SAMPLE SIZE DETERMINATION

🔹 Definition

Sample size determination is the process of calculating the minimum number of subjects or observations required in a study to:

• Detect a true effect or difference,

• Achieve statistical significance, and

• Ensure adequate power and precision of results.

It prevents both underpowered studies (too few participants → false negatives) and wasteful studies (too many participants → unnecessary cost/time).

⚖️ Key Factors Affecting Sample Size

🧩 1. For Estimating a Mean

n=(Z×SDE)2n = \left( \frac{Z \times SD}{E} \right)^2n=(EZ×SD​)2

Where:

• n = required sample size

• Z = Z-score for desired confidence (1.96 for 95%)

• SD = estimated standard deviation

• E = margin of error (precision)

Example 1 (Mean)

A study aims to estimate mean fasting blood sugar.

• SD = 15 mg/dL

• Desired precision (E) = ±5 mg/dL

• Confidence = 95% (Z = 1.96)

n=(1.96×155)2=(5.88)2=34.6n = \left( \frac{1.96 \times 15}{5} \right)^2 = (5.88)^2 = 34.6n=(51.96×15​)2=(5.88)2=34.6

✅ Minimum sample size = 35 participants

🧩 2. For Estimating a Proportion

n=Z2×p(1−p)E2n = \frac{Z^2 \times p(1-p)}{E^2}n=E2Z2×p(1−p)​

Where:

• p = estimated proportion (from pilot or literature)

• E = acceptable margin of error

• Z = Z-score for desired confidence level

Example 2 (Proportion)

Prevalence of hypertension estimated at 30% (p = 0.3).
Margin of error = 5% (E = 0.05).
Confidence = 95% (Z = 1.96).

n=(1.96)2×0.3(1−0.3)(0.05)2=3.8416×0.210.0025=323.1n = \frac{(1.96)^2 \times 0.3(1 - 0.3)}{(0.05)^2} = \frac{3.8416 \times 0.21}{0.0025} = 323.1n=(0.05)2(1.96)2×0.3(1−0.3)​=0.00253.8416×0.21​=323.1

✅ Minimum sample size = 324 participants

🧩 3. For Comparing Two Groups (Means)

n=2×(Zα/2+ZβΔ/SD)2n = 2 \times \left( \frac{Z_{\alpha/2} + Z_{\beta}}{\Delta/SD} \right)^2n=2×(Δ/SDZα/2​+Zβ​​)2

Where:

• Δ = expected difference between group means

• Zα/2 = 1.96 for 95% confidence

• Zβ = 0.84 for 80% power

Example 3 (Two Means)

Comparing two antihypertensive drugs:

• Expected difference = 10 mmHg

• SD = 15 mmHg

• α = 0.05, β = 0.20 (power 80%)

n=2×(1.96+0.8410/15)2=2×(2.8×1.5)2=2×(4.2)2=2×17.64=35.28n = 2 \times \left( \frac{1.96 + 0.84}{10/15} \right)^2 = 2 \times (2.8 \times 1.5)^2 = 2 \times (4.2)^2 = 2 \times 17.64 = 35.28n=2×(10/151.96+0.84​)2=2×(2.8×1.5)2=2×(4.2)2=2×17.64=35.28

✅ n ≈ 36 per group

📊 4. Adjustments

• For Non-response or Attrition:
  nadjusted=n(1−expected dropout rate)n_{adjusted} = \frac{n}{(1 - \text{expected dropout rate})}nadjusted​=(1−expected dropout rate)n​
  e.g., If dropout = 10%, then multiply by 1.11.

• For Finite Population (N 10,000):
  nadj=n1+n−1Nn_{adj} = \frac{n}{1 + \frac{n - 1}{N}}nadj​=1+Nn−1​n​

💡 Practical Notes for Medical Research

• Use pilot studies or previous literature to estimate SD or p.

• For unknown proportions, assume p = 0.5 (gives largest sample).

• Always justify sample size in proposals (Ethics & IRB requirement).

• Many use software like OpenEpi, Raosoft, or Epi Info for precise calculations.