The Education Production Function Empirical evidence indicates that: increasing school spending has a modest impact on achievement. Urban schools spend more per student than suburban schools yet the achievement gap persists Questions: What is the nature of the relationship between school spending and achievement? What variables determine achievement? Do scores capture educational output?
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12 26 11 38 09 29 13 24 33 15 52 35 30 47 32 41 10 48 02 F F M M F F M M F F F M F M F F M M M Suburban Urban Urban Suburban Rural Suburban Rural Urban Urban Rural Suburban Suburban Rural Suburban Rural Suburban Rural Suburban Rural Other Republican Democrat Other Republican Democrat Independent Democrat Democrat Other Republican Other Independent Republican Republican Other Other Republican Democrat 2 2 4 3 6 4 9 3 1 2 0 1 3 6 2 3 3 2 8 34 24 62 44 44 38 47 44 45 56 32 54 41 50 59 44 62 53 59 34,000 30,000 78,000 68,000 29,000 40,000 39,000 60,000 49,000 39,000 33,000 65.000 25,000 61,000 41,000 44,000 45,000 64,000 39,000 5 2 1 7 9 9 6 2 3 8 5 3 7 5 8 3 10 1 8 8 11 18 17 8 9 8 27 9 6 11 14 5 15 8 10 10 13 9 10 15 11 12 27 25 20 4 12 27 15 14 25 15 27 18 23 15 24
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5. Mean projections and mean student scores are calculated. Student Projection1 Student Score 1 Student Projection 2 Student Score 2 Student Projection 3 Student Score 3 Student Projection 4 Student Score 4 Student Projection 5 Your School Student Score 5 Student Projection 6 Student Score 6 Student Projection 7 Student Score 7 Student Projection 8 Student Score 8 Student Projection 9 Student Score 9 Student Projection 10 Student Score 10 Student Projection 11 Student Score 11 Student Projection 12 Student Score 12 Student Projection 13 Student Score 13 Student Projection 14 Student Score 14 Student Projection 15 Student Score 15 Student Projection 16 Student Score 16 Student Projection 17 Student Score 17 Student Projection 18 Student Score 18 Student Projection 19 Student Score 19 Student Projection 20 Student Score 20 Mean Projected Score Mean Student Score Copyright © 2003. Battelle for Kids
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VOMmean w F=(DPP-MN)/4 Concrete4150(C, W, FA, Ag) 0 1 1 1 5 1 6 1 7 1 8 4 med=14 9 1 10 1 11 2 12 1 13 5 14 1 15 3 med=18 16 3 17 4 18 1 19 3 20 9 21 4 22 3 23 7 24 2 med=40 25 4 26 8 27 7 28 7 med=56 29 10 30 3 31 1 32 3 33 6 med=61 34 4 35 5 37 2 38 2 40 1 42 3 43 1 44 1 45 1 46 4 ______ CLUS 4 gap=7 49 1 56 1 [52,74) 0L 7M 0H CLUS_3 58 1 61 1 65 1 66 1 69 1 ______ gap=6 71 1 77 1 [74,90) 0L 4M 0H CLUS_2 80 1 83 1 ________ gap=14 86 1[0.90) 43L 46 M 55H 100 1 [90,113) 0L 6M 0H CLUS_1 103 1 105 1 108 2 112 1 _____________At this level, FinalClus1={17M} 0 errors C1 C2 C3 C4 med=10 med=9 med=17 med=21 med=23 med=34 med=33 med=57 med=62 med=71 med=71 med=86 CLUS 4 (F=(DPP-MN)/2, Fgap2 _______ 0L 0M 3H CLUS 4.4.1 gap=7 0 3 =0 0L 0M 4H CLUS 4.4.2 gap=2 7 4 =7 9 1 [8,14] 1L 5M 22H CLUS 4.4.3 1L+5M err H 10 12 11 8 gap=3 12 7 ______ 0L 0M 4H CLUS 4.3.1 gap=3 15 4 =15 18 10 0L 0M 10H CLUS 4.3.2 gap=3 21 3 =18 22 7 ______ 23 2 [20,24) 0L 10M 2H CLUS 4.7.2 gap=2 25 2 [24,30) 10L 0M 0H CLUS_4.7.1 26 3 27 1 28 2 gap=2 29 1 31 3 CLUS 4.2.1 gap=2 32 1 [30,33] 0L 4M 0H Avg=32.3 34 2 0L 2M 0H CLUS 4.2.2 gap=6 40 4 =34 ______ 0L 4M 0H CLUS_4.2.3 gap=7 47 3 =40 52 1 0L 3M 0H CLUS_4.2.4 gap=5 53 3 =47 54 3 55 4 56 2 57 3 ______ gap=2 58 1 [50,59) 12L 1M 4H CLUS 4.8.1 L60 2 8L 0M 0H CLUS_4.8.2 61 2 [59,63) gap=2 62 4 ______ =64 2L 0M 2H CLUS 4.6.1 gap=3 64 4 [66,70) 10L 0M 0H CLUS 4.6.2 67 2 gap=3 68 1 71 7 ______ gap=7 72 3 [70,79) 10L 0M 0H CLUS_4.5 79 5 5L 0M 0H CLUS_4.1.1 gap=6 85 1 =79 87 2 [74,90) 2L 0M 1H CLUS_4.1 1 Merr in L Median=0 Avg=0 Median=7 Avg=7 Median=11 Avg=10.7 Median=15 Avg=15 Median=18 Avg=18 Median=22 Avg=22 2H errs in L Median=26 Avg=26 Median=31 Median=34 Avg=34 Median=40 Avg=40 Median=47 Avt=47 Accuracy=90% Median=55 Avg=55 1M+4H errs in Median=61.5 Avg=61.3 Median=64 Avg=64 2 H errs in L Median=67 Avg=67.3 Median=71 Avg=71.7 Median=79 Avg=79 Median=87 Avg=86.3 Suppose we know (or want) 3 clusters, Low, Medium and High Strength. Then we find Suppose we know that we want 3 