Sampling Distributions for Differences in Sample Proportions - AP Statistics

SECTION 1

Historical Context & Motivation

Comparing two groups is one of the oldest and most fundamental questions in statistics: Is a new treatment more effective than an old one? Do voters in one region favor a policy more than voters in another? The formal machinery for answering such questions took shape over several centuries, building on the theory of probability, the Central Limit Theorem, and the development of sampling distributions. Before statisticians established the sampling distribution of the difference in sample proportions, practitioners had no rigorous way to decide whether an observed difference between two groups was real or merely the product of random variation.

1713 • Bernoulli's Ars Conjectandi

Jacob Bernoulli published the first version of the law of large numbers, establishing that sample proportions converge to the true population proportion as the sample size grows. This was the conceptual seed for all future work on sampling distributions.

1733 • De Moivre's Normal Approximation

Abraham de Moivre showed that the binomial distribution could be approximated by the normal curve when sample sizes are large, providing the foundation for using z-scores with proportions.

1810 • Laplace's Central Limit Theorem

Pierre-Simon Laplace generalized the normal approximation to sums and means of arbitrary random variables. His Central Limit Theorem guaranteed that linear combinations of independent statistics—including differences—are approximately normal.

1900s • Pearson & Fisher Formalize Hypothesis Testing

Karl Pearson and Ronald Fisher developed the framework of significance testing and confidence intervals. Two-sample tests for proportions became standard tools in biostatistics, social science, and quality control.

2000s • Modern AP Statistics Curriculum

The AP Statistics course codified the two-sample z-test for proportions as a core inference procedure, embedding the sampling distribution of p̂₁ − p̂₂ as essential knowledge for every statistics student.

The central question this topic addresses is deceptively simple: if we draw an independent random sample from each of two populations and compute the difference in sample proportions, what does the distribution of all possible such differences look like? Once we characterize that distribution—its center, spread, and shape—we have the tools to build confidence intervals and conduct hypothesis tests for comparing two proportions.

SECTION 2

Core Principles & Definitions

Before diving into formulas, it is essential to be precise about what we mean by a sampling distribution in the two-sample context. Suppose Population 1 has a true proportion p₁ of individuals with some characteristic, and Population 2 has a true proportion p₂. We independently draw a random sample of size n₁ from the first population and a random sample of size n₂ from the second, then compute the sample proportions p̂₁ and p̂₂. The quantity of interest is the difference in sample proportions, p̂₁ − p̂₂. If we were to repeat this process infinitely many times, the collection of all resulting differences would form the sampling distribution of p̂₁ − p̂₂.

Center (Mean)

The mean of the sampling distribution equals the true difference: μ(p̂₁ − p̂₂) = p₁ − p₂. This follows from the linearity of expectation. The statistic is an unbiased estimator of the population parameter.

Spread (Standard Deviation)

Because the two samples are independent, the variances add: σ(p̂₁ − p̂₂) = √[p₁(1 − p₁)/n₁ + p₂(1 − p₂)/n₂]. Larger samples shrink this standard deviation, making the estimate more precise.

Shape (Normality)

When n₁p₁ ≥ 10, n₁(1 − p₁) ≥ 10, n₂p₂ ≥ 10, and n₂(1 − p₂) ≥ 10, the sampling distribution is approximately normal. This Large Counts condition ensures the CLT applies.

Independence

The two samples must be independent of each other, and within each sample, observations must be independent. For sampling without replacement, we invoke the 10% condition: n₁ ≤ 0.10N₁ and n₂ ≤ 0.10N₂.

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KEY TAKEAWAY

Think of each sample proportion as a dart thrown at a target centered on its true population proportion. Each dart lands with some random scatter. When you compute the difference p̂₁ − p̂₂, you are measuring the gap between two independently scattered darts. The variability in that gap is greater than the variability in either dart alone—precisely because independent variances add. This additive property is what makes the standard deviation formula for the difference larger than either single-proportion standard deviation.

SECTION 3

Visual Explanation

The following diagram illustrates how two individual sampling distributions for p̂₁ and p̂₂ combine to produce the sampling distribution of their difference. Each bell curve is centered on its respective population proportion, and the resulting difference distribution is centered on p₁ − p₂ with a spread determined by the sum of the two variances.

