Brenda King
Tree Problem
The
problem:
Using data from the
lumber industry which gives the approximate number of board feet of lumber per
tree in a forest of a given age, find a function that will fit the data. Predict the harvest for ages other than those
given.
Age of Tree 
100s of Board Feet 
20 
1 
40 
6 
60 

80 
33 
100 
56 
120 
88 
140 

160 
182 
180 

200 
320 
Introduction
In many realworld
problems there exist patterns or relationships between sets of numerical data. The relationship among the data can be
influenced by many variables. In the
case of trees, the variables may include age, soil condition, weather, insects,
and human intervention. In this
investigation, the relationship between Age of Trees and number
of 100s of board feet will be modeled.
How can a model be fit to
data? How can the “best” model be
selected? Can the model be used to
predict values other than those given? Are there limits on the use of the
model?
Examining Relationships
An effective way to see a
relationship in data is to display the information in a scatter plot. A scatter plot shows how two variables
relate to each other by looking for patterns in the data. Strong relationships
show data following a specific pattern or trend and weak relationships show data
widely dispersed or with no pattern at all. The scatter plot for the tree data is
shown in diagram 1.
Diagram 1
Scatterplots show the form and strength of the relationship
between sets of data. For the tree
problem, the data moves in a positive direction from the lower left corner to
upper right corner of the graph. The
form of the relationship is slightly curved.
The strength of a relationship is determined by how closely the points
follow the curve. The tree scatterplot
seems to indicate a strong relationship. A model can be constructed to represent this
data.
When the points in a
scatter plot are represented by a line of best fit, the line can be used to
predict values other than the ones given. Linear relationships are quite common
and simple to use. Since the
scatterplot displayed for the tree data appears to be curved, the function
model will probably not be linear. Many
statistical measurements, such as correlation r, are based on the strength of a
straightline relationship. In order to use these measures to determine the
“best” model, a transformation will be necessary to achieve linearity.
Curve fitting
The process of fitting a
set of data with a model can be done in many ways.
For example, given three
noncollinear points, a quadratic model can be fit to the data. To find a, b,
and c in f(x)=ax^{2}+bx+c, write and solve a system of three linear
equations using the three unknowns.
Three data points were
selected from the tree problem and setup in equations as shown below.
Point 
Substitution 
Equation 
(20,1) 
a(20)^{2}+b(20)+c
= 1 
400a+20b+c=1 
(80,33) 
a(80)^{2}+b(80)+c
= 33 
6400a+80b+c=33 
(160,182) 
a(160)^{2}+b(160)+c
= 182 
25600a+160b+c=182 
Using a matrix equation,
X=a^{1} b, to solve for a, b, and c, gives the following output:
These results produce the
quadratic model of f(x) = .009494x^{2}. .416071x + 5.52381.
The graph of the tree data
and quadratic model are shown in diagram 2.
Diagram 2
Although this model is
not an exact fit, the last point is clearly not on the line, it does look like
a nice fit. The equation produced from
these three randomly selected points will not produce the only quadratic model
for the tree problem. The model is
dependant on the points selected for the calculation. If a different set of points are used, the
equation would change.
The degree of the
polynomial can increase if more points are used. Given four noncollinear points, a quartic
model can be fit to the data.
Another way to produce
function models is to use graphing calculators or excel spreadsheets. Diagram 3
 8 are sample models produced in this way.
Diagram 3 Linear
The linear model does not
seem to be a good fit, only two points are close to the line.
Diagram 4 Exponential
The exponential model
does not seem to be a good fit either, however, points are closer to this curve
than the linear model.
Diagram 5 Quadratic
For the models with
degree greater than 1 (as shown in diagram 58), the domain must be restricted
to positive values (tree ages). These diagrams included negative values to
better display the shape of the model.
The quadratic model
created from the calculator produces a tighter fit, to all the points, than the
3 point model. The calculator has more
data to use in the equation.
3 point model: f(x) =
.009494x^{2}. .416071x + 5.52381
7 point model: f(x) =
.016203x^{2}. .475431x + 2.43061
Diagram 6 Cubic
With the increase of each
power, the curve appears to be passing through and closer to more and more data
points.
Diagram 7 Quartic
The changes shown in the
correlation of determination, R^{2}, confirms the improvement of each
graph.
Diagram 8 Power
The power model seems to
have slipped in the measure of fit as determined by R^{2}, but still a
very strong relationship is modeled.
Selecting the best model
A correlation
coefficient, r. indicates how closely the data points cluster around a linear
model. The coefficient of determination, r^{2}, is a numerical quantity
that tells how well the leastsquares line does at predicting values.
When data is nonlinear, a
transformation can be done to use correlation measures. One of the transformations that will be
described here (to flatten out nonlinear models) is for power equations y=ax^{p}.
Taking the logarithm of
both sides of the power equation gives log y = log a + p log x. The
results is a linear relationship between log x and log y. The power p in the power equation becomes the
slope of the straight line that links log y to log x. If by taking the log of both variables
produces a linear scatterplot, then the results is a reasonable model for the
original data.
With the linear model
just created, the leastsquare regression analysis and the measures of good
fit, such as correlation, can be used.
Diagram 9 shows the linearized Tree data and the graph of the results.
Log (AGE) 
Log (Board) 
2.995732274 
0 
3.688879454 
1.791759469 
4.094344562 
2.785608595 
4.382026635 
3.496507561 
4.605170186 
4.025351691 
4.787491743 
4.477336814 
4.941642423 
4.911544711 
5.075173815 
5.204006687 
5.192956851 
5.511128522 
5.298317367 
5.768320996 
With the correlation of
determination, r^{2}, set at .999864, we see we have a good fit to the
original data. The graphing calculator
does provide a correlation of determination with the models shown in diagrams
38, but it is also nice to see how a nonlinear curve is “straighten” out for
the calculation just done and shown in diagram 9.
Predicting values
Using the model produced in the transformation,
f(x)=2.4952x7.44665, and plugging in log(60), log(140), and log(180) values
not given can be predicted. To get the results
needed the data is converted back to the original form by using them as an
exponent. The table below has been
filled in with the missing values.
Age of Tree 
100s of Board Feet 
20 
1 
40 
6 
60 
15.95 
80 
33 
100 
56 
120 
88 
140 
132.12 
160 
182 
180 
247.35 
200 
320 
There are some limits to fitting
data to models. Predictions should stay
within the bounds of the original data if possible. The model summarizes the relationship between
two variables only when one of the variables helps explain or predict the
other. Another problem is data outliers.
A deviation that falls outside the overall pattern of the relationship, such
as an outlier, will distort the model.
Summary
Models can be fit to data
using systems of equations, matrices, built in calculator programs, excel
spreadsheets and computer programs. When
necessary, nonlinear models can be transformed to linear models to take
advantage of leastsquare analysis. Goodness of fit measures, like correlation,
can be used to select the “best” model for the data . By using models, unknown values can be predicted. Caution should be taken when trying to make predictions
outside the range of the original data.