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I feel like this is kind of an elementary question but w/all my reading in and around the subject I have yet to arrive at a conclusive answer. I'm testing an ecological (site*species) data set as part of an investigation into an invasive species (Rosa rugosa). I have grouped my data according to the quadrant they were part of, I don't how relevant this is, but the species is only seen at certain quadrants along the transect.

Is there a test that I can run to test the null hypothesis that this specific species has no impact on the entire assemblage. I believe that I am looking for some kind of multivariate T-test, for which I'd be looking for Hotelling's T^2 test, although many of the functions in the vegan package seem that they test along similar lines. Any of envfit, betadisper, adonis, or even anosim seem to allow me to group a sort of

>nta.bray ## bray-curtis matrix w/wisconsin transformation
>test(nta.bray, Rosa_rugosa ~ Quadrant, data=nta)

Does anyone know of a test that will allow me to test along these lines, or even a paper that performed a similar hypothesis test?

  • How are you defining "the entire assemblage" or how would you like to measure the change in that? Are you planning on defining composition as the combined presence/absence of all taxa other than Rosa rugosa, for example. Then you could envisage a test where the response is that matrix (data frame) of presence absence of all species (minus Rosa rugosa) and then fit a `cca(spp_data ~ Rosa-rugosa, data = nta)`, the idea being that you treat `Rosa_rugosa` as an "environmental" variable to explain composition in the remaining species. 1/2 – Gavin Simpson Sep 16 '22 at 09:22
  • Or you could use the actual composition (+/- or abundance, or relative abund, etc) and define a new variable that is a factor indicating if *R. rugosa* is present or not at each qudrant. Then you'd do a `rda()` (on transformed composition), `cca()`, `dbrda()` or `adonis2()` (but most likely the former three, not `adonis2()`) with this factor as the sole predictor variable and the species data matrix as the response. Even here you'd need to think whether you include *R. rugosa* in the response spp data as this is what the factor predictor already indicates. 2/2 – Gavin Simpson Sep 16 '22 at 09:25
  • We really need more information about how you want to measure "impact" and what "entire assemblage" means to you/your study. – Gavin Simpson Sep 16 '22 at 09:26
  • My data is relative abundance data, and what I mean to say by 'entire assemblage' is the entire species matrix including *R. rugosa*. By impact, I'm slightly less sure: the results of similarity percentage tests & species indicator tests do indicate that the presence of the species is a significant marker of the quadrant that they're in. To test that it has a significant impact on the Bray-Curtis dissimilarity index seems reasonable. – Dean Bayliss Sep 16 '22 at 14:15
  • I had attempted to run an adonis2, and an envfit between *R. rugosa* as a variable and the species matrix, although I think I knew at the time that they are not testing the exact hypothesis I was hoping to test. The other idea that I had, that I no longer think is feasible, it to test it directly against the Shannon diversity index for a direct evaluation of its effect on biodiversity, but like I said, I don't think that this test is possible in any meaningful way. – Dean Bayliss Sep 16 '22 at 14:21
  • To run a cca for this - the *R. rugosa* variable would need to be transformed into a binary 0/1 factor? – Dean Bayliss Sep 16 '22 at 14:23
  • But if *R. rugosa* is in the assemblage, and predictor variable you generate from *R. rugosa* will have it's importance in any test *inflated* because it will be able to "explain" the differences between communities that is due to presence of itself, even if it doesn't affect composition of the other species at all. That's why I suggested conversion to +/- for the other species, exclude *R. rugosa* from the response spp 0/1 matrix & test for differential composition between sites with/without *R. rugosa* using a factor variable `Rosa_rugosa > 0` to test if mere presence has an effect, or 1/2 – Gavin Simpson Sep 16 '22 at 18:47
  • Include `Rosa_rugosa` as a continuous explanatory variable with the same binary spp response matrix as above (excluding *R.rugosa*) to test if composition depends not only on presence but also on abundance of the invasive. 2/2 – Gavin Simpson Sep 16 '22 at 18:48
  • Using reproducible example, let's see if presence of _Trifolium repens_ has an "effect" on the whole community in terms of Bray-Curtis: `anova(dbrda(vegdist(wisconsin(dune[, names(dune) != "Trifrepe"])) ~ Trifrepe>0, dune))`. Naturally, you see only correlation and not causality. Other species may respond in similar way to hidden environmental variables, but _T. repens_ has no impact. You see that in the `dune` data if you cycle over species: _any_ species of moist places is "influential" because all these species occur together in moist places. – Jari Oksanen Sep 18 '22 at 10:34
  • Sorry for the delay in getting back to yous, I thought to talk it over w/someone who knows the data. We are going to test the matrix against presence/absence, and against abundance, while acknowledging the inherent flaws in the tests only really proving correlation. – Dean Bayliss Sep 21 '22 at 14:46
  • @JariOksanen , you have made a Bray-Curtis matrix of a wisconsin transformation and run that against a binary list of the target species, with a anova.dbRDA as the permutation test. – Dean Bayliss Sep 21 '22 at 14:57
  • @GavinSimpson , whereas you have suggest that I convert to plus minus, which I have tried using a couple of methods that I took from flutterby's to transform the data to a mean of one, although running that against an anova.dbRDA returned an error. Was that the method you intended that I use to transform the data? – Dean Bayliss Sep 21 '22 at 14:57
  • To compare said matrix against abundance I would simply remove the >0 ? That would make it continuous, or should there be another transformation that I perform on it as a singular variable? – Dean Bayliss Sep 21 '22 at 14:58

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