The top row shows the individual sampling distributions of p̂₁ (cyan, centered at 0.60) and p̂₂ (violet, centered at 0.45). The bottom curve (emerald) shows the resulting sampling distribution of p̂₁ − p̂₂, centered at 0.15, with a wider spread reflecting the addition of variances from both component distributions.

Notice that the bottom curve is visibly wider than either individual curve. This is not an accident; it reflects a fundamental property of independent random variables. When you subtract two independent quantities, the uncertainties do not cancel—they compound. The variance of a difference equals the sum of the variances, not the difference. This is one of the most common sources of student error on the AP exam, where test-takers sometimes mistakenly subtract variances.

SECTION 4

Mathematical Framework

We now formalize the three properties—center, spread, and shape—of the sampling distribution. Recall that p̂₁ and p̂₂ are independent random variables because the two samples are drawn independently from their respective populations.

MEAN OF THE DIFFERENCE

μ_{p̂₁ − p̂₂} = p₁ − p₂

By linearity of expectation, E[p̂₁ − p̂₂] = E[p̂₁] − E[p̂₂] = p₁ − p₂. The statistic p̂₁ − p̂₂ is an unbiased estimator of the true difference in population proportions.

STANDARD DEVIATION OF THE DIFFERENCE

σ_{p̂₁ − p̂₂} = √[ p₁(1 − p₁)/n₁ + p₂(1 − p₂)/n₂ ]

Since p̂₁ and p̂₂ are independent, Var(p̂₁ − p̂₂) = Var(p̂₁) + Var(p̂₂). We then take the square root. Here p₁, p₂ are population proportions and n₁, n₂ are the respective sample sizes.

STANDARD ERROR (ESTIMATED)

SE_{p̂₁ − p̂₂} = √[ p̂₁(1 − p̂₁)/n₁ + p̂₂(1 − p̂₂)/n₂ ]

In practice, the population proportions are unknown. We substitute sample proportions to obtain the standard error (SE), which estimates the standard deviation. This version is used for confidence intervals.

NORMALITY (LARGE COUNTS) CONDITION

n₁p₁ ≥ 10, n₁(1 − p₁) ≥ 10, n₂p₂ ≥ 10, n₂(1 − p₂) ≥ 10

When all four products meet or exceed 10, the Central Limit Theorem guarantees that p̂₁ − p̂₂ is approximately normally distributed. In practice, when the true proportions are unknown, we check with the observed counts: n₁p̂₁, n₁(1 − p̂₁), n₂p̂₂, n₂(1 − p̂₂).

⚠

AP Exam Tip: Pooled vs. Unpooled

When constructing a confidence interval, use the unpooled SE formula above (with p̂₁ and p̂₂ separately). When conducting a hypothesis test under H₀: p₁ = p₂, use the pooled proportion p̂_c = (X₁ + X₂)/(n₁ + n₂) to compute the SE. Mixing these up is a common error on the exam.

SECTION 5

Conditions & Checking Assumptions

For the normal model to apply to p̂₁ − p̂₂, three sets of conditions must be verified. On the AP exam, you are expected to name each condition, check it using the problem's data, and state whether it is satisfied. Skipping this step costs points on every inference FRQ. The following diagram and table summarize the full checklist.

The three condition boxes—Random, 10% Condition, and Large Counts—must all be satisfied before you can assert that p̂₁ − p̂₂ follows an approximately normal distribution.

Condition	What to Check	Why It Matters
Random	Both samples were selected using a random mechanism (SRS, stratified random sample, or random assignment in an experiment).	Ensures the sample statistics are unbiased estimators of the population parameters and that the sampling distribution theory applies.
10% (Independence)	n₁ ≤ 0.10 × N₁ and n₂ ≤ 0.10 × N₂, where N₁ and N₂ are the population sizes.	When sampling without replacement, observations are not perfectly independent. The 10% condition ensures the dependence is negligible so the variance formula remains accurate.
Large Counts (Normality)	n₁p̂₁ ≥ 10, n₁(1 − p̂₁) ≥ 10, n₂p̂₂ ≥ 10, n₂(1 − p̂₂) ≥ 10.	Guarantees each individual sampling distribution is approximately normal, which in turn ensures the difference is approximately normal.

