Just do the Mann-Whitney on the untransformed data, that's probably the simplest and most honest way to analyse it.

Bojim se da Vam poduzorci nisu izbalansirani po broju sudionika. u prvom je 39, a u drugom 281 ispitanik. U opvom slučaju ne bi trebalo raditi t-test, jer su, pretpostavljam, i varijance dobi ovih poduzoraka statistički značajno različite.

Just do the Mann-Whitney on the untransformed data, that's probably the simplest and most honest way to analyse it.

Haris, this kind of variable does not follow a normal distribution. As it assumes two situation (not smoke/yes smoke) this kind of data adjust to the Binomial distribution. So, if you use a SAS System you can perform a PROC GENMOD to obtain a ANOVA and analyze your data to verify the average differences between the group.

This is a methodology to use. Otherwise, you can use a non-parametric statistic like a Mann-Whitney test. However, this is the last option in Experimental Design

Best regards

Nonsense, smoke or no-smoke is a grouping variable, group 1 or group 2, of course it doesn't have a normal distribution, it's effectively the independent variable.

If your hypothesis is look at the mean age difference between the smokers and non smokers means you can just run non parametric test called Mann-Whitney test is enough (Because your data does't follow normal distribution). In this case you should represent your result table as median,interquartile range and your p-value.

Usually we will go ANOVA for to compare more than two mean differences. Hence you cannot run ANOVA in your case because you have only two groups (Smoker and Non-smoker).

Coming to your log transformation, you can do t-test after log transformation in your dataset. If you look at the P-value of your log transformed t-test and non-parametric Mann-Whitney test both gives nearly same value. When you do the log transformation your t-test will take geometric mean for run the analysis behalf of the arithmetic mean. So better you can just run the non-parametric Mann-Whitney U test.

Haris,

It is not clear from what you say whether you performed a normality test on the whole distribution of ages (i.e., smokers and non-smokers all together) or otherwise. I just wanted to point out that performing a normality test on the whole distribution of ages irrespective of the fact that this confounds smokers with non-smokers does not make any sense. The normality test should bear on the ages *conditional on smoker status*.

Indeed, imagine 1°) the distribution is normal for each group (i.e., conditionally on the smoker status) and 2°) there is an effect of the smoker status. If you look at the whole distribution (smokers and non-smokers all together), it will have two bumps and thus will not be a normal distribution.

So, I suggest you first read my contribution to this thread (sorry for the self-promotion):

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

If after you log-transformed age, the distribution of log-transformed age (again, conditional on smoker status!) is not different from the normal distribution (i.e., a normality test (e.g., Shapiro-Wilk) yields a nonsignificant p-value), you can go on and interpret the result of your t-test. If its p-value is < 0.05 you are entitled to reject the null hypothesis!

HTH,

cheers,

SCM

The mann whitney test is telling you something useful, but is is not testing the same hypothesis as the t-test.

It is calculating the probability that a smoker will be older than a non-smoker. This is a rather useful statistic. If you're using Stata, add the -porder- option to see the calculated probability. If you have SPSS, you will have to do this by hand. The U statistic is the probability if you divide it by the maximum value of U, which is n1 x n2.

This statistic has many names: the Mann Whitney statistic, the common language effect size, Harrell's C (though C is a special case, applied to censored data), and the area under the ROC curve. But it has a simple interpretation: the probability that a member of one group will have a higher value than a member of the other group.

The real question is whether the difference you detected is of any real importance.

Thank you all for the answers. The entire age distribution is positively skewed as well as the distribution for smokers and non-smokers separately. I am aware of the non-parametric procedures (Mann-Whitney test) but was wondering about the interpretation of the results. Median age of smokers is 34 (almost same as the mean) and nonsmokers 29. Clearly the smokers group is somewhat older, but is the difference due to chance or is there really a difference in age of smokers and nonsmokers (in this study)? The problem is that I have two different conclusions and need to be careful in making interpretations, transformed data and MW test p.05). And last concern regards the question of which type of error I am more likely to make when analyzing nonnormal data by using parametric statistics!? Of course, the best thing would be to take another sample of the population of interest and test the age difference again among smokers and non-smokers but will leave that possibility out (for the time being) as we want to elaborate on stat. concepts.

p.s. the interpretation of log data is also problematic, mean smokers 1.52, nonsmokers 1,47, p=.037. And variances are the same in both groups.

As both distributions are not normal, then you must use either the transformed data t-test or the MW, and both of them give the same result, there is a significant difference. So you should conclude that the difference is not due to chance. In terms of error type, it depends on what happens. In this case, if you used the t-test on the untransformed data it gives you p>.05, so you are making a false negative interpretation, a type ii error. If on the other hand it agreed with the alternatives, MW and transformed t-test, there would likely be no error.

For the issue in general, which error type is more likely when using a t-test with non normally distributed data, I think it probably depends on the reasons for the non normality, a skew in one direction will increase the chance of one type of error and a skew in the other, the other type of error.

Finally, it does sound a bit like you are reluctant to accept that there is a significant difference in age in your groups. If this is the case, remember that p values are only part of the story, the effect size here is about .33 (Cohen's d), and would be called a 'small' effect.

Haris,

I pretty much agree with Graham (except for looking at the size effect, which I do not believe in, but this is another story): like it or not, you've got an age difference between smokers and non-smokers.

I'm not a fan of t-tests in general but in this case if on the logtransformed data i) variances are homogeneous between the 2 age groups, and ii) the distributions of age conditional on the smoking status are normal, then a conclusion based on a t-test on the logtransformed data is warranted.

And you may interpret the logtransformed age data very simply: smokers are older.

I have nothing again nonparametric tests per se, but people often forget that there are assumptions that have to be met even for a MW test, namely that the distributions that you compare are the same. And this is what I find difficult to warrant. I mean, if the distributions are skewed and not normal, how can someone warrant they are *the same*? This, plus the fact that nonparametric tests generally are less powerful than parametric tests, would make me want to give a try to a t-test comparison. As with your data the conditions are met for a t-test comparison (on the logtransformed data), stick with it and do believe in what it yields as a result! My five cents.

Cheers,

SCM

Haris, ou said: "The problem is that I have two different conclusions" - That's not correct. You have different p-values. What conclusions you draw from these p-values is your choice.

I presume that you have not planned the test (otherwise you would have known the effect size to look for and you would have known the distribution and you would have known the test and you would have fixed the power). To at best you are controlling the type-I error rate but not the type-II error rate. This in turn means that you do not have ANY conclusion for a "non-significant" result. Such a result is just that you should not reject H0 based on your data when you wish to keep the type-I error rate. That's all.

Regarding your concern about the type-I error: this error is not done by data analysis, nor by any transfromation or by any test. This error is done when you wrongly decide to reject the H0. By no means you can find out how likely your decision is right or wrong. Rejecting H0 only when p