Geography 215: QUANTITATIVE METHODS
Dr. Rodrigue
Graded Lab 8: Nearest-Neighbor Analysis and Bi-Variate Hypothesis Testing
with Nominal Data
This lab covers two techniques applicable to hypothesis testing with spatial
data. One is nearest-neighbor analysis and the other is Chi-squared in
quadrat analysis.
For all questions, please do your calculations at the full capacity of your
spreadsheet or calculator, but round your answers to three decimal
places of accuracy (i.e., 0.000).
LAB EXERCISE A: Nearest-Neighbor Analysis
Examine the distribution of adult fetid lilac shrubs, Ceanothus
foetidus (which I made up, by the way, so don't go running to the Munz
flora in the excitement of learning about a new chaparral plant!) in Figure 1. The area mapped is a hillside in the Sulphur
Mountain area of Ventura County (which does exist and, ooooo, is it
stinky!). Employ nearest-neighbor analysis to ascertain whether, at this
scale, this "species" has a clumped, uniform, or random distribution.
Calculating the nearest-neighbor co-efficient (R) entails the tedious process
of measuring the distance between each point in a given space and the point
that is its nearest neighbor. It should be noted that point a may well have
point b as its nearest neighbor, but point b may have another point entirely,
say, c, as its nearest neighbor. Anyhow, having measured all those
nearest neighbor distances, you figure out the mean nearest-neighbor distance
and then create an expected mean nearest-neighbor distance from the
density of the points in your study area. You then create R as the ratio of
the mean observed nearest-neighbor distance to this expected mean
nearest-neighbor distance.
R can vary from 0 to 2.149 (I've always liked that extra .149 bit!). A score
of 0 means perfect clustering: All points are found at the exact same point
in space (which is, of course, a physical impossibility if your study entails
point data collected at one time). A score of 2.149 means perfect uniformity.
A score of 1 represents perfectly random distribution of points in space. So,
you can use this descriptive statistic to characterize a distribution as more
clustered or more uniform or just random. Unfortunately, I know of no
significance test for the R ratio, so there's no way to test whether a given
distribution departs significantly from randomness.
- What is n (i.e., how many plants are there in the area shown in Figure 1?
n = ________
- What is A (the size of the study area, measured in square meters)?
A = ________
- What is p (the density of C. foetidus in numbers per unit of area,
or n/A, or, in this case, number of plants divided by the size of the study
area in square meters)?
p = ________
- The tedious part: Measure and record the distance from each plant to its
nearest-neighbor (remember, one plant can wind up the nearest-neighbor for
more than one other plant). Record your measurements in the worksheet below.
"#" stands for the individual plant identification number shown on the map,
while "r" is the distance between each plant and its nearest neighbor.
# r # r # r
1) ________ 11) ________ 21) ________
2) ________ 12) ________ 22) ________
3) ________ 13) ________ 23) ________
4) ________ 14) ________ 24) ________
5) ________ 15) ________ 25) ________
6) ________ 16) ________ 26) ________
7) ________ 17) ________ 27) ________
8) ________ 18) ________ 28) ________
9) ________ 19) ________ 29) ________
10) ________ 20) ________ 30) ________
_
- What is the ro (the observed mean nearest-neighbor distance)?
_
ro = ________
_
- What is the re (the expected mean nearest-neighbor distance)?
_ 1 1
re = ______ _____
2\/ n/A 2\/ p
_
re = ________
- What is R (the ratio of observed to expected mean nearest-neighbor
distances)?
_
ro
R = _____
_
re
= ________
- On the basis of the R you calculated and the guidelines for interpreting
it given at the top of Lab A, how would you characterize the local
distribution of C. foetidus, at least at this particular scale of
analysis?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
LAB EXERCISE B: Chi-Squared Quadrat Analysis
You may recall that quadrat-based techniques of spatial analysis involve the
division of an area into equal-sized plots, usually through a grid of squares.
This permits the use of statistical techniques to analyze quantitative data
with no more measurement sophistication than mere frequencies by category
(nominal data). Nearest-neighbor analysis, by contrast, required the
collection of data at the ratio level of measurement, which is the highest
level.
For your reference pleasure, the definitional formula for Chi-squared is:
r k
__ __ (Oij - Eij)2
X2 =\ \ ____________
/_ /_ Eij
i=1 j=1
You'll be comforted to know I'll walk you through a much easier computational
process.
- The first order of business, as usual, is to set up null and alternative
hypotheses and an appropriate standard to judge whether the null hypothesis
can be rejected for your purposes. So, eyeballing the map in Figure 2, formulate your hunch about the relationship
between the distributions of the two plant species described below. Now,
please state the null version of that hypothesis:
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
- Now, this is a classic physical geography kind of study, so it will be
driven mainly by some sort of scientific theoretical concern, testing the
validity of some argument about niche-sharing or perhaps allelopathy. So, is
the consequence of a Type I error likely to be life-threatening?
