Difference of Proportions

When researchers want to compare two groups — say, the proportion of students who pass an exam in two different schools, or the fraction of voters who support a policy in two different cities — they need a way to reason about the difference between two sample proportions. The sampling distribution of the difference of proportions is the probability distribution of the statistic p^1−p^2\hat{p}_1 - \hat{p}_2 computed from all possible pairs of random samples drawn from two separate populations. Just as a single sample proportion p^\hat{p} has its own sampling distribution (approximately normal under certain conditions), the difference p^1−p^2\hat{p}_1 - \hat{p}_2 also follows an approximately normal distribution when sample sizes are large enough. This powerful result lets us calculate probabilities, build confidence intervals, and conduct hypothesis tests to decide whether two population proportions are truly different or whether the observed gap is just random chance. Understanding this concept connects directly to many real-world questions: Does a new drug reduce side-effect rates more than the old one? Do boys and girls differ in the proportion who prefer a certain activity? The mathematics of the difference of proportions gives us a rigorous, quantitative way to answer these questions rather than relying on intuition.

Key takeaways
  • Define the two populations
  • Draw independent random samples
  • Find the mean of the sampling distribution
  • Find the standard deviation (standard error)
  • Check the Normal approximation conditions
  • Standardize to find probabilities

Define the two populations

Start with two distinct populations. Population 1 has a true proportion p1p_1 of individuals with some characteristic, and Population 2 has a true proportion p2p_2. For example, p1p_1 might be the proportion of left-handed students at School A, and p2p_2 the proportion at School B. These are fixed but usually unknown parameters.

Draw independent random samples

Take a simple random sample of size n1n_1 from Population 1 and an independent simple random sample of size n2n_2 from Population 2. Compute the sample proportions p^1=x1/n1\hat{p}_1 = x_1 / n_1 and p^2=x2/n2\hat{p}_2 = x_2 / n_2, where x1x_1 and x2x_2 are the counts of successes in each sample.

Find the mean of the sampling distribution

Because E[p^1]=p1E[\hat{p}_1] = p_1 and E[p^2]=p2E[\hat{p}_2] = p_2, the expected value of the difference is μp^1−p^2=p1−p2\mu_{\hat{p}_1 - \hat{p}_2} = p_1 - p_2. The statistic is unbiased: on average across all possible samples, the difference of sample proportions equals the true difference of population proportions.

Find the standard deviation (standard error)

Since the two samples are independent, the variances add. The variance of p^1\hat{p}_1 is p1(1−p1)/n1p_1(1-p_1)/n_1 and the variance of p^2\hat{p}_2 is p2(1−p2)/n2p_2(1-p_2)/n_2. Therefore, the standard deviation of the sampling distribution (called the standard error) is σp^1−p^2=p1(1−p1)n1+p2(1−p2)n2\sigma_{\hat{p}_1 - \hat{p}_2} = \sqrt{\dfrac{p_1(1-p_1)}{n_1} + \dfrac{p_2(1-p_2)}{n_2}}.

Check the Normal approximation conditions

The sampling distribution of p^1−p^2\hat{p}_1 - \hat{p}_2 is approximately normal provided: (1) Both samples are randomly selected and independent of each other. (2) The success-failure condition holds for each sample: n1p1≥10n_1 p_1 \geq 10, n1(1−p1)≥10n_1(1-p_1) \geq 10, n2p2≥10n_2 p_2 \geq 10, and n2(1−p2)≥10n_2(1-p_2) \geq 10. (3) Each sample is less than 10% of its population (the 10% condition) to ensure approximate independence. When these conditions hold, we can apply the Central Limit Theorem.

