Interval for a Difference of Proportions

Compare support in two library populations

• Independent random samples find 60 supporters among 100 users in library 1 and 50 among 100 in library 2. Estimate $p_1-p_2$.
• The point estimate is $\hat p_1-\hat p_2=0.60-0.50=0.10$, or ten percentage points. It is not a 10% relative increase.

Check both groups and their independence

• Check the random design and independence between groups. For sampling without replacement, check $n_1\le0.10N_1$ and $n_2\le0.10N_2$.
• The observed counts are 60, 40, 50 and 50; all are at least ten. Surveys of the same people before and after are paired, not independent samples.

Combine the two variance estimates

• The two-sample z-interval 双样本z区间 uses $SE=\sqrt{\hat p_1(1-\hat p_1)/n_1+\hat p_2(1-\hat p_2)/n_2}$.
• Here $SE=\sqrt{0.60(0.40)/100+0.50(0.50)/100}=0.07$. The variance terms add; the standard errors do not simply add.

Build the difference interval

• Use $(\hat p_1-\hat p_2)\pm z^*SE$. At 95%, this is $0.10\pm1.96(0.07)$, giving $(-0.0372,0.2372)$.
• The interval uses separate sample proportions, because it does not assume $p_1=p_2$. Do not replace them with a pooled estimate.

Keep the group order throughout

• We are 95% confident that population support in library 1 minus library 2 lies between -3.72 and 23.72 percentage points.
• Reversing the subtraction reverses the endpoints: the interval for $p_2-p_1$ is $(-0.2372,0.0372)$.

Explain the remaining uncertainty

• The positive estimate favours library 1, but this interval also includes zero and negative differences. It does not establish which population has higher support.
• Both samples contribute uncertainty. A large first sample does not excuse a second sample with too few successes or failures.

Write from an interval that is entirely positive

• A confidence interval for a difference of proportions is 0.02 to 0.18.
• The difference was group 1 minus group 2. The lesson interval that includes negative 0.2372 is another study.