Lab 3-1
Population ecology
Population distribution patterns
I. Introduction to population distribution patterns
The distribution of individuals in a population describes their spacing relative to each other.
Different species and different populations of the same species can exhibit drastically different
distribution patterns. Generally, distributions can follow one of three basic patterns: random, uniform
(evenly spaced or hyper-dispersed), or clumped (aggregated or contagious; see Figure 2). Species traits
such as territoriality, other social behaviors, dispersal ability, and allelochemistry will shape individual
dispersal (i.e., movements within a population), emigration, and immigration, all of which affect
population distribution patterns. In addition to species traits, the distribution of resources or microhabitats
links population distribution patterns to the surrounding abiotic environment.
II. Measuring population distribution
Population distribution is commonly quantified by population ecologists. With mobile organisms,
this requires intensive sampling. Therefore, we will measure the distribution patterns of less mobile
species. Analyses of population distribution patterns usually follow a standard method in which observed
distribution patterns are compared to predicted, random distribution patterns. As such, the following
steps are followed:
1. A particular method of field sampling is chosen, which will allow the ecologist to quantify the
distribution of individuals within a population of a given species (e.g. Quadrat method or Pointto-Plant method). – This is the observed distribution pattern.
2. A standardized technique is used to determine what values would have been obtained if the same
sampling techniques was used for a randomly distributed population (e.g. the Poisson
distribution). – This is the predicted, random distribution pattern.
3. A statistical test is used to evaluate the hypotheses:
HO: There is no significant difference between the observed distribution pattern and the
predicted, random distribution pattern.
Ha: There is a significant difference between the observed distribution pattern and the
predicted, random distribution pattern.
4. If p > 0.05, then we fail to reject the null hypothesis and therefore state that the observed
population is randomly distributed.
If p < 0.05, then we reject the null hypothesis and therefore state that the observed population is
not randomly distributed. It is therefore uniform or clumped.
5. If the population is not randomly distributed, we then use a method to determine whether it is
uniform or clumped.
In today’s lab exercise, we will utilize two different techniques to characterize the distribution pattern of
our focal species: (A) a modified quadrat-based method and (B) a point-to-plant method.
Lab 3-2
A. The Quadrat Method
The quadrat method involves counting the frequency of
occurrences of the species of interest in each of the 100 individual
sub-quadrats that compose the quadrat. The example on the right
shows a quadrat made up of 36 sub- quadrats. Each sub-quadrat
contains 1 individual.
If the individuals within the population are randomly
dispersed, there will be a random number of individuals in each
quadrat, centered about the mean (Figure 2A, D). If the individuals
in the population are uniformly dispersed, there will be the same
number of individuals in each sub- quadrat (Figure 2B, E). If the
individuals in the population are clumped in dispersion, there will
be a few quadrats with many individuals, and many quadrats with no
individuals (Figure 2C, F).
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Figure 1: Example of a quadrat,
divided into sub-quadrats.
A
B
C
D
E
F
Figure 2: Common dispersion patterns are represented above. Figures A, B, and C represent the spacing
of individuals within a population relative to each other. The entire square indicates the entire quadrat,
and each small square indicates one sub-quadrat. Figures D, E, and F indicate the number of individuals
within each sub-quadrat. Note that Figure D is derived from a randomly dispersed population, and that it
indicates a Poisson distribution (example data and figure from S. Whitfield).
In today’s lab, we will modify the quadrat method. We will be assessing the distribution of epiphytes
growing on trees on campus. Each tree will represent one quadrat. Within each tree, all epiphytes,
Lab 3-3
Tillandsia usneoides and Tillandsia recurvata, will be counted. To analyze the data from this we will use
a chi-square test to evaluate our proposed hypotheses:
HO: There is no significant difference between the observed distribution pattern and the predicted,
random distribution pattern.
Ha: There is a significant difference between the observed distribution pattern and the predicted,
random distribution pattern.
The chi-square test compares a given (observed) distribution to the Poisson (random)
distribution. The Poisson distribution basically allows us to determine the probability of obtaining “x”
number of individuals within a tree, if the population is randomly distributed. For example, if the
population of interest is randomly distributed, what is the probability of obtaining 3 individuals in a tree?
To do this, follow these steps:
1. Determine the observed distribution pattern.
This information is collected in the field and then used to complete the table below, where “x” is
the number of individuals found in a tree and “f” represents the number of trees that contained x number
of individuals (This column is also labeled “O” meaning “observed distribution”). For example, suppose
we sample 40 trees to determine the distribution of the individual epiphyte species. We may get the
following results: Nine trees have 0 individuals each, 22 trees have 1 individual each, 6 trees have 2
individuals each, 2 trees have 3 individuals each, 1 tree has 4 individuals and none of the trees have 5
individuals or more (Table I).
