I'm sorry if this is the wrong place to post this question. I'm an undergraduate with fairly limited statistical education up to this point.
I have a research project examining the contents of fox scat. Within each unit of scat we gather whatever animal parts we can find record the following data: 1) the taxa those parts belong to 2) the likely mass and life stage of that organism.

For example: One scat contains a teeth attributed to a juvenile cottontail weighing 50-150 g. Therefore 1 occurrence of cottontail is counted for the 50-150g column within the cottontail category.

This study was fairly opportunistic, so control for variables is limited, but as far as I can tell we can still use a chi-squared statistic to measure the significance of: 1) which taxa was dominant by overall frequency of occurrence 2) which taxa was dominant by amount of biomass 3) which biomass category was overall dominant.

Am I correct in applying chi-squared here and that our degrees of freedom would be determined with n = # of categories for each test?

  • $\begingroup$ Did you do any DNA analysis to identify which scat belongs to which fox? It looks like your data may not be independent if you sample many times a same fox. $\endgroup$
    – Nakx
    Jan 20, 2018 at 3:34
  • $\begingroup$ Hi Nakx, unfortunately we did not identify individual foxes. Would this violate independence because a single prey item consumed by a single fox could appear in multiple scats, which would inflate their frequency of occurrence? If so, is there any statistic to be salvaged from such a data set? I would consider the mission of this opportunistic study as one of salvage. Sequencing DNA from each of the 300+ scat would be too expensive for a study that really had no experimental design to begin with. At the very least we have a list of prey for a poorly understood species of fox. $\endgroup$ Jan 20, 2018 at 3:58
  • $\begingroup$ Yes the bias you mention could appear in your results, but without sequencing or identifying individuals there is no way around it. This happens all the time in biology, it is a limitation of your study that you have to accept and simply report and discuss when you give the results. $\endgroup$
    – Nakx
    Jan 20, 2018 at 23:26

1 Answer 1


1) which taxa was dominant by overall frequency of occurrence

Depending on how many taxa you have doing a Chi-squared Test might be tricky. It is often recommended to have at least five expected count per category, if you have a lot of taxa you will have a lot of small numbers of occurrence. If this is the case, simply consider grouping some of the taxa together (by how they are related for example). Then you can perform your chi-squared test and you will obtain a p-value to report.

This is a question you will have to discuss anyway: if your fox ate 1 individual each time from 10 different species of mice that are very closely related, and 5 brown rats from the same species, will you consider that the fox prefers rats or mice? Once you have decided how you want answer to this question, it should help you to decide how you want to regroup your taxa.

2) which taxa was dominant by amount of biomass 3) which biomass category was overall dominant.

I do not see how these two questions differ, it looks like you just want to describe your data and do not actually need a test for this. The chi-squared test is not appropriate for this as these data are not count data.

You could consider your scats as replicates, and estimate the weight of each taxa per scat, then you could use an ANOVA to compare the average weight of each taxa in the scats. The difficulty here is that you will most likely get zero inflated data and the data will likely not be homoscedastic and normal. A Kruskal–Wallis would not be appropriate either because you will obtain a lot of ex-aequo values (a zero every time one of the taxa is not found in the scats).

Have a look at your data and if you are very lucky you may be able to perform the ANOVA, otherwise making good figures should be enough for you to describe and interpret your results.


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