6 Applying Logistic Regression Models - Nc State Universitydzhang2/st744/chap6.pdf · 6 Applying...

CHAPTER 6 ST744, D. Zhang

6 Applying Logistic Regression Models

I Model Selection and Diagnostics

I.1 Model Selection

• # of x’s can be entered in the model:

Rule of thumb: # of events (both [Y = 1] and [Y = 0]) per x ≥ 10.

• Need to be aware of collinearity in x’s.

• Use traditional model selection procedures (used when p << n)

1. Forward selection (simple one + variant)

2. Backward elimination

• Use modern model selection procedures, usually in the form of

penalized likelihood (can handle p > n); New research area.

Slide 344


• Use LRT for nested models (e.g., Table 6.2)

• Use AIC (Akaike information criterion) or BIC (Bayesian information

criterion) for model selection (not necessarily nested models)

AIC = −2{ℓmax − p}BIC = −2{ℓmax − 0.5 log(n)p}

Smaller AIC/BIC, the better.

Note: BIC tends to yield a simpler model than AIC.

• Use common sense in model building (e.g. time ordering, etc. Table

6.3).

Slide 345


I.2 Model Diagnostics

• Use standardized residuals to check model fit and identify outliers:

yi|xiind∼ Bin(ni, πi)

logit(πi) = xTi β

πi =exT

ibβ

1 + exTi

bβ

1. Standardized Pearson residual:

ei =yi − πi√niπi(1− πi)

esti =

ei√1− hi

Slide 346


2. Standardized deviance residual:

di = 2

(yi log

yi

niπi+ (ni − yi) log

ni − yi

ni − niπi

)

di =√disign(yi − πi)

dsti =

di√1− hi

• If |esti | (or |dst

i |) > 2, 3 ⇒ outliers.

• Plots of esti (or dst

i ) v.s. xi or xTi β may detect lack of fit.

• When ni = 1, esti (or dst

i ) not very informative.

• Note: Proc Logistic does not report esti and dst

i . Need to use

Proc GenMod to get esti and dst

i .

Slide 347


• Example 1: Residual plot for the crab data:

Model: logit(P [Y = 1|x, c]) = β0 + β1c1 + β2c2 + β3c3 + β4xdata crab;

input color spine width satell weight;weight=weight/1000;color=color-1;satbin=(satell>0);c1 = (color=1);c2 = (color=2);c3 = (color=3);c4 = (color=4);s1 = (spine=1);s2 = (spine=2);datalines;

3 3 28.3 8 30504 3 22.5 0 15502 1 26.0 9 23004 3 24.8 0 21004 3 ...

proc genmod data=crab descending;model satbin = width c1 c2 c3 / dist=bin link=logit;output out=resid ResRaw=ResRaw ResChi=ResChi StdReschi=StdReschi;

run;

data _null_; set resid;file "crab_res";put stdreschi width;

run;

Slide 348


22 24 26 28 30 32 34

−3

−2

−1

01

23

Carapace Width

Sta

ndar

dize

d P

ears

on R

esid

ual

Standardized Pearson Residual Plot for Crab Data

Slide 349


• Example 2: Heart disease and bloop pressure (Table 6.5, P. 217)data HD;

input bp $ n y;if bp="<117" thenx=111.5;

else if bp="117-126" thenx=121.5;

else if bp="127-136" thenx=131.5;

else if bp="137-146" thenx=141.5;

else if bp="147-156" thenx=151.5;

else if bp="157-166" thenx=161.5;

else if bp="167-186" thenx=176.5;

elsex=191.5;

cards;<117 156 3117-126 252 17127-136 284 12137-146 271 16147-156 139 12157-166 85 8167-186 99 16>186 43 8

;

proc genmod;model y/n = x /dist=bin link=logit residual;

run;

Slide 350


Criteria For Assessing Goodness Of Fit

Criterion DF Value Value/DF

Deviance 6 5.9092 0.9849Scaled Deviance 6 5.9092 0.9849Pearson Chi-Square 6 6.2899 1.0483Scaled Pearson X2 6 6.2899 1.0483

Analysis Of Maximum Likelihood Parameter Estimates

Standard Wald 95% Confidence WaldParameter DF Estimate Error Limits Chi-Square

Intercept 1 -6.0820 0.7243 -7.5017 -4.6624 70.51x 1 0.0243 0.0048 0.0148 0.0338 25.25

