# IMPROVING RESULTS WITH A DIFFERENT KIND OF CRUISE

by Kim Iles, Kim Iles & Associates

Cruises do not have to be of the same type to be combined. This general process is useful whenever you happen to have (or need to have) two different kinds of cruises on the same property. Here is an example.

The worst case nightmare with a 3P cruise is when you have not chosen any "insurance trees" beforehand and fail to meet your sampling error. Perhaps you wanted ±10% and got ±12% and there is a legal requirement for the ±10%. You have to get it.

We have suggested some ways in past articles to insure that extra trees are available, but suppose someone just messed up royally and these are not in place. Traditionally your only option using 3P sampling is to estimate all of the trees again over the whole area to choose the extra samples and get that 2% improvement. Not an attractive option on a 3P cruise of 100 acres.

Could you install a few variable plots? Yes, you can – just enough to improve the answer to what you require (perhaps that would only be about a dozen or so). That would be a practical solution to the problem even though you have to spread those plots over the entire area.

There is a very powerful principle when dealing with several unbiased samples that says "you can always combine them and come up with another unbiased sample. Not only that, you can weight them differently, and thereby recognize the different quality of each contribution."  If you want to avoid almost all the math here you can go directly to the last section titled "Let’s look at this a simpler way."

Combining two answers for the SAME thing.
Suppose that you had a 3P cruise with a ±10% sampling error (on an average of 30,000 bf) and a VP cruise with a ±30% sampling error (on an average of 25,000 bf). How would you combine them?

The standard way to do this is to "weight inversely proportional to the variance." This means that you compute the total SE in units for the cruise, square it, divide into 1.0, and use that as a multiplier. We will use the volumes in thousands to keep the number of zeros minimized. In this case the weights are:

{1 / (±10%*30 mbf)2 }
= { 1 / 9} = 0.11111

and

{1 / (±30%*25 mbf)2 }
= { 1 / 56.25} = 0.017778

The sum of the two weights is:

(1/9 + 1/56.25) = 0.12889

So the relative weights for the two cruises are:

(1/9) /0 .12889 = 0.86207 = 86.2%

(1/56.25) / 0.12889 = 0.13793 = 13.8%

We can now get an average of the two samples by multiplying their relative weights by the two individual averages:

(86.2% * 30 mbf) + (13.8% * 25 mbf)

= (25,862 bf + 3,449 bf) = 29,310 bf

so this is the proper estimate for the area, recognizing the quality of each of the two samples.

This is pretty close to the 30,000 bf from the 3P cruise alone. Since the sampling error of ±10% was small, that average was weighted heavily. That makes sense, because you are saying that the ±10% average is a good one compared to a sample with a ±30% sampling error and you want to pay more attention to the better quality answer.

How do we get a combined sampling error? That is pretty simple. Let’s combine the sampling error of each of the two parts (we have to square them before adding, and then take the square root of the total):

(25,862 bf * ±10%)2 + (3,449 bf * ±30%)2

= ±7,759,034.5 bf2

taking the square root, we get:

±2,785 bf as a sampling error, which is ±9.5% of 29,310 bf.

The variable plot sample we used did reduce the sampling error from 10%, but only by 1/2% because there were very few plots in that sample.

So this process can give us the answer when we combine two very different samples on the same area, but what if we have one sample already and want to know how hard we have to work to improve it?

Calculating sample size:
Let’s use our example. We are working with 2 averages. One has a ±12% error, the other (x%) must be calculated, and we want a combined result of ±10%. Assuming that the average of both of the samples will be approximately the same, we can compute the sampling error needed for the additional sample with this formula:

(1 / 102) - (1 / 122) = (1 / x2)

Doing the math, we get:

0.01 - 0.006944 = 0.003056 = (1 / x2)

(1 / 0.003056) = 327, and the square root of that is 18% which is the sampling error needed on the additional sample.

OK, now we know what SE% the variable plot sample must produce. How many plots would that be? Suppose the CV of the variable plot was 55%. To get an 18% error would require:

(55/18)2 plots = 9.3 or about 10 plots.

So, you have your choice. Do you want to do a 100 acre 3P cruise again, or put in 10 variable plots and average the answers? For heavens sake put in a few extra plots on the second visit so we do not have to go out there a third time!

This kind of computation is easy to put on a spreadsheet, and the calculations are a snap.

Let’s look at this a simpler way.
The math was a little involved back there in the last section. All statistical and mathematical work is really LOGICAL work when you look at it the right way. There is another way to view this whole process, and you might find it more satisfying.

Suppose you were going to do a variable plot sample in the first place and wanted to get a 10% error from a population having a CV of 55%. That would require a sample size of:

(55/10)2 = 30.25 or 31 plots

Now, you have a 3P sample already which you are going to use. Here is the critical question -- how many variable plots is that 3P sample "worth"?

Well, to have produced a sample of ±12% you would have had to install:

(55%/±12%)2 = 21 plots

so the sampling equivalent of that 3P sample you are going to use is 21 variable plots, and you just have to install the remaining 10.

Simple.

Originally published October 1998