2. Decision Trees A “divisive” method (splits)
Start with “root node” – all in one group
Get splitting rules
Response often binary
Result is a “tree”
Example: Loan Defaults
Example: Framingham Heart Study
Example: Automobile Accidentsnts

3. Recursive Splitting Pr{default} =0.008 Pr{default} =0.012
X1=Debt To
Income
Ratio
Pr{default} =0.0001
Pr{default} =0.003
X2 = Age
No default
Default

4. Some Actual Data Framingham Heart Study
First Stage Coronary Heart Disease
P{CHD} = Function of:
Age - no drug yet! 
Cholesterol
Systolic BP
Import

5. Using Default Settings
Example of a “tree” 
Using Default Settings
Copyright © 2010, SAS Institute Inc. All rights reserved.

6. Example of a “tree”  Using Custom Settings
(Gini splits, N=4 leaves)

7. How to make splits? Contingency tables 95 5 55 45 75 25 100 100 150 50
Heart Disease
No Yes
Heart Disease
No Yes 95 5 55 45
100 75 25
100
Low BP
High
180 ?
240?
DEPENDENT (effect)
INDEPENDENT (no effect)

8. How to make splits? Contingency tables = 2(400/75)+ 2(400/25) = 42.67
Heart Disease
No Yes
Observed - Expected = 95 75 5 25 55 45
100
Expected
Low BP
High
180 ?
240?
2(400/75)+ 2(400/25) =
42.67
Compare to c2 tables –
Significant!
(Why do we say “Significant” ???)
DEPENDENT (effect)

9. Beyond reasonable doubt
H0:
H1:
H0: Innocence
H1: Guilt
Beyond reasonable doubt
P<0.05 95 75 5 25 55 45
H0: No association
H1: BP and heart disease
are associated P=
Framingham Conclusion: Sufficient evidence against the (null) hypothesis of no relationship.

10. Demo 1: Chi-Square for Titanic Survival (SAS)
Alive Dead
Crew
3rd Class
2nd Class
1st Class
“logworth” =
all possible splits c P-value -log10(p-value)
Crew vs. other x
Crew&3 versus 1& x
First versus other x
Copyright © 2010, SAS Institute Inc. All rights reserved.

11. Demo 2: Titanic (EM) Open Enterprise Miner File  New  Project
Specify workspace
Startup Code – specify data library
New Data Source  Titanic
Data has class (1,2,3,crew) & MWC
Bring in data, Decision Tree (defaults)
Copyright © 2010, SAS Institute Inc. All rights reserved.

12.  3 possible splits for M,W,C
10 possible splits for class
(is that fair?)

13. Demo 3: Framingham (EM) Open Enterprise Miner File  New  Project
Specify workspace
Startup Code – specify data library
New Data Source  Framingham
Firstchd = target, reject obs
Explore Firschd (pie) vs. Age (histogram)
Bring in data, Decision Tree (defaults)
Copyright © 2010, SAS Institute Inc. All rights reserved.

14. Other Sp lit Criteria Gini Diversity Index Shannon Entropy
(1) { A A A A B A B B C B}
Pick 2, Pr{different} = 1-Pr{AA}-Pr{BB}-Pr{CC}
= (sampling with replacement)
(2) { A A B C B A A B C C }
= 0.66  group (2) is more diverse (less pure)
Shannon Entropy
Larger  more diverse (less pure)
-Si pi log2(pi)
{0.5, 0.4, 0.1} 
{0.4, 0.2, 0.3}  (more diverse)

15. How to make splits? Which variable to use? Where to split?
Cholesterol > ____
Systolic BP > _____
Idea – Pick BP cutoff to minimize p-value for c2
Split point data-derived!
What does “significance” mean now?

16. Multiple testing a = Pr{ falsely reject hypothesis 2} a a a =
Pr{ falsely reject one or the other} < 2a
Desired: probability or less
Solution: Compare 2(p-value) to 0.05

17. Validation
Traditional stats – small dataset, need all observations to estimate parameters of interest.
Data mining – loads of data, can afford “holdout sample”
Variation: n-fold cross validation
Randomly divide data into n sets
Estimate on n-1, validate on 1
Repeat n times, using each set as holdout.
Titanic and Framingham examples did not use holdout.

18. Pruning Grow bushy tree on the “fit data”
Classify validation (holdout) data
Likely farthest out branches do not improve, possibly hurt fit on validation data
Prune non-helpful branches.
What is “helpful”? What is good discriminator criterion?