Raw Pearson DevianceObservation Residual Residual Residual

Std Deviance Std Pearson LikelihoodResidual Residual Residual

1 -2.194866 -0.979434 -1.061683-1.198648 -1.105788 -1.179257

2 6.3932374 2.0057053 1.85010722.1903838 2.3745999 2.2447199

3 -3.072737 -0.813338 -0.841966-0.978546 -0.945274 -0.970016

4 -2.081617 -0.50673 -0.51623-0.583485 -0.572747 -0.581169

5 0.3836399 0.1175816 0.11700160.1254648 0.1260868 0.1255461

6 -0.856987 -0.304247 -0.308775-0.330927 -0.326074 -0.330303

7 1.791237 0.5134723 0.50496570.6411542 0.651955 0.6452766

8 -0.361958 -0.139464 -0.140243-0.178337 -0.177346 -0.177959

Slide 351


• Example 3: Admission to Graduate School at UF in 1997-1998

(Table 6.6)

Let π(k, g) = P [admission|D = k,G = g] for department D = k

and gender G = g. We consider three models:

1. π(k, g) = Dk: Admission is independent of gender at each

department.

2. π(k, g) = Dk +Gg: Admission-Gender association is the same

across departments.

3. π(k, g) = Gg: Get the marginal Admission-Gender association

collapsed over departments.

options ls=75 ps=100;

data admit;input dept $ gender y yno;n = y+yno;male=gender-1;cards;anth 1 32 81anth 2 21 41astr 1 6 0astr 2 3 8

Slide 352


chem 1 12 43chem 2 34 110...

title "Model 1: Logistic model assuming gender and admission are";title2 "conditional independent given department";proc genmod;

class dept;model y/n = dept /dist=bin link=logit;output out=resid Resraw=Resraw Reschi=Reschi StdReschi=StdReschi;

run;

data resid; set resid;keep dept male Resraw Reschi StdReschi;

run;

title "Residuals from Model 1";proc print data=resid;run;

title "Model 2: Logistic model with homogeneous GA and DA association";proc genmod data=admit;

class dept;model y/n = dept male;

run;

title "Model 3: Logistic model for marginal GA association";proc genmod data=admit;

model y/n = male;run;

Slide 353


Part of the output:Model 1: Logistic model assuming gender and admission are 1

conditional independent given department




StdObs dept male Reschi Resraw Reschi

1 anth 0 -0.45509 -2.22286 -0.764572 anth 1 0.61438 2.22286 0.764573 astr 0 2.30940 2.82353 2.870964 astr 1 -1.70561 -2.82353 -2.870965 chem 0 -0.22824 -0.71357 -0.268306 chem 1 0.14105 0.71357 0.268307 clas 0 -0.75593 -0.50000 -1.069048 clas 1 0.75593 0.50000 1.069049 comm 0 -0.16670 -1.04167 -0.6326010 comm 1 0.61024 1.04167 0.6326011 comp 0 0.85488 1.63636 1.1575212 comp 1 -0.78040 -1.63636 -1.1575213 engl 0 0.67452 3.32130 0.9420914 engl 1 -0.65769 -3.32130 -0.9420915 geog 0 1.79629 2.75000 2.1664116 geog 1 -1.21106 -2.75000 -2.1664117 geol 0 -0.21822 -0.30000 -0.2608218 geol 1 0.14286 0.30000 0.2608219 germ 0 0.89974 0.77273 1.8873020 germ 1 -1.65903 -0.77273 -1.88730

