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Dear all, Here's a list of all the replies I got to my question. Many thanks to all those who responded! All the Best, Kim. **************************** My original question: Hello everyone Just a small query about non parametric statistics..... It has been indicated to me that, when testing for differences between 2 groups, the Wilcoxon Signed Ranks Test and the Mann Whitney Test assume that the distribution of the two groups have the same shape. Even though these distributions do not have to be normal..do they have to have only 'one-peak'? Also, for the Mann Whitney test and Wilcoxon Signed Ranks Test it is stated in one text that the variables can be ordinal and in another text (when talking about the Mann Whitney test)it is also stated that the variable does not have to be on an interval scale; an ordinal scale is sufficient. However, I have been in conversation with various bods who say that a continuous scale is assumed. Can anyone shed some light? Many thanks, Kim. ************************** For formal assumptions of the tests, I would check a text like Hollander. But more broadly, if one or both distributions have more than one peak then what difference between the two distributions do you want to test? ************************************ Kim, > -----Original Message----- > From: K F Pearce [mailto:[log in to unmask]] > Sent: Thursday, May 06, 2004 3:49 PM > To: [log in to unmask] > Subject: Non parametric assumptions > > > Hello everyone > > Just a small query about non parametric statistics..... > > It has been indicated to me that, when testing for > differences between 2 > groups, the Wilcoxon Signed Ranks Test and the Mann Whitney > Test assume > that the distribution of the two groups have the same shape. Even > though these distributions do not have to be normal..do they have to > have only 'one-peak'? Indeed, the distributions need to have the same shape. But nothing else is required, i.e. they may even have more than one peak. > Also, for the Mann Whitney test and Wilcoxon Signed Ranks Test it is > stated in one text that the variables can be ordinal and in another > text (when talking about the Mann Whitney test)it is also stated that > the variable does not have to be on an interval scale; an ordinal > scale is sufficient. However, I have been in conversation with > various bods who > say that a continuous scale is assumed. Can anyone shed some light? From a theoretical point of view, continuous scales are required as ties are not allowed (even though corrections for ties are available). However, in applications this can be weakend in case the "underlying" distribution is continous. Assume the sweetness of something to be rated on an ordinal scale 1 : 10. Then, in fact, you may observe more than just 10 different values of the "true sweetness", i.e. sweetness is continuous, you only measured it ordinal. In that case, Wilcoxon is okay. However, you will have to assure that the number of ties remains small, otherwise this does not hold anymore. I.e. with just 4 categories and 20 observations, say, I would not recommend the use of Wilcoxon anymore. Best, **************************************************** Hello Kim The Wilcoxon test does not assume that the 2 distributions are unimodal. Also, there are ways of correcting for ties if the Y-variable is not continuous. The main problem with the Wilcoxon test is that it is only a test, and does not provide confidence intervals for a parameter. There are 2 possible confidence intervals corresponding to the Wilcoxon test. These are for Somers' D and for the Hodges-Lehmann median difference. It is possible to calculate confidence intervals for these without even assuming that the 2 groups have the same shape. More about this can be found in Newson (2002). A pre-publication draft of this can be downloaded from my website (see below), and so can a few more documents on rank statistics. I hope this helps. References Newson R. Parameters behind "nonparametric" statistics: Kendall's tau, Somers' D and median differences. The Stata Journal 2002; 2(1): 45-64. *********************************************** Kim, This one comes up regularly. It all depends on how you state your null hypothesis. If your null hypothesis is that the probablity that a member of one population will exceed a member of the other is equal to the probablity that a member of one population will be less than a member of the other, then you need make no distributional assumptions. This is sometimes written as the two populations are not schochastically different. If your null hypothesis is that the probablity that a member of one population will exceed a member of the other is equal to zero, then you need only assume that there are no ties, i.e. that distribution is continuous (always an approximation, of course). In practice, the exact distributions for U and T are worked out for no ties and we approximate when there are ties, which is usually the case in my experience. If your null hypothesis is that the means of the two populations are the same, you must assume that the distributions have the same shape, differing only in location. These tests are sometimes described as tests of inequality of medians because under this assumption the difference between the medians is equal to the difference between the means. If the assumption is true, the variances must be the same. This is unlikely if the disribution is not Normal. In my book An Introduction to Medical Statistics, 3rd ed., I have an example of a Mann Whitney U test which is significant even though the medians are equal. Nearly all the observations were zero. Note that the means, medians, and shape of the distribution do not appear in the calculations for these tests. We only need assumptions about them if we want to draw conclusions about them. I hope this helps, *********************************************************** Kim My understanding is as follows: The Mann-Whitney test assumes the two samples are drawn from identical populations (i.e. that's the assumption of the null hypothesis). This 'common' population can be of any shape. The 'simplest' alternative hypothesis is therefore that the two populations have the same shape, but are 'shifted' relative to one