19. Goals
Split (or keep split) if diversity in parent “node” > summed diversities in child nodes
Prune to optimize
Estimates
Decisions
Ranking
in validation data

20. Assessment for:
Decisions Minimize incorrect decisions (model versus realized)
Estimates Error Mean Square (average error)
Ranking C (concordance) statistic = proportion concordant + ½ (proportion tied)
Obs number (2, 3)
Probability of ( from model)
Actual (0,1)
Concordant Pairs: (1,3) (1,4) (2,4)
Discordant Pairs: (3,5) (4,5)
Tied (2,3)
6 ways to get pair with 2 different responses
C = 3/6 + (1/2)(1/6) = 7/12=0.5833

21. Accounting for Costs Pardon me (sir, ma’am) can you spare some change?
Say “sir” to male +$2.00
Say “ma’am” to female +$5.00
Say “sir” to female -$1.00 (balm for slapped face)
Say “ma’am” to male -$10.00 (nose splint)

22. Including Probabilities
Leaf has Pr(M)=.7, Pr(F)= You say:
Sir Ma’am
Expected profit is
2(0.7)-1(0.3) = $1.10
if I say “sir”
= -$5.50 (a loss)
if I say “Ma’am”
Weight leaf profits by leaf size (# obsns.) and sum.
Prune (and split) to maximize profits.
True Gender M F
0.7 (2)
0.7 (-10)
0.3 (-1)
0.3 (5)
+$ $5.50

23. Regression Trees Continuous response Y
Predicted response Pi constant in regions i=1, …, 5
Predict 80
Predict 50 X2
Predict
130
Predict 20
Predict 100 X1

24. Regression Trees Predict Pi in cell i. Yij jth response in cell i.
Split to minimize Si Sj (Yij-Pi)2

25. Real data example: Traffic accidents in Portugal*
Y = injury induced “cost to society”
* Tree developed by Guilhermina Torrao, (used with permission)
NCSU Institute for Transportation Research & Education
Help - I ran
Into a “tree”
Help - I ran
Into a “tree”

26. Demo 4: Chemicals (EM) Hypothetical potency versus chemical content
(proportions of A, B, C, D)
New Data Source  Chemicals
{Sum, Dead} rejected, PotencyTarget, Partition node: training 75%, validation 25%
Connect and run tree node, look at results
Tree Node, Results.
View  Model Variable Importance
Copyright © 2010, SAS Institute Inc. All rights reserved.

27. Demo 4 (cont.) Only A and B amounts contribute to splits.
Variable Number of Importance Validation Ratio of Validation
Name Splitting Rules Importance to Training Importance B A
C NaN
D NaN
Only A and B amounts contribute to splits.
B splits reduce variation more.
A reduces it by about 2/3 as much as B.
Properties panel  Exported data
Explore3D graph
X=A Y=B Z=P_Potency
Copyright © 2010, SAS Institute Inc. All rights reserved.

28. B A A B A A

29. 3 5 1 6 2 4

30. Logistic Regression Logistic – another classifier
Older – “tried & true” method
Predict probability of response from input variables (“Features”)
Linear regression gives infinite range of predictions
0 < probability < 1 so not linear regression.

31. Logistic Regression
Three Logistic Curves
ea+bX
(1+ea+bX) p X

32. Example: Seat Fabric Ignition
Flame exposure time = X
Y=1  ignited, Y=0 did not ignite
Y=0, X= 3, 5, , ,
Y=1, X = , , 15, , 25, 30
Q=(1-p1)(1-p2)(1-p3)(1-p4)p5p6(1-p7)p8p9(1-p10)p11p12p13
p’s all different pi = f(a+bXi) = ea+bXi/(1+ea+bXi)
Find a,b to maximize Q(a,b)

33. Given temperature X, compute L(x)=a+bX then p = eL/(1+eL)
Logistic idea:
Given temperature X, compute L(x)=a+bX then p = eL/(1+eL)
p(i) = ea+bXi/(1+ea+bXi)
Write p(i) if response, p(i) if not
Multiply all n of these together, find a,b to maximize this “likelihood”
Estimated L = ___+ ____X
maximize Q(a,b)
-2.6
-2.6 a b
0.23
0.23