Slide 354


21 hist 0 -0.14639 -0.31034 -0.1762722 hist 1 0.09820 0.31034 0.1762723 lati 0 1.22493 3.25676 1.6456424 lati 1 -1.09895 -3.25676 -1.6456425 ling 0 0.78403 2.13043 1.3729826 ling 1 -1.12711 -2.13043 -1.3729827 math 0 1.00845 3.30631 1.2884428 math 1 -0.80193 -3.30631 -1.2884429 phil 0 1.22474 1.00000 1.3416430 phil 1 -0.54772 -1.00000 -1.3416431 phys 0 1.17573 2.57576 1.3245832 phys 1 -0.61005 -2.57576 -1.3245833 poli 0 -0.18041 -0.68707 -0.2331834 poli 1 0.14772 0.68707 0.2331835 psyc 0 -1.16905 -2.41176 -2.2722236 psyc 1 1.94841 2.41176 2.2722237 reli 0 0.63246 0.75000 1.2649138 reli 1 -1.09545 -0.75000 -1.2649139 roma 0 0.05868 0.17647 0.1397040 roma 1 -0.12677 -0.17647 -0.1397041 soci 0 0.17272 0.56164 0.3012342 soci 1 -0.24679 -0.56164 -0.3012343 stat 0 -0.00960 -0.02439 -0.0122944 stat 1 0.00768 0.02439 0.0122945 zool 0 -1.23400 -3.10769 -1.7587346 zool 1 1.25314 3.10769 1.75873

Model 2: Logistic model with homogeneous GA and DA association 4




Slide 355



Standard Wald 95% WaldParameter DF Estimate Error Confidence Limits Chi-Square

Intercept 1 -2.0323 0.2877 -2.5962 -1.4685 49.91dept anth 1 1.2585 0.3277 0.6162 1.9008 14.75dept astr 1 2.2622 0.5631 1.1586 3.3659 16.14

...

male 1 -0.1730 0.1123 -0.3932 0.0472 2.37

Model 3: Logistic model for marginal GA association 6





Standard Wald 95% Confidence WaldParameter DF Estimate Error Limits Chi-Square

Intercept 1 -0.6455 0.0637 -0.7703 -0.5207 102.77male 1 0.0662 0.0921 -0.1142 0.2467 0.52

Models 2 & 3 show Simpson’s Paradox.

Slide 356


II Inference on The Conditional Association in 2× 2×K Tables

Example: Multi-center clinical trial evaluating a cream in curing skin

infection (Table 6.9, P.226)

S F

trt 11 25

control 10 27

Z = 1

S F

16 4

22 10

Z = 2

S F

14 5

7 12

Z = 3

S F

2 14

1 16

Z = 4

S F

trt 6 11

control 0 12

Z = 5

S F

1 10

0 10

Z = 6

S F

1 4

1 8

Z = 7

S F

4 2

6 1

Z = 8

What we observed: There is a lot of variation in success probabilities

among centers.

Slide 357


If we collapse the tables over centers, we got:

Y

S F

X trt 55 75

control 47 96

⇒θXY =

96× 55

47× 75≈ 1.5

The above estimate θXY may not be very useful since this is not a

random sample, so we cannot use the famous formula for calculating the

variance of log θXY :

var(log θXY ) 6= 1

55+

1

75+

1

47+

1

96.

⇒ Should focus on conditional association!

Slide 358


II.1 Testing Conditional Independence between X and Y Given Z

(H0 : X ⊥ Y |Z)

1. Method 1: Use logistic model with ML inference (good when K is

fixed, small ∼ moderate)

• Let

Y = 1 for success, 0 for failure

x = 1 for treatment, 0 for control

z = 1, 2, ..., 8 for centers

π(x, z) = P [Y = 1|x, z]

and consider the (homogeneous) model:

logitπ(x, z = k) = βx+ βzk −−−−(∗)

• ⇒ common odds-ratio model:

π(x = 1, z = k)/{1− π(x = 1, z = k)}π(x = 0, z = k)/{1− π(x = 0, z = k)} = eβ

Slide 359


π(x = 0, z = k)/{1− π(x = 0, z = k)} = eβZk

• Under this model, H∗0 : β = 0 ⇔ H0 : X ⊥ Y |Z.

data table6_9;input center trt y y0;n=y+y0;cards;1 1 11 251 0 10 272 1 16 42 0 22 10

...

title "Use homogeneous model to test no treatment effect at each center";proc logistic;

class center / param=ref;model y/n = center trt / selection=f include=1 slentry=1;

run;

Use homogeneous model to test no treatment effect at each center 1

The LOGISTIC Procedure

The following effects will be included in each model:

Intercept center

Step 0. The INCLUDE effects were entered.

