WB University – MPH (1st Semester)

Epidemiology & Biostatistics – Fully Solved Paper | Theory + Stepwise Calculations

Q1 (a) Disease control, elimination, eradication + Outbreak investigation steps

Disease Control: Reduction to acceptable level by continuous intervention (e.g., Malaria control).
Disease Elimination: Zero incidence in defined region, measures continue (e.g., Leprosy elimination).
Disease Eradication: Permanent global zero, no intervention needed (e.g., Smallpox).

Steps of Outbreak Investigation

  1. Prepare field work
  2. Confirm outbreak
  3. Verify diagnosis
  4. Define & identify cases
  5. Describe by time/place/person
  6. Develop hypothesis
  7. Evaluate hypothesis
  8. Refine hypothesis
  9. Implement control
  10. Communicate findings

Q1 (b) Index case vs First case + Risk Ratio (Fish fry)

Index case = first reported; First case = actual earliest case.

ExposureIllNot IllTotal
Ate fish fry404080
Did not eat103040

Attack rate exposed = 40/80 = 0.5 ; Attack rate unexposed = 10/40 = 0.25.
RR = \( \frac{0.5}{0.25} = 2 \) → fish fry eaters had 2× higher risk → suspected food item.

Q2 (a) Sequential testing + Net Specificity

Test A (Se=80%, Sp=70%), Test B (Se=90%, Sp=90%), Pop=3000, Prev=5% (D=150, ND=2850).
After A: TP=120, FN=30 ; TN=1995, FP=855. Test B on positives (975):
TP₂=108, FN₂=12 ; TN₂=769.5, FP₂=85.5.
Final TN = 1995+769.5 = 2764.5 ; Final FP=85.5.
Net Specificity = \( \frac{2764.5}{2764.5+85.5} = \frac{2764.5}{2850} = 0.97 \) (97%).

Q2 (b) Reliability + Percentage & Positive agreement

TB YesTB NoTotal
MO₁ TB Yes201030
MO₁ TB No51520
Total252550

Percentage agreement = \( \frac{20+15}{50}=70\% \).
Positive agreement = \( \frac{20}{20+10+5}= \frac{20}{35} \approx 57.14\% \).

Q3 (a) Case-control vs Cohort + Type I/II error + PAR

Case-control: starts with disease, OR, rare diseases. Cohort: starts with exposure, RR, rare exposure.
Type I error (α): reject true H₀ (false positive). Type II error (β): accept false H₀ (false negative).
PAR = \( I_t - I_0 \) (excess risk in population due to exposure).

Q3 (b) AR and PAR (smoking & lung cancer)

Iₑ=95/10000, Iᵤ=15/10000, Pₑ=0.4.
AR = \( \frac{95-15}{10000} = \frac{80}{10000} \) = 80 per 10,000.
PAR = \( 0.4 \times \frac{80}{10000} = \frac{32}{10000} \) = 32 per 10,000.

Q4 (a) Correlation ≠ causation (Hill’s Criteria)

Confounding may explain correlation. Hill’s criteria: Strength, Consistency, Specificity, Temporality (most important), Biological gradient, Plausibility, Coherence, Experiment, Analogy.

Q4 (b) Stratified analysis & confounding

Crude OR = \( \frac{60\times164}{140\times36}=1.95 \).
Age <40: OR=1 ; Age ≥40: OR=1 → crude OR differs → Age is a confounder.

Q5 (a) Mean, Median, Mode (Grouped data)

Classes: 100–150(25),150–200(30),200–250(35),250–300(10). Midpoints:125,175,225,275.
Mean = \( \frac{19000}{100}=190 \).
Median class 150–200: Median = \( 150 + \frac{50-25}{30}\times50 = 191.67 \).
Mode = \( 200 + \frac{35-30}{70-30-10}\times50 = 208.33 \).

Q5 (b) Corrected mean & SD

n=100, old mean=65, wrong 60 → correct 55. Corrected mean = \( \frac{6500-60+55}{100}=64.95 \) kg.
Corrected SD ≈ 5.07 kg (after Σx² correction).

Q6 (a) Binomial mean/SD & birth weight probability

n=60, success p=0.4 → mean = 24, SD = \( \sqrt{60\times0.4\times0.6}=3.79 \).
Birth weight: μ=2.8, σ=0.4, P(X<2.5): z=-0.75 → prob≈0.2266 → expected newborns = \( 0.2266\times500 \approx 113 \).

Q6 (b) Bayes theorem

P(correct)=0.6, P(death|correct)=0.4, P(death|wrong)=0.7. P(death)=0.52.
P(correct|death) = \( \frac{0.24}{0.52} \approx 0.4615 \) (46.15%).

Q7 (a) Correlation coefficient & regression

x: 60,65,70,75,80 ; y:140,145,150,155,160 → perfect linear → r = +1.
Regression y on x: slope b=1, intercept a=80 → y = x + 80.

Q7 (b) Direct standardization

Standard population: young=13000, old=14000. X std rate = \( \frac{0.008625\times13000 + 0.01045\times14000}{27000} = 0.00957 \) (9.57/1000).
Y std rate = \( \frac{0.0096\times13000 + 0.02\times14000}{27000}=0.01499 \) (14.99/1000).

Q8 (a) Chi-square test (cars crossing bridge)

Observed: 13804,13930,13863,14023,14345,14944,15044 ; total=99953, expected=14279.
χ² = Σ(O−E)²/E ≈ 113.30, critical (df=6, α=0.05)=12.59 → significant → cars not equally distributed.

Q8 (b) One-sample t-test

n=10, mean=66, SD=3, H₀: μ=64. t = \( \frac{66-64}{3/\sqrt{10}} = 2.108 \).
Tabulated t₀.₀₅,₉ = 2.262 → 2.108 < 2.262 → not significant; cannot conclude mean height >64 inches.

📌 FINAL SHORT SUMMARY OF ANSWERS
1(b) RR
2
2(a) Net Specificity
97%
2(b) % agreement
70%
2(b) Positive agreement
57.14%
3(b) AR
80/10,000
3(b) PAR
32/10,000
4(b) Crude OR
1.95
4(b) Stratified ORs
1 and 1 → confounding
5(a) Mean
190
5(a) Median
191.67
5(a) Mode
208.33
5(b) Corrected mean
64.95
5(b) Corrected SD
≈5.07
6(a) Binomial mean
24
6(a) Binomial SD
3.79
6(a) Newborns <2.5kg
≈113
6(b) Bayes (Correct|Death)
46.15%
7(a) r
+1
7(a) Regression
y = x + 80
7(b) Std rate X
9.57/1000
7(b) Std rate Y
14.99/1000
8(a) χ²
113.30 (significant)
8(b) t
2.108 (not signif.)