34. Demo 5: Space Shuttle Challenger

35. Example: Shuttle Missions
O-rings failed in Challenger disaster
Prior flights “erosion” and “blowby” in O-rings (6 per mission)
Feature: Temperature at liftoff
Target: (1) - erosion or blowby vs. no problem (0)
L= – (temperature)
p = eL/(1+eL)

37. Neural Networks Very flexible functions “Hidden Layers”
(0,1) H1
Output = Pr{1}
inputs X2
Very flexible functions
“Hidden Layers”
“Multilayer Perceptron” X3 H2
Logistic function of
Logistic functions **
Of data X4
** (note: Hyperbolic tangent functions are just reparameterized logistic functions)

38. Example:
Y = a + b1 H1 + b2 H2 + b3 H3
Y = H H H3
“bias”
-1 to 1
“weights” H1 b1 H2 b2 X Y b3 H3
Arrows on right represent linear
combinations of “basis
functions,” e.g. hyperbolic tangents
(reparameterized logistic curves)

39. A Complex Neural Network Surface
(-10)
-0.4 3
0.8 X1 X1
(-1)
(-13) -1 P P
0.25 X2 X2
-0.9
2.5
0.01
(20)
(“biases”)

40. Demo 6: EM part Model comparison node picks Neural Net
From results: view->score code
SAS program uses score code (copy & paste)

41. Note: No validation data used – surfaces are unnecessarily complex
Demo 6 (EM & SAS) Framingham Neural Network
Note: No validation data used – surfaces are unnecessarily complex
“Slice” of 4-D surface at age=25
Code for graphs in NNFramingham.sas
Age 25
Nonsmoker
Age 25
Smoker
.057
.001
.057
.001
Cholest
Cholest
Sbp
Sbp

42. “Slice” at age 55 – are these patterns real?
Framingham Neural Network Surfaces
“Slice” at age 55 – are these patterns real?
Age 55
Nonsmoker
Age 55
Smoker
.262
.009
.261
.004
Cholest
Cholest
Sbp
Sbp

43. Handling Neural Net Complexity
Use validation data, stop iterating when fit gets worse on validation data. Problems for Framingham (50-50 split) Fit gets worse at step 1 no effect of inputs Algorithm fails to converge (likely reason: too many parameters) (2) Use regression to omit predictor variables (“features”) and their parameters. This gives previous graphs. (3) Do both (1) and (2).
Run Demo 6A in SAS to see the results

44. * Cumulative Lift Chart - Go from leaf of most to least predicted
3.3 1
* Cumulative Lift Chart
- Go from leaf of most to least predicted
response.
- Lift is
proportion responding in first p%
overall population response rate
Predicted response
high   low

45. Receiver Operating Characteristic Curve
Cut point 1
Logits of 1s Logits of 0s
red black
Logits of 1s Logits of 0s
red black
Select most likely p1% according to model.
Call these 1, the rest 0. (call almost everything 0)
Y=proportion of all 1’s correctly identified.(Y near 0)
X=proportion of all 0’s incorrectly called 1’s (X near 0)
(or any model predictor
from most to least likely
to respond)

46. Receiver Operating Characteristic Curve
Cut point 2
Logits of 1s Logits of 0s
red black
Logits of 1s Logits of 0s
red black
Select most likely p2% according to model.
Call these 1, the rest 0.
Y=proportion of all 1’s correctly identified.
X=proportion of all 0’s incorrectly called 1’s

47. Receiver Operating Characteristic Curve
Cut point 3
Logits of 1s Logits of 0s
red black
Logits of 1s Logits of 0s
red black
Select most likely p3% according to model.
Call these 1, the rest 0.
Y=proportion of all 1’s correctly identified.
X=proportion of all 0’s incorrectly called 1’s

48. Receiver Operating Characteristic Curve
Cut point 3.5
Logits of 1s Logits of 0s
red black
Select most likely 50% according to model.
Call these 1, the rest 0.
Y=proportion of all 1’s correctly identified.
X=proportion of all 0’s incorrectly called 1’s

49. Receiver Operating Characteristic Curve
Cut point 4
Logits of 1s Logits of 0s
red black
Select most likely p5% according to model.
Call these 1, the rest 0.
Y=proportion of all 1’s correctly identified.
X=proportion of all 0’s incorrectly called 1’s

