Timeline for Test two groups with only sample statistics or without distributional assumptions
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2015 at 0:00 | history | edited | Glen_b | CC BY-SA 3.0 |
added 124 characters in body
|
| Mar 30, 2015 at 23:35 | history | edited | Glen_b | CC BY-SA 3.0 |
added 15 characters in body
|
| Mar 30, 2015 at 23:26 | history | edited | Glen_b | CC BY-SA 3.0 |
added 233 characters in body
|
| Mar 30, 2015 at 23:20 | history | edited | Glen_b | CC BY-SA 3.0 |
added 233 characters in body
|
| Mar 30, 2015 at 23:11 | history | edited | Glen_b | CC BY-SA 3.0 |
added 1413 characters in body
|
| Mar 30, 2015 at 22:26 | comment | added | Glen_b | If we go to second order stochastic dominance, I think the restriction about the integrals above isn't needed and the expectations will automatically be ordered. However, this stuff isn't my forte, so I hope I haven't made some glaring error in there. | |
| Mar 30, 2015 at 22:18 | comment | added | Glen_b | @gung There's not stochastic dominance at first order there -- the F's will cross. If you specify something similar (that's not sample dependent), you can just draw the difference in F and see it crosses the axis. [I'm not talking about the even more general P(X>Y)>1/2 case that the most general take on the MW gives you; stochastic dominance is more restrictive.] | |
| Mar 30, 2015 at 22:05 | comment | added | gung - Reinstate Monica |
I'm not sure I'm following all of this. Here's what I tried: set.seed(1656); x=rlnorm(10000); x=x-mean(x); y=-1*x; wilcox.test(x,y). That is significant w/ identical means. I like the MW suggestion & the note that you have to assume the shapes are the same in order for this to be a test of the means is a good one, but I wonder if that subtlety might slip past casual readers.
|
|
| Mar 30, 2015 at 22:04 | history | edited | Glen_b | CC BY-SA 3.0 |
added 377 characters in body
|
| Mar 30, 2015 at 22:00 | comment | added | Glen_b | So anyway, the W-MW is useful in a pretty broad range of situations for testing difference in means (more broadly useful than some of my early answers here might have suggested). I should probably edit. | |
| Mar 30, 2015 at 21:51 | comment | added | Glen_b | er, that should read "... started with the CDF..." and the "(as under the alternative)" should be later in that last sentence, since it includes the second case. For anyone reading along, the relationship between survival function and expectation is discussed here | |
| Mar 30, 2015 at 21:45 | comment | added | Glen_b | @gung I suspect that I was being unclear. I mean stochastic dominance at first order; $X$ stochastically dominates $Y$ at first order if $F_Y\geq F_X$ (I should have started the CDF relationship). Because of the relation between expectation and the survival function ($1-F$), if that inequality is strict (as under the alternative), or even strict in some interval and elsewhere equal, such that the integral of the survival functions differ, that implies $E(X) > E(Y)$. | |
| Mar 30, 2015 at 21:36 | comment | added | gung - Reinstate Monica | Hmm, maybe I'm misunderstanding. Can't you have 2 distributions, 1 positively skewed & 1 negatively skewed, but w/ identical means, st NS is stochastically dominant over PS & hence MW would be significant w/ identical means? I thought that was the case, & I read your answer as suggesting MW only w/ identical shape b/c of that potential problem. I may be confused / misunderstanding something. | |
| Mar 30, 2015 at 21:13 | comment | added | Glen_b | @gung While your comment is true (and a consideration to keep in mind), I didn't recommend against W-MW anywhere in my answer. I offered an alternative which may be useful. In fact with some care, one can extend the applicability of the W-MW from the equal shape location-shift case (e.g. if both means exist, we could extend it to the fairly general situation under the alternative of stochastic dominance, since it implies unequal mean). The W-MW is ... a very handy test. If you face heavy tails, it should have better power than the permutation test. | |
| Mar 30, 2015 at 16:11 | comment | added | gung - Reinstate Monica | Is your recommendation against the MW when the distributions differ b/c the MW can be significant w/ equal means but different shapes? | |
| Mar 30, 2015 at 14:52 | history | answered | Glen_b | CC BY-SA 3.0 |