Problem Description:
The regression analysis assignment involves the application of logistic regression to analyze the impact of various factors on a binary outcome related to patient outcomes in an intensive care unit (ICU). The factors include age, cancer presence, infection status, cardiopulmonary resuscitation (CPR) prior to admission, and race. The goal is to interpret the model, assess significance, and provide confidence intervals for key variables.
Question 1: Data Transformation
RECODE RACE (2=1) (ELSE=0) INTO Race_01.
VARIABLE LABELS Race_01 'Race_01'.
EXECUTE.
RECODE RACE (3=1) (ELSE=0) INTO Race_02.
VARIABLE LABELS Race_02 'Race_02'.
EXECUTE.
Design Matrix:
| RACE | Race_01 | Race_02 |
|---|---|---|
| White | 0 | 0 |
| Black | 1 | 0 |
| Other | 0 | 1 |
Question 2: Logistic Regression Model
π(X)=e^(β0+β1(AGE)+β2(CAN)+β3(CPR)+β4(INF)+β5(Race_01)+β6(Race_02))/(1-e^(β0+β1(AGE)+β2(CAN)+β3(CPR)+β4(INF)+β5(Race_01)+β6(Race_02)) )X=vector of covariates or predictors i.e.,AGELikelihood Function: l(β)=∏_(i=1)^200▒〖ζ(Xi〗),CAN,CPR,INF,Race_01,and Race_02
Logit Transformation: g(X)=β0+β1(AGE)+β2(CAN)+β3(CPR)+β4(INF)+β5(Race_01) +β6(Race_02)
There are total 7 parameters, including the intercept term of the model.
Question 3: Likelihood and Log-Likelihood
Likelihood Function: l(β)=∏_(i=1)^200▒〖ζ(Xi〗)
In the above equation, the Xi represents the set of covariates
ζ(Xi)=π(X_i )^(y_i ) (1-π〖(X_i ))〗^(〖1-y〗_i )
y_i=0 ,1
Question 4: Logistic Regression Model Fitting
LOGISTIC REGRESSION VARIABLES STA
/METHOD=ENTER RACE CAN INF CPR AGE
/CONTRAST (RACE)=Indicator
/SAVE=PRED
/PRINT=GOODFIT CORR ITER(1) CI(95)
/CRITERIA=PIN(0.05) POUT(0.10) ITERATE(20) CUT(0.5).
Model Summary:
- Variables Entered: RACE, CAN, INF, CPR, AGE
β ̂_0=-3.512,β ̂_1=0.027,β ̂_2=0.245,β ̂_3=1.646,β ̂_4=0.681,β ̂_5=1.217,β ̂_6=-0.260
Parameters: g ̂(X)=-3.512+0.027(AGE)+0.245(CAN)+1.646(CPR)+0.681(INF)-0.260(Race_01) -1.217(Race_02)
Logistic Regression Model
π ̂(X)=e^(-3.512+0.027(AGE)+0.245(CAN)+1.646(CPR)+0.681(INF)-0.260(Race_01 )-1.217(Race_02))/(1-e^(-3.512+0.027(AGE)+0.245(CAN)+1.646(CPR)+0.681(INF)-0.260(Race_01 )-1.217(Race_02)) )
Question 5: Likelihood Ratio Test
- Null Hypothesis: 6=0β1=β2=β3=β4=β5=β6=0
- Likelihood Ratio Test: -value=0.002<0.05p-value=0.002<0.05
- Deviance: 179.3179.3 with 66 degrees of freedom
Question 6: Wald Statistics
| Test for a predictor | df | Wald | Sig. |
|---|---|---|---|
| AGE | 1 | 5.478 | 0.019 |
| CAN | 1 | 0.157 | 0.692 |
| INF | 1 | 3.201 | 0.074 |
| CPR prior to ICU admission | 1 | 6.975 | 0.008 |
| Constant | 1 | 9.373 | 0.002 |
| Race | 2 | 0.902 | 0.637 |
| Race 1 | 1 | 0.089 | 0.766 |
| Race 2 | 1 | 0.812 | 0.368 |
Table 1: Full model statistics
Test for a predictor df Wald Sig. AGE 1 5.478 0.019 CAN 1 0.157 0.692 INF 1 3.201 0.074 CPR prior to ICU admission 1 6.975 0.008 Constant 1 9.373 0.002 Race 2 0.902 0.637 Race 1 1 0.089 0.766 Race 2 1 0.812 0.368
Reduced Model:
- Variables Entered: AGE, CPR prior to ICU admission
| Test for a predictor | df | Wald | Sig. |
|---|---|---|---|
| AGE | 1 | 7.052 | 0.008 |
| CPR prior to ICU admission | 1 | 8.630 | 0.003 |
| Constant | 1 | 20.216 | .000 |
Table 2: Test for a predictor
Question 7: Confidence Intervals
- 95% CI for Age: 0.008≤1≤0.051560.008≤β1≤0.05156
- 95% CI for CPR: 0.59428≤2≤2.973720.59428≤β2≤2.97372
Question 8: Covariance Matrix and Model Application
Covariance Matrix: Cov=[.556−.091−.008−.091.369.001−.008.001.000]Cov(β)=⎣⎡.556−.091−.008−.091.369.001−.008.001.000⎦⎤
Estimated Logit Model: =3.355−1.784(AGE)−0.030(CPR)g(X)=3.355−1.784(AGE)−0.030(CPR)
Prediction for a 60-year-old patient with CPR: =−103.72g^(X)=−103.72
- 95% CI: −104.832≤≤−102.598−104.832≤g^(X)≤−102.598
- Estimated Probability: =0.000π^(X)=0.000