I am currently calculating weighted means of the DBH and height of my forest inventory data. I used 3 different approaches to investigate this data: I used nested plots like so:Plot Design

  • The smallest circle includes only the smallest trees(< 22cm Diameter [futher on referred as DBH])
  • the medium circle only medium trees(22cm $\le$ DBH $\le$ 42cm)
  • the biggest circle contains only the biggest trees($\ge$ 42cm DBH) (ignore the small 2m circle).

My data looks like that: Data_Table

Now I want to calculate the overall DBH of all three circles combined (is that even possible?). I know I could just include all trees in the smallest circle and calculate their mean (which I did) and would get the proper DBH of all trees in the forest, however that excludes a lot of data. So I asked a professor of mine and he told me to weight the different means of DBH/Height according to the number of their appearance (which is in this case Number of trees per hectare) like so (the result is in the picture above in the column "DBH_plotwise", thus i caluclated this row-wise):

$\frac{\mathrm{Nr}_\mathrm{small} \cdot \mathrm{DBH}_\mathrm{small}+\mathrm{Nr}_\mathrm{medium} \cdot \mathrm{DBH}_\mathrm{medium} + \mathrm{Nr}_\mathrm{large} \cdot \mathrm{DBH}_\mathrm{large}}{\mathrm{Nr}_\mathrm{small} + \mathrm{Nr}_\mathrm{medium} + \mathrm{Nr}_\mathrm{large}} = \mathrm{DBH}_\mathrm{plotwise}$

The mean of the column "DBH_plotwise" is 18.26. However, during my calculations I calculated this value in a different manner by taking the mean of DBH_s/m/l and the mean of Nr_s/m/l and using the same formula as above I get the following numbers:

$\frac{10.05\cdot613+26.2\cdot 105+52.9\cdot 33.2}{613+105+33.2}= 14.20$

Shouldn't there be no difference? 14 and 18 is a huge difference.

So to sum it up:

  • Is it possible to calculate the overall DBH/Height with all of the three circles included?
  • Why is there a difference in the result of the two approaches above to calculate the mean DBH?
  • 1
    $\begingroup$ for an explanation of the acronym DBH, that is maybe well known in some circles but never heard in others, see (en.wikipedia.org/wiki/Diameter_at_breast_height) $\endgroup$
    – Jean Marie
    Jan 4, 2017 at 19:14
  • $\begingroup$ Don't use mean of Nr and DBH, use sums $\endgroup$
    – zzz
    Jan 5, 2017 at 15:56

1 Answer 1


I suppose, your $\mathrm{Nr}$ is actually a density (number/area) and $\mathrm{DBH}$ ist already a mean.


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:

  1. 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 $\tilde n_i$ ($i=m,l$)
  2. 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:

$$\begin{align}\tilde\varnothing &= \frac{1}{n_s + \tilde n_m + \tilde n_l}\left(n_s \cdot \varnothing_s + \tilde n_m \cdot \tilde \varnothing_m + \tilde n_l \cdot \tilde \varnothing_l\right)\\ &= \frac{\frac{n_1}{A_s}\cdot \varnothing_s + \varrho_m \cdot \tilde\varnothing_m + \varrho_l \cdot \tilde\varnothing_l}{\frac{n_1}{A_s}+\varrho_m + \varrho_l}\\ &= \frac{\varrho_s \cdot \varnothing_s + \varrho_m \cdot \tilde\varnothing_m + \varrho_l \cdot \tilde\varnothing_l}{\varrho_s +\varrho_m + \varrho_l}\end{align}$$


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).

  • $\begingroup$ thanks a lot mate! That answered my question in all detail, it's so well explained, almost as if you explained it to me face to face ;) $\endgroup$
    – Lukas
    Jan 6, 2017 at 12:37

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