Standardize to find probabilities

Once we know the distribution is approximately N(p1−p2,  σp^1−p^22)N(p_1 - p_2,\; \sigma_{\hat{p}_1 - \hat{p}_2}^2), we standardize using a zz-score: z=(p^1−p^2)−(p1−p2)p1(1−p1)n1+p2(1−p2)n2z = \dfrac{(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)}{\sqrt{\dfrac{p_1(1-p_1)}{n_1} + \dfrac{p_2(1-p_2)}{n_2}}}. This zz-score follows the standard normal distribution N(0,1)N(0,1), enabling us to look up probabilities in a standard normal table or use technology.

The Key Equation

σp^1−p^2=p1(1−p1)n1+p2(1−p2)n2\sigma_{\hat{p}_1 - \hat{p}_2} = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}
  • \sigma_{\hat{p}_1 - \hat{p}_2} — Standard error of the difference (dimensionless): Measures the typical spread of the statistic p^1−p^2\hat{p}_1 - \hat{p}_2 across all possible pairs of samples.
  • p_1 — Population proportion 1 (dimensionless (0 to 1)): The true proportion of successes in Population 1; a fixed parameter.
  • p_2 — Population proportion 2 (dimensionless (0 to 1)): The true proportion of successes in Population 2; a fixed parameter.
  • n_1 — Sample size from Population 1 (count): Number of individuals randomly sampled from Population 1.
  • n_2 — Sample size from Population 2 (count): Number of individuals randomly sampled from Population 2.

Worked Example

Problem. A survey finds that 55% of adults in City A support a new transit policy (p1=0.55p_1 = 0.55) and 48% in City B support it (p2=0.48p_2 = 0.48). Independent random samples of n1=200n_1 = 200 from City A and n2=150n_2 = 150 from City B are taken. (a) Describe the sampling distribution of p^1−p^2\hat{p}_1 - \hat{p}_2. (b) What is the probability that the sample from City A shows a proportion at least 0.10 higher than the sample from City B, i.e., P(p^1−p^2≥0.10)P(\hat{p}_1 - \hat{p}_2 \geq 0.10)?

P(p^1−p^2≥0.10)≈0.289P(\hat{p}_1 - \hat{p}_2 \geq 0.10) \approx 0.289

Answer. The sampling distribution of the difference is approximately Normal with mean 0.07 and standard error ≈ 0.0539. The probability that the City A sample proportion exceeds the City B sample proportion by at least 0.10 is approximately 28.9%.

Figure
Diagram
Sampling Distribution of p̂₁ − p̂₂N(p₁−p₂, σ²) with p₁−p₂ = 0.07, σ ≈ 0.054μ = 0.070.10P ≈ 0.2890−0.070.0160.100.178f(x)xBell curvecentered at 0.07
Origins · 1 minAbraham de Moivre and how this was discovered
AD
Abraham de Moivre
1667–1754 · French-British

De Moivre was the first mathematician to derive the normal distribution as an approximation to the binomial distribution (1733), laying the theoretical foundation that allows us to treat sampling distributions of proportions — and their differences — as approximately normal for large samples. His work predated Gauss and Laplace and was published in his landmark text 'The Doctrine of Chances.' De Moivre reputedly predicted the date of his own death by noticing he was sleeping 15 minutes longer each night, extrapolating arithmetically until he calculated he would sleep a full 24 hours — and he died on that exact predicted day in 1754.

  • Sampling Distribution of a Proportion — The difference of proportions builds directly on how a single sample proportion p^\hat{p} is distributed; mastering one makes the other straightforward.
  • Central Limit Theorem — The CLT is the theoretical reason why the sampling distribution of p^1−p^2\hat{p}_1 - \hat{p}_2 is approximately normal for large sample sizes.
  • Confidence Interval for Difference of Proportions — The sampling distribution derived here is exactly what is used to construct and interpret confidence intervals for p1−p2p_1 - p_2.
  • Two-Proportion Z-Test — Hypothesis testing for whether p1=p2p_1 = p_2 relies on the sampling distribution of p^1−p^2\hat{p}_1 - \hat{p}_2 under the null hypothesis.