another. The Wilcoxon signed ranks test assumes that the distribution of differences (between the data pairs) is symmetrical (of any shape). The Mann-Whitney test assumes ordinal data, and that the data can be ranked without ties -- so, not necessarily an interval scale, but capable of an unambiguous ranking. Hence many stats packages apply a correction for 'ties' when estimating the significance level. The Wilcoxon signed rank test makes similar assumptions about the differences (between the data pairs) -- which implies that the original measurements, from which the differences are calculated, must be at least interval in scale. (So, all in all, Wilcoxon signed ranks is not as 'assumption-free' as many people think.) ************************************************* Hi Kim I had until now always assumed that one used Mann-Whitney and Wilcoxon tests for ordinal data as well as non-normal continuous data, given that the tests use ranks not absolute values (thus in my mind retaining the features of an ordinal scale). I would be extremely interested in any advice you receive on this topic, in case this assumption turns out to be misguided... Thanks *********************************** Hi Kim, It does not matter what the shape of the distribution is so long as (in the null hypothesis being tested) it is the same distribution for both samples. The continuity consideration comes in because the standard computation of the distribution of the M-W/W test assumes that there will be no ties (equal observations); this is assured for continuous distributions (though observations rounded to a relatively small number of significant figures, as most are, could give a few ties). For ordinal variables, which typically do not have many categories, the probability of a tie may be considerable, and in that case the standard distribution will not be right (though how far it is wrong will depend on the probability of ties). There are various approaches to handling ties. One which makes sense in these days when computer simulation is easy and quick is to break the tie-clusters randomly (e.g. a cluster [ABB] could be A<B<B, B<A<B, B<B<A) and re-compute the M-W statistic each time. You will end up with a simulated distribution of P-values, which will indicate the loss of information due to ties. The rationale for this approach, if adopted, is that the data record a more or less coarse grouping of an underlying continuous variable which has not been directly observed. If you could see inside the bin, you would be able to separate the results; but you can't. So you simulate it by random tie-breaking. OK if the grouping is not really coarse. Such a rationale, however, is not applicable to categorical variables which, even if ordered, do not naturally correspond to an underlying continuous variable. For instance, you may have decided that listening to a classical concert on the radio is "more cultured" than listening to "Moneybox" which is "more cultured" than listening to pop music. So you go and sample people from two neighbourhoods A and B and ask, for these categories, which one they most recently listened to. Then you can test whether A is "more cultured" than B. You will have a lot of ties; there is no obvious underlying variable; and, even if you use a Mann-Whitney type of statistic (sum of all (A,B) pairs in which the B response is "more cultured" than the A response), you certainly shouldn't be referring it to the Mann-Whitney distribution, and tie-breaking for the purpose would be a very dubious thing to do. You're in contingency-table territory here. So there's a line to be drawn. Hope this helps, ****************************************************** Kim, Distributions for the Mann-Whitney and Wilcoxon matched-pairs signed-ranks test do need to be the same for both groups, as you say, but they do not need to be unimodal. The minimum level of measurement for both tests is ordinal. The data do NOT have to be continuous. (They are often used for ranks, which are definitely discrete). ***************************************************************** Hi Kim Not sure about the answer to your 'one peak' question (Though the tests are for location of the median, which may not be particularly helpful for multimodal distributions). I suspect the answer to the issue of whether variables have to be continuous or just ordinal is tied up with ties; the tests work by ranking observations, so if several observations have the same value and so the same rank, you loose a bit of precision (? not sure that's right technical term). If the scales only have a small number of categories, you'll be getting a lot of ties,and this may become a serious issue I'd be inetersted to hear what else people have to say on this ******************************************************** Kim You ask 'can anyone shed some light?'. I suspect we may spread more fog, but I will have a go. The 'Mann- Whitney' (and Wilcoxon rank sum) test does not as far as I can tell make any assumptions about the shapes of the distributions. Nor does it assume that they are the same shape- this is often assumed but the test will detect a wide range of alternatives. It is not a test for differences in mean, or median, of the distributions; rather it is a test that the median of differences is zero. Usually this is associated with differences in the 2 medians (and means) but not necessarily. Therefore it can be applied meaningfully to ordinal variables and discrete distributions; however you will get a lot of ties and you need to know how to allow for these. The best thing is probably to get some decent software and rely upon the methods it uses. Personally I am not convinced that there is much difference between 'ordinal' and 'interval' scales; the real difference is between discrete and continuous scales, and all actual measurements are discrete. The Wilcoxon Signed Ranks Test on the other hand is a test of differences: like a 'paired- t-test' it is actually a one sample test. The only assumptions it makes I think is that the differences are independent and symmetrically distributed. How often one comes across situations where these are valid but the data are too far from normal to use a 'paired- t' is another matter. I am sure you will get lots of different answers as well. Regards