_____ yes _____ no
Given that, would you set alpha at the high end (larger prob-value) or low end
(smaller prob-value) of the scientific continuum of common alphas?
_____ larger alpha _____ smaller alpha
On the other side, this is clearly not a marketing type of study, where a Type
II error (missing a significant and exploitable relationship, if it exists)
would have the more serious consequences. So, there is not much pressure to
increase alpha: It is more important for theory that you not delude yourself
into seeing significant relationships where there might be none. So, you want
to select an alpha that is on the smaller side compared to, say, the marketing
continuum and yet on the larger end of the scientific continuum. Balancing
these concerns, which of the commonly used alpha levels is
best able to suit your purposes?
_____ 0.10 _____ 0.05 _____ 0.01
- Figure 2 shows the distribution of two (real) plant
species, Salvia apiana (white sage) and Avena barbata (slender
oat). Characterize each of the larger quadrats (the ones labeled A1 or F9 or
J5, for example) as belonging to one of the four quadrat types listed below.
Be sure that you've accounted for 100 quadrats (10 x 10), and that no quadrat
fits in more than one category. This can be a little tedious to keep track
of, so be really careful from the "git-go" in creating your observed counts.
- (a) containing both Avena and Salvia;
- (b) containing Avena but no Salvia;
- (c) containing Salvia but no Avena; OR
- (d) containing neither Avena nor Salvia.
- Now, with all 100 quadrats accounted for, each in no more than one
category, fill in the following spreadsheet with quadrat counts in each of the
four categories. Put your count numbers by category in the upper part of the
appropriate cell. These are your observed or real-world frequencies.
| SALVIA |
| | |
| present | absent | row totals
_________________________________________________________________________
|(a) |(b) |-e-
present | | |
| | |
AVENA _________________________________________________________________
|(c) |(d) |-f-
absent | | |
| | |
_________________________________________________________________________
|-g- |-h- |-i-
column totals | | |
| | | n =
- Compute the marginal totals. That is, sum the observed frequencies in
each row and put those sums in the appropriate row total (e or f). Do the
same for the frequencies in each column and put those sums in the appropriate
column total (g or h). The sum of row totals should equal the sum of column
totals. If so, put the total number or n (which had better equal 100)
in cell i.
- Create the expected frequencies for each data cell (a through d). This
is the distribution of cell counts you would expect from your data if there
were no association between the two plant species (i.e., random processes were
allocating them among the cells). To do this for each data cell, a through d,
multiply the row total to its right by the column total below it and then
divide the answer by n. Put the answer, rounded to three decimal
places of accuracy, in its cell below the actual observed frequency.
-
- Still lost? Okay, okay. In other words, multiply cells e and g and
divide the answer by cell i. Put the answer, properly rounded, in the lower
part of cell a. Similarly, multiply cells e and h and divide by i, and put
that answer in cell b. Multiply cells f and g and divide by i, and plop that
answer in cell c. Lastly, multiply cell f by cell h, divide by i again, and
put the result in cell d.
That done, examine the expected frequencies. Chi-square should not be used if
any expected frequencies are below 2 (or, irrelevantly in this case, if more
than 20 percent of the data cells have fewer than 5 actual cases). You will
note that there are no such problems with your contingency table, so you can
safely proceed through Chi-square.
- Now, move on to the worksheet below for calculating Chi-squared. In the
first column, enter the observed frequencies for each data cell (the number in
the upper part of cells a through d).
- In the second column, square those frequencies.
- In the third column, divide each squared frequency by the corresponding
expected frequency in the bottom of the appropriate data cell (a through d).
- Now, sum the third column and put the answer near the bottom of the
spreadsheet (sum(O2/E). Show your work here to three
decimal places of accuracy.
- Finally, subtract n (from cell i) from that sum. This answer is your
calculated Chi-squared (X2). Put it at the bottom of the whole
spreadsheet, also rounded to three decimal places of accuracy.
________________________________________________________________________
DATA CELL | O | O2 | O2/E
________________________________________________________________________
(a) | | |
________________________________________________________________________
(b) | | |
________________________________________________________________________
(c) | | |
________________________________________________________________________
(d) | | |
________________________________________________________________________
| sum(O2/E) =
________________________________________________________________________
| sum(O2/E) - n = X2 =
________________________________________________________________________
- Now, to interpret this hard-gained number, your
X2calc, you need to compare it with a critical
X2. To do this, you will need the Chi-squared table I distributed
in class, the one suited to the classical approach to hypothesis testing. You
need your pre-selected alpha level to pick the right
column and the degrees of freedom for your 2 x 2 contingency table to choose
the right row to enter the table. Degrees of freedom in Chi-squared can be
defined as:
DF = (r - 1)(k - 1)
where r = number of rows and k = number of columns
So, you will enter the table at the intersection of:
the column headed ________
and the row corresponding to ________ degrees of freedom.
