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As noted earlier, quadrat sampling is the term used to describe a procedure of sampling and recording point-based data within regularly shaped regions (typically a grid of square cells). Since grid data are already in this form, it is possible to analyse grids where the contents of grid cells (or blocks of cells) are regarded as counts of point objects (Figure 5‑12). In the example illustrated a 5x5 grid has been used to collect data on point events shown by the symbol x (the larger * symbol in this diagram shows the position of the MAT point under the L2 metric). The event distribution has then been coded as counts in the grid below. Simple statistics may then be computed, such as the mean number of points per cell/cell block (4 in this example), and the variance of this measure (4.59 in this case).
If the distribution of points across the set of grid cells is random, it can be modelled using the Poisson distribution. The Poisson distribution is applicable where events (points in our case) are independent, there are a large number of events (typically 100+), and the probability of an individual event occurring (e.g. a point falling in a particular location) is small. It is derived as an approximation to the Binomial distribution by applying these conditions.
As noted in Table 1‑4, the Poisson distribution has the form:
; x=0,1,2,…
where m is the mean and x is the count of events. In our example the (sample) mean is 4, so the individual terms of the distribution may be computed (i.e. for x=0,1,2,3…) and used as a set of “expected” values, under the null hypothesis that the observed frequency distribution is random. The set of n observed frequency values may then be compared to the set of expected values using a simple Chi-square test, to obtain an estimate of the probability that the data reflects a random distribution of events.
Figure 5‑12 Quadrat counts

|
3 |
2 |
6 |
2 |
2 |
|
2 |
4 |
3 |
7 |
3 |
|
2 |
6 |
6 |
9 |
4 |
|
5 |
6 |
3 |
5 |
5 |
|
3 |
7 |
3 |
2 |
0 |
For the example shown above a frequency analysis of the form shown in Table 5‑5 may be drawn up. The sum row shows the total number of observations made (grid cells) and the value of the χ2 statistic. The degrees of freedom in this case are 11‑1‑1=9, because there are 11 frequency classes, the total count is known (‑1DF), and the mean (m) has been estimated from the sample (‑1DF). The 5% probability level from tables or computed value of the Chi-square distribution is χ20.05,9=16.9, thus a value of 9.3 is well within the expectation for a random pattern and thus we cannot reject the null hypothesis on the basis of this information. Aggregating rows to ensure most table cell counts are greater than 5 — e.g. by grouping frequencies into four classes (0,1), (2,3), (4,5) and (6+) — gives a χ2 statistic of 4.6 with 3 degrees of freedom, and χ20.05,2=6, confirming the previous result.
Table 5‑5 Simple Chi-square frequency table computation
|
Freq |
Obs, O |
Exp, E |
|O-E| |
|O‑E|2/E |
|
0 |
1 |
.5 |
.5 |
0.64 |
|
1 |
0 |
1.8 |
1.8 |
1.83 |
|
2 |
6 |
3.7 |
2.3 |
1.49 |
|
3 |
6 |
4.9 |
1.1 |
0.25 |
|
4 |
2 |
4.9 |
2.9 |
1.7 |
|
5 |
3 |
3.9 |
.9 |
0.21 |
|
6 |
4 |
2.6 |
1.4 |
0.75 |
|
7 |
2 |
1.5 |
.5 |
0.18 |
|
8 |
0 |
.7 |
.7 |
0.74 |
|
9 |
1 |
.3 |
.7 |
1.35 |
|
10 |
0 |
.1 |
.1 |
0.13 |
|
Sum |
25 |
|
|
χ2=9.3 |
In principle the observed and expected frequency distributions could also be compared using the maximum absolute difference in their cumulative probability distributions, using the Kolmogorov-Smirnov test statistic. However, this procedure should really be reserved for expected distributions that are continuous (e.g. the Normal distribution and other, non-Normal continuous distributions), and there may be difficulties in estimation of the appropriate mean value for the expected distribution. Testing procedures of this type are rarely supported directly within GIS and related packages, but are widely supported in statistical packages such as SPSS and STATA and may be readily computed programmatically or by use of a generic tool such as Excel.
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