Mean and variance for unequal samples 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?
 A: 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.