What, then, is your critical Chi-squared value?
X2crit = ________
- Is your X2calc ________ greater than or ________
less than the X2crit?
- If your actual, calculated Chi-square value is greater than the critical
Chi-square, you may safely conclude that your pattern is not just a random
one. In other words, there is a statistically significant probability that
there is a real association of some sort between your variables (in this case,
between the two plant species). If the calculated Chi-square value is less
than the critical test value, the relationship probably is random. Can the
null hypothesis of random association between these two plant species in this
study area be rejected in this case?
_____ reject Ho _____ do not reject Ho
- It's always good etiquette, whenever possible, to calculate the
prob-value of a Type I error, to express your faith in the null hypothesis,
however, in the off chance that a reader may have compelling reasons to use a
different standard of alpha than you chose. Unfortunately, M&M somehow messed
up and did not put the oh-so-important 1 DF column in their prob-value table
for Chi-squared. In a rare attack of generosity, I have decided to provide
you the missing column by spending a lot of time with the probability
calculator within Statistica, a very nice full-featured statistics package.
Consult Figure 3 to get the missing column and tell me
the probability that you could have gotten results as extreme as yours if
there is but a random association between the two plant species.
________ prob-value of Ho
- Plot complication. Chi-squared is notoriously sensitive to sample size.
That is, the same percentages in each cell can appear significant in a big
sample (large n) or insignificant in a small sample. It might help to assess
the strength of a significant relationship, should the Chi-squared test
find one. For that, you can use Yule's Q. Yule's Q, however, can only be
calculated for contingency tables with no more than two rows and two columns
(bigger tables can sometimes be collapsed into a 2 x 2 format, by combining
rows and columns in some sort of logical way). Conveniently, this lab just
happens to feature a 2 x 2 table.
To calculate Yule's Q, multiply cells a and d and also cells b and c. Then,
enter these multiplications into the following formula:
ad - bc
Q = _______
ad + bc
So, what is the Q value for this lab? ________
- Now, what does it all MEAN? Basically, Yule's Q can vary from -1 to +1.
The closer it is to 0, the weaker the relationship is. The closer it is to -1
or +1, the stronger the relationship is, whether inverse (negative) or direct
(positive).
Please interpret the results of Lab B, taking into consideration both
Chi-squared and Yule's Q. What sort of ecological relationship, if any,
exists between Salvia apiana and Avena barbata at this scale of
analysis?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
And that's that for another lab, folks!
Figure 1 -- Map of Ceanothus foetidus Plants
(or why they don't let me teach cartography!)
![[ map of ceanothus plants ]](http://www.csuchico.edu/~lapaloma/geog215/lab8fig1.jpg)
Figure 2 Map of Oats and Sage (you might want to
recopy these figures at 120 percent or so)
![[ map of oats and sage ]](http://www.csuchico.edu/~lapaloma/geog215/lab8fig2.jpg)
Figure 3: p-Values for
X2
X2 1 DF X2 1 DF X2 1 DF X2 1 DF
3.2 .0736 4.4 .0359 5.6 .0180 6.8 .0091
3.3 .0692 4.5 .0339 5.7 .0170 6.9 .0086
3.4 .0652 4.6 .0320 5.8 .0160 7.0 .0082
3.5 .0614 4.7 .0302 5.9 .0151 7.1 .0077
3.6 .0578 4.8 .0285 6.0 .0143 7.2 .0073
3.7 .0544 4.9 .0268 6.1 .0135 7.3 .0669
3.8 .0513 5.0 .0254 6.2 .0128 7.4 .0065
3.9 .0483 5.1 .0239 6.3 .0121 7.5 .0062
4.0 .0455 5.2 .0226 6.4 .0114 7.6 .0058
4.1 .0429 5.3 .0213 6.5 .0108 7.7 .0055
4.2 .0404 5.4 .0201 6.6 .0102 7.8 .0052
4.3 .0381 5.5 .0190 6.7 .0096 >7.8 <.0050
-
- data collected by Dr.
Rodrigue from the Probability Calculator within Statistica®
at ALST, Inc., of Northridge, CA, 11/98.
first placed on the web: 11/26/98
last revised: 11/30/98
© Dr. Christine M.
Rodrigue