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Modeling Ordinal Associations

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Opinions on Premarital Sex and Teenage Birth Control. S mij. i. Teenage Birth Control pj (column) ... Premarital Sex. pi (row) 9. Pearson Residuals = -0.8 ... – PowerPoint PPT presentation

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Title: Modeling Ordinal Associations


1
Modeling Ordinal Associations
Section 9.4
Roanna Gee
2
1991 General Social Survey
National Opinion Survey
  • Opinions were asked about a man and a woman
    having sexual
  • relations before marriage.
  • Always Wrong
  • Almost Always
  • Wrong Only Sometimes
  • Not Wrong At All
  • Opinions were also asked whether methods of birth
    control should be available to teenagers between
    the ages of 14 and 16.
  • Strongly Disagree
  • Disagree
  • Agree
  • Strongly Agree

3
Opinions on Premarital Sex and Teenage Birth
Control
4
Independence Model
log mij ? ?i ?j
X
Y
mij Expected count ? Mean log cell
count ?i Adjustment for Row i ?j Adjustment
for Column j
X
Y
Degrees of Freedom n (r 1)(c 1)
5
Sample Calculation of mij
log m23 log (row total) log(column total)
log(table total) log 93 log 324 log
926 3.48 so m23 exp(3.48) 32.5 or
P(A ? B) n P(A) P(B) n (93/926)(324/926)(9
26)
m23 can be calculated as (93)(324)/(926) 32.5
log means natural logarithm
6
Calculate m and li
X
m is the mean of the logs of the expected all the
cell counts. m (log 42.4 log 51.2 . . .
log 111.4)/16 3.8836
X
?i is the adjustment to m for row iits mean
less m. l2 (log 16.0 log 19.3 log 32.5
log 25.2)/4 3.8836 -0.7734 ?j is the
adjustment to m for column j. l3 0.3692
X
Y
Y
7
Degrees of Freedom
There are 4 rows and 4 columns giving us a total
of 16 cells and therefore 16 degrees of freedom.
For each parameter we add to the model, we lose
one degree of freedom. We lose one degree for
m. We lose 3 degrees for the li s. (Since S li
0, l4 l1 l2 l3.) We also lose 3 degrees
for the lj s.
X
X
X
X
X
X
Y
n 16 1 3 3 9 n rc 1 (r 1) (c
1) (r 1)(c 1)
8
Data and Independence Model
9
Pearson Residuals
-0.8
A standardized Pearson residual that exceeds 2 or
3 in absolute value indicates a lack of fit.
10
Data and Pearson Residuals
11
SAS Code Independence Model
data sex input premar birth u v count _at__at_
linlin uv datalines 1 4 1 4 38 1 3 1 3 60 1
2 1 2 68 1 1 1 1 81 2 4 2 4 14 2 3 2 3 29 2 2 2 2
26 2 1 2 1 24 3 4 3 4 42 3 3 3 3 74 3 2 3 2 41 3 1
3 1 18 4 4 4 4 157 4 3 4 3 161 4 2 4 2 57 4 1 4 1
36 proc genmod class premar birth model
count premar birth / distpoi linklog run
12
SAS Output Independence Model
Criteria For Assessing Goodness Of Fit
Criterion DF
Value Value/DF Deviance
9 127.6529
14.1837 Scaled Deviance
9 127.6529 14.1837
Pearson Chi-Square 9 128.6836
14.2982 Scaled Pearson
X2 9 128.6836 14.2982
Log Likelihood
2983.6850 Algorithm converged.
Analysis Of
Parameter Estimates
Standard Wald 95 Confidence
Chi- Parameter DF Estimate
Error Limits Square Pr gt
ChiSq Intercept 1 4.7132
0.0731 4.5700 4.8564 4162.07
lt.0001 premar 1 1 -0.5092
0.0805 -0.6670 -0.3514 40.00
lt.0001 premar 2 1 -1.4860
0.1148 -1.7111 -1.2609 167.47
lt.0001 premar 3 1 -0.8538
0.0903 -1.0307 -0.6769 89.48
lt.0001 premar 4 0 0.0000
0.0000 0.0000 0.0000 .
. birth 1 1 -0.4565 0.1014
-0.6552 -0.2579 20.29 lt.0001
birth 2 1 -0.2680 0.0959
-0.4559 -0.0800 7.81 0.0052
birth 3 1 0.2553 0.0841
0.0905 0.4201 9.22 0.0024
birth 4 0 0.0000 0.0000
0.0000 0.0000 . . Scale
0 1.0000 0.0000 1.0000
1.0000
13
Flat Plane
14
Saturated Model
log mij ? ?i ?j ?ij
Y
X
XY
?ij Adjustment for Cell ij
XY
Degrees of Freedom n 0
?23 log n23 m l2 l3 log 29 3.8836
(-0.7734) .3692 2.0728
XY
Y
X
15
Linear-by-Linear Model
X
Y
log mij ? ?i ?j ?uivj ? linear-by
linear association ui row scores vj column
scores
The Linear-by-Linear model adds a parameter so we
lose a degree of freedom n (r 1)(c 1) 1
8
16
SAS Code Linear-by-Linear Model
data sex input premar birth u v count _at__at_
linlin uv datalines 1 4 1 4 38 1 3 1 3 60 1
2 1 2 68 1 1 1 1 81 2 4 2 4 14 2 3 2 3 29 2 2 2 2
26 2 1 2 1 24 3 4 3 4 42 3 3 3 3 74 3 2 3 2 41 3 1
3 1 18 4 4 4 4 157 4 3 4 3 161 4 2 4 2 57 4 1 4 1
36 proc genmod class premar birth model
count premar birth linlin / distpoi
linklog run
17
SAS Output Linear by Linear Model
Criteria For Assessing Goodness Of Fit
Criterion DF
Value Value/DF Deviance
8 11.5337
1.4417 Scaled Deviance
8 11.5337 1.4417
Pearson Chi-Square 8 11.5085
1.4386 Scaled Pearson
X2 8 11.5085 1.4386
Log Likelihood
3041.7446 Algorithm converged.
Analysis Of
Parameter Estimates
Standard Wald 95 Confidence
Chi- Parameter DF Estimate
Error Limits Square Pr gt
ChiSq Intercept 1 0.4735
0.4339 -0.3769 1.3239 1.19
0.2751 premar 1 1 1.7537
0.2343 1.2944 2.2129 56.01
lt.0001 premar 2 1 0.1077
0.1988 -0.2820 0.4974 0.29
0.5880 premar 3 1 -0.0163
0.1264 -0.2641 0.2314 0.02
0.8972 premar 4 0 0.0000
0.0000 0.0000 0.0000 .
. birth 1 1 1.8797 0.2491
1.3914 2.3679 56.94 lt.0001
birth 2 1 1.4156 0.1996
1.0243 1.8068 50.29 lt.0001
birth 3 1 1.1551 0.1291
0.9021 1.4082 80.07 lt.0001
birth 4 0 0.0000 0.0000
0.0000 0.0000 . . linlin
1 0.2858 0.0282 0.2305
0.3412 102.46 lt.0001 Scale
0 1.0000 0.0000 1.0000
1.0000
18
Sample Calculation in the Linear-by-Linear Model
log m23 m l2 l3 bu2v3 0.4735 1.7537
1.1551 0.2858(2)(3) 3.4511 m23
exp(3.4511) 31.5
Y
X
18
19
Data and Linear-by-Linear Model
19
20
Constant Odds Ratio by Uniform Association Model
21
Odds Ratio
Example
22
Saddle
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