Spatial Correspondence of Areal Distributions Quadrat and nearest-neighbor analysis deal with a single distribution of points

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• Proposed by Court (1970)

• Limits are –1 to +1 with a perfect negative correspondence given a value of –1
• Sampling distribution is roughly normal, so you can test for statistical significance

• Class Example

• Relationship between wheat yield and precipitation
• Two maps showing high and low yields and high and low precipitation

• In this example we are using a grid

• The finer the grid, the more precise the measurement
• Four possibilities exist

• Low rainfall, low yield
• Low rainfall, high yield
• High rainfall, low yield
• High rainfall, high yield

• Create a table of expected frequencies using probability statistics (% High rain * # of high yield cells)

• Row total * column total / table total

• So, the overall map comparison is really a function of:

• Total cells on the diagonal / total number of cells.
• (2 + 2 + 1) / (2 + 2 + 0 +0 + 2 + 1 + 0 + 1+1) = 5/9 = .55% agreement

• The total correspondence of our example is 55%. But, that only tells us part of the story. What if we were really interested in classification B? Where there changes in classification B? Even here, there are two different ways of interpreting that question:

• If I were interested in mapping all the areas of B, how well did I get them all? This is called the map Producer’s Accuracy. That is, how well did we produce a map of classification B.
• If I were to use the map to find B, how successful would I be? This is called the Map User’s Accuracy. That is, much confidence should a user of the map have for a given classification.
• To compute the map user’s accuracy, we would divide the total number correct within a row with the total number in the whole row. Staying with our example of classification B:

• We said that we had two cells where B was correct. However, we actually said that there were three cells that contained B (in other words, we incorrectly called a cell B, when it should have been C). Therefore, we have:
• 2 correct B values / 3 total values = .66 user’s accuracy.
• This means that if we were to use this map and look for the classification of B, we would be correct 66% of the time.
• To compute the map producer’s accuracy, we would divide the total number of correct within a column with the total number in the whole column. Staying with our example of classification B:

• We said that we had two cells where B was correct. However, we actually said that there were five cells that should have been B. Therefore, we have:
• 2 correct B values / 5 total values that should be B = .4 producer’s accuracy
• This means that the map produced only 40% of all the B’s that were out there.

• This also gives us some indication of the nature of the errors. For instance, it appears that we confused classification A with classification B (we said on two occasions that B was A). By understanding the nature of the errors, perhaps we can go back, look over our process and correct for that mistake.

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