I'm running a biological experiment with rodents, have two groups (each consists of 26 animals), where one is treated with a chemical, and one is control (saline).

In one variable, there doesn't seem to be much of a difference between groups. However, when I identify outliers, the data becomes very significant ($p < .01$). I am wondering are my criteria to exclude those data points valid.

I've used two methods to identify outliers. First of all, I was identifying outliers separately for control and treatment, as the data seems very different (check histogram attached for the treatment group)—is this wrong or not? To me, it sounds logical, as the groups are different (in the sense they are chemically treated differently).

Histogram for treated group

First method I mentioned—I checked for data that is 2.5 STDEV from the mean, and excluded those points. The second one is similar: http://www.wikihow.com/Calculate-Outliers—this one gives the same result, and identifies the outliers in the treatment group.

However, as I would like to prepare a publication, I need a real "peer-accepted" test for outliers. I have found some tests in StatSoft Statistica, but the box whisker plot does not show outliers there. However, in the Grubbs' test, I get $p < .05$, so it is analytically detected as an outlier, right?

So, my questions are:

1) Is it ok to calculate outliers differently for each group, or should I collapse the sample into one group ($N = 52$) and calculate it then, for both groups together?

2) Which method should I use to identify (and exclude) outliers in order for it to be accepted by a critical review?

I attach the data points here: https://www.mediafire.com/?0qdsifib0hugd9u

To me, it seems that any animal with the score higher than 54 is an outlier.


1 Answer 1


It is most definitely not OK to remove outliers just because they happen to be far from the mean. The fact that there are many of them just confirms that they are not flukes. The values above 50 seem very much in line with the long-tailed distribution of the control samples. Depending on what is the outcome, a test that appropriately adds to information about the distribution of the outcomes might have more power than a t-test (you don't mention what test you use), which specifically compares the means. Or you might want to consider whether you really care about the mean, or perhaps some other aspect of the distribution.

  • $\begingroup$ Yes, the distribution is long-tailed in the control sample, but not so much in the treatment one. This is why I ask is it ok to consider them outliers (one of the animals in the treatment sample actually died after the test, others did not, and it was doing poorly in other tests as well)... I used t-test, but I do not think that I can consider this sample to follow a normal distribution, so I used the Mann-Whitney U Test as well, which might be a better choice for this dataset. Correct? $\endgroup$
    – praznin
    Commented Jan 17, 2014 at 22:37
  • $\begingroup$ So which one should I use? T-test assumes normality and I don't have that either. :/ $\endgroup$
    – praznin
    Commented Jan 17, 2014 at 22:45
  • 1
    $\begingroup$ Mann-Whitney U test does not assume a symmetrical distribution. If you want to interpret it as a test of medians, then you have to assume that the two distributions have the same shape. Otherwise it is a test of $P(X < Y)=0.5$, where $X$ and $Y$ are random observations from the two distributions. It seems a reasonable test in this case, but I know nothing about your hypothesis. $\endgroup$
    – Aniko
    Commented Jan 17, 2014 at 23:30
  • $\begingroup$ @Iliasfl this is wrong. There is no such assumption. $\endgroup$
    – Michael M
    Commented Jan 17, 2014 at 23:33
  • $\begingroup$ @Aniko, the hypothesis is that the treatment animals will have a lower score in the variable measured, which is indicative for anxious behaviour (like an anxyolitic effect). I don't know if this answers your question. So, Mann-Whitney U test could be fine in the end? $\endgroup$
    – praznin
    Commented Jan 17, 2014 at 23:37

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