strength clusters, Low, Medium and High. We can use an antichain that gives us exactly 3 subclusters two ways, one show in brown and the other in purple Which would we choose? The brown seems to give slightly more uniform subcluster sizes. Brown error count: Low (bottom) 11, Medium (middle) 0, High (top) 26, so 96/133=72% accurate. The Purple error count: Low 2, Medium 22, High 35, so 74/133=56% accurate. What about agglomerating using single link agglomeration (minimum pairwise distance? Agglomerate (build dendogram) by iteratively gluing together clusters with min Median separation. Should I have normalize the rounds? Should I have used the same Fdivisor and made sure the range of values was the same in 2nd round as it was in the 1st round (on CLUS 4)? Can I normalize after the fact, I by multiplying 1st round values by 100/88=1.76? Agglomerate the 1st round clusters and then independently agglomerate 2nd round clusters? CONCRETE
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Hypotheses Result Description 1. x 0 L An increase in population will increase urban extent and urban expansion. 2. x 0 y An increase in household income will increase urban extent and urban expansion. 3. x 0 t An increase in transportation costs will reduce urban extent and limit urban expansion. 4. x 0 rA 5. x 0 H l 6. x 0  7. x 0 f l 8. x 0 w An increase in the opportunity cost of non-urban land will reduce urban extent and limit urban expansion. An increase in the marginal productivity of land in housing production will cause urban expansion. An increase in the share of land available for housing development will increase urban extent and urban expansion. An increase in marginal productivity of land in production of the export good will increase urban extent and urban expansion. An increase in the world price of the export good will increase urban extent and urban expansion.
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"Gap Hill Climbing": mathematical analysis 1. To increase gap size, we hill climb the standard deviation of the functional, F (hoping that a "rotation" of d toward a higher StDev would increase the likelihood that gaps would be larger since more dispersion allows for more and/or larger gaps. This is very heuristic but it works. 2. We are more interested in growing the largest gap(s) of interest ( or largest thinning). To do this we could do: F-slices are hyperplanes (assuming F=dotd) so it would makes sense to try to "re-orient" d so that the gap grows. Instead of taking the "improved" p and q to be the means of the entire n-dimensional half-spaces which is cut by the gap (or thinning), take as p and q to be the means of the F-slice (n-1)-dimensional hyperplanes defining the gap or thinning. This is easy since our method produces the pTree mask of each F-slice ordered by increasing F-value (in fact it is the sequence of F-values and the sequence of counts of points that give us those value that we use to find large gaps in the first place.). The d2-gap is much larger than the d1=gap. It is still not the optimal gap though. Would it be better to use a weighted mean (weighted by the distance from the gap - that is weighted by the d-barrel radius (from the center of the gap) on which each point lies?) In this example it seems to make for a larger gap, but what weightings should be used? (e.g., 1/radius2) (zero weighting after the first gap is identical to the previous). Also we really want to identify the Support vector pair of the gap (the pair, one from one side and the other from the other side which are closest together) as p and q (in this case, 9 and a but we were just lucky to draw our vector through them.) We could check the d-barrel radius of just these gap slice pairs and select the closest pair as p and q??? 0 1 2 3 4 5 6 7 8 9 a b c d e f 1 0 2 3 4 5 6 7 8 =p 9 d 2-gap d2 d 1-g ap j d k c e m n r f s o g p h i d1 l q f e d c b a 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 a b c d e f 1 2 3 4 5p 6 7 8 9d 1-g ap d 2-gap a q=b d2 f e d c b a 9 8 7 6 5 4 3 2 1 0 a b d d1 j k qc e q f
