# T-test / ANOVA on Box-Cox transformed non-normal data

Suppose I apply a Box-Cox transformation to my data and now it looks rather like a normal distribution. I then add another dataset, transform it by Box-Cox with the same lambda and run a t-test to compare the means. Would this approach make sense if my data is non-normal by its nature? In other words, is the fact that a Box-Cox transform produces a Gaussian-like distribution sufficient to then use standard methods for normally distributed data such as t-test and ANOVA?

Update - to formulate this question a bit more specifically: I want to test whether there are significant differences between the means of two samples. I can see that the distributions in each sample are very much non-normal. My question is: if I force them to look normal by using a transformation, will this be enough to essentially forget about their "original" non-normal nature for testing this hypothesis?

Update 2 - I suppose my question is similar in spirit to this one, which asked the same thing about log-transformation.

• ANOVA does not require normally distributed data it requires normally distributed residuals. Apr 9, 2013 at 19:23
• That is correct, but in my example above, isn't this essentially the same thing? I.e., if I then merge these two datasets into one and use their labels as explanatory variables? Please do correct me if I'm wrong. Apr 9, 2013 at 19:37
• I just wanted to clarify that. Apr 9, 2013 at 19:45

If you're interested in comparing means, once you transform you end up with a comparison of things that are not means. If the right assumptions hold you can still test for a difference, but the alternative won't be location-shift.

I didn't want the details to detract form the general point.

On the other - and more important - hand, if you omit essential details you'll be more likely to end up with less useful - or even potentially misleading - answers that you won't even realize aren't the answers you need.

By leaving out the fact that you were dealing with count data, you were risking exactly that. While leaving out unnecessary detail is probably useful, knowing it's count data is pretty much central to the problem.

There are techniques for comparing means that are suitable for count data. With some more information about the kind of analysis/information you were after (even if it's what you would have done if the data were normal), we may be able to guide you better.

Transformation is less useful than doing something suited to your actual data.

What do you mean "not normal by its nature"? A variable is distributed a certain way, that is its nature. Now, you might get an outlier by chance, but that is also part of the nature of the variable - e.g. if you sample people's heights and get one person who is 7 feet tall, well, there ARE such people.

Is the fact that you can make a variable normal sufficient to use a t-test? No. The key question should be the research question. Sometimes you don't want to transform the data. Sometimes the outliers are vital. Sometimes you want to transform data even if it is already normal.

It depends on the situation.

• Thanks for your answer. My research question is whether there are significant differences between the means of two samples. I can see that the distributions in each sample are very much non-normal. My question is: if I force them to look normal by using a transformation, will this be enough to essentially forget about their "original" non-normal nature for the purposes of testing the differences in means? Apr 9, 2013 at 20:36
• Two samples of what? Incomes? Height? Mohs scale of hardness? IQ? Mass? Answering statistical questions without context is like boxing while blindfold. Apr 9, 2013 at 21:15
• I share your frustration about the vagueness of the question - it's just that the dataset is quite complex and I didn't want the details to detract form the general point. The data are sequence counts coming out of a slightly unconventional genomic analysis. Classically next-gen sequencing data are treated as Poisson or negative binomial, but this doesn't seem to be appropriate in my case. Apr 9, 2013 at 22:57
• Why are Poisson and NB not appropriate? If you have count data, they are likely to be better than the Normal, unless the mean is high and there are many different values. Apr 9, 2013 at 22:59
• They don't seem appropriate, because the data looks consistently underdispersed compared to Poisson, let alone NB. Apr 10, 2013 at 8:18