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I'm looking at two plant categories (all species fall into one type or the other) at several different types of locations. Within each type of location, I have several samples, so I have an average of each type of plant as it is found in each type of location.

Within each type of location, I want to compare the average number of one type to the average number of the other type. The number of one type of plant is not necessarily independent of the other (there is only a certain amount of resources after all).

I feel as though there must be a simple test to look at this, but I can't remember what it is. For the difference in average number of ONE type across the different locations, I used an ANOVA; if there were only two locations I could have used a t-test. If I had one population type and one type of plant, and looked at it before and after some manipulation, then I would use a paired/dependent t-test...

Any suggestions would be phenomenally helpful.

Thanks!

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A location (Between) x Plant Type (Within) mixed ANOVA may do what you want. The disadvantage of this approach is that there is no clear post-hoc to perform at each location, so you'll have to do a bonferroni correction or take some other approach (e.g. question 1 or question 2). Since your data is sourced from percentages and is bounded at the upper and lower extremities you may want to transform the raw data first (see this issue/approach).

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russellpierce
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Assuming that you are looking at proportions you need to do a One-proportion z-test. The link takes you to the wiki which has common test statisitics used in various situations including the one-proportion z-test.

If I have understood your situation correctly you have the following set-up. Let:

$n$ be the total number of trees in a location,

$n_a$ be the number of trees in that location of type 'a'.

Thus, your hypothesis potentially is:

$H_0: \pi_a = \pi_0$,

$H_A: \pi_a \ne \pi_0$,

where

$\pi_a$ is the true proportion of trees of type $a$,

You are approximating $\pi_a$ with $p_a$ where:

$p_a = \frac{n_a}{n}$

and

$\pi_0$ is your hypothesis about the true proportion for that type.

  • I've been looking at my data, and I misrepresented the question a little bit. –  Oct 23 '10 at 21:47
  • @Jane If the added context makes my answer irrelevant I will delete my answer or perhaps edit to fit your context better. –  Oct 23 '10 at 21:49
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For a particular plot we estimate "% cover" per plant species (summed across all species within a type to give the % cover for the plant type); in a forest canopy (as one example of the locations we used) this means that the total will definitely be greater than 100%, so a statistic using proportions wont work...

When I say 'plants', I mean all the plants there (trees, herbs, shrubs, etc). The two categories were "native" and "introduced/invasive" so all plants will fall into one or the other.

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    @Jane Please add the above information to your question using the 'edit' link below the tags and delete this 'answer' as it really does not answer your question. –  Oct 23 '10 at 22:26
  • @Srikan Vadali For some reason the site wont let me edit my own question, but it does give me an edit button for the question, so there seems to be something strange going on... Sorry about that. –  Oct 24 '10 at 15:40