Summary of conditions for inference on p̂₁ − p̂₂

SECTION 6

Worked Example

A researcher wants to compare the proportion of adults who support a new environmental regulation in two states. In State A, a random sample of 400 adults finds that 248 support the regulation. In State B, a random sample of 500 adults finds that 265 support it. Describe the sampling distribution of p̂_A − p̂_B.

Describing the Sampling Distribution of p̂_A − p̂_B

1
Step 1 — Identify Given Values

From the problem: $n_A$ = 400, $X_A$ = 248, so p̂_A = 248/400 = 0.62. Also $n_B$ = 500, $X_B$ = 265, so p̂_B = 265/500 = 0.53. The observed difference is p̂_A − p̂_B = 0.62 − 0.53 = 0.09.

p̂_A = 0.62, p̂_B = 0.53, p̂_A − p̂_B = 0.09
2
Step 2 — Check the Random Condition

The problem states that both samples are random samples of adults from their respective states. The two samples are selected independently. The Random condition is satisfied.
3
Step 3 — Check the 10% Condition

Each state has millions of adults, so 400 < 0.10 × $N_A$ and 500 < 0.10 × $N_B$ . The 10% condition is satisfied for both samples, and we can treat individual observations as approximately independent within each sample.
4
Step 4 — Check the Large Counts Condition

Compute the four counts: $n_A$ × p̂_A = 400 × 0.62 = 248 ≥ 10 ✓, $n_A$ × (1 − p̂_A) = 400 × 0.38 = 152 ≥ 10 ✓, $n_B$ × p̂_B = 500 × 0.53 = 265 ≥ 10 ✓, $n_B$ × (1 − p̂_B) = 500 × 0.47 = 235 ≥ 10 ✓. All four counts exceed 10, so the Large Counts condition is satisfied.
5
Step 5 — Compute the Mean

The mean of the sampling distribution is μ_{p̂_A − p̂_B} = $p_A$ − $p_B$ . We estimate this with our observed p̂_A − p̂_B = 0.62 − 0.53 = 0.09. If we are describing the distribution under known population parameters, the mean would be $p_A$ − $p_B$ .

Mean = $p_A$ − $p_B$ (estimated by 0.09)
6
Step 6 — Compute the Standard Deviation

Using the observed proportions as estimates: SE = √[p̂_A(1 − p̂_A)/ $n_A$ + p̂_B(1 − p̂_B)/ $n_B$ ] = √[(0.62)(0.38)/400 + (0.53)(0.47)/500] = √[0.2356/400 + 0.2491/500] = √[0.000589 + 0.000498] = √0.001087 ≈ 0.03297.

SE ≈ 0.033
7
Step 7 — State the Sampling Distribution

Since all conditions are met, the sampling distribution of p̂_A − p̂_B is approximately normal with mean $p_A$ − $p_B$ and standard deviation ≈ 0.033. In notation:

p̂_A − p̂_B ~ Normal( $p_A$ − $p_B$ , 0.033)

SECTION 7

Common Errors & Pitfalls

The sampling distribution of p̂₁ − p̂₂ is a topic where students frequently lose points on the AP exam—not from a lack of formula knowledge, but from procedural missteps and conceptual confusions. The table below catalogs the most common errors alongside the correct approach.

Common Error	Why It's Wrong	Correct Approach
Subtracting standard deviations instead of adding variances	Standard deviations are not additive. The variance of a difference of independent variables equals the sum of the variances, not the difference.	Always add the variances first, then take the square root: σ = √(σ₁² + σ₂²).
Using the pooled proportion when constructing a confidence interval	Pooling assumes p₁ = p₂, which is the null hypothesis. A CI does not assume H₀ is true.	Use separate sample proportions p̂₁ and p̂₂ in the SE formula for confidence intervals.
Forgetting to check conditions	Without verified conditions, the normal model may be invalid, and the resulting inference is unsupported.	Always state and verify Random, 10%, and Large Counts conditions with specific numbers from the data.
Treating dependent samples as independent	If the same individuals are measured twice (e.g., before/after), the samples are paired, not independent, and this formula does not apply.	Use the two-sample z-procedure only for genuinely independent samples. For paired data, analyze the single set of differences.
Confusing standard deviation with standard error	The standard deviation uses population parameters (p₁, p₂); the standard error uses sample statistics (p̂₁, p̂₂). Terminology matters on the AP rubric.	Use 'standard deviation' when referencing theory with known p₁ and p₂. Use 'standard error' when estimating with p̂₁ and p̂₂.

