Problem 1a) Create a Moran correlogram for the number of woodpeckers. Copy the graph into your Word
document.
b) Interpret the Moran correlogram. What can be said about the spatial autocorrelation in the
data? What can be said about the range of autocorrelation in the data?
–
The correlograms created by 3 different approaches in defining neighbors are similarly
displaying a positive spatial autocorrelation (> 0).
A weak negative spatial autocorrelation (< 0) is noted at one of the lags in Queen’s Case.
The negative slope indicates the positive spatial autocorrelations between the quadrant and its
neighbors become weaker with increasing lag.
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c) Create a Moran scatterplot for the number of woodpeckers. Copy the graph into your Word
document.
d) Interpret the Moran scatterplot for the number or woodpeckers. What can be said about the
spatial autocorrelation and the local structure in the data? Please write a short paragraph in
the Word document to answer this.
The positive slope of the scatterplot implies a positive spatial autocorrelation among the
neighboring cell with a mean of woodpecker count around 6-7.
Most of the quadrants fell at upper-right and lower-left corners implies that most of them are
surrounded by neighbors with a similar count of woodpeckers. The scatters are more
concentrated at the lower-left of the plot, which suggests that the most of the quadrants have
low count of woodpeckers.
Some of the quadrants fell into the upper-left and lower-right corners, which implies the
quadrants share a negative spatial autocorrelation with their neighbors.
There were 12 influential points identified (16 ,24 ,25 ,26 ,27 ,28 ,34 ,35 ,36 ,44 ,45 , and 46)
based on the matrix of influence measure. 26 is the most influential one since it is the furthest
away from the mean.
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Problem 2
c) Table 1: Results for Moran’s I and Geary’s c tests for the number of woodpeckers (per 1ha)
using point data (spatial weights = binary).
Neighbor
Observed Expected
Method
Statistic
Variance
p-value
selection
Value
Value
Euclidean
I
0.7601
-0.0101
0.0053
< 2.2e-16
Normality
k=4
I
0.7051
-0.0101
0.0046
< 2.2e-16
Euclidean
I
0.7601
-0.0101
0.0051
< 2.2e-16
Randomization
k=4
I
0.7051
-0.0101
0.0044
< 2.2e-16
Euclidean
I
0.7601
0
---0.000999
Monte Carlo
k=4
I
0.7051
0
---0.000999
Euclidean
c
0.3146
1.0000
0.0057
< 2.2e-16
Normality
k=4
c
0.2917
1.0000
0.0047
< 2.2e-16
Euclidean
c
0.3146
1.0000
0.0065
< 2.2e-16
Randomization
k=4
c
0.2917
1.0000
0.0051
< 2.2e-16
Monte Carlo
Euclidean
c
0.3146
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---0.000999
e) Table 2: Results for Moran’s I and Geary’s c tests for the number of woodpeckers (per 1ha) using
polygon data with Rook’s and Queen’s case neighborhood definitions (spatial weights=binary).
Method
Normality
Randomization
Monte Carlo
Normality
Randomization
Monte Carlo
Neighbor
selection
Rook’s
Queen’s
Rook’s
Queen’s
Rook’s
Queen’s
Rook’s
Queen’s
Rook’s
Queen’s
Rook’s
Queen’s
Statistic
I
I
I
I
I
I
c
c
c
c
c
c
Observed
Value
0.7601
0.7012
0.7601
0.7012
0.7601
0.7012
0.3146
0.4143
0.3146
0.4143
0.3146
0.4143
Expected
Value
-0.0101
-0.0101
-0.0101
-0.0101
0
0
1.0000
1.0000
1.0000
1.0000
1
1
Variance
p-value
0.0053
0.0027
0.0051
0.0026
------0.0057
0.0037
0.0065
0.0058
-------
< 2.2e-16
< 2.2e-16
< 2.2e-16
< 2.2e-16
0.000999
0.000999
< 2.2e-16
< 2.2e-16
< 2.2e-16
8.731e-15
0.000999
0.000999
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f) Are there differences (if so, how do values differ?) in the observed values and variances of the
statistics (Moran’s I, Geary’s c) depending on
i.
Method used to select neighbors (Euclidean vs. k = 4 neighbors vs. Rook’s vs. Queen’s
case)?
The observed value and variance of the statistics are slightly different if we adopt
different rules to select neighbors. The Queen’s case always gives a smaller observed
values and variance. Yet, all of them share a very small p-valve and imply that we are
confident to reject the null hypothesis.
ii.
Test used to determine significance (normality, randomization, Monte Carlo
simulation)?
The variance is slightly different under different assumption but the observed values are
nearly the same under different assumptions and in the simulation. Nonetheless, all of
them share a very small p-valve and imply that we are confident to reject the null
hypothesis.
iii.
Point vs. polygon data?
In Moran’s I test, the observed value and variance do not change much if we change the
data type.
In Geary’s c test, the observed value and variance are also similar after we change the
data type. Adopting a polygon data and using Queen’s case to define the neighbor to
run Geary’s c test under normality assumption would give a lower variance than the
others.
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g) Is spatial autocorrelation present in the data? Write out all four steps of the hypothesis test
using one of the above tests (Moran’s I or Geray’s c; just pick one option to define neighborhoods
and weights.
We are going to use point data and analyze with Moran’s I for this question.
1. Neighborhood is defined by the Euclidean distance at 100 m.
> nlistd W.dW moran.test(WP_Counts,W.dW,randomisation = TRUE, alternative
= “greater”)
Moran I test under randomization
data:
WP_Counts
weights: W.dW
Moran I statistic standard deviate = 9.9299, p-value < 2.2e-16
alternative hypothesis: greater
sample estimates:
Moran I statistic
Expectation
Variance
0.709917057
-0.010101010
0.005257677
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4. Monte Carlo simulation is used to simulate the “experiment”.
> moran.mc(WP_Counts,W.dW,nsim = 1000, alternative =
“greater”)
Monte-Carlo simulation of Moran I
data:
WP_Counts
weights: W.dW
number of simulations + 1: 1001
statistic = 0.70992, observed rank = 1001, p-value = 0.000999
alternative hypothesis: greater
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