Model Fit Statistics

InterceptIntercept and

Slide 360


Criterion Only Covariates

-2 Log L 360.828 283.689

Residual Chi-Square Test

Chi-Square DF Pr > ChiSq

6.5583 1 0.0104

Step 1. Effect trt entered:

Model Fit Statistics

InterceptIntercept and

Criterion Only Covariates

-2 Log L 360.828 277.020

Analysis of Maximum Likelihood Estimates

Standard WaldParameter DF Estimate Error Chi-Square Pr > ChiSq

Intercept 1 0.8859 0.6755 1.7201 0.1897center 1 1 -2.2079 0.7195 9.4166 0.0022center 2 1 -0.1525 0.7381 0.0427 0.8363center 3 1 -1.0550 0.7457 2.0015 0.1571center 4 1 -3.6264 0.9071 15.9813 <.0001center 5 1 -2.7278 0.8184 11.1104 0.0009center 6 1 -4.3548 1.2293 12.5499 0.0004center 7 1 -3.0056 1.0200 8.6836 0.0032trt 1 0.7769 0.3067 6.4174 0.0113

Slide 361


• Three Tests for H0 : β = 0:

(a) Score test: χ2 = 6.5583, df = 1, P = 0.0104.

(b) LRT: G2 = 283.689− 277.020 = 6.669, df = 1, P = 0.0098.

(c) Wald test: χ2 = 6.4174, P = 0.0113.

Strong evidence to reject H0 : β = 0.

• β = 0.7769, ebβ = 2.17 ⇒ At each center, the odds of success

(infection is cured) for treated patients is 2.17 times the odds of

success for untreated patients.

• Note 1: The above test results are based on the homogeneous

model (*). When β = 0, model (*) reduces to

logitπ(x, z = k) = βzk

⇔ to H0 : X ⊥ Y |Z, can be tested by conducting the GOF test

for this model.

Slide 362


title "Use goodness-of-fit statistics to test conditional independence";Proc genmod;

class center;model y/n = center;

run;

***************************************************************************

Use goodness-of-fit statistics to test conditional independence 3

The GENMOD Procedure

Response Profile

Ordered Binary TotalValue Outcome Frequency

1 Event 1022 Nonevent 171




χ2 = 13.71, df = 16− 8 = 8, P = 0.089.

G2 = 16.42, df = 8, P = 0.037. Less powerful.

Slide 363


• Note 2: We can also test the adequacy of the homogeneous

model (*) using its GOF statistics:title "Use goodness-of-fit statistics to test homogeneity";Proc genmod;

class center;model y/n = center trt;

run;

***************************************************************************

Use goodness-of-fit statistics to test homogeneity 4

The GENMOD Procedure




χ2 = 8.0256, df = 7, P = 0.33

G2 = 9.7463, df = 7, P = 0.20; adequate fit.

Slide 364


2. Method 2. Use Cochran-Mental-Haenszel (CMH) test for

H0 : X ⊥ Y |Z (good when K →∞ or K is fixed but n++k →∞)

• The above analysis assuming N = 2×K = 2× 8 = 16 is fixed

may be problematic in many situations.

• One way to test X ⊥ Y |Z is to use the CMH test:

Y

S F

X trt n11k n12k n1+k

control n21k n22k n2+k

n+1k n+2k

Z = k

Slide 365


• Under H0 : X ⊥ Y |Z, n11k|n1+k, n+1k ∼ hypergeometric

distribution:

E(n11k|H0, n1+k, n+1k) =n1+kn+1k

n++k= µ11k,

var(n11k|H0, n1+k, n+1k) =n1+kn2+kn+1kn+2k

n2++k(n++k − 1)

.

⇒χ2 =

[∑K

k=1(n11k − µ11k)]2∑K

k=1 var(n11k|H0, n1+k, n+1k)

H0∼ χ21.

• CMH with correction:

χ2c =

{|∑Kk=1(n11k − µ11k)| − 0.5}2

∑Kk=1 var(n11k|H0, n1+k, n+1k)

H0∼ χ21.

• The CMH does not require the homogeneous model.

