How to calculate proportions of individuals from an area based sample? I have a sample with 4 randomly located 1/10 acre circular plots. In all plots we measure the diameters of all trees (real trees with leaves and bark not regression trees). The data look like
Plot 1: 1 trees 12 inches diameter
Plot 2: 0 trees 
Plot 3: 0 trees
Plot 4: 3 trees 5, 8, 11 inches in diameter
What percentage of trees in the population are less than 6 inches in diameter. I feel like it should be (1/3+0/1)/2 = 16.5% but I need some backup because my colleagues want 1/4 = 25%. It would be nice to get variance of estimate but not completely necessary.
 A: I'm sorry to tell you that your colleagues are correct.
Let the population size be $N$; $X_i$ be the number of trees in plot  $i$  and $Y_i$ be the number of trees in plot $i$  Let there be $n$ plots in the sample (here $n = 4$) and let $x_i$ $y_i$ be the sample counts of trees and of <6 inch-diameter trees, respectively in the i-th plot. Call the population and sample totals $X,\thinspace x,\thinspace Y,\thinspace y$.
Then the fraction of all trees in the population with diameter $<$ 6 inches is:
$$
R = \frac{\sum_{i=1}^NY_i}{\sum_{i=1}^NX_i}= \frac{Y}{X}
$$
Plots with no trees contribute zero to both numerator and denominator and do not affect the calculation. This is a ratio. 
The sample totals of trees and trees $<$ 6 inches in diameter are $y = 1$ and $x = 4$. Your colleague's estimate of the population proportion $R$  is:
 $$
\widehat{R} = \frac{y}{x}= \frac{1}{4}
$$
This is known as a ratio estimator, 
Your solution does not estimate $R$. Let $M$ be the number of plots in the population with trees; and let $m$ be the corresponding number in the sample. You are estimating the following average of plot-specific fractions:
$$
R^* = \frac{1}{M} \sum_{i=1}^M \frac{Y_i}{X_i}
$$
This will not be equal to $R$ unless all plots have the same number of trees.
To illustrate the difference in what the two solutions estimate: suppose only two plots in the population have trees.  In one plot, $X_1 = 1000$  trees and $Y_1 = 500$; in the second $X_2 = 10$ and $Y_2 = 0$. The actual fraction of trees with diameter $<$ 6 inches in the population is $R = .495$.  The quantity estimated by your solution is $R^* =.25 $.
Note that $R$ can be expressed as an average of the
$\dfrac{Y_i}{X_i}$, but it is a weighted average, with weights proportional to the the number of trees in the plot; i.e if  $W_i =\dfrac{X_i}{\sum_i X_i}$, then
$$
R =  \sum_{i=1}^N W_i \frac{Y_i}{X_i} 
$$
Ratio estimators are covered in every sampling text e.g. Lohr, 2009, Chapter 4, Cochran, 1977. Chapter 6. There are formulas for standard errors, but they require a sample size (number of plots with trees) that exceeds 30 (Cochran, 1977, p. 153).
Added: There are situations in which $R^*$ would be the proper target parameter.  Suppose that the sampling units are people and that the target parameter is the average fraction of decayed teeth per person. This is $R^*,$ with $X_i$ the number of person $i$'s teeth and $Y_i$ the number decayed.
For a related question about a forestry sample see: Mean and variance for unequal samples
References
Cochran, WG. 1977. Sampling Techniques, Wiley, NY
Lohr, Sharon L. 2009. Sampling: Design and Analysis. Boston, MA: Cengage Brooks-Cole.