50. Receiver Operating Characteristic Curve
Cut point 5
Logits of 1s Logits of 0s
red black
Logits of 1s Logits of 0s
red black
Select most likely p6% according to model.
Call these 1, the rest 0.
Y=proportion of all 1’s correctly identified.
X=proportion of all 0’s incorrectly called 1’s

51. Receiver Operating Characteristic Curve
Cut point 6
Logits of 1s Logits of 0s
red black
Select most likely p7% according to model.
Call these 1, the rest 0.(call almost everything 1)
Y=proportion of all 1’s correctly identified. (Y near 1)
X=proportion of all 0’s incorrectly called 1’s (X near 1)

52. Why is area under the curve the same as proportion of concordant pairs + (number of ties)/2 ??
Tree
Leaf 1: 20 0’s, 60 1’s Leaf 2: 20 0’s, 20 1’s Leaf 3: 60 0’s 20 1’s
Cut pt
Left of cut, call it  |  and right of cut, call it 0.
X = number of incorrect 0 decisions , Y= number of correct 1 decisions
X = number of 0’s to the left of cut point, Y = number of 1’s to the left of cut point.
(X,Y) on ROC curve
Cut 0 X=0 (all points called 0 so none wrong) Y=0 (no 1 decisions) (0, 0)
Cut 1 (20)(60) = 1200 ties, 60(20+60)=1200 concordant (20, 60)
(20,60)
60 1’s
 Areas are
number of tied pairs in leaf 1
plus
number of concordant pairs from 0’s in leaves 2 and 3
ROC curve (line) splits ties in two.
(0,0)
Ties
Concordant

53. Why is area under the curve the same as number of concordant pairs + (number of ties)/2??
Tree
Leaf 1: 20 0’s, 60 1’s Leaf 2: 20 0’s, 20 1’s Leaf 3: 60 0’s 20 1’s
Cut pt
Left of cut, call it  |  and right of cut, call it 0.
Cut point 2: 20 more 1’s, 60 0’s to the right, 20x20=400 ties in leaf point on ROC curve: (40.80)
Cut point 3: call everything a 1. All 100 1’s are correctly called, all 0’s incorrectly called.
point on ROC curve:(100,100)
20 1’s
ROC curve is in terms of proportions so join points are
(0,0), (.2,.6), (.4,.8), and (1,1).
For Logistic Regression or Neural Nets, each point is a potential
cut point for the ROC curve.
20 1’s
60 1’s
(1,1) ’s
----(0.4,0.8)
-(0.2,0.6)
Ties
Concordant
(0,0)

54. Misclassification Rate Avg. Squared Error Area Under ROC
 Framingham ROC for 4 models
and baseline 45 degree line
Neural net highlighted, area 0.780
Sensitivity: Pr{ calling a 1 a 1 }=Y coordinate Specificity; Pr{ calling a 0 a 0 } = 1-X.
Selected Model
Model Node
Model Description
Misclassification Rate
Avg. Squared Error
Area Under ROC Y
Neural
Neural Network
0.780
Tree
Decision Tree
0.720
Tree2
Gini Tree N=4
0.675
Reg
Regression
0.734

55. A Combined Example Cell Phone Texting Locations Black circle:
Phone moved > 50 feet in first two
minutes of texting.
Green dot:
Phone moved < 50 feet. .

56.  Three Models Training Data  Lift Charts Validation Data Lift Charts
Tree Neural Net Logistic Regression
 Three Models
Training Data
 Lift Charts
Validation Data
Lift Charts
Resulting
Surfaces

57. Demo 7 (EM): Three Breast Cancer Models
Features are computed from a digitized image of a fine needle
aspirate (FNA) of a breast mass. They describe characteristics
of the cell nuclei present in the image.
1. Sample code number id number
2. Clump Thickness
3. Uniformity of Cell Size
4. Uniformity of Cell Shape
5. Marginal Adhesion
6. Single Epithelial Cell Size
7. Bare Nuclei
8. Bland Chromatin
9. Normal Nucleoli
10. Mitoses
11. Class: (2 for benign, 4 for malignant)

58. Use Decision Tree to Pick Variables

59. Decision Tree Needs Only BARE & SIZE

60. Can save scoring code from EM
Data A; do bare=0 to 10; do size=0 to 10;
(score code from EM)
Output; end; end;
Model Comparison Node Results
Cancer Screening Results: Model Comparison Node
Selected Model
Model Node
Model Description
Train Misclassification Rate
Train Avg. Squared Error
Train Area Under ROC
Validation Misclassification Rate
Validation Avg. Squared Error
Validation Area Under ROC Y
Tree
Decision Tree
0.975
0.933
Neural
Neural Network
0.994
0.985
Reg
Regression
0.992
0.987