We can then calculate the total number of individuals recorded within the tree. This is done in the “fx”
column.
Table I: Example data -- there are 40 total trees, 44 total individuals, and a mean of 1.1 individuals per
tree.
Number of
Individuals tree
(xi)
0
1
2
3
4
5
∑
µ
Number of
trees
(fi)
(O)
fixi
9
22
6
2
1
0
40
1.1
9*0=0
1* 22 = 22
2 * 6 = 12
3*2=6
4*1=4
5*0=0
44
-
2. Calculate the mean number of individuals per tree.
Lab 3-4
To calculate the mean value for data in this format use the following equation…
…in which, f is the number of trees and x is the number of individuals per tree for each row in Table I. In
the given example, µ = 44/40 = 1.1
3. Calculate the Poisson probability values.
Once the mean is calculated, we can then calculate the Poisson probability values. To do this we use the
‘Poisson expression’ (Cox 2001)…
…where e = the base of the natural log = 2.7182818, µ = mean, and x = the number of individuals per
tree.
You can do this calculation in Excel, by using the following formula:
=(µ^x)/((EXP(µ))*(FACT(x)))
We use the Poisson probabilities to determine the probability of obtaining “x” number of individuals
within a tree, if the population is randomly distributed. We can calculate these probabilities for each row
in Table I as shown below:
P(x0) =
(1.1)0
=
1
= 0.3329
1.1
(2.718) (0!)
3.0038
P(x1) =
(1.1)1
= 1.10 = 0.3662
(2.718)1.1(1!)
3.0038
P(x2) =
(1.1)2
= 1.21 = 0.2014
(2.718)1.1(2!) 6.0076
P(x3) =
(1.1)3
= 1.31 = 0.0739
1.1
(2.718) (3!) 18.0228
P(x4) =
(1.1)4
= 1.4641 = 0.0203
(2.718)1.1(4!) 72.0912
P(x5) =
(1.1)5
= 1.6105 = 0.0045
(2.718)1.1(5!) 360.456
Lab 3-5
These values can now be used to calculate the expected number of trees that would obtain “x” individuals,
if the population was randomly distributed. For example, the results above show that, if the population
was randomly distributed, then 33% of the trees should each have 0 individuals, 7% of the trees should
each have 3 individuals and 0.4% of the trees should each have more than 5 individuals.
4. Calculate the expected values.
We can now use these probabilities to calculate expected values using the equation below (essentially,
multiply each probability above by the total number of trees, in this case, 40.):
E(xi) = P(xi)∑fi
E(x0) = P(x0)∑fi = 0.3329*40 = 13.316
E(x1) = P(x1)∑fi = 0.3662*40 = 16.646
E(x2) = P(x2)∑fi = 0.2014*40 = 8.055
E(x3) = P(x3)∑fi = 0.0739*40 = 2.954
E(x4) = P(x4)∑fi = 0.0203*40 = 0.812
E(x5) = P(x5)∑fi = 0.0045*40 = 0.217
Based on the results above, we can see that if our population was randomly distributed, we would expect
13.3 of the 40 trees to have 0 individuals, 2.9 of the trees to have 3 individuals and .2 of the trees to have
more than 5 individuals.
5. Compare the Observed values to the Expected values using the Chi-squared test (χ2).
We can now compare our observed values and our expected values to see whether or not there is a
significant difference between them. To calculate our Chi-square value, we use the following equation:
χ2 = ∑ (O – E)2
E
Please note that you must calculate this equation for each row, and THEN sum all the values to obtain
your Chi-square value.
In this example, the expected values for trees with three or more individuals are combined for the Chisquare analysis because those trees have expected values that are less than 1. If the expected frequency for
any category or set of categories is less than 1, you must combine the values from this category and the
one preceding it, and then use the combined fi value to calculate Chi-square (see Table 2).
In the example below, because the Expected values for rows 4 and 5 are less than 1, the values for these
two rows were added to the row preceding them. Therefore:
The combined Observed frequency = 3
The combined Probability = 0.0987
The combined Expected frequency = 3.983
The combined Chi-square value is therefore: (3-3.983)2/3.983 = 0.243
Lab 3-6
Table II. Example of a completed Poisson Table
Number of
individuals
per tree
(xi)
Observed
Frequency
(O)
P(x)
Expected
Frequency
E
0
1
2
3
4
5
∑
9
22
6
2
1
0
40
0.3329
0.3662
0.2014
0.0739
0.0203
0.0045
1
13.315
16.646
8.055
2.954
0.812
0.217
-
(O-E)2/E
1.398
3.693
0.525
5.859
Table IIb. Example of a Poisson Table -- Asterisk (*) denotes that the categories for 3, 4, and 5
individuals per tree were combined into one class in order to better meet assumptions of the Chi-square
test.