Frequent mistakes on the AP Statistics exam related to inference for two proportions

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REMEMBER: VARIANCES ADD

Imagine two machines on an assembly line, each introducing its own random error to a product's length. If one machine's error has variance 4 and the other's has variance 9, the total variability in the combined error is 4 + 9 = 13, regardless of whether the errors are added or subtracted. In the same way, the variance of p̂₁ − p̂₂ equals the sum of the individual variances. This principle—that independent uncertainties always compound—is a cornerstone of statistical inference.

SECTION 8

Connection to Inference Procedures

The sampling distribution of p̂₁ − p̂₂ is not an end in itself; it is the theoretical engine that powers two major inference procedures you will encounter on the AP exam. Understanding the sampling distribution allows you to construct confidence intervals for p₁ − p₂ and to perform two-proportion z-tests. The table below contrasts the two procedures and highlights how the SE formula changes between them.

Feature	Confidence Interval	Two-Proportion z-Test
Goal	Estimate the true difference p₁ − p₂ with a range of plausible values.	Test whether p₁ − p₂ equals a hypothesized value (usually 0).
SE Formula	Unpooled: √[p̂₁(1−p̂₁)/n₁ + p̂₂(1−p̂₂)/n₂]	Pooled: √[p̂_c(1−p̂_c)(1/n₁ + 1/n₂)] where p̂_c = (X₁+X₂)/(n₁+n₂)
Why the difference?	We make no assumption about the relationship between p₁ and p₂, so each is estimated separately.	Under H₀: p₁ = p₂, both populations share a common proportion—pooling gives a better estimate of this shared value.
Output	(p̂₁ − p̂₂) ± z* × SE	z = (p̂₁ − p̂₂ − 0) / SE, then find the p-value
Large Counts check uses	Individual sample proportions p̂₁ and p̂₂	Pooled proportion p̂_c for both samples

Confidence Interval vs. Hypothesis Test for two proportions

Beyond the AP course, the ideas behind the sampling distribution of a difference extend naturally to more advanced settings. In logistic regression, differences in proportions are modeled through log-odds, and the sampling distributions of regression coefficients follow similar normal approximation logic. In Bayesian inference, the posterior distribution for p₁ − p₂ can be computed directly by simulating differences from the posterior distributions of p₁ and p₂. The frequentist z-based approach you learn now is the simplest entry point into a rich family of comparison methods.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Two independent random samples are drawn from two populations. The sampling distribution of p̂₁ − p̂₂ has a standard deviation that depends on both p₁(1 − p₁)/n₁ and p₂(1 − p₂)/n₂. Why are the variances added rather than subtracted when computing the variance of the difference?

A. Because Var(X − Y) = Var(X) + (−1)²Var(Y), and squaring the coefficient makes both terms positive, so independent variances always add regardless of whether the variables are summed or differenced.
B. Because the positive covariance between the two sample proportions cancels the negative sign, effectively converting the subtraction of variances into an addition.
C. Because the Central Limit Theorem requires that all component variances be summed before the normal approximation can be applied.
D. Because larger sample sizes always produce larger combined variances, so the variances must be added to account for total sample size.
E. Because the addition of variances only applies when both sampling distributions are exactly normal, and the Large Counts condition guarantees exact normality.

Show answer (A)

(A) is correct. When two random variables are independent, the variance of their difference equals the sum of their variances: Var(X − Y) = Var(X) + Var(Y). This is because Var(X − Y) = Var(X) + (−1)²Var(Y) = Var(X) + Var(Y). The squaring of the −1 coefficient is what converts the subtraction into addition. Intuitively, subtracting two uncertain quantities does not reduce uncertainty—it increases it, because both sources of randomness contribute to variability in the result. (B) is incorrect because the covariance is zero for independent samples; the addition of variances does not depend on the sign of a covariance term. (C) is incorrect because the CLT does not dictate whether variances add or subtract—the addition of variances is a property of independence, not normality. (D) is incorrect because sample size affects the magnitude of each variance, not whether they are added or subtracted. (E) is incorrect because the formula holds for any difference of independent random variables, not just those that are normal.

Question 2: BASIC CALCULATION

A random sample of 200 voters in City X found that 120 favor a ballot measure (p̂₁ = 0.60). An independent random sample of 300 voters in City Y found that 135 favor the measure (p̂₂ = 0.45). What is the standard error of p̂₁ − p̂₂?