Slide 366


data y1; set table6_9;count=y;drop y0;y=1;

run;

data y0; set table6_9;count=y0;drop y0;y=0;

run;

data new; set y1 y0;run;

title "MH test for conditional independence and MH common OR";proc freq data=new order=data;

weight count;tables center*trt*y/nopercent norow nocol cmh;

run;

*****************************************************************************

MH test for conditional independence and MH common OR 8

The FREQ Procedure

Summary Statistics for trt by yControlling for center

Cochran-Mantel-Haenszel Statistics (Based on Table Scores)

Statistic Alternative Hypothesis DF Value Prob---------------------------------------------------------------

1 Nonzero Correlation 1 6.3841 0.01152 Row Mean Scores Differ 1 6.3841 0.01153 General Association 1 6.3841 0.0115

Slide 367


Estimates of the Common Relative Risk (Row1/Row2)

Type of Study Method Value 95% Confidence Limits-------------------------------------------------------------------------Case-Control Mantel-Haenszel 2.1345 1.1776 3.8692

(Odds Ratio) Logit ** 1.9497 1.0574 3.5949

Cohort Mantel-Haenszel 1.4245 1.0786 1.8812(Col1 Risk) Logit ** 1.2194 0.9572 1.5536

Cohort Mantel-Haenszel 0.8129 0.6914 0.9557(Col2 Risk) Logit 0.8730 0.7783 0.9792

** These logit estimators use a correction of 0.5 in every cellof those tables that contain a zero.

Breslow-Day Test forHomogeneity of the Odds Ratios------------------------------Chi-Square 7.9955DF 7Pr > ChiSq 0.3330

CMH χ2 = 6.3841, df = 1, P = 0.0115.

MH Common odds-ratio estimate θMH = 2.1345 with 95% CI

[1.1776, 3.8692].

Breslow-Day Test for common odds-ratio: χ2 = 7.9955, df = 7,

P = 0.3330, similar to the GOF test.

Slide 368


3. Method 3: Use a conditional logistic regression under homogeneous

model (*) (good even when K →∞):

logitπ(x, k) = xβ + βk.

• Problem: # of βk’s may →∞; want to get rid of them.

• Idea: find out sufficient statistics of βk and conduct inference on

β based on the conditional distribution of the data given those

sufficient statistics.

• Data from center k:

Y

S F

X trt n11k n12k n1+k

control n21k n22k n2+k

Z = k

Slide 369


• Given

n11k|n1+k ∼ Bin(n1+k, π(1, k)),

n21k|n2+k ∼ Bin(n2+k, π(0, k)),

we got the likelihood function of β and (β1, ..., βK):

L(β, β1, ..., βK) =K∏

k=1

Lk(β, βk)

where Lk(β, βk) is the likelihood contributed by the data from

center Z = k:

Lk(β, βk) = {π(1, k)}n11k{1− π(1, k)}n12k

{π(0, k)}n21k{1− π(0, k)}n22k ,

π(1, k) =eβ+βk

1 + eβ+βk

π(0, k) =eβk

1 + eβk

Slide 370


⇒

Lk(β, βk) =

(eβ+βk

1 + eβ+βk

)n11k ( 1

1 + eβ+βk

)n12k

(eβk

1 + eβk

)n21k ( 1

1 + eβk

)n22k

=eβn11k+βk(n11k+n21k)

(1 + eβ+βk)n11k+n12k(1 + eβk)n21k+n22k

=eβn11k+βkn+1k

(1 + eβ+βk)n1+k(1 + eβk)n2+k

Since n1+k and n2+k are fixed already, so n+1k = n11k + n21k

(total # of successes in center k) is a sufficient statistic for βk.

⇒ Lk(β, βk|n+1k) should be free of βk – noncentral

hypergeometric dist.

Slide 371


• The conditional logistic inference (on β) is based on the

conditional likelihood:

Lc(β|{n+1k}) =K∏

k=1

Lk(β, βk|n+1k),

which only has one parameter β no matter how large K is!

Treat this as a regular likelihood function, we can estimate β by

maximizing Lc(β|{n+1k}). We can also conduct the Wald, score

and LRT for testing H0 : β = 0.

Slide 372


• SAS program and output:title "Use a conditional logistic regression to assess treatment effect";proc logistic;

class center;model y/n = trt;strata center;

run;

***************************************************************************

Use a conditional logistic regression to assess treatment effect 5


Conditional Analysis

Model Information

Data Set WORK.TABLE6_9Response Variable (Events) yResponse Variable (Trials) nNumber of Strata 8Model binary logitOptimization Technique Newton-Raphson ridge

Testing Global Null Hypothesis: BETA=0

Test Chi-Square DF Pr > ChiSq

Likelihood Ratio 6.4914 1 0.0108Score 6.3841 1 0.0115Wald 6.2464 1 0.0124

Slide 373




trt 1 0.7563 0.3026 6.2464 0.0124

• β = 0.7563(SE = 0.3026), ebβ = 2.13, similar to before since

K = 8 is small.

