Notes on Hypothesis Tests for 2 Proportions

Comparing Two Proportions

• Estimating Difference Between Two Proportions:
  • Example: Comparing admission rates at ages 17 and 18.
  • Statistic used: Difference in sample proportions
  • Population proportions:
    • Population 1: $p<em>1$, size $n</em>1$, sample proportion $\bar{p}_1$
    • Population 2: $p<em>2$, size $n</em>2$, sample proportion $\bar{p}_2$

Sampling Distribution of Proportions

• For independent samples of sizes $n<em>1$ and $n</em>2$ from populations with parameters $p<em>1$ and $p</em>2$:
  • Mean:
  • $ext{Mean of the sampling distribution } \bar{p} = p$
  • $ext{Mean of } (\bar{p}<em>1 - \bar{p}</em>2) = p<em>1 - p</em>2$
  • Standard Deviation:
  • $ext{Std. deviation of } \bar{p} = \frac{p(1 - p)}{n}$
  • $ext{Std. deviation of } (\bar{p}<em>1 - \bar{p}</em>2) = \sqrt{\frac{p<em>1(1 - p</em>1)}{n<em>1} + \frac{p</em>2(1 - p<em>2)}{n</em>2}}$
  • Normality Conditions:
  • $n p ext{ and } n(1 - p) ext{ should be } ext{≥} 10$ for both samples

Assumptions and Conditions

• Independence Observations:
  • Randomization Condition:
  • Data drawn independently and randomly from a homogeneous population.
  • 10% Condition:
  • Sample should not exceed 10% of population when sampled without replacement.
• Independent Groups:
  • Two groups must be independent.
• Sample Size Condition:
  • Each group must be sufficiently large.
  • Success/Failure Condition:
  • $n<em>1 p</em>1 ext{ and } n<em>1(1 - p</em>1) ext{ ≥ } 10$
  • $n<em>2 p</em>2 ext{ and } n<em>2(1 - p</em>2) ext{ ≥ } 10$

Confidence Interval for 2 Population Proportions

• Formula:
  • $\bar{p}<em>1 - \bar{p}</em>2 ext{ ± } z^* \sqrt{\frac{\bar{p}<em>1(1 - \bar{p}</em>1)}{n<em>1} + \frac{\bar{p}</em>2(1 - \bar{p}<em>2)}{n</em>2}}$
• Example:
  • Smokers (n1=150): 95 with prominent wrinkles, $\bar{p}_1 = 0.63$
  • Nonsmokers (n2=250): 105 with prominent wrinkles, $\bar{p}_2=0.42$
  • 95% CI for smokers:
  • $0.63 ± 1.96 × 0.0394 = (0.55, 0.71)$
  • 95% CI for nonsmokers:
  • $0.42 ± 1.96 × 0.0312 = (0.36, 0.48)$
  • Check for overlap in intervals: indicates proportion differences.

Two-Proportion z Test

• Hypotheses:
  • Two-tailed: $H<em>0: p</em>1 - p<em>2 = p</em>0$ vs $H<em>a: p</em>1 - p<em>2 \neq p</em>0$
  • Upper-tailed: $H<em>0: p</em>1 - p<em>2 = p</em>0$ vs $H<em>a: p</em>1 - p<em>2 > p</em>0$
  • Lower-tailed: $H<em>0: p</em>1 - p<em>2 = p</em>0$ vs $H<em>a: p</em>1 - p<em>2 < p</em>0$
• Assumption & Conditions:
  • Random samples, independent observations, large sizes
  • Normal Conditions: $n<em>1 p</em>1, n<em>1(1 - p</em>1), n<em>2 p</em>2, n<em>2(1 - p</em>2) ext{ ≥ } 10$
• Test Statistic:
  • $z<em>0 = \frac{\bar{p}</em>1 - \bar{p}<em>2 - (p</em>1 - p<em>2)}{\sqrt{\frac{p</em>1(1 - p<em>1)}{n</em>1} + \frac{p<em>2(1 - p</em>2)}{n_2}}}$
  • If null is true, pool the proportions:
  • $\bar{p}<em>{pooled} = \frac{n</em>1 \bar{p}<em>1 + n</em>2 \bar{p}<em>2}{n</em>1 + n_2}$
• Modified test statistic:
  • $z<em>0 = \frac{\bar{p}</em>1 - \bar{p}<em>2}{\bar{p}</em>{pooled}(1 - \bar{p}<em>{pooled})(\frac{1}{n</em>1} + \frac{1}{n_2})}$

Decision Making

• p-value Criteria:
  • For two-tailed: p-value = 2 × P(Z > z_0)
  • For upper-tail: p-value = P(Z > z_0)
  • For lower-tail: p-value = P(Z < z_0)
• Decision rule:
  • If $p-value ≤ \alpha$, then reject $H_0$
  • If p-value > \alpha, do not reject $H_0$

Example Continuation

• Proportions of smokers and non-smokers with wrinkles at $\alpha=0.05$:
  • Test statistic:
  • $z<em>0 = \frac{\bar{p}</em>1 - \bar{p}<em>2}{\bar{p}</em>{pooled}(1 - \bar{p}<em>{pooled})(\frac{1}{n</em>1} + \frac{1}{n_2})} = 4.2$
  • Pooled proportion:
  • $\bar{p}_{pooled} = 0.5$
  • p-value calculation:
  • $p-value = 0$
• Conclusion:
  • Reject $H_0$, indicating a significant difference in proportions of smokers and non-smokers with wrinkles.