A. 0.033
B. 0.040
C. 0.045
D. 0.052
E. 0.060

Show answer (C)

(C) is correct. The standard error formula is SE = √[p̂₁(1 − p̂₁)/n₁ + p̂₂(1 − p̂₂)/n₂] = √[(0.60)(0.40)/200 + (0.45)(0.55)/300] = √[0.24/200 + 0.2475/300] = √[0.00120 + 0.000825] = √0.002025 = 0.045. (A) 0.033 is incorrect—this would result from using incorrect proportions or from accidentally subtracting variances. (B) 0.040 is incorrect—this might arise from a rounding error or from only computing one of the two variance components. (D) 0.052 is incorrect—this could result from using n − 1 denominators or from incorrectly squaring the proportions. (E) 0.060 is incorrect—this is the value of p̂₁ itself, not the SE of the difference.

Question 3: INTERMEDIATE

A researcher collects two independent random samples: 150 adults from Population A (62 successes) and 180 adults from Population B (90 successes). She wishes to describe the sampling distribution of p̂_A − p̂_B. Which of the following is the best description?

A. Approximately Normal with mean $p_A$ − $p_B$ and standard deviation 0.045
B. Approximately Normal with mean $p_A$ − $p_B$ and standard deviation 0.055
C. Approximately Normal with mean 0.087 and standard deviation 0.055
D. Approximately Normal with mean $p_A$ − $p_B$ and standard deviation 0.110
E. Cannot be determined because the Large Counts condition is not satisfied

Show answer (B)

(B) is correct. First, compute p̂_A = 62/150 ≈ 0.413 and p̂_B = 90/180 = 0.500. The mean is p̂_A − p̂_B ≈ −0.087. Check Large Counts:

n_A

× p̂_A = 62 ≥ 10 ✓;

n_A

(1 − p̂_A) = 88 ≥ 10 ✓;

n_B

× p̂_B = 90 ≥ 10 ✓;

n_B

(1 − p̂_B) = 90 ≥ 10 ✓. SE = √[(0.413)(0.587)/150 + (0.500)(0.500)/180] = √[0.001617 + 0.001389] = √0.003006 ≈ 0.055. So the sampling distribution is approximately Normal with mean

p_A

−

p_B

and SD ≈ 0.055. (A) is incorrect because the SD of 0.045 is too small, likely resulting from a computation error. (C) is incorrect because the mean is listed as 0.087 (positive), but the correct direction is p̂_A − p̂_B which is negative. (D) is incorrect because a standard deviation of 0.110 is roughly double the correct value. (E) is incorrect because the Large Counts condition IS satisfied, so the normal approximation is valid.

Question 4: APPLIED

4 pts

A pharmaceutical company runs a randomized controlled trial to compare the effectiveness of a new drug versus a placebo for relieving symptoms. In the treatment group, 84 out of 250 patients experienced symptom relief (p̂₁ = 0.336). In the placebo group, 51 out of 250 patients experienced relief (p̂₂ = 0.204). (a) Verify that the conditions for inference are met. (b) Describe the sampling distribution of p̂₁ − p̂₂, assuming the true proportions are p₁ and p₂. (c) Compute and interpret a 95% confidence interval for p₁ − p₂. (d) Based on the confidence interval, is there convincing evidence that the drug is more effective than the placebo? Justify your answer.

Show answer

Rubric (4 points): Point 1 — Conditions (a): Random: Patients were randomly assigned to treatment and placebo groups in a randomized experiment, so the random condition is met. 10% Condition: 250 is less than 10% of all potential patients with this condition. Large Counts: n₁p̂₁ = 84 ≥ 10 ✓, n₁(1 − p̂₁) = 166 ≥ 10 ✓, n₂p̂₂ = 51 ≥ 10 ✓, n₂(1 − p̂₂) = 199 ≥ 10 ✓. All conditions are satisfied. Point 2 — Sampling Distribution (b): Because all conditions are met, the sampling distribution of p̂₁ − p̂₂ is approximately Normal with mean μ = p₁ − p₂ and standard deviation σ = √[p₁(1 − p₁)/250 + p₂(1 − p₂)/250]. Point 3 — Confidence Interval (c): SE = √[(0.336)(0.664)/250 + (0.204)(0.796)/250] = √[0.000893 + 0.000650] = √0.001543 ≈ 0.03928. The 95% CI is (p̂₁ − p̂₂) ± 1.96 × SE = 0.132 ± 1.96(0.03928) = 0.132 ± 0.077 = (0.055, 0.209). We are 95% confident that the true difference in proportions (drug minus placebo) is between 0.055 and 0.209. Point 4 — Conclusion (d): Because the entire 95% confidence interval lies above 0, we have convincing evidence that p₁ > p₂—that is, the drug produces a higher proportion of symptom relief than the placebo. Since the study was a randomized experiment, we can conclude that the drug causes a higher rate of symptom relief, by somewhere between 5.5 and 20.9 percentage points.

