Timeline for How to choose the test statistic in Mann-Whitney test?
Current License: CC BY-SA 3.0
11 events
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| Jan 13, 2023 at 3:53 | history | edited | User1865345 |
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| Oct 6, 2018 at 1:59 | vote | accept | r_31415 | ||
| Oct 5, 2018 at 13:37 | answer | added | ely | timeline score: 1 | |
| Nov 17, 2013 at 15:47 | comment | added | r_31415 | use the corresponding tables for that statistic Great, that was the part I was missing. I thought those statistics could potentially give different decisions. Would you add an answer in order to accept it? | |
| Nov 17, 2013 at 14:28 | comment | added | Glen_b | All of the statistics mentioned are equally correct statistics, yielding equivalent tests. As long as you're clear which one you're using, and use the corresponding tables for that statistic, they all reject or fail to reject the same cases. There's a little bit of relevant discussion in this answer. | |
| Nov 17, 2013 at 8:25 | comment | added | ttnphns | A single answer to your confusion cannot be done because different programs (implementations) differ in details. The fact is that whether to rely on U1 or U2, there always a due move is done to compute the unique correct Z or the exact p-value. Don't bother your brains. | |
| Nov 17, 2013 at 8:15 | comment | added | r_31415 | Correct. However, I don't think it's impossible that a test statistic $U_{1}$ gives a value that rejects the null hypothesis $H_{0}$ and $U_{2}$ fails to reject $H_{0}$. Otherwise, why to suggest to take the smaller value, the biggest value or the first one? That seems a bit pointless. | |
| Nov 17, 2013 at 8:11 | comment | added | ttnphns | Because since you know the sum of ranks, U1+U2, then if you get to know U1 you automatically know U2, and vice versa. | |
| Nov 17, 2013 at 8:08 | comment | added | r_31415 | Could you elaborate a bit more? Obviously there are two test statistic to choose from and those have different values depending on sum of ranks and possibly sample size, so why using min() or max() makes no difference? | |
| Nov 17, 2013 at 7:58 | comment | added | ttnphns | Because the two samples of known sizes are combined first and then ranked (and the overall sum of ranks of is therefore fixed), it makes then no difference whether you base the test on min() or max(). | |
| Nov 17, 2013 at 7:38 | history | asked | r_31415 | CC BY-SA 3.0 |