SME
Geographic Information Systems (GIS)
Spatial statistics
Use this for “50 wells stand in a 100 km² basin, each 0.9 km on average from the nearest other well. Are they clustered, random or dispersed?”
Diagram
Expected
mean
distance
for
50
random
points
in
100
km²:
0.5
÷
√(
50
100
)
=
0.707
km.
R
=
0.9
km
0.707
km
=
1.27:
dispersed.
SE
=
0.26136
÷
√(
50
2
100
)
=
0.0523
km;
z
=
0.9 − 0.707
0.0523
=
3.69.
The drawn points’ own index is 1.27.
This is an example. Type your own values to start a new problem.
n
Number of points
example
A
Study area
example
d̄
Observed mean distance
example
d̄ₑ
Expected mean distance
worked out (not typed)
R
Nearest-neighbor index
worked out (not typed)
SE
Standard error
worked out (not typed)
z
z-score
worked out (not typed)
Formulas
d̄ₑ
=
0.5
÷
√(
n
A
)
0.7071
=
0.5
÷
√(
50
100
)
R
=
d̄
d̄ₑ
1.2728
=
0.9
0.7071
SE
=
0.26136
÷
√(
n
2
A
)
0.05227
=
0.26136
÷
√(
50
2
100
)
z
=
d̄
− d̄ₑ
SE
3.6902
=
0.9 − 0.7071
0.05227
Assumptions
•
Nearest-neighbor index R near 1 is random; toward 0, clustered; up to 2.15, dispersed (a perfect triangular lattice).
•
Constants for random points (Clark and Evans): expected distance 0.5 ÷ √density, standard error 0.26136 ÷ √(n² ÷ A).
•
Edge effects are ignored: points near the border have fewer neighbors, so the observed mean runs a little high.
•
The z-score past ±1.96 is significant at the 5% level; the drawn points are one seeded pattern with this R.
Step-by-step
WE KNOW
n = 50, A = 100 km², d̄ = 0.9 km
FIND
expected mean distance (d̄ₑ), nearest-neighbor index (R), standard error (SE), z-score (z)
Step 1 · Find expected mean distance (d̄ₑ)
Use d̄ₑ = 0.5 ÷ √(n ÷ A)
Random points at this density sit, on average, half the square root of the area per point apart.
d̄ₑ
=
0.5
÷
√(
50
100
)
d̄ₑ
=
0.5
÷
√
0.5
d̄ₑ
=
0.5
0.7071
d̄ₑ = 0.707107 km
Step 2 · Find nearest-neighbor index (R)
Use R = d̄ ÷ d̄ₑ
The observed mean distance over the random one: below 1 the points crowd, above 1 they spread.
R
=
0.9
0.707107
R
=
1.2728
Step 3 · Find standard error (SE)
Use SE = 0.26136 ÷ √(n² ÷ A)
How much the mean distance varies from one random pattern of this many points to the next.
SE
=
0.26136
÷
√(
50
2
100
)
50
2
100
=
25
√
25
=
5
0.26136
5
=
0.05227
SE = 0.052272 km
Step 4 · Find z-score (z)
Use z = (d̄ − d̄ₑ) ÷ SE
How many standard errors the observed distance lies from random; past ±1.96 it is significant at 5%.
z
=
0.9 − 0.707107
0.052272
z
=
0.1929
0.05227
z
=
3.6902
Check
Put the numbers back in. Both sides should match.
0.707107
=
0.5
÷
√(
50
100
)
✓
1.2728
=
0.9
0.707107
✓
0.052272
=
0.26136
÷
√(
50
2
100
)
✓
3.6902
=
0.9 − 0.707107
0.052272
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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