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I want to understand whether the differences in the means of two independent samples (distributed lognormally) are statistically significant. In order to use Student's t-test, the data must be distributed normally. Since I have lognormal distribution, do I need to test the data for normality? (Using for example Shapiro-Wilk/Kolmogorov–Smirnov tests).
I know the differance between normal and lognormal distributions (normal distribution is symmetric and lognormal distribution is right-skewed. And lognormal random variable is positive, whereas normal has also negative values).
But I don't really understand whether I have to do test for normality at all. Thank you :)

Mary
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    Question (meant to be bit rhetorical): If you have samples that you already know/assumed are log-normally distributed, what information will a test for normality provide? – B.Liu Apr 25 '21 at 14:41
  • @B.Liu Hmm yes, it must be pointless. But can I use t-test for lognormal distribution? Is it appropriate? – Mary Apr 25 '21 at 14:44
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    Theoretically, you will incur some bias and some will consider it inappropriate. Practically, depending on the skewness of the lognormal distribution and the number of samples you have, people may use t-tests anyway, along with various amount of salt they bring with them. See [this question](https://stats.stackexchange.com/questions/69898), [this question](https://stats.stackexchange.com/questions/492750), and [this question](https://stats.stackexchange.com/questions/504262) for some of the discussions related to this topic that I am aware of. – B.Liu Apr 25 '21 at 14:55
  • @B.Liu Thank you for your help! I have just one more question :) Is it ok to use ks-test (two-sample Kolmogorov-Smirnov test) in this case? Or are there any other better/more appropriate options? Thank you!!! – Mary Apr 25 '21 at 15:01
  • If you _know_ data are lognormal, then take logs of both datasets and compare the resulting normal samples using a valid t test. – BruceET Apr 25 '21 at 15:31
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    @Bruce Unfortunately, that approach tests a different hypothesis: it compares geometric means, not arithmetic means. – whuber Apr 25 '21 at 17:45
  • @whuber. Yes, right. Also, it's difficult to make sense of the resulting CI. Is there something more powerful than nonparametric tests? – BruceET Apr 25 '21 at 18:03

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