Number of
individuals
per tree
(xi)
0
1
2
3
4
5
∑
Observed
Frequency
(O)
9
22
6
P(x)
Expected
Frequency
E
(O-E)2/E
0.3329
0.3662
0.2014
13.315
16.646
8.055
1.398
3.693
0.525
0.243*
3
40
0.0987
1
3.983
-
5.859
6. Determine the adjusted number of categories and calculate the degrees of freedom.
Once you have calculated the Chi-square statistic, you will then need to calculate one other value in order
to obtain your p-value, the degrees of freedom.
The degrees of freedom (df) is used to locate the χ2statistic on a Chi-square table:
df = k-2
…where k is the number of categories remaining after you perform any necessary adjustments to the
number of rows in the table. In the example above, after adjusting the table, we end up with 4 categories
(instead of the original 6). Therefore the degrees of freedom = 4 – 2 = 2.
Lab 3-7
7. Use the Chi-square statistic and the degrees of freedom to obtain the associated p-value.
The χ2 statistic for our example is 5.841. You can locate this value on the Chi-square table and then find
the associated p-value, or use the following Excel formula to get a precise p-value:
=CHIDIST(χ2,df)
…where χ2 is the test statistic you calculated and df are the degrees of freedom.
Using the following equation in Excel: =CHIDIST(5.859,2), we get an associated p-value of 0.054.
8. Based on the p-value, determine whether to reject or fail to reject your null hypothesis.
If p > 0.05, then we fail to reject the null hypothesis and therefore state that the observed population is
randomly distributed.
If p < 0.05, then we reject the null hypothesis and therefore state that the observed population is not
randomly distributed. It is therefore uniform or clumped.
Using our example, the p-value was 0.054 which is greater than 0.05. Therefore the population is
randomly distributed.
If our p-value was less than 0.05, meaning that the population is not randomly distributed, we would then
have to plot our observed data on a graph to determine whether it is uniform or clumped, based on the
pattern of the graph (See Figure 2).
B. Point-to-Plant method
The point-to-plant distance method utilizes a ratio to detect deviation from a random distribution
pattern. We use the point-to-plant method to sample the distribution of organisms that cannot easily be
sampled using a 1 m2 quadrat (like trees or species that occur much more spaced out).
To analyze the data from the point-to-plant method we will use a z-test to evaluate our proposed
hypotheses:
HO: There is no significant difference between the observed distribution pattern and the predicted,
random distribution pattern.
Ha: There is a significant difference between the observed distribution pattern and the predicted,
random distribution pattern.
To do this, follow these steps:
1. Determine the observed distribution pattern.
To collect the appropriate data, you will haphazardly select a point of origin (by throwing an object of
some sort), and then measure the distance from that point to the two nearest individuals of the species of
interest.
A total of at least 40 haphazardly selected points must be used, to end up with a data set as shown below:
Lab 3-8
Sample
point
d1
(m)
d2
(m)
d12
d22
d12 / d22
1
2
3
……
……
……
……
38
39
40
4.3
1.2
3.7
……
……
……
……
1.1
0.8
3.8
5.8
4.3
4.6
……
……
……
……
2.3
1.5
4.2
18.49
1.44
13.69
……
……
……
……
1.21
0.64
14.44
33.64
18.49
21.16
……
……
……
……
5.29
2.25
17.64
0.549643
0.07788
0.646975
……
……
……
……
0.228733
0.284444
0.818594
∑ = 17.35
This table will then be expanded to allow the calculation of the coefficient of aggregation (A), which is
our observed distribution measure for this population.
…where n = the number of sample points, d1 = the distance from the selected location to the closest tree
and d2= the distance from the selected location to the second closest tree.
This coefficient of aggregation will always be between 0 and 1.
The expected value of A for a randomly dispersed population is 0.5.
If A is significantly less than 0.5, the dispersion is uniform.
If A is significantly greater than 0.5, the dispersion is aggregated.
Once A is calculated, we must then determine whether A is significantly different to 0.5 or not.
Essentially, this is the same as asking whether the observed population distribution pattern is significantly
different to a random distribution pattern. To do this we must use a statistical test.
NOTE: just because A is less than 0.5, does not mean it is automatically uniform. For example if A=0.48,
then it is less than 0.5, but you must use a statistical test to determine whether or not 0.48 is significantly
less than 0.5.