• LRT G2 = 6.4914, Score χ2 = 6.3841, Wald χ2 = 6.2464.

Reject H0 : β = 0.

• Note 1: Score χ2 stat using Lc(β|{n+1k}) is equivalent to

CMH χ2,

• Note 2: We can make exact conditional inference for a

regression coefficient in a regular regression model using the

same idea.

Yi = 1/0 for success/failure, covariates: xi1, xi2, ..., xip.

π(xi) = P [Yi = 1|xi]

Slide 374


Model:

logit{π(xi)} = β1xi1 + β2xi2 + · · ·+ βpxip

We can find out suff. stat. for each βk, denoted by Tk. Suppose

we would like to make exact conditional inference on, βp, say,

then the exact inference can be based on

f(y1, y2, ..., yn|T1, T2, ..., Tp−1) = L(βp).

For exact test of H0 : βp = 0, the cond. dist. of data

(Y1, Y2, ..., Yn) given T1, T2, ..., Tp−1 is completely known. We

can do exact score test based on L(βp).

We can also construct an exact CI for βp based on L(βp).

Software:Proc Logistic descending;

model y = x1 x2 x3 / link=logit;exact x3;

run;

Slide 375


Warning: It is usually very time consuming to conduct the

exact inference, especially for non-sparse data, in which case no

exact inference is needed.

• Note 3: If we apply the above procedure to our homogeneous

model (*)

logitπ(x, k) = xβ + βk,

we can make exact conditional inference on the treatment effect

β. In this case L(β) is the conditional likelihood we got before

using the conditional logistic approach. Therefore, we will get

exact CMH test for H0 : β = 0.

title "Exact p-value for MH test of no treatment effect at each center";proc logistic data=table6_9;

class center / param=ref;model y/n = center trt;exact trt;

run;

***************************************************************************

Exact p-value for MH test of no treatment effect at each center 9


Slide 376




Intercept 1 0.8859 0.6755 1.7201 0.1897center 1 1 -2.2079 0.7195 9.4166 0.0022center 2 1 -0.1525 0.7381 0.0427 0.8363center 3 1 -1.0550 0.7457 2.0015 0.1571center 4 1 -3.6264 0.9071 15.9813 <.0001center 5 1 -2.7278 0.8184 11.1104 0.0009center 6 1 -4.3548 1.2293 12.5499 0.0004center 7 1 -3.0056 1.0200 8.6836 0.0032trt 1 0.7769 0.3067 6.4174 0.0113

--- p-Value ---Effect Test Statistic Exact Mid

trt Score 6.3841 0.0134 0.0110Probability 0.00469 0.0134 0.0110

We can see that 6.3841 is the CMH χ2, which is the score stat.

based on L(β) (row 1). We can also conduct Fisher exact test

on H0 : β = 0 using table prob. (row 2).

Slide 377


4. Method 4. Use mixed model approach (good when K →∞ as

n→∞):

logitπ(x, k) = xβ + βk.

• Data from center k:

Y

1 0

X 1 n11k n12k n1+k

0 n21k n22k n2+k

Z = k

• Here 8 centers is probably a random sample of centers drawn

from a large population of centers. Then the analysis should take

this into account – clustered data.

• βk – log odds of being a “success” for patients in center k if

they all receive the control treatment. It reflects the general

healthy status of patients in center k.

Slide 378


Since center k is randomly sampled, it is reasonable to assume

βk is a random variable and has a distribution.

A commonly used dist. is βk ∼ N(µ, σ2).

Let bk = βk − µ, then bk ∼ N(0, σ2) and our model becomes:

logitπ(x, k) = µ+ xβ + bk.

Only 3 model parameters: µ, β and σ2.

• The likelihood function of (µ, β, σ2):

L(µ, β, σ2) =K∏

k=1

f(n11k|bk)f(n21k|bk)f(bk)dbk.

The inference on β is based on L(µ, β, σ2).

