Why shift the Mann Whitney distribution?


by fadecomic
Tags: distribution, mann, shift, whitney
fadecomic
fadecomic is offline
#1
Mar19-09, 03:59 PM
P: 10
I'm reading up on the Mann Whitney test, and I can't wrap my head around one thing. Most of the test makes perfect sense. If two samples come from populations with similar medians, then the sum of ranks of both of those populations should hover around some expected value. The "T" or "U" statistic, depending on what you're reading, is determined, and one determines whether or not U falls within a certain interval. Fine. U is defined as the difference of the observed sum of ranks and either the minimum or maximum possible value of the sum of ranks (doesn't matter). The U distribution is parameterized by the two sample sizes and nothing more, as is the maximum possible U. That means the max possible U is nothing but a shift of the distribution. Why bother? Is there some advantage to doing it that way? Why not tabulate the distribution based on the sum of ranks only (which is done--it's called the "Wilcoxan test")?

Thanks.
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fadecomic
fadecomic is offline
#2
Mar19-09, 04:37 PM
P: 10
Incidentally, the expected value for the U statistic is [tex](n_an_b)/2[/tex], which is easy to prove. So why does the Mann-Whitney test require a statistic that is shifted to a central value of [tex](n_an_b)/2[/tex]?


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