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Logistic

Logistic Regression — Predict Pass/Fail from Study Hours

Binary logistic regression with OR=3.06 (95% CI [1.19, 7.87]). Model χ²=14.5, Nagelkerke R².

📊 OR = 3.06, 95% CI [1.19, 7.87]
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⚙️ 15 · Logistic Regression

Predict Student Pass/Fail from Study Hours
Logistic Regression
Research Question: Does weekly study hours predict the probability of passing (Y=1) vs failing (Y=0) an examination? n=20 students.
Logistic Function — Probability of Success
P ( pass ) = 1 1 + e − ( b 0 + b 1 X ) = e b 0 + b 1 X 1 + e b 0 + b 1 X
  1. 1
    MLE estimates (obtained via software):
    b 0 = − 8.76 ( S E = 3.42 ) , b 1 = 1.12 ( S E = 0.48 ) Y ^ = 1 1 + e − ( − 8.76 + 1.12 X )
  2. 2
    Predicted probabilities at key values:
    Study Hours678910
    P(pass).115.283.550.798.922
    At  X = 8 : P ( pass ) = 1 1 + e − ( − 8.76 + 8.96 ) = 1 1 + e − 0.20 = 1 1.819 = 0.55
  3. 3
    Odds Ratio — how much the odds multiply per extra hour:
    O R = e b 1 = e 1.12 = 3.06 95 [ 1.19 , 7.87 ] Each additional study hour multiplies the odds of passing by 3.06×
  4. 4
    Model significance — Omnibus test:
    Overall model:  χ 2 ( 1 ) = 14.5 , p < .001 , Nagelkerke  R 2 = .72 Wald test:  χ 2 ( 1 ) = 5.44 , p = .020 (for  b 1 ) Correctly classified:  85 %
3.06
OR
<.001
Model p
.72
Nagelkerke R²
85%
Classified
✅ Study hours significantly predicts pass/fail, OR=3.06 — each additional study hour triples the odds of passing.
APA-7
Logistic regression was performed to examine whether study hours predicted exam outcome. The overall model was significant, χ²(1) = 14.5, p < .001, Nagelkerke R² = .72, correctly classifying 85% of cases. Study hours was a significant predictor, b = 1.12, SE = 0.48, Wald χ²(1) = 5.44, p = .020, OR = 3.06, 95% CI [1.19, 7.87].

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