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I have a sampling of variable sized plots. Each plot contains the number of trees present on the plot. Given:

  • $n=$ the number of plots
  • $s_i=$ the size of the $i^{th}$ plot
  • $y_i=$ the number of trees on the $i^{th}$ plot

What method should I use to determine the mean and the variance of the sample?

Because the plots are variable sized, I'm hesitant to use the standard mean and variance formulas. My intuition tells me that larger plots should be weighted more than smaller ones.

A colleague of mine has suggested to use the density of each plot as the sampling statistic and the average of the densities as the mean to calculate variance using the standard variance formula. In my opinion this method weights smaller plots equal to larger plots, thus in effect giving smaller plots more weight.

I'm not a statistician, but an idea I had was to weight each plot by the number of averaged sized plots in the sample and to inversely weight the number of trees sampled, then use the weighted average formula to calculate the mean and the weighted sample variance formula for frequency based weights to calculate the variance. E.g.,

\begin{align} w_i &= \frac{s_i}{\bar{s}} \\ \ \\ x_i &= y_i\frac{\bar{s}}{s_i} \\ \ \\ \bar{x} &= \frac{\sum{w_ix_i}}{\sum{w_i}} \\ \ \\ s^2 &= \frac{w_i(x_i-\bar{x})^2}{n-1} \end{align}

It should be noted that: $w_ix_i=y_i$ and $\sum{w_i}=n$ thus $\bar{x}=\bar{y}$ where $\bar{y}$ is the number of trees for an averaged sized plot.

Is this method statistically valid or is there another formula I should be using?

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  • $\begingroup$ Some questions: When you say a "sampling" of plots, what do you mean? Exactly how were the plots selected for study? Were they a random sample of some kind from a target population of trees? If yes, is the mean you are estimating supposed to estimate the mean of that population? If no, then exactly what target parameter are you trying to estimate? $\endgroup$ Commented Aug 4, 2015 at 22:41
  • $\begingroup$ The plots are randomly selected from the population. The goal is to estimate the number of trees in the population based upon the sample and state the associated standard error of that estimate. There are two ways to calculate this value. $\endgroup$
    – deAtog
    Commented Aug 5, 2015 at 16:32
  • $\begingroup$ 1. Assume the density of the sample represents the density of the population. Eg: total_area * number_of_trees_sampled / area_sampled 2. Calculate the total number of averaged sized plots in the population area and multiply it by the total number of trees per averaged sized plot (total_area / average_plot_size) * (number_of_trees_sampled / number of samples) Both are essentially the same. Eq2 = Eq1 $\endgroup$
    – deAtog
    Commented Aug 5, 2015 at 16:40
  • $\begingroup$ Just two more questions: 1) When you say "randomly selected"; do you mean a "simple random sample (without replacement)", where each plot had the same chance of selection, or something else? 2) Can you categorize the sizes of the plots in the population? That would make it possible to "post-stratify" the sizes of tghe sampled plots. $\endgroup$ Commented Aug 5, 2015 at 19:06
  • $\begingroup$ 1) simple random with replacement. A random point for each plot is chosen within the population area and designated as the center of the plot. 2) It's hard to categorize the size of a plot. The majority of the plots should all be of the same size, but it's hard to enforce that constraint if the user of our software generates additional plots at a later time. 3) To complicate things, a user may choose to only sample a portion of a plot due to physical limitations. Eg: the plot overlaps the boundary of the study area (Note: I realize this causes bias, but it has been done this way for years) $\endgroup$
    – deAtog
    Commented Aug 5, 2015 at 19:41

1 Answer 1

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It looks like you are doing a version of FIREMON sampling (https://www.frames.gov/partner-sites/firemon/sampling-methods/). I know little about such point smpling, and I think you would benefit from consultation with someone knowledgeable about such methods. I'll give you my best guess, but I could be wrong.

I recommend a ratio estimate of the total number of trees. Larger plots are automatically represented in proportion to their size.

Denote $s_i$ = area for plot $i$ and $y_i$ = the number of trees in the plot. $S$ and $Y$ are the totals in the population. You know $S$ and want to estimate $Y$. The average density in the population is: $$ R = \frac{Y}{S} $$

You know the total area in the population is $S$. So the number of trees in the population $Y$ obeys the equation:

$$ Y = R\,S $$

The sample will permit you to estimate $R$, so that your estimated total will be:

$$ \hat{Y}_R = \hat{R} S $$

The natural estimate of $R$ is the ratio of sample totals, which is the same as the ratio of sample means for number of trees and area:

$$ \hat{R} = \frac{y}{s} = \frac{\overline{y}}{\overline{s}} $$

Calculation of estimated Standard Error

Methods for estimating the standard error include: linearization, the jackknife, and the bootstrap. All are presented in standard sampling texts. I'll present the linearization method, which requires a "large" n. (Cochran, 1977, pp; 32; 160-163).

The approximate standard error of $\hat{Y}_R$ under sampling with replacement is (Cochran, 1977, $v_2$, page 155):

\begin{equation} SE(\hat{Y}_R) = S\sqrt{\frac{\sum_{i=1}^n (y_i - \hat{R}s_i)^2}{n(n-1) \overline{s}^2}} \end{equation}

If, in fact, the process is not "simple random sampling" of plots (see my comment), then this might still be valid. I would speculate that for the point sampling method, one could multiply the standard error above by a "finite population correction" $\sqrt{1-{\dfrac{s}{S}}}$.

References

Cochran, William G. 1977. Sampling techniques. New York: Wiley.

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  • $\begingroup$ Thank you for the excellent response. A quick comparison of this method against the standard variance formula shows that if all the plots are the same size then it converges appropriately. Once it has been peer reviewed, I'll mark it as the accepted solution if it's found acceptable. $\endgroup$
    – deAtog
    Commented Aug 10, 2015 at 17:55

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