Sample Size Determination

Overview

Sample size determination involves calculating how many observations are needed to achieve a desired level of precision or statistical power.

For Estimating a Mean

When $\sigma$ is Known

$$n = \left(\frac{z_{\alpha/2} \times \sigma}{ME}\right)^2$$

Where $ME$ = desired margin of error

When $\sigma$ is Unknown

Use a pilot study or estimate $\sigma$, or use:

$$n = \left(\frac{t_{\alpha/2, df} \times s}{ME}\right)^2$$

For Estimating a Proportion

$$n = \hat{p}(1-\hat{p}) \times \left(\frac{z_{\alpha/2}}{ME}\right)^2$$

Conservative Estimate

When $p$ is unknown, use $p = 0.5$ (maximizes variance):

$$n = 0.25 \times \left(\frac{z_{\alpha/2}}{ME}\right)^2$$

For Hypothesis Testing (Power Analysis)

$$n = \frac{(z_\alpha + z_\beta)^2 \times \sigma^2}{\delta^2}$$

Where:

• $z_\alpha$ = critical value for significance level
• $z_\beta$ = critical value for power ($1 - \beta$)
• $\delta$ = minimum detectable effect size

Common Values

Confidence	$z_{\alpha/2}$
90%	1.645
95%	1.96
99%	2.576

Power	$z_\beta$
80%	0.84
90%	1.28
95%	1.645

Practical Considerations

Factor	Effect on Required $n$
Smaller ME	Larger $n$
Higher confidence	Larger $n$
Higher power	Larger $n$
Larger $\sigma$ or variance	Larger $n$
Smaller effect size	Larger $n$

Examples

Example 1: Sample Size for Mean

Estimate mean with $ME = 5$, $\sigma = 20$, 95% confidence:

$$n = \left(\frac{1.96 \times 20}{5}\right)^2 = \left(\frac{39.2}{5}\right)^2 = 7.84^2 = 61.5 \Rightarrow n = 62$$

Example 2: Sample Size for Proportion

Estimate proportion within $\pm 3\%$ ($ME = 0.03$), 95% confidence, unknown $p$:

$$n = 0.25 \times \left(\frac{1.96}{0.03}\right)^2 = 0.25 \times 65.33^2 = 0.25 \times 4268 = 1067$$

Example 3: With Prior Estimate of $p$

If we expect $p \approx 0.20$, $ME = 0.03$, 95% confidence:

$$n = (0.20)(0.80) \times \left(\frac{1.96}{0.03}\right)^2 = 0.16 \times 4268 = 683$$

Less than conservative estimate since $p(1-p) < 0.25$

Example 4: For Power Analysis

Detect difference of $\delta = 5$, $\sigma = 15$, $\alpha = 0.05$, power = 80%:

$$n = \frac{(1.96 + 0.84)^2 \times 15^2}{5^2} = \frac{7.84 \times 225}{25} = \frac{1764}{25} = 70.6 \Rightarrow n = 71 \text{ per group}$$

Adjustments

Finite Population Correction

If sampling from population of size $N$:

$$n_{\text{adj}} = \frac{n}{1 + \frac{n-1}{N}}$$

Adjusting for Nonresponse

If expected response rate is $r$:

$$n_{\text{needed}} = \frac{n}{r}$$

Sample Size Quick Reference

For 95% CI for mean ($\sigma = 10$):

Desired ME	Required $n$
5.0	16
2.0	97
1.0	385
0.5	1537

For 95% CI for proportion (conservative):

Desired ME	Required $n$
10%	97
5%	385
3%	1068
1%	9604