modeling ordinal associations
DESCRIPTION
Modeling Ordinal Associations. Section 9.4. Roanna Gee. 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 - PowerPoint PPT PresentationTRANSCRIPT
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Modeling Ordinal Associations
Section 9.4
Roanna Gee
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National Opinion Survey
1991 General Social Survey
Opinions were asked about a man and a woman having sexualrelations 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
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Opinions on Premarital Sex and Teenage Birth Control
Premarital Sex
i (row)
Teenage Birth Control j (column)
ij i
Strongly Disagree Disagree Agree Strongly Agree
Always Wrong
81(42.4)
7.6(80.9)
68(51.2)
3.1(67.6)
60(86.4)-4.1
(69.4)
38(67.0)-4.8
(29.1)
247
Almost Always
Wrong
24(16.0)
2.3(20.8)
26(19.3)
1.8(23.1)
29(32.5)-0.8
(31.5)
14(25.2)-2.8
(17.6)
93
Wrong Only
Sometimes
18(30.1)-2.7
(24.4)
41(36.3)
1.0(36.1)
74(61.2)
2.2(65.7)
42(47.4)-1.0
(48.8)
175
Not Wrong at All
36(70.6)-6.1
(33.0)
57(85.2)-4.6
(65.1)
161(143.8)
2.4(157.4)
157(111.4)
6.8(155.5)
411
ij j
159 192 324 251 926
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log ij = + i + j
Independence ModelX Y
ij =Expected count
=Mean log cell count
i =Adjustment for Row i
j =Adjustment for Column j
X
Y
Degrees of Freedom = (r – 1)(c – 1)
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Sample Calculation of ij
log†23 = log (row total) + log(column total) – log(table total) = log 93 + log 324 – log 926≈ 3.48
so 23 ≈ exp(3.48) ≈ 32.5
or
† log means natural logarithm
23 can be calculated as (93)(324)/(926) ≈ 32.5
P(A B) • n = P(A) P(B) • n = (93/926)(324/926)(926)
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Calculate and i
is the mean of the logs of the expected all the cell counts.
= (log 42.4 + log 51.2 + . . . + log 111.4)/16 = 3.8836
i is the adjustment to for row i—its mean less .
2 = (log 16.0 + log 19.3 + log 32.5 + log 25.2)/4 – 3.8836= -0.7734
j is the adjustment to for column j.
3 = 0.3692
X
X
Y
X
Y
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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 .
We lose 3 degrees for the i ’s.
(Since i = 0, 4 = – 1 – 2 – 3.)
We also lose 3 degrees for the j ’s.
X
X X X X X
Y
= 16 – 1 – 3 – 3 = 9
= rc – 1 – (r – 1) – (c – 1) = (r – 1)(c – 1)
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Data and Independence Model
Premarital Sex
i (row)
Teenage Birth Control j (column)
ij i
Strongly Disagree Disagree Agree Strongly Agree
Always Wrong81
(42.4)68
(51.2)60
(86.4)38
(67.0) 247
Almost Always
Wrong
24(16.0)
26(19.3)
29(32.5)
14(25.2) 93
Wrong Only
Sometimes
18(30.1)
41(36.3)
74(61.2)
42(47.4) 175
Not Wrong at All36
(70.6)
57(85.2)
161(143.8)
157(111.4) 411
ij j
159 192 324 251 926
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eij nij ˆ ij
ˆ ij (1 pi)(1 p j )
Pearson Residuals
= -0.8
A standardized Pearson residual that exceeds 2 or 3 in absolute value indicates a lack of fit.
e23 29 32.5
32.5 1 93
926
1 324
926
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Data and Pearson Residuals
Premarital Sex
i (row)
Teenage Birth Control j (column)
ij i
Strongly Disagree Disagree Agree Strongly Agree
Always Wrong817.6
683.1
60-4.1
38-4.8 247
Almost Always
Wrong
242.3
261.8
29-0.8
14-2.8 93
Wrong Only
Sometimes
18-2.7
411.0
742.2
42-1.0 175
Not Wrong at All36
-6.157
-4.61612.4
1576.8 411
ij j
159 192 324 251 926
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SAS Code Independence Modeldata sex;input premar birth u v count @@; linlin = u*v ;datalines;1 4 1 4 381 3 1 3 601 2 1 2 681 1 1 1 812 4 2 4 142 3 2 3 292 2 2 2 262 1 2 1 243 4 3 4 423 3 3 3 743 2 3 2 413 1 3 1 184 4 4 4 1574 3 4 3 1614 2 4 2 574 1 4 1 36;proc genmod; class premar birth; model count = premar birth / dist=poi link=log;run;
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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 > ChiSq
Intercept 1 4.7132 0.0731 4.5700 4.8564 4162.07 <.0001 premar 1 1 -0.5092 0.0805 -0.6670 -0.3514 40.00 <.0001 premar 2 1 -1.4860 0.1148 -1.7111 -1.2609 167.47 <.0001 premar 3 1 -0.8538 0.0903 -1.0307 -0.6769 89.48 <.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 <.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
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Flat Plane
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log ij = + i + j + ij
Saturated Model:
X Y
ij = Adjustment for Cell ijXY
Degrees of Freedom = 0
XY
23 = log n23 – – 2 – 3
= log 29 – 3.8836 – -0.7734) – .3692
= 2.0728
XY X Y
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log ij = + i + j + uivj
:linear-by linear associationui : row scoresvj : column scores
Linear-by-Linear Model
X Y
The Linear-by-Linear model adds a parameter so we lose a degree of freedom:
= (r – 1)(c – 1) – 1 = 8
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SAS Code Linear-by-Linear Modeldata sex;input premar birth u v count @@; linlin = u*v ;datalines;1 4 1 4 381 3 1 3 601 2 1 2 681 1 1 1 812 4 2 4 142 3 2 3 292 2 2 2 262 1 2 1 243 4 3 4 423 3 3 3 743 2 3 2 413 1 3 1 184 4 4 4 1574 3 4 3 1614 2 4 2 574 1 4 1 36;proc genmod; class premar birth; model count = premar birth linlin / dist=poi link=log;run;
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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 > 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 <.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 <.0001 birth 2 1 1.4156 0.1996 1.0243 1.8068 50.29 <.0001 birth 3 1 1.1551 0.1291 0.9021 1.4082 80.07 <.0001 birth 4 0 0.0000 0.0000 0.0000 0.0000 . . linlin 1 0.2858 0.0282 0.2305 0.3412 102.46 <.0001 Scale 0 1.0000 0.0000 1.0000 1.0000
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Sample Calculation in the Linear-by-Linear Model
log 23 = + 2 + 3 + u2v3
= 0.4735 + 1.7537 + 1.1551 + 0.2858(2)(3)
= 3.4511
23 = exp(3.4511) = 31.5
X Y
18
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Data and Linear-by-Linear Model
Premarital Sex
i (row)
Teenage Birth Control j (column)
ij i
Strongly Disagree Disagree Agree Strongly Agree
Always Wrong81
(80.9)68
(67.6)60
(69.4)38
(29.1) 247
Almost Always
Wrong
24(20.8)
26(23.1)
29(31.5)
14(17.6) 93
Wrong Only
Sometimes
18(24.4)
41(36.1)
74(65.7)
42(48.8) 175
Not Wrong at All36
(33.0)57
(65.1)161
(157.4)157
(155.5) 411
ij j
159 192 324 251 926
19
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Constant Odds Ratio by Uniform Association Model
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Odds Ratio
logabcd
adcb
uc ua vd vb
7283
7382
e 11
72
73
e 82
83
72
82
e 73
83
and
Example:
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Saddle
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