Slide 379


• SAS program and output:

title "Proc glimmix treating center effect as random";proc glimmix method=quad data=table6_9;

class center;model y/n = trt / s dist=bin;random int / subject=center type=vc;

run;

******************************************************************

Proc glimmix treating center effect as random 12

The GLIMMIX Procedure

Model Information

Data Set WORK.TABLE6_9Response Variable (Events) yResponse Variable (Trials) nResponse Distribution BinomialLink Function LogitVariance Function DefaultVariance Matrix Blocked By centerEstimation Technique Maximum LikelihoodLikelihood Approximation Gauss-Hermite QuadratureDegrees of Freedom Method Containment

Class Level Information

Class Levels Values

center 8 1 2 3 4 5 6 7 8

Number of Observations Read 16

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Number of Observations Used 16Number of Events 102Number of Trials 273

Iteration History

Objective MaxIteration Restarts Evaluations Function Change Gradient

0 0 4 76.921833212 . 5.40261 0 2 75.07979884 1.84203437 5.3622692 0 3 74.240313174 0.83948567 1.3338353 0 3 74.055295009 0.18501817 0.3159534 0 2 74.044936835 0.01035817 0.0341825 0 2 74.044537699 0.00039914 0.0139666 0 3 74.044495142 0.00004256 0.0046597 0 3 74.044493119 0.00000202 0.0004538 0 3 74.044493114 0.00000000 2.59E-6

Convergence criterion (GCONV=1E-8) satisfied.

Fit Statistics

-2 Log Likelihood 74.04AIC (smaller is better) 80.04AICC (smaller is better) 82.04BIC (smaller is better) 80.28CAIC (smaller is better) 83.28HQIC (smaller is better) 78.44

Fit Statistics for ConditionalDistribution

-2 log L(y | r. effects) 49.29

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Pearson Chi-Square 8.37Pearson Chi-Square / DF 0.52

Covariance Parameter Estimates

StandardCov Parm Subject Estimate Error

Intercept center 1.9591 1.1903

Solutions for Fixed Effects

StandardEffect Estimate Error DF t Value Pr > |t|

Intercept -1.1974 0.5561 7 -2.15 0.0683trt 0.7385 0.3004 7 2.46 0.0436

From the output, we see µ = −1.1974

β = 0.7385(SE = 0.3004), ebβ = 2.1.

σ2 = 1.9591, variation in log odds of “success” among centers.

Huge variation.

• Since the success prob. for patients receiving control at center k

is

π0k = π(0, k) =

eµ+bk

1 + eµ+bk

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and the success prob. for patients receiving treatment at center

k is

π1k = π(1, k) =

eµ+β+bk

1 + eµ+β+bk,

we can generate a random sample {bk}’s to get a feeling on the

distributions of π0k and π1

k

π0 = E(π0k) = 0.29, π1 = E(π1

k) = 0.42 ⇒ θXY = 1.77.R function:postscript(file="cream-prob.ps", horizontal = F)par(mfrow=c(1,2), pty="s")

b <- rnorm(10000, 0, sqrt(1.9591))expeta0 <- exp(-1.1974 + b)expeta1 <- exp(-1.1974 + 0.7385 + b)

pi0 <- expeta0/(1+expeta0)pi1 <- expeta1/(1+expeta1)

mean0 <- mean(pi0)mean1 <- mean(pi1)

hist(pi0, main="Histogram of pi_0")hist(pi1, main="Histogram of pi_1")dev.off()

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Histogram of pi_0

pi0

Fre

quen

cy

0.0 0.2 0.4 0.6 0.8 1.0

020

060

010

00

Histogram of pi_1

pi1F

requ

ency

0.0 0.2 0.4 0.6 0.8 1.0

020

040

060

0

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II.2 Estimation of The Common Odds-ratio in 2× 2×K Tables

• Each of the above methods provides an estimate of the common

odds-ratio in 2× 2×K tables, except the CMH method (Method 2).

• There is also an MH estimate of the common odds-ratio

θMH =

∑Kk=1

n11kn22k

n++k∑Kk=1

n12kn21k

n++k

• Motivation of θMH : We could estimate θ using the data from the

kth table as:

θ =n11kn22k

n12kn21k

Slide 385


⇒ Estimating equation:

θn12kn21k = n11kn22k

⇒ θn12kn21k/n++k = n11kn22k/n++k

⇒ θ

K∑

k=1

n12kn21k/n++k =

K∑

k=1

n11kn22k/n++k

⇒ θMH =

∑Kk=1

n11kn22k

n++k∑Kk=1

n12kn21k

n++k

.