61. Demo 7 (SAS): Three Breast Cancer Models
Copyright © 2010, SAS Institute Inc. All rights reserved.

62. A B Association Analysis is just elementary probability with new names
0.3
Pr{B}=0.3 A
A: Purchase Milk
B: Purchase Cereal
Pr{A and B}
= 0.2 B
0.1
Pr{A} =0.5
0.4
= 1.0

63. Cereal=> Milk
Rule B=> A “people who buy B will buy A”
Support:
Support= Pr{A and B} = 0.2
Independence means that Pr{A|B} = Pr{A} = 0.5
Pr{A} = 0.5 = Expected confidence if there is no
relation to B..
Confidence:
Confidence = Pr{A|B}=Pr{A and B}/Pr{B}=2/3
??- Is the confidence in B=>A the same as the
confidence in A=>B?? (yes, no)
Lift:
Lift = confidence / E{confidence} = (2/3) / (1/2) = 1.33
Gain = 33% A B
0.2
0.1
0.3
0.4 B
Marketing A to the 30%
of people who buy B will
result in 33% better sales
than marketing to a random
30% of the people.

64. Demo 8 (EM) : Grocery cart items (hypothetical)
Sort Criterion: Confidence
Maximum Items: 2
Minimum (single item) Support: 20%
item cart
bread 1
milk 1
soap 2
meat 3
bread 3
bread 4
cereal 4
milk 4
soup 5
bread 5
cereal 5
milk 5
Association Report
Expected
Confidence Confidence Support Transaction
Relations (%) (%) (%) Lift Count Rule
cereal ==> milk
milk ==> cereal
meat ==> milk
bread ==> milk
soup ==> milk
milk ==> bread
bread ==> cereal
cereal ==> bread
soup ==> cereal
cereal ==> soup
milk ==> meat
milk ==> soup

65. Link Graph 
Sort Criterion: Confidence
Maximum Items: 3
e.g. milk & cerealbread
Minimum Support: 5%
Slider bar at 62%

66. Unsupervised Learning
We have the “features” (predictors)
We do NOT have the response even on a training data set (UNsupervised)
Another name for clustering EM
Large number of clusters with k-means (k clusters)
Ward’s method to combine (less clusters , say r<k)
One more k means for final r-cluster solution.

67. Demo 9 (EM) : Cluster the breast cancer data (disregard target)
Plot clusters and actual target values:

68. Segment Profiler

69. Text Mining Hypothetical collection of news releases (“corpus”) :
release 1: Did the NCAA investigate the basketball scores and
vote for sanctions?
release 2: Republicans voted for and Democrats voted against
it for the win.
(etc.)
Compute word counts:
NCAA basketball score vote Republican Democrat win
Release
Release

70. Text Mining Mini-Example: Word counts in 16 e-mails
 words 
R B T
P e a o
d E r p s D u
o l e u k e r S S
c e s b e m V n S c c
u c i l t o o a p o o
m t d i b c t N L m e W r r
e i e c a r e C i e e i e e
n o n a l a r A a n c n _ _
t n t n l t s A r t h s V N
#SASGF13
Copyright © 2013, SAS Institute Inc. All rights reserved.

71. Variable Prin1 Basketball -.320074 NCAA -.314093 Tournament -.277484
Eigenvalues of the Correlation Matrix
Eigenvalue Difference Proportion Cumulative
(more)
Variable Prin1
Basketball
NCAA
Tournament
Score_V
Score_N
Wins
Speech
Voters
Liar
Election
Republican
President
Democrat
55% of the variation in
these 13-dimensional
vectors occurs in one
dimension.
Prin 2
Prin 1

72. Variable Prin1 Basketball -.320074 NCAA -.314093 Tournament -.277484
Eigenvalues of the Correlation Matrix
Eigenvalue Difference Proportion Cumulative
(more)
Variable Prin1
Basketball
NCAA
Tournament
Score_V
Score_N
Wins
Speech
Voters
Liar
Election
Republican
President
Democrat
55% of the variation in
these 13-dimensional
vectors occurs in one
dimension.
Prin 2
Prin 1

