GEOG 3020 Lecture 23-9 Spatial Autocorrelation

let's work out an example of calculating more enzyme annually for this smallmap pattern that we have in front of us in this case each of our polygons A through F have a an x-value indicated by this number over here these are our data values and these are our labels for our different zones in the map this l is the number of links each polygon has when we have a when we use a rook connectivity structure so neighborhood c has a value of 2.4 and we see that it's connected to one two three four neighboring regions the first thing that we're going to do is construct a binary rook matrix for this connectivity structure so let's look at neighborhood a in which zones are a neighbor too we see that a shares a boundary with b and c so we're going to put ones in these two locations to denote that neighborhood a is neighbors with b and c and will put zeros everywhere else on that row B is neighbors to a C and D C is neighbors to a B D and E D is neighbors with b c e and f these neighbors with C D and F and F is neighbors with D and E and again we have to put zeros everywhere else so now we have our weight matrix and in the formula for Moran's I we have to know the sum over all eyes and all J's of W IJ in other words we have to add up all the values in this matrix and when we do that we see that the sum of all of these ones is equal to 18 so that sum of sum of W appears say over here in the numerator in the denominator of this first fraction the next thing that we're going to have to do is calculate all of our deviations so in order to calculate the denominator we need to know the sum over X I minus X bar squared so for each X we are going to calculate X minus X bar we found that X bar is one point seven and we just start going to do the normal calculation of the deviance in the deviance squared so 2.6 minus one point seven is zero point nine and we square that to get point eight one so this denominator over here is just the sum of this column which we find to be ten point three to this n over here is just the number of observations we had we had one two three four five six so this was six the sum of sum of W was 18 in here we had ten point three two and now we need to calculate all of the cross product terms and multiply them by the weights in order to do that we can visualize the numerator as a matrix so let's visualize it as a matrix where we are keeping track of the eyes down the rows and the J's across the columns so what we have is the sum over all i and all J of W IJ times the two cross products and W IJ is that binary weight matrix so we know right away that wherever W IJ equals zero this entire cross product term for that specific IJ will equal zero so we're going to put zeros in all the same locations where we had zeros in our original W matrix so everywhere we have zeros here we're going to have zeros there where we have ones in the W matrix we're going to place the cross product term or we're going to put one times the cross product term so for example area and before I had these as a through F whoops c d e f a just so that you don't get confused it's very uncommon that we use letters instead of numbers to to refer to locations but I like the letters down here so a was neighbors with B so what we are going to have in this cell of the matrix is X a minus X bar which is now just d1 minus which is d1 and xB minus X bar which is d2 and the weight of one with two was equal to one so we don't actually write it down but we know we also have a one multiplying this thing so this D 1 D 2 is just X 1 minus X bar times X 2 minus X bar over here as another example we had locate the location E and D being neighbors so for that pair of neighbors we need to calculate its cross product term which is just going to be X 5 minus X bar so that's D 5 the deviation for point 5 and XJ minus X bar which is D for the deviation for location 4 or location D so we're going to go ahead and fill out this entire matrix and on this page we're just showing you the schematic of what all of these entries are going to contain and the next step is actually to fill in and replace these D ones these deviances with the actual values that we have from our table of deviances over here so this is D 1 this is d2 d3 d5 and d6 so over here where I call for d1 times d2 we're going to multiply 0.9 times minus 1.2 d1 times d2 and place it in this cell so that is equal to minus one point zero eight and we do that for each of the cross product terms that have nonzero weights so we're calculating the cross products for neighboring locations and now we have to sum them all up and when we sum them all up we find that the sum of the cross product terms is minus five point six six now what does that mean it means that more often than not when X I and XJ are neighbors X I and XJ don't have the same aren't similar values so more often than not when X I is above the mean XJ is below the mean and we end up with negative cross product terms and we if we add up all of the cross product terms over all we find that the sum is negative so right away we know that Marans I is also going to be negative or Anne's I is going to be negative so we'll have negative autocorrelation let's just calculate the rest of the statistic so we had n which was 6 we had n over the sum of sum of W which was 18 6 over 18 then we had the sum of the cross product terms minus 5 point 6 6 and then in the sum of squared DV INSAS which is 10 point 3 2 and that ends up within Marans I of minus 0.183 just some weak negative spatial autocorrelation if we wanted to know whether or not this moran's I was justit istic aliy different to 0 we would have to standardize it into a z-score using Z equals e minus the expected value of I over the square root of the variance of I now the expected value of I if the data were random we only have 6 observations Yi of I equals minus 1 over n minus 1 equals minus 1 v equals minus 0.2 so the expected value in such a small sample size is minus 0.2 we have an eye of minus 0.183 so even though we don't we aren't going to compute the full z test because we don't have an estimate of the variance in this case I'm pretty sure that this isn't going to be significantly different to this random case and in fact we would conclude in this case that despite I being less than 0 it's not significantly different to this random eye that we would expect to see and therefore we would fail to reject a null hypothesis of randomness in this case here we have the Moran scatter plot of X versus the neighborhood average of X and here we see that the best fitting line through the cloud of points is a negative sloping line so we do have a negative level of autocorrelation but we also see that the scattering of points around this line is really really loose we don't have a nice tight linear pattern which would give us a Moran's I close to minus 1 instead we have this really loose pattern which gives us somewhere ends I close to 0 0.18

Channel: Steven Farber

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Keywords: Spatial Autocorrelation, Moran's I

Id: vhuh1v3UPvA

Length: 11min 40sec (700 seconds)

Published: Tue Jun 11 2013