AP Statistics · Unit 3 · Topic 3.9

Sampling Distribution of a Difference in Proportions Study Lab

When you compare two groups, the difference in sample proportions p̂₁ − p̂₂ changes from sample to sample. Set two true proportions and two sample sizes, take many pairs of random samples, and compare the simulated differences with the normal model.

1. Simulate the sampling distribution of p̂₁ − p̂₂

Set the two true population proportions and the two sample sizes. Each simulated trial takes an independent random sample from each population and records the difference in sample proportions.

μ = p₁ − p₂ σ = √( p₁(1 − p₁)/n₁ + p₂(1 − p₂)/n₂ )
Preset situations

Populations and samples

Population 1
Population 2

Simulated values of p̂₁ − p̂₂

Normal model N(μ, σ)μ = p₁ − p₂Latest difference
Mean μ (theory)
Standard deviation σ (theory)
Mean of simulated values
SD of simulated values

Center and spread

Is a normal model OK?

2. Key ideas

Center

Centered at the true difference

The sampling distribution of p̂₁ − p̂₂ has mean μ = p₁ − p₂. That makes p̂₁ − p̂₂ an unbiased estimator of the difference in population proportions.

Spread

Variances add, not SDs

For independent samples, σ² = p₁(1 − p₁)/n₁ + p₂(1 − p₂)/n₂. You add the two variances, then take the square root. The variances add even though you subtract the proportions.

Shape

Large counts → normal

The shape is approximately normal when n₁p₁, n₁(1 − p₁), n₂p₂, and n₂(1 − p₂) are all at least 10.

Independence

Two conditions

The samples must be independent of each other (two separate random samples or random assignment), and each sample should be at most 10% of its population when sampling without replacement.

Sample size

Bigger n, less variability

σ shrinks as n₁ and n₂ grow. To cut σ in half you need about four times as many observations in each group.

Using it

Probability of a difference

With the normal model, find P(p̂₁ − p̂₂ > c) by computing z = (c − μ)/σ and using a normal table or normalcdf.

3. Quick reference

The sampling distribution of p̂₁ − p̂₂, and what each condition checks.

FeatureFormula or ruleCondition needed
Meanμ = p₁ − p₂None: always true for random samples
Standard deviationσ = √( p₁(1 − p₁)/n₁ + p₂(1 − p₂)/n₂ )Independent samples; each n ≤ 10% of its population
ShapeApproximately normaln₁p₁, n₁(1 − p₁), n₂p₂, n₂(1 − p₂) all ≥ 10
z-scorez = ((p̂₁ − p̂₂) − (p₁ − p₂)) ÷ σNormal shape
Bigger samplesσ gets smaller; center does not move—
Equal populationsμ = 0, but individual differences still vary—

4. On the FRQ

How to earn the points

  • Describe all three features with numbers and context: "The sampling distribution of p̂₁ − p̂₂ is approximately normal with mean 0.10 and standard deviation 0.069."
  • Check the large counts condition with all four products, and show the numbers.
  • State the 10% condition for each population when sampling without replacement.
  • For a probability, show the z-score: z = (0.15 − 0.10) ÷ 0.069 = 0.72, then P(Z > 0.72) = 0.235. Label normalcdf inputs if you use a calculator.
  • Say why the samples are independent of each other: two separate random samples, or random assignment to two groups.

Common mistakes

  • Adding the standard deviations instead of the variances.
  • Subtracting the variances because the proportions are subtracted. Variability always adds.
  • Using sample proportions p̂ in the mean or SD formulas when the problem gives the true p.
  • Checking large counts only for successes. Failures count too.
  • Thinking a difference of 0 is guaranteed when p₁ = p₂. The center is 0, but samples vary.

5. Practice questions

AP-style multiple choice. Pick an answer to see the explanation.

0 / 0 correct

6. Key terms

The vocabulary for topic 3.9. Read the list, flip through flashcards, or quiz yourself. For every term in the unit, see the Unit 3 terminology guide.