CDA provides a variance formula of log(θMH) on P. 229, can be

used to construct CI’s for the common odds-ratio θ.

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• For our cream example, we have

θMH =11×27

73 + 16×1052 + 14×12

38 + 2×1633 + 6×12

29 + 1×1021 + 1×8

14 + 4×113

25×1073 + 4×22

52 + 5×738 + 14×1

33 + 11×029 + 10×0

21 + 4×114 + 2×6

13

= 2.1345

See Method 2 in the previous section for SAS program and output.

Slide 387


III Summarizing Predictive Power, Classification Tables and ROC

Curves (P. 223)

• Suppose we have binary response Yi = 1/0 (success/failure), xi a

vector of covariates.

π(xi) = P [Yi = 1|xi]

logit{π(xi)} = xTi β

After we fit the model, we got β ⇒ we got πi as

πi =exT

ibβ

1 + exTi

bβ.

• Choose a known value π0 (e.g., π0 = 0.5), and conduct prediction

Yi as

Yi =

1 if πi > π0

0 otherwise

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and then construct the table (classification table)

Y

1 0

Y 1 n11 n12

0 n21 n22

The following two quantities tell us how good the prediction is:

sensitivity = n11

n11+n12

specificity = n22

n21+n22

• Using only one table with one π0 loses information.

• Solution: use many different values of π0 ⇒ many classification

tables ⇒ many pairs of sensitivity and specificity ⇒ plot sensitivity

v.s. 1− specificity ⇒ ROC (receiver operating characteristic

curve⇒ Area under the ROC curve summarizes the predictive power

of the model, often called the c-index.

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• An example:

Y bπ bY0.3−bY0.4−

bY0.5−bY0.6−

bY0.7−bY0.8−

bY0.9−

1 0.8 1 1 1 1 1 1 0

1 0.6 1 1 1 1 0 0 0

1 0.4 1 1 0 0 0 0 0

0 0.7 1 1 1 1 1 0 0

0 0.5 1 1 1 0 0 0 0

0 0.3 1 0 0 0 0 0 0

bY

Y 1 0

1 3 0

0 3 0

se =3

3

sp =0

3

3 0

2 1

se =3

3

sp =1

3

2 1

2 1

se =2

3

sp =1

3

2 1

1 2

se =2

3

sp =2

3

1 2

1 2

se =1

3

sp =2

3

1 2

0 3

se =1

3

sp =3

3

0 3

0 3

se =0

3

sp =3

3

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ROC curve for the example

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

1 − Specificity

Sen

sitiv

ity

Slide 391


• The AUC for the above ROC curve:

1− 3

9=

2

3

= proportion of concordant pairs in (Yi, πi) among all pairs with

different outcome Yi.

# of pairs with different outcomes: 3× 3 = 9.

# of concordant pairs: 3 + 2 + 1 = 6.

• If there are ties in πi’s, need to do some adjustment. For example,suppose two πi’ for a Yi = 1 and a Yi = 0 are the same (0.4):

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Y bπ bY0.4−bY0.5−

bY0.6−bY0.7−

bY0.8−bY0.9−

1 0.8 1 1 1 1 1 0

1 0.6 1 1 1 0 0 0

1 0.4 1 0 0 0 0 0

0 0.7 1 1 1 1 0 0

0 0.5 1 1 0 0 0 0

0 0.4 1 0 0 0 0 0

The corresponding classification tables are:

bY

Y 1 0

1 3 0

0 3 0

se =3

3

sp =0

3

2 1

2 1

se =2

3

sp =1

3

2 1

1 2

se =2

3

sp =2

3

1 2

1 2

se =1

3

sp =2

3

1 2

0 3

se =1

3

sp =3

3

0 3

0 3

se =0

3

sp =3

3

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ROC curve when there are tied predictive probs

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

1 − Specificity

Sen

sitiv

ity

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• AUC = 5.59

9 = # of pairs with diff outcomes

5.5 = # of concordant pairs (5) + 0.5 × # of ties in πi’s with diff.

outcomes (1).

Slide 395

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