73. Sports
Cluster
Politics
Cluster

74. R B T
P e a o
d E r p s D u
o C l e u k e r S S
c L e s b e m V n S c c
u U P c i l t o o a p o o
m S r t d i b c t N L m e W r r
e T i i e c a r e C i e e i e e
n E n o n a l a r A a n c n _ _
t R n t n l t s A r t h s V N
(biggest gap) P r i n 1
Sports
Documents
Politics
Documents

75. PROC CLUSTER (single linkage) agrees !

76. Demo 10: (EM) Memory Based Reasoning
(optional)
Usual name: Nearest Neighbor Analysis
Probe Point
9 blue
7red
Classify as
BLUE
Estimate Pr{Blue}
as 9/16

77. Note:
Enterprise Miner -> Exported Data -> Explore
Chooses colors 

78. Score Data (grid)
scored
by Enterprise
Miner Nearest
Neighbor
method
Original Data
Original Data
scored
by Enterprise
Miner Nearest
Neighbor
method

79. Fisher’s Linear Discriminant Analysis
- an older method of classification
(optional)
Assumes multivariate normal distribution
of inputs (features)
Computes a “posterior probability” for each
of several classes, given the observations.
Based on statistical theory

80. Example: One input, (financial stress index) Three Populations:
(pay all credit
card debt,
pay part,
default).
Normal distributions with same variance (9). Classify a credit card holder by
financial stress index: pay all (stress<20), pay part (20<stress<27.5),
default(stress>27.5) Data are hypothetical. Means are 15, 25, 30.

81. Example: Differing variances. Not just closest mean now.
Normal distributions with different variances. Classify a credit card holder by financial
stress index: pay all (stress<19.48), pay part (19.48<stress<26.40),
default(26.40<stress<36.93), pay part (stress>36.93)
Data are hypothetical. Means are 15, 25, 30. Standard Deviations 3,6,3.

82. Example: 20% defaulters, 40% pay part 40% pay all
Normal distributions with “priors.” Classify a credit card holder by financial stress index:
pay all (stress<19.48), pay part (19.48<stress<28.33), default(28.33<stress<35.00),
pay part (stress>35.00) Data are hypothetical.
Means are 15, 25, 30. Standard Deviations 3,6, Population size ratios 2:2:1

83. m = 15, , or 30, s =3 here.
The part that changes from one population to another is Fisher’s linear
discriminant function. Here it is (mx-m2/2)/s2, that is, (mx-m2/2)/9 . The
bigger this is, the more likely it is that X came from the population with
mean m.
The three probability densities are in the same proportions as their
values of exp((mx-m2/2)/9).

84. Example: Stress index X=21. Classify as someone that pays only
part of credit card bill because 21
is closer to 25 than to 15 or 30.
Probabilities 0.24, 0.74, 0.02
give more detail.
= 15, , or 30, s =here.
59.11* * *108
probability

85. Fisher for credit card data was (mx-m2/2)/9 when variance was 9
in every group.
(mx-m2/2)/9 = 1.67 X for mean 15
(mx-m2/2)/9 = 1.67 X for mean 25
(mx-m2/2)/9 = 1.67 X for mean 30
For 2 inputs it again has a linear form ___+___X1+___X2 where the
coefficients come from the bivariate mean vector and the 2x2
matrix is assumed constant across populations.
When variances differ, formulas become more complicated,
discriminant becomes quadratic, and boundaries are not
linear.

86. Example:
Two inputs,
three
populations
Multivariate
normals, same
2x2 covariance
matrix.
Viewed from
above, boundaries
appear to be linear.

87. Defaults
Pays
part
only
debt
Pays in
full
Income

88. Generated sample: 1000 per group
Debt
index
Income index
Copyright © 2010, SAS Institute Inc. All rights reserved.

89. proc discrim data=a; class group; var X1 X2; run; _ -1 _ -1 _
_ _ _
Constant = -.5 X' COV X Coefficient Vector = COV X
j j j
Linear Discriminant Function for group
Variable
Constant X X

90. Number of Observations and Percent Classified into group
From group Total
Total
Priors
Error Count Estimates for group
Total
Rate
Priors

91. Generated sample: 1000 per group
Sample Discriminants Differ Somewhat from Theoretical
Defaults
Pays
part
only
Debt
index
Pays in
full
Income index
Copyright © 2010, SAS Institute Inc. All rights reserved.