In our example A = 17.35/40 = 0.434
2. Calculate the z statistic which will be used in the Z-test to obtain a p-value.
To test our hypotheses, the z-equation is used…
Lab 3-9
…where n = the number of sample points, 0.2887 = the standard deviation of A values for a randomly
dispersed population.
Note that this is very similar to the equation for a t-test from the first lab. The z-equation is simply a
special version of the t-equation, except that the degrees of freedom are irrelevant because the number of
sample points (n) must always be greater than 30.
The calculated z is looked up on the z table to find an associated p-value. You can use excel to look up a
precise p-value from the z-table (z is your calculated z-value) using the following equation: =1NORMSDIST(z)
In our example z = (0.5 – 0.434) / (0.2887/v40) = 1.446
The associated p-value (obtained from Excel) = 1-NORMSDIST(1.446) = 0.074
9. Based on the p-value, determine whether to reject or fail to reject your null hypothesis.
If p > 0.05, then we fail to reject the null hypothesis and therefore state that the observed population is
randomly distributed. (i.e. there is no significant difference between our observed coefficient of
aggregation (A) and the expected coefficient of aggregation (0.5)).
If p < 0.05, then we reject the null hypothesis and therefore state that the observed population is not
randomly distributed. (i.e. there is a significant difference between our observed coefficient of
aggregation (A) and the expected coefficient of aggregation (0.5)). The population is therefore uniform or
clumped.
If you reject the null hypothesis, you must then look at the observed coefficient of aggregation and
determine if it is less than 0.5 (in which case the population is uniform) or greater than 0.5 (in which case
the population is clumped).
Using our example, the p-value was 0.074 which is greater than 0.05. Therefore the population is
randomly distributed.
If our p-value was less than 0.05, meaning that the population is not randomly distributed, we would then
have to look at the A value to determine if the population is uniform or clumped.
III. Objective
The field portion of today’s lab will involve collecting data on the dispersion pattern of
populations of epiphytic species and a large tree species chosen by your TA. The objective of Lab 3 is to
determine if the dispersion patterns of the populations you investigate are random, uniform, or clumped.
IV. Instructions
Before setting out to sample your focal species, complete the following pre-field instructions.
Lab 3-10
1. Generate several testable hypotheses as a class that you can test with today’s exercise.
2. Discuss how to record the 2 different kinds of dispersion data. Set up field data sheets for your
sampling procedures.
3. Be sure that you have all the field sampling equipment that you will need.
4. All field teams should participate in sampling all habitats. Your TA will pool data from all teams
to generate larger datasets for each population that you investigated. Use these complete datasets
for your analysis.
EPIPHYTE DISTRIBUTION SAMPLING
You will be collecting data on all trees assessing the epiphytic species distribution of the entire
population. Count the number of individual epiphytes within each tree. Then, to test the accuracy of
randomly sampling a population, you will randomly select 30 individual trees, assess the epiphytic
population distribution, and compare that analysis to that of the entire population.
Record data as shown below, using a tally for each row:
Number of Individuals per Tree
(xi)
Tally of Trees (fi)
(O)
Number of Trees (fi)
(O)
0-10
11-20
21-30
31-40
41-50
51-60
61-70
71-80
81-90
91-100
100-more
11111
111
1111
11
11111111
111
11
1111
1
0
1
5
3
4
2
8
3
2
4
1
0
1
You will then use this data to complete your chi-square calculations.
POINT-TO-PLANT METHOD S
To perform the point-to-plant distance method, locate the required number of random points in
the population of interest. Measure the distance from each random point to the two nearest trees of the
focal species. Once you are finished, you will have two distances (in meters) for each random point: the
distance from the point to the nearest tree and the distance from the point to the next nearest tree.
Organize your data as shown below:
Lab 3-11
Sample
point
d1
(m)
d2
(m)
1
2
3
4
5
6
7
8
9
10
You will use these data to calculate the coefficient of aggregation (A) for your focal population in order
carry out the z-test to determine the dispersion pattern of the population.
Literature Cited
Cox, G. W. 2001. General Ecology Laboratory Manual, 8th edition. McGraw-Hill, New York.
Further Reading
Cornell, H. V. 1982. The notion of minimum distance or why rare species are clumped. Oecologia
52(2):278-280.
Zavala-Hurtado, J. A., P. L. Valverde, M. C. Herrera-Fuentes, A. Diaz-Solis. 2000. Influence of leafcutting ants (Atta mexicana) on performance and dispersion patterns of perennial desert shrubs in an
inter-tropical region of Central Mexico. Journal of Arid Environments 46(1):93-102.