it means that $\ln x_i$ and $\ln y_i$ are from normal distribution.

As @whuber noted, this would be a test on medians of $x,y$, but if you assume that $\sigma_x=\sigma_y$, then equal medians will imply equal means. On the other hand if you don't assume the equal variances, then you may end up with the same means and different variances. In this regard having the same mean is less informative. So, it's better to analyze medians.

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    $\begingroup$ This is a test of medians, not means. Especially when the coefficient of variation is not very small (greater than 40% or so), the two parameters differ substantially. $\endgroup$
    – whuber
    Dec 29, 2014 at 21:31
  • $\begingroup$ @whuber, right, but I'm assuming she's interested in medians, although the question is about means. I think what she's really interested is whether the samples are from the same distribution. $\endgroup$
    – Aksakal
    Dec 29, 2014 at 21:38
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    $\begingroup$ In many applications involving the lognormal distribution this is not a valid interpretation. For instance, in many countries environmental standards are based on human health risk assessments that (1) are concerned about arithmetic means (because they reflect total exposure to a pollutant over a period of time) and (2) assume lognormal distributions of concentrations. Thus, without further justification, it would be incorrect--and possibly quite erroneous--to recommend analyzing medians instead. $\endgroup$
    – whuber
    Dec 29, 2014 at 22:01
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What about doing a likelihood ratio test where under the null, the location and scale parameters of the lognormal are equal, and under the alternative they are distinct? In that case data from sample 1 would have $E(X) = \exp(\mu_1+.5\sigma_1^2)$ and data from sample 2 would have $E(Y) = \exp(\mu_2+.5\sigma_2^2)$. Under the null, set $\mu_1 = \mu_2$ and $\sigma_1 = \sigma_2$ then you would have equal means and the difference in the -2log(likelihood) values under each hypothesis would have a Chi-square distribution.

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