1

Modeling Ordinal Associations

Section 9.4

Roanna Gee

2

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

3

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

4

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)

5

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)

6

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

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 .

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)

8

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

9

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

10

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

11

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;

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 > 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

13

Flat Plane

14

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

15

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

16

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;

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 > 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

18

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

19

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

20

Constant Odds Ratio by Uniform Association Model

21

Odds Ratio

logabcd

adcb

uc ua vd vb

7283

7382

e 11

72

73

e 82

83

72

82

e 73

83

and

Example:

22

Saddle

Movie