# Logistic Regression Analysis of Patient Outcomes in Intensive Care: Model Development, Interpretation, and Inference

In this comprehensive logistic regression analysis, we delve into the intricacies of patient outcomes within an intensive care setting. From the transformation of categorical variables to the application of likelihood ratio tests, the study provides a detailed exploration of the factors influencing outcomes. The model interpretation, confidence intervals, and covariance matrices further contribute to a thorough understanding of the complex interplay between age, medical conditions, and race in predicting critical care outcomes.

## 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