GEOG 371 Lecture Notes - Lecture 12: Chi-Squared Test, Soil Type, Null Hypothesis
5 May 2018
School
Department
Course
Professor
Week 12
4/9 – Spatial Association
Spatial Association
• Relationship between two or more spatial variables
• Map comparison and overlay
• Is variable X associated with variable Y?
• How can we quantitatively analyze the relationships b/w maps?
Maps as Outcomes of Spatial Stochastic Processes
• How would the maps look if there was no association between them?
• Presence of X at a location is not associated with presence of Y
• Null hypothesis = presence of x at location not associated w/ y presence
Analyzing Spatial Association Between Two Variables
Variable 1
Variable
2
Nominal
Ordinal
Interval
Nominal
Chi Square
Mann-
Whitney, etc.
T-test
ANOVA
Ordinal
Rank
correlation
Rank
correlation
Interval
Correlation
Chi Square Test
• Two sample – analyze association between two nominal variables based on counts/frequencies
• Compare two maps, each containing nominal data
• Based on contingency tables
• Ex – x: soil type, y: dominant flower species
o X: climate type, y: terrain type
o X: industry type, y: voting outcome
• Ex – is crop type associated with soil type?
o Data: Counts of grid points with crop type i and soil type j
o Oij = # of grid points with crop type i and soil type j
o Null Hypothesis (Ho): Soil type is not related to crop type\
▪ No relationship or no difference
o Alternative Hypothesis (Ha): Soil type is related to crop type
find more resources at oneclass.com
find more resources at oneclass.com
Document Summary
Is the actual relationship: data in contingency table: observed frequencies (oij, oij = frequency of values in category i, j. Soil type a soil type b soil type c total. 100: what would the contingency table look like if crop type and soil type were statistically independent (ho), multiplication rule for independent events, expected frequency under the ho: Cr i n j: ri = row i total, cj = column j total, n = number of observations (table total, expected contingency table under ho: 100: p (corn a) = p (corn) x p(a) = 0. 7 x 0. 25 = 0. 125, p (corn) = 70/100 = 0. 7, p (soil a) = 25/100 = 0. 25, 0. 175 x n = 0. 175 x 100 = 17. 5. Cr i n j: ecorn (a) = (70x25)/100 = 17. 5, ecorn (b) = (70x45)/100 = 31. 5, ecorn (c) = (70x30)/100 = 21, chi square test statistic:
