MLR & Spatial Data Lab Report

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. 1 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. 2 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 1 ---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 3 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. 4 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 5 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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