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Lets imagine 5 plots of different size are sampled for a target species:

plot#  count  area(m^2)  plot_density
  1      1       5         0.2
  2      3       2         1.5
  3      0      10         0.0
  4      5       1         5.0
  5      2       6         0.33

What is this species' density? I see two ways to calculate the density that give completely different values.

The first way averages the density at each plot:

$$ \frac{\Sigma(\frac{count_i}{area_i})}{5} = 1.41/m^2 $$

This seems OK but it doesn't do much to control for changing plot sizes. For example if plot 3 above was 10 times larger the density would still be the same. In a situation where plot sizes are determined by the environment (say, under natural cover objects) this seems less than ideal.

The second way totals the counts and the search area and divides them:
$$ \frac{\Sigma(count_i)}{\Sigma(area_i)}=0.46/m^2 $$

I prefer the second method because it seems to describe the animals' density more accurately. And I would be happy to use the second method, but my issue is that I am unsure how to calculate a summary statistic for method two since a mean was never calculated. Method one gives a standard deviation of 2.1. What is the standard deviation for method two??

One possible solution I've come up with involves breaking the larger plots into 1m^2 plots and dividing the number of animals across those smaller plots. So now I have 24 1m^2 plots with the following "counts":

plot#  count  area(m^2)  plot_density
1-5      0.2       1         0.2
6-7      1.5       1         1.5
8-17     0.0       1         0.0
18       5.0       1         5.0
19-24    0.33      1         0.33 

Now, using the first equation above I get: $$ \frac{\Sigma(\frac{count_i}{area_i})}{24} = 0.73/m^2 $$ With a standard deviation of 1.26.

Is this a reasonable approach? Is there an established solution to this problem?

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To me, there are two issues that are mixed up here (if I understand you correctly). First, do you want to estimate the mean and variance for a statistical population (i.e. to characterize a larger population by independent samples), or do you want to calculate the actual density for a particular area, where you have counted all occurences in that entire area (but maybe divided the area into subareas out of convenience when counting)? This is not clear from your question.

In the second case, your second option of pooling counts and areas is suitable. However, then you have only calculated the actual density in that particular area (ignoring issues of detectability of the organism when counting), and you cannot draw any inferences about a larger statistical population.

If your aim is to draw inferences for a larger statistical population, a start is to calculate the mean and standard deviation (sd) of your sample. In that case, I assume that your samples are chosen randomly and independently from a larger statistical population. Your first option is then the right approach. However, since your samples have different sizes, you might want to attach more weight to larger samples, since they can be assumed to better describe the population average than smaller samples. This is called a weighted mean.

Generally, the weighted arithmetic mean is defined as:

$$\bar{x}_w = \dfrac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i}$$

where $w_i$ are the weights for each sample ($x_i$) and $n$ are the number of samples. Using this formula, you will arrive at exactly the same value for the weighted mean as you calculated in your second attempt (0.458), if you use the areas as weights.

The weighted standard deviation is a bit more problematic, in the sense that there doesn't exist one single standard way of calculating this. However, a commonly used formula is:

$$ \sigma_{w_1} = \sqrt{ \frac{ \sum_{i=1}^n w_i (x_i - \bar{x}_w)^2 }{ \frac{(M-1)}{M} \sum_{i=1}^n w_i } },$$

where $M$ are the number of nonzero weights. Other definitions of the weighted standard deviation can be found at Wikipedia: weighted sample variance.

Another version is defined as (called "reliability weights" on the wiki page):

$$ \sigma_{w_2} = \sqrt{ \frac{ \sum_{i=1}^n w_i (x_i - \bar{x}_w)^2 }{ \sum_{i=1}^n w_i - \frac{\sum_{i=1}^n w_i^2}{\sum_{i=1}^n w_i}}},$$

If I haven't made a mistake, these will give the standard deviations 1.15 and 1.22, using your data.

As for the biological interpretation, all calculations of average densities indicate a clumped distribution, since the coefficient of variation (CV) is larger than 1 (CV = sd/mean), but more so if you use a weighted mean. The CV using the arithmetric mean is 1.5, while it is 2.5-2.65 for the weighted mean, which is reasonable since you are giving more weight to the large area sample with a zero count. However, I should also note that you should be cautious about using a weighted mean when you have a strongly clumped distribution, since your run the risk that e.g. a large sample plot land on an area with no occurences, which might bias your density estimate. Generally, when you have a clumped distribution, you need to sample more intensively to get a good estimate of average density, and many small samples is often better than a few large ones.

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    $\begingroup$ Great answer, thank you! If anyone else comes along and finds your answer helpful they might also be interested to learn the r package SDMTools has the functions wt.mean() wt.sd() and wt.var(). wt.sd() uses the more common equation that you described above. $\endgroup$ – Adam C Jun 10 '15 at 23:51
  • $\begingroup$ After finding more time to consider your answer I have minor correction on your answer and a follow-up question. The weighted mean for the above example is 1.16 not 0.458 like you say. And now consider these three plots: Count: 1, 1, 20; area(m2): 1,1,20. I would say the density is 1animal/m2 and the three methods in my original post agree. But the weighted mean is 18.2. That doesn't seem right. What are your thoughts on this? $\endgroup$ – Adam C Jun 11 '15 at 21:41
  • $\begingroup$ If you get a weighted mean of 1.16 you are calculating a mean based on counts, not densities, and the same goes your your new example. It also doesn't make much sense to calculate the average number of occurences (counts) for plots of different sizes, which is what you have done in your example in the comment. If you calculate a mean (weighted or otherwise) based on densities you will get 1animal/m2. $\endgroup$ – fileunderwater Jun 11 '15 at 22:45
  • $\begingroup$ Yep you're exactly right. Of course you would use the densities when calculating the weighted density. How I didn't see that before is beyond me. Thanks again. $\endgroup$ – Adam C Jun 11 '15 at 23:10
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    $\begingroup$ @AdamC You're welcome. Note that I've added a bit on the biological interpretation, and some issues when using the weighted mean for clumpted distributions. $\endgroup$ – fileunderwater Jun 12 '15 at 11:42

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