I suppose, your $\mathrm{Nr}$ is actually a density (number/area) and $\mathrm{DBH}$ ist already a mean.
1
Observation: The inner circle is the only one with information about small trees. Therefore just there you have the proper ration in diameters but with a big error.
Say you have $n_{Pi}$ trees in the inner circle of your area $Pi$. These $n_{Pi}$ trees consists of
- $n^s_{Pi}$ trees with small diameter
- $n^m_{Pi}$ trees with middle diameter
- $n^l_{Pi}$ trees with llarge diameter
so that $n_{Pi} = n^s_{Pi} + n^m_{Pi} + n^l_{Pi}$. Now you can calculate the mean DBHs in the inner circles for small, middle and large diameters separately (Note: This is just the usual mean over all trees in the inner circle if you put it together but writing it this way, you can later on replace the different means for m and l). Then your (total) mean DBH (over all trees in the inner circle) is:
$$\mathrm{DBH} = \frac{1}{n_{Pi}}\left(n^s_{Pi} \cdot \mathrm{DBH}_s + n^m_{Pi} \cdot \mathrm{DBH}_m + n^l_{Pi} \cdot \mathrm{DBH}_l\right)$$
Now, lets change the notation to prolong the life of the keyboard (compare yourself).
$$\varnothing = \frac{1}{n}\left(n_s \cdot \varnothing_s + n_m \cdot \varnothing_m + n_l \cdot \varnothing_l\right) \qquad n = n_s + n_m + n_l$$
What information can we get from the additional circles? For the middle one, we lack the information about the small trees.
Therefore my approach:
- Replace the number of trees found in the inner circle $n_i$ ($i=m,l$) with what you would expect for the small circle/area using the density you get from the bigger ones. Let's call them $\tilde n_i$ ($i=m,l$)
- Replace the mean diameters $\varnothing_i$ ($i=m,l$) with the more exact ones calculated using the larger circles/areas (and therefore more trees and a more precise mean). Let's call them $\tilde\varnothing_i$ ($i=m,l$). In other words: $\tilde\varnothing_m$ is the mean DBH of all middle sized trees in the middle sized circle and $\tilde\varnothing_l$ is the mean DBH of all large sized trees contained in the large sized circle. You can directly calculate both.
Then you get:
$$\tilde\varnothing = \frac{1}{n_s + \tilde n_m + \tilde n_l}\left(n_s \cdot \varnothing_s + \tilde n_m \cdot \tilde \varnothing_m + \tilde n_l \cdot \tilde \varnothing_l\right)$$
Now to $\tilde n_i$. For this let's call:
- $N_i$ ($i=s,m,l$) the total number of small/medium/large trees in the small/medium/large circle (Note: $N_s=n_s$ but $N_i\neq n_i$ for $i=m,l$).
- $A_i$ ($i=s,m,l$) the surface area of the small/medium/large circle.
Then your tree densities $\varrho_i$ ($i=m,l$) (you called them Nr) in the outer circles are
$$\varrho_i = \frac{N_i}{A_i},\quad i=m,l$$
and the therefore expected number of trees in the inner circle:
$$ \tilde n_i = \varrho_i \cdot A_s = N_i \cdot \frac{A_s}{A_i},\quad i=m,l$$
In other words: You take the larger amount of trees on the larger area to calculate a more precise trees/surface ratio and calculate what you would actually expect on $A_s$.
Putting this together is exactly what the professor said:
$$\begin{align}\tilde\varnothing &= \frac{1}{n_s + \tilde n_m + \tilde n_l}\left(n_s \cdot \varnothing_s + \tilde n_m \cdot \tilde \varnothing_m + \tilde n_l \cdot \tilde \varnothing_l\right)\\
&= \frac{\frac{n_1}{A_s}\cdot \varnothing_s + \varrho_m \cdot \tilde\varnothing_m + \varrho_l \cdot \tilde\varnothing_l}{\frac{n_1}{A_s}+\varrho_m + \varrho_l}\\
&= \frac{\varrho_s \cdot \varnothing_s + \varrho_m \cdot \tilde\varnothing_m + \varrho_l \cdot \tilde\varnothing_l}{\varrho_s +\varrho_m + \varrho_l}\end{align}$$
2
If you simply average means, you ignore that these means contain different numbers of trees. You therefore need to use a weighted sum -- weighted by the number of trees.
So, if you now have not just $\tilde\varnothing$ but $\tilde\varnothing_{P1}, \dots, \tilde\varnothing_{P28}$ and lets say $\tilde N_i$ ($i=P1,\dots ,P28$) is the number of trees involved in the mean of every P (compare: $n_s + \tilde n_m + \tilde n_l$ for the single one above), then your sum should be:
$$\tilde\varnothing_{\mathrm{total}} = \frac{1}{N_{P1} + \dots + N_{P28}} \cdot \left( N_{P1} \cdot \tilde\varnothing_{P1} + \dots + N_{P28} \cdot \tilde\varnothing_{P28} \right)$$
This is the total mean of the DBH of all the trees in the inner circles of all sample areas (enhanced with the results of the outer circles).
