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I have a dataset of six variables that represent respondents' ranking to six different stimuli. Each respondent ranked this stimuli in order of preference from 1 to 6. This means that each row of a data matrix with these 6 variables can't have two of the same values. I also have a variable that represents respondent's gender.

I would like to examine whether there are gender differences in terms of their ranking preferences for each of these 6 stimuli (variables). I am not sure if ordinal regression is an appropriate statistical analysis, since a person's response on one of the variables depends on the his/her responses on the other variables.

Any suggestions or insights on what could be an appropriate analysis for such rank ordered data is appreciated!

tvl
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  • If I understand it right, your 6 variables are the 6 stimuli, and data values are the ranks. Because your task is to test for each stimulus separately, the situation is not different any fundamentally from when the data are ratings (see pt. 1 http://stats.stackexchange.com/a/141669/3277). You may confidently do Mann-Whitney test for each variable. – ttnphns Jun 02 '16 at 19:26
  • @ttnphns thank you for referring to the earlier post. Are you saying that, if I were to analyze each variable separately, then the Mann-Whitney test is an appropriate test, regardless of that we know the responses for each of the six are not independent? – tvl Jun 02 '16 at 20:20
  • And if were to analyze these 6 variables simultaneously, what would be the appropriate procedure? – tvl Jun 02 '16 at 20:21
  • That will be a repeated-measures compositional data (compisitional data are what sum to a constant "horizontally"). Special methods exist for them, including some modelings available through generalized estimating equations (GEE). However, as I've noted in my link, traditional RM-ANOVA could be used also, without being a gross violation. – ttnphns Jun 02 '16 at 20:39
  • regardless of that we know the responses for each of the six are not independent? They are "not independent" in case of free rating as well. You may see the rankings by individual as just ratings with two constraints: (1) imposed sum, same for all individuals; (2) no ties (equal values). Neither of the constraints is important in the analysis of each variable separately. They are not very dramatically important even in the analysis of all the variables at once. – ttnphns Jun 02 '16 at 20:47
  • @ttnphns thank you very much for clarifying, this was very helpful. Could you please explain: 1) Why "neither of the two constraints are important in the analysis of each variable separately" and "it is not dramatically important" if we analyze all variables at once? Is there a "statistical" explanation for why these constraints can be "ignored" and what consequences does it have for the model estimates/interpretations? – tvl Jun 04 '16 at 00:24
  • @ttnphns 2) If I were to analyze all 6 variables at once and use RM-ANOVA, should I be using a 'multivariate approach' to RM-ANOVA, instead of a univariate. I read your post here http://stats.stackexchange.com/questions/13197/differences-between-manova-and-repeated-measures-anova?rq=1 and it seems like multivariate approach is more appropriate, since I have six different stimuli (variables). I am also wondering if running MANOVA on the these variables could be also appropriate or not? – tvl Jun 04 '16 at 00:25
  • Respondents are independent (they don't "know" each other's answers) - so you may use Mann-Whitney and even might use t-test (some would say), for univariate testing of rankings. The circumstance that the data are compositional (equal sum for each respondent) makes it clear that out of the p variables only p-1 tests are independent, that's true and that bears on interpretation; but it doesn't make respondents dependent. – ttnphns Jun 04 '16 at 08:53
  • As I've said, for tests which consider all p variables at once it is important that rankings in not the same as ratings. Special procedures do exist for compositional data. Still, it is not a deadly sin to use things such as unual RM-ANOVA for them. You could use both "RM univariate" and "MANOVA" approaches. – ttnphns Jun 04 '16 at 09:10

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