Sample Size Determination

Goal: Estimate mean household income within $2,000 (margin of error)

Given:

• 95% confidence → z* = 1.96
• Estimated σ = $15,000
• E = $2,000

Solution: $n = \left(\frac{1.96 \times 15000}{2000}\right)^2 = \left(\frac{29400}{2000}\right)^2 = 14.7^2 = 216.09$

Required: n = 217 (always round UP)

Goal: Estimate voter support within ±3 percentage points

Given:

• 95% confidence → z* = 1.96
• E = 0.03
• p unknown → use 0.5

Solution: $n = 0.5(0.5)\left(\frac{1.96}{0.03}\right)^2 = 0.25 \times 65.33^2 = 0.25 \times 4268.4 = 1067.1$

Required: n = 1068 (round UP)

That’s why polls often survey about 1000 people!

If a previous poll showed 60% support:

$n = 0.6(0.4)\left(\frac{1.96}{0.03}\right)^2 = 0.24 \times 4268.4 = 1024.4$

Required: n = 1025

Slightly smaller because p(1-p) is smaller when p ≠ 0.5.

Goal: Detect a 5-point improvement in blood pressure

Given:

• α = 0.05 (two-tailed) → zα/2 = 1.96
• Power = 80% → zβ = 0.84
• σ = 12 mmHg
• δ = 5 mmHg

Solution: $n = \frac{2(1.96 + 0.84)^2 (12)^2}{5^2} = \frac{2(7.84)(144)}{25} = \frac{2258}{25} = 90.3$

Required: n = 91 per group (182 total)

To detect a medium effect (d = 0.5) with 80% power at α = 0.05:

n per group ≈ 64

For small effect (d = 0.2): n per group ≈ 393

For large effect (d = 0.8): n per group ≈ 26

Calculated n = 400 Expected response rate = 60%

$n_{adjusted} = \frac{400}{0.60} = 667$

Need to initially contact 667 people to get ~400 responses.

Calculated n₀ = 400, Population N = 2000

$n_{adjusted} = \frac{400}{1 + \frac{399}{2000}} = \frac{400}{1.20} = 333$

Need only 333 instead of 400.

Note: If N is very large, this correction is negligible.