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I have GPS collar data that I have used to extract values to points through ArcGIS Pro on MODIS landcover data. From this, I have a record of the number of times the animal has been in each landcover class. My data looks like this:

Landcover Selection Frequency
Evergreen 105
WoodySavanna 327
Savanna 30
Grassland 2
Croplands 2

It looks like there is a pattern of selection there so I am wondering what statistical test I can run to identify if there is a significant difference between the selection frequencies of the landcover classes.

Ideally would be running this in R.

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2 Answers 2

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Your problem can be rephrased in terms of testing if the cell probabilities of a Multinomial distribution follow a given pattern.

In particular, given the sample $(X_1,\ldots,X_5)\sim \text{Mn}(466,\theta_1,\ldots,\theta_5)$ the problem is to test

$$H_0: \theta_1=\cdots=\theta_5=1/5$$ against $$H_1:\theta_i\neq\theta_j\, \text{for at least one pair } i,j, \text{with } i\neq j.$$

Note that $466$ is the sum of cell counts.

There are several ways to implement this test, and in R the simplest way is perhaps this

counts = c(105, 327, 30, 2, 2)
expected = rep(1/5,5)
chisq.test(counts, p = expected)
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    $\begingroup$ Thank you so much that makes more sense now! $\endgroup$
    – Chelsea_VictoriaS
    Dec 7, 2022 at 10:29
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I think there are issues with the multinomial test proposed by @utobi, the null hypothesis of equal probabilities for the five landcover types and with the resulting "pattern of selection" interpretation.

  • Is the multinomial distribution justified? The counts are number of times the tagged animal has been in each landcover area. How reasonable it is to assume that the observations (sightings) are independent? With movement data, we would expect that if the animal is in the Evergreen now, it is more likely to stay in that area than to move to another area.
  • Along those lines, if the animal wants to move to Grassland but Grassland doesn't directly adjoin Evergreen, it would have to cross other landcovers to reach it. Again, the independence of sightings required by the multinomial distribution would be violated.
  • Is the null hypothesis $H_0:\theta_1=\cdots=\theta_5=\frac{1}{5}$ justified? The availability of the landcover types in the animal's habitat matters. If the habitat consists of mostly Savanna, why would be interested in the testing the naive null hypothesis that the animal spends equal amount of time in each landcover type? The null hypothesis that the probabilities are all equal sounds unrealistic.

We don't need a formal statistical test to reject the null hypothesis that the animal been in each landcover class equal number of times; it's enough to look at the table of counts. However, rejecting this specific null hypothesis may in fact say little about the behavior of the animal.

References

A quick search for "habitat selection" turns up a number of relevant articles and even an R package. These could be a starting point for looking into how to analyze GPS tracking data in a meaningful way.

[1] William, G., Jean-Michel, G., Sonia, S. et al. Same habitat types but different use: evidence of context-dependent habitat selection in roe deer across populations. Sci Rep 8, 5102 (2018). https://doi.org/10.1038/s41598-018-23111-0
[2] Fattorini, L., Pisani, C., Riga, F. et al. A permutation-based combination of sign tests for assessing habitat selection. Environ Ecol Stat 21, 161–187 (2014). https://doi.org/10.1007/s10651-013-0250-7
[3] phuasses: Proportional Habitat Use Assessment

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    $\begingroup$ This is not an answer. It's an extended comment as to the reasonability of the question. As a stand alone question, it is reasonable and there is no need to read further into it with assumptions. $\endgroup$
    – Mari153
    Mar 12 at 8:34
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    $\begingroup$ _We don't need a formal statistical test to reject the null hypothesis that the animal been in each landcover class equal number of times; it's enough to look at the table of counts. _ it seems you are suggesting performing statistical inference "by eye".... Anyway, I do see your points and I do agree with some of them (that's why my +1). However, given the meagre information at hand, an approximate but correct answer is better than nothing. $\endgroup$
    – utobi
    Mar 12 at 13:29
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    $\begingroup$ Sure, domain-specific knowledge is paramount to the correct statistical implementation. Good to see that we agree to disagree :-) $\endgroup$
    – utobi
    Mar 12 at 13:38
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    $\begingroup$ So agree that we don't have domain knowledge, but suggesting equal probabilities in this scenario strikes me as not reasonable @utobi on its face. Can you imagine drawing a grid anywhere in the world and the cover proportions are near equal for these land classes? (Also there are probably other land classes in the sample area with 0 counts.) $\endgroup$
    – Andy W
    Mar 12 at 13:57
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    $\begingroup$ @AndyW, yes I agree with you. In my answer I've adopted $H_0$ of uniform cell probabilities because of the lack of concrete extra information, adopting thus the principle of insufficient reasons (P. S. Laplace). The OP can easily change $H_0$ at will. $\endgroup$
    – utobi
    Mar 12 at 14:14

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