We focus on the difference $\hat{p}_1 - \hat{p}_2$.

Sampling Distribution of $\hat{p}_1 - \hat{p}_2$

When conditions are met, $\hat{p}_1 - \hat{p}_2$ is approximately Normal with:

• Mean: $\mu_{\hat{p}_1 - \hat{p}_2} = p_1 - p_2$
• Standard deviation: Depends on context (see below)

Two-Sample z-Interval for $p_1 - p_2$ (Unpooled SE)

For a confidence interval, we use the unpooled standard error:

$SE_{\text{unpooled}} = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$

Confidence Interval: $(\hat{p}_1 - \hat{p}_2) \pm z^* \cdot SE_{\text{unpooled}}$

Where $z^*$ is the critical value (e.g., $z^* = 1.96$ for 95% confidence).

Two-Sample z-Test for $p_1 = p_2$ (Pooled SE)

For a hypothesis test (usually testing $H_0: p_1 = p_2$), we use a pooled standard error that assumes the null hypothesis:

$\hat{p}_{\text{pool}} = \frac{\text{successes}_1 + \text{successes}_2}{n_1 + n_2}$

$SE_{\text{pooled}} = \sqrt{\hat{p}_{\text{pool}}(1 - \hat{p}_{\text{pool}}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}$

Test Statistic: $z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{SE_{\text{pooled}}}$

Compare to the standard Normal distribution to find the p-value.

Conditions for Two-Sample z-Procedures

Both methods require:

Worked Example 1: Two-Sample z-Interval

In a clinical trial:

• Treatment group: 35 successes out of 100 patients → $\hat{p}_1 = 0.35$
• Control group: 20 successes out of 100 patients → $\hat{p}_2 = 0.20$

Construct a 95% confidence interval for $p_1 - p_2$.

Check conditions:

• $n_1 \hat{p}_1 = 100(0.35) = 35 \geq 10$ ✓
• $n_1(1-\hat{p}_1) = 100(0.65) = 65 \geq 10$ ✓
• $n_2 \hat{p}_2 = 100(0.20) = 20 \geq 10$ ✓
• $n_2(1-\hat{p}_2) = 100(0.80) = 80 \geq 10$ ✓

Calculate unpooled SE: $SE = \sqrt{\frac{0.35(0.65)}{100} + \frac{0.20(0.80)}{100}} = \sqrt{\frac{0.2275}{100} + \frac{0.16}{100}} = \sqrt{0.003875} \approx 0.0622$

95% CI (z = 1.96):* $(0.35 - 0.20) \pm 1.96(0.0622) = 0.15 \pm 0.122 = (0.028, 0.272)$

We are 95% confident that $p_1 - p_2$ is between 0.028 and 0.272 (2.8% to 27.2% difference).

Worked Example 2: Two-Sample z-Test

Test $H_0: p_1 = p_2$ vs. $H_a: p_1 \neq p_2$ at $\alpha = 0.05$ using the data above.

Calculate pooled proportion: $\hat{p}_{pool} = \frac{35 + 20}{100 + 100} = \frac{55}{200} = 0.275$

Calculate pooled SE: $SE = \sqrt{0.275(0.725) \left( \frac{1}{100} + \frac{1}{100} \right)} = \sqrt{0.199375 \times 0.02} = \sqrt{0.00399} \approx 0.0631$

Test statistic: $z = \frac{0.35 - 0.20}{0.0631} = \frac{0.15}{0.0631} \approx 2.378$

p-value (two-tailed): $p\text{-value} = 2 \times P(Z \geq 2.378) \approx 2(0.0087) \approx 0.0174$

Since $p\text{-value} = 0.0174 < 0.05$, we reject $H_0$. Significant evidence that $p_1 \neq p_2$.

Common Pitfalls

⚠️ Pooled vs. Unpooled: Use unpooled SE for confidence intervals (we estimate both $p_1$ and $p_2$ separately). Use pooled SE for hypothesis tests under $H_0: p_1 = p_2$ (assumes they are equal).

⚠️ Forgetting the 10% Condition: Always check $n < 0.1N$ within each population to ensure independence of observations.

⚠️ Large Counts with Pooled Proportion: For tests, check counts using $\hat{p}_{pool}$, not individual $\hat{p}_i$.

Calculator Tip

💡 TI-84 / TI-Nspire: Use 2-PropZInt for confidence intervals and 2-PropZTest for hypothesis tests. Input $x_1$, $n_1$, $x_2$, $n_2$ and the alternative hypothesis. The calculator will compute the interval or test statistic automatically.

Pooled proportion: $\hat{p}_{pool} = \frac{48 + 35}{150 + 150} = \frac{83}{300} \approx 0.2767$

Check Large Counts (with pooled):

• $n_1 \hat{p}_{pool} = 150(0.2767) \approx 41.5 \geq 10$ ✓
• All four counts pass.

Pooled SE: $SE = \sqrt{0.2767(0.7233) \left( \frac{1}{150} + \frac{1}{150} \right)} = \sqrt{0.2 \times 0.00133} \approx 0.0516$

Test statistic: $z = \frac{0.32 - 0.233}{0.0516} \approx 1.68$

p-value (one-tailed, $H_a: p_1 > p_2$): $P(Z > 1.68) \approx 0.0465$

Since $p\text{-value} \approx 0.0465 < 0.05$, we reject $H_0$. Significant evidence that treatment is more effective.

Values:

• $z_{\alpha/2} = 1.96$ (two-tailed, $\alpha = 0.05$)
• $z_{\beta} = 0.84$ (80% power)
• $p_1 = 0.60$, $p_2 = 0.50$
• $p_1 - p_2 = 0.10$

Compute: $p_1(1-p_1) + p_2(1-p_2) = 0.60(0.40) + 0.50(0.50) = 0.24 + 0.25 = 0.49$

$n \approx \frac{(1.96 + 0.84)^2 \times 0.49}{0.10^2} = \frac{(2.80)^2 \times 0.49}{0.01}$ $= \frac{7.84 \times 0.49}{0.01} = \frac{3.84}{0.01} = 384$

Answer: Approximately 384 participants per group (768 total) are needed.