Question 5: CRITICAL THINKING

4 pts

Suppose p₁ = 0.50 and p₂ = 0.50, and you draw independent random samples of sizes n₁ and n₂ from the two populations. (a) Derive a simplified expression for σ_{p̂₁ − p̂₂} in terms of n₁ and n₂ when p₁ = p₂ = 0.50. (b) Show that if n₁ = n₂ = n, the standard deviation simplifies to 1/√(2n). Explain intuitively why doubling the sample size in both groups does not cut the standard deviation in half. (c) Suppose you have a total budget of N = n₁ + n₂ observations. Using calculus or algebraic reasoning, show that σ_{p̂₁ − p̂₂} is minimized when n₁ = n₂ = N/2. Explain why equal allocation is optimal. (d) Discuss how this result informs practical study design when comparing two proportions.

Show answer

Rubric (4 points): Point 1 — Derivation (a): When p₁ = p₂ = 0.50, each term p(1 − p) = 0.25. So σ_{p̂₁ − p̂₂} = √[0.25/n₁ + 0.25/n₂] = 0.5 × √[1/n₁ + 1/n₂]. Point 2 — Simplification and Interpretation (b): If n₁ = n₂ = n, then σ = 0.5 × √[1/n + 1/n] = 0.5 × √[2/n] = 0.5 × √2/√n = 1/√(2n). Doubling the sample size in both groups (replacing n with 2n) gives σ = 1/√(4n) = 1/(2√n), which is the original σ divided by √2 ≈ 1.414, not by 2. This is because the standard deviation decreases with the square root of total observations, a diminishing-returns property inherent to all sampling distributions. Point 3 — Optimization (c): We wish to minimize f(n₁) = 1/n₁ + 1/(N − n₁). Taking the derivative: f'(n₁) = −1/n₁² + 1/(N − n₁)². Setting f'(n₁) = 0 gives n₁² = (N − n₁)², so n₁ = N − n₁, hence n₁ = N/2. The second derivative is positive, confirming this is a minimum. Alternatively, by the AM-HM inequality, 1/n₁ + 1/n₂ ≥ 4/(n₁ + n₂) = 4/N, with equality when n₁ = n₂. Equal allocation minimizes the sum of reciprocals. Point 4 — Practical Implication (d): When comparing two proportions, researchers should allocate equal sample sizes to both groups whenever feasible. Unequal allocation increases the standard error and reduces the precision of the estimated difference. In practice, equal allocation may not always be possible (e.g., if one group is rarer or more expensive to sample), but any departure from balance incurs a cost in statistical power. This result also shows why well-designed clinical trials and A/B tests typically aim for 1:1 randomization ratios.

SUMMARY

Summary

The sampling distribution of p̂₁ − p̂₂ describes the behavior of the difference in sample proportions across all possible pairs of independent random samples. Its mean equals p₁ − p₂ (making the statistic an unbiased estimator), and its standard deviation is √[p₁(1 − p₁)/n₁ + p₂(1 − p₂)/n₂], reflecting the critical fact that independent variances add. When the Random, 10%, and Large Counts conditions are all satisfied, this distribution is approximately normal, enabling z-based inference procedures.

In practice, the standard error (using p̂₁ and p̂₂ in place of the unknown population proportions) is used for confidence intervals, while a pooled proportion is used when testing H₀: p₁ = p₂. Always verify conditions with specific numerical values from the data, and remember that the most common exam errors involve subtracting variances instead of adding them, or using the wrong SE formula for the procedure at hand. Mastering this sampling distribution unlocks the entire family of two-proportion inference methods that appear frequently on the AP Statistics exam.