You haven't said much about your variable. For example, it it discrete? You also haven't said what aspect of the variable that you want to compare the two groups by? You might even show histograms or qqnorm plots to help commentators provide help.

Did you try to transform your second group and calculate again the normal distribution?

"Normality test using Shapiro-wilk shows normal distribution in one group while non normal in another. ( p value of 0.014 and 0.362)"

This is an awfully wrong conception of what these tests do and what they are good for. What you actually have shown is that your data is sufficient to recognize a statistical difference from the (idealized) normal distribution model in the first group, but not so much in the second. The "conclusion" is that more data would be needed to also reject the hypothesis for the second group.* It does NOT mean that the data from that group are sampledfrom a normal distribution! If you think about it you will find that these formal normality-tests (where you eventually use a p-value to make a dichotomous decision [normal or not]) are really useless (if not even harmful).

You should not start with automatically assuming a normal distribution and then panic if the data turns out not to quite follow this assumption. Instead you should start with understanding the variable, its "data generating process" to come up with a sensible, reasonable assumption about a distribution model. This is some kind of idealization, and maybe you understand that the distribution should actually be rather complicated, but that some simplification is tolerable. Then you would use this model. - However, this does not mean that you could not cross-check that the observed data does not considerably contradict your distributional model. But formal tests are not helpful. You need to visualize the discrepancy between the data and the model (as Daniel said: QQ-plots are a good tool for that) and judge the relevance of the observed discrepancy for your case. If it is unclear what effect a particular discrepancy has one can use simulations to get an impression.

In your case the normal model may still be an appropriate idealization. If the sample sizes in both groups are equal, the nominal significance level is kept but the power is compromized. So if the test lets you reject the null hypothesis (mean age difference is zero), you have what you want (i.e. demonstrated that your data are sufficient to interpret the sign of the mean age difference). If you cannot reject the hypothesis, there is no conclusion, so this is also not a problem from a scientific prespective (but it is usually a career problem, because it's difficult to sell a paper making a story about inconclusive data). A better distribution model would be the Weibull distribution.

I would not transform age unless you come up with a resonable explanation what the practical meaning of the transformed variable should be.

Non-parametric tests are hard to interpret when the distribution shapes are different (as in your case). These are no tests of median differences! There are tests for median differences, but they have really low power. And it's still hard to interpret the meaning. Medians don't represent expectations (=means),and I am usually puzzled what it tells me to know that the probability of sampling a higer value from one group is higher than from the other group. However, you can always bootstrap the sampling distribution of any statistic you want, what is a non-parametric approach (as you don't need a parametric formulationof that sampling distribution). But this works well only for larger sample sizes, and just in these cases (larger sample sizes), most discrepancies between the idealized (parametric) distribution models and the real world don't matter much.

* Your variable is strictly positive, but the normal distribution assigns some probability to negative values. So for simple theoretical reasons, is is really only a matter of sample size to "see" that the model isn't perfect. This is more evident (and for smaller sample sizes) when the ages are close to zero (-> distribution must be right-skewed). Since age can not be infinitively large, this is also the case if the ages are close to the maximum life span of the species (-> distribution must be left-skewed).

Daniel Wright : The variable was age, if I understood correctly :)

"formal normality-tests (where you eventually use a p-value to make a dichotomous decision [normal or not]) are really useless (if not even harmful)"

- Jochen Wilhelm

Thanks Jochen Wilhelm . I misread and thought that was the predictor variable (i.e., two groups defined by age). Nice answer btw.

It is my pleasure and an honor to offer a comment to this expert discussion on an important, timely, and interesting question.

A sea of researchers have compared mean ages between two or more groups in a unimaginably broad spectrum of applications: the majority which I've read haven't even considered assumptions or inferences, and many made nary a comment!

Yet, among those who aim to get it right, this particular issue has been a big deal for a very long time. Anyway, here's how I do this sort of thing, these days:

• https://odajournal.com/2020/06/28/what-is-novometric-data-analysis/

Here's an example involving age.

• https://odajournal.com/2016/09/19/novometrics-vs-multiple-regression-analysis-age-and-clinical-measures-of-pcp-survivors/

Regarding "Normality test", the above explanations are exhaustive.

Regarding the opportunity to transform variables, I agree that “I would not transform age unless you come up with a resonable explanation what the practical meaning of the transformed variable should be” ( Jochen Wilhelm).

In addition,

https://www.researchgate.net/post/How_can_we_trust_in_the_back-transformed_confidence_interval_and_statistical_analysis

follow the answers to a similar question including my answer

https://www.researchgate.net/post/Transform_and_use_a_parametric_test_or_use_non_parametric_test

Good luck,

Hamid