Topic 3.10 Notes – Constructing a Confidence Interval for the Difference Between Two Population Proportions

What a Two-Proportion z-Interval Estimates

A two-proportion $z$-interval estimates the difference between two population proportions.

The parameter is

$$p_1-p_2$$

where $p_1$ and $p_2$ are the true proportions of successes in group 1 and group 2.

That means you need:

• a categorical response variable
• exactly two outcomes, or a way to turn it into success/failure
• two independent groups being compared on that same response

The point estimate comes from the samples:

• $\hat p_1 = \dfrac{x_1}{n_1}$
• $\hat p_2 = \dfrac{x_2}{n_2}$
• estimated difference $= \hat p_1-\hat p_2$

What you must define in context:

• the success category
  (example: “plans to attend a four-year college”)
• who group 1 and group 2 are
  (example: District A seniors and District B seniors)
• the order of subtraction
  (example: District A minus District B)

If your estimate or interval is:

• positive, group 1 likely has the higher proportion
• negative, group 2 likely has the higher proportion

Reversing the order flips the sign but keeps the same width.

This method fits two independent groups only. It does not fit:

• one proportion
• matched pairs or repeated measures
• multi-category distribution comparisons like chi-square

Conditions for a Two-Proportion z-Interval

You need the same three-condition logic you’ve used before, but checked for both groups.

Randomization

The data should come from:

• two independent random samples, or
• a randomized experiment with two treatment groups

You have to justify this from the description. You cannot just assume randomness because the problem wants a confidence interval.

The groups also must be independent of each other. If the same people are measured twice, this is not the right procedure.

10% condition

For sampling without replacement, check each sample separately:

• $n_1 \le 0.10N_1$
• $n_2 \le 0.10N_2$

This condition is not needed for randomized experiments.

Normality condition

All four observed counts must be at least 10:

• $n_1\hat p_1 \ge 10$
• $n_1(1-\hat p_1) \ge 10$
• $n_2\hat p_2 \ge 10$
• $n_2(1-\hat p_2) \ge 10$

Same idea, written as counts:

• $x_1 \ge 10$, $n_1-x_1 \ge 10$
• $x_2 \ge 10$, $n_2-x_2 \ge 10$

These are observed counts, not expected counts from a hypothesis test.

Building the Interval

Here’s the full build. For a confidence interval, you keep the middle area and split the remaining area evenly into the two tails.

1. Identify the procedure
   Two-sample $z$-interval for $p_1-p_2$

2. Compute sample proportions
   $$\hat p_1=\frac{x_1}{n_1}, \qquad \hat p_2=\frac{x_2}{n_2}$$

3. Compute the point estimate
   $$\hat p_1-\hat p_2$$

4. Compute the standard error
   $$SE=\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}$$

5. Choose $z^*$

   • 90% → 1.645
   • 95% → 1.96
   • 99% → 2.576

6. Compute margin of error
   $$ME=z^*\cdot SE$$

7. Build the interval
   $$(\hat p_1-\hat p_2)\pm ME$$

Quick example with $x_1=48, n_1=80, x_2=36, n_2=90$:

• $\hat p_1=0.60$
• $\hat p_2=0.40$
• point estimate $=0.20$

For 95% confidence, use the middle 95% of the normal curve, which gives $z^*=1.96$. The interval is centered at 0.20 using the SE formula above. On a calculator, use 2-PropZInt with $x_1, n_1, x_2, n_2$, and the confidence level.

What the Formula Is Doing

Every confidence interval is point estimate ± margin of error.

Here, the center is the observed difference $\hat p_1-\hat p_2$. Even though you are subtracting proportions, the variability adds through the variances, which is why both terms appear with a plus sign inside the square root.

The width changes in predictable ways:

• higher confidence level → bigger $z^*$ → wider interval
• larger sample sizes → smaller SE → narrower interval

One of the biggest AP traps is pooling. For a confidence interval, you do not combine the proportions. You use $\hat p_1$ and $\hat p_2$ separately because you are estimating an unknown difference, not assuming $p_1=p_2$.

Report differences as proportions or percentage points. A difference of $0.08$ means 8 percentage points.

Common Mistakes and Exam Traps

• Parameter vs statistic gets mixed up a lot. The parameter is $p_1-p_2$. The point estimate is $\hat p_1-\hat p_2$.
• Order problems cost points fast. Your parameter, arithmetic, interval, and interpretation all need the same group 1 minus group 2 order.
• Pooled $\hat p$ belongs to a two-proportion significance test, not this interval.
• All four counts must be checked. A large total sample size does not fix one group having too few successes or failures.
• Observed counts are the ones used here. Do not switch to expected counts from hypothesis testing.
• Dependent groups mean wrong method. Paired data needs a different procedure.
• Early rounding can change endpoints. Keep calculator values as long as possible.
• Context matters on AP Stats. Name the success, both groups, and the subtraction order in words.