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I am trying to compare financial investment of 2 groups, however, my data is not normally distributed AND has unequal variance between the two groups. I thought of using the Mann-Whitney on SPSS but I read that to use the Mann-Whitney data must have equal variance, is there any exception to this rule? I have decent sample sizes with my smallest sample size being 85, would this large sample size allow me to use a t-test with unequal variance?

Any help would be appreciated.

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    $\begingroup$ Where did you read that the Mann-Whitney U-test requires equal variance? Can you site the source and quote where this was claimed? $\endgroup$
    – gung - Reinstate Monica
    Commented Aug 12, 2016 at 1:13
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    $\begingroup$ Data are neither parametric nor nonparametric; those are adjectives that apply to models or techniques. If you mean "not normally distributed" that's not at all the same thing as "nonparametric" and similarly "parametric" is not at all the same thing as "normally distributed" -- one can fit parametric non-normal models and correspondingly, one can happily use nonparametric procedures on data drawn from normal distributions. Please amend your post to more clearly express the actual situation. $\endgroup$
    – Glen_b
    Commented Aug 12, 2016 at 3:29
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    $\begingroup$ There's no such rule. If you want to make a particular kind of interpretation of the results, you might assume equal shapes and equal variance (e.g. if you explicitly want to be able to interpret rejection as a mean-shift). But the test has no such requirement. Who says this is a rule? Can we have the context? $\endgroup$
    – Glen_b
    Commented Aug 12, 2016 at 3:31
  • $\begingroup$ Hi Gung, I do not have the original source, however, here is a powerpoint I got off of the University of Ohio statistics site which has the same information; ohio.edu/plantbio/staff/mccarthy/quantmet/lectures/Nonparm.pdf ; also by nonparametric I meant not normally distributed. $\endgroup$
    – E.Brogdon
    Commented Aug 13, 2016 at 17:09
  • $\begingroup$ @E.Brogdon, the information in that document is at best misleading. You do not need identical distributions, unless you want to interpret the result as showing the means differ. (Note that, for me to be 'pinged', you need to put "@" before my username.) $\endgroup$
    – gung - Reinstate Monica
    Commented Aug 13, 2016 at 17:15

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In general, the reason we use non-parametric analyses is that we don't want to have to make (or count on) any distributional assumptions. In other words, you might choose the Mann-Whitney U-test because you think the variances are unequal, for example. In short, non-parametric analyses do not require equal variance.

Let me speculate on what may have been the source of confusion. The Mann-Whitney U-test is a test of one distribution is stochastically larger. That means that if you drew a single value from each population, it is likely that the value from the one distribution will be higher than the other value. Now if the two distributions are the same, except that one is shifted up relative to the other, then stochastically larger implies the higher population has a higher mean. This is only true if the distributions are identical except for shifted. Note further that identical distributions entails equal variances.

If your goals would only be satisfied by a test of the equality of means, and your distributions are neither sufficiently normal nor identically shaped, you could try bootstrapping.

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  • $\begingroup$ Yes, the test sees if one value is stochastically larger than the other. (I've learned to avoid "stochastic dominance" since it is formally used to mean something a bit different from this -- though a number of people have used it in this sense) $\endgroup$
    – Glen_b
    Commented Aug 12, 2016 at 3:32
  • $\begingroup$ @Glen_b, I changed the phrasing. Let me know if you think it needs more. $\endgroup$
    – gung - Reinstate Monica
    Commented Aug 12, 2016 at 18:45
  • $\begingroup$ It looks very good to me. $\endgroup$
    – Glen_b
    Commented Aug 13, 2016 at 2:09
  • $\begingroup$ @gung this explanation is so fantastic, helps me immensely, you are much more clear than much of what I have found! Again, thank you for your help. $\endgroup$
    – E.Brogdon
    Commented Aug 13, 2016 at 17:37
  • $\begingroup$ You're welcome, @E.Brogdon. $\endgroup$
    – gung - Reinstate Monica
    Commented Aug 13, 2016 at 17:52

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