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"Gap Hill Climbing": mathematical analysis One way to increase the size of the functional gaps is to hill climb the standard deviation of the functional, F (hoping that a "rotation" of d toward a higher STDev would increase the likelihood that gaps would be larger ( more dispersion allows for more and/or larger gaps). This is very general. We are more interested in growing the one particular gap of interest (largest gap or largest thinning). To do this we can do as follows: F-slices are hyperplanes (assuming F=dotd) so it would makes sense to try to "re-orient" d so that the gap grows. Instead of taking the "improved" p and q to be the means of the entire n-dimensional half-spaces which is cut by the gap (or thinning), take as p and q to be the means of the F-slice (n-1)-dimensional hyperplanes defining the gap or thinning. This is easy since our method produces the pTree mask of each F-slice ordered by increasing F-value (in fact it is the sequence of F-values and the sequence of counts of points that give us those value that we use to find large gaps in the first place.). The d2-gap is much larger than the d1=gap. It is still not the optimal gap though. Would it be better to use a weighted mean (weighted by the distance from the gap - that is weighted by the d-barrel radius (from the center of the gap) on which each point lies?) In this example it seems to make for a larger gap, but what weightings should be used? (e.g., 1/radius2) (zero weighting after the first gap is identical to the previous). Also we really want to identify the Support vector pair of the gap (the pair, one from one side and the other from the other side which are closest together) as p and q (in this case, 9 and a but we were just lucky to draw our vector through them.) We could check the d-barrel radius of just these gap slice pairs and select the closest pair as p and q??? 0 1 2 3 4 5 6 7 8 9 a b c d e f 1 0 2 3 4 5 6 7 8 =p 9 d 2-gap d2 d 1-g ap j d k c e m n r f s o g p h i d1 l q f e d c b a 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 a b c d e f 1 2 3 4 5p 6 7 8 9d 1-g ap d 2-gap a q=b d2 f e d c b a 9 8 7 6 5 4 3 2 1 0 a b d d1 j k qc e q f
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"Gap Hill Climbing": mathematical analysis One way to increase the size of the functional gaps is to hill climb the standard deviation of the functional, F (hoping that a "rotation" of d toward a higher STDev would increase the likelihood that gaps would be larger ( more dispersion allows for more and/or larger gaps). We can also try to grow one particular gap or thinning using support pairs as follows: F-slices are hyperplanes (assuming F=dotd) so it would makes sense to try to "re-orient" d so that the gap grows. Instead of taking the "improved" p and q to be the means of the entire n-dimensional half-spaces which is cut by the gap (or thinning), take as p and q to be the means of the F-slice (n-1)-dimensional hyperplanes defining the gap or thinning. This is easy since our method produces the pTree mask of each F-slice ordered by increasing F-value (in fact it is the sequence of F-values and the sequence of counts of points that give us those value that we use to find large gaps in the first place.). The d2-gap is much larger than the d1=gap. It is still not the optimal gap though. Would it be better to use a weighted mean (weighted by the distance from the gap - that is weighted by the d-barrel radius (from the center of the gap) on which each point lies?) In this example it seems to make for a larger gap, but what weightings should be used? (e.g., 1/radius2) (zero weighting after the first gap is identical to the previous). Also we really want to identify the Support vector pair of the gap (the pair, one from one side and the other from the other side which are closest together) as p and q (in this case, 9 and a but we were just lucky to draw our vector through them.) We could check the d-barrel radius of just these gap slice pairs and select the closest pair as p and q??? 0 1 2 3 4 5 6 7 8 9 a b c d e f 1 0 2 3 4 5 6 7 8 =p 9 d 2-gap d2 d 1-g ap j d k c e m n r f s o g p h i d1 l q f e d c b a 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 a b c d e f 1 2 3 4 5p 6 7 8 9d 1-g ap d 2-gap a q=b d2 f e d c b a 9 8 7 6 5 4 3 2 1 0 a b d d1 j k qc e q f
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