When you compare two groups by subtracting their sample means, the answer changes from sample to sample. The sampling distribution of x̄₁ − x̄₂ describes that variation: where it centers, how spread out it is, and when it is approximately normal. Simulate it yourself and check every formula.
Take a random sample from each of two populations, find each sample mean, and subtract. Repeat many times. The distribution of all those differences is the sampling distribution of x̄₁ − x̄₂.
The mean of the sampling distribution of x̄₁ − x̄₂ is exactly μ₁ − μ₂. On average, the difference in sample means hits the true difference, so x̄₁ − x̄₂ is an unbiased estimator of μ₁ − μ₂.
Each sample mean has its own variability, σ²/n. When you subtract two independent sample means, the variances add: σ₁²/n₁ + σ₂²/n₂. Take the square root to get the standard deviation. Never subtract the standard deviations.
Larger n₁ and n₂ shrink the standard deviation of x̄₁ − x̄₂. Multiplying both sample sizes by 4 cuts the standard deviation in half.
If both populations are normal, the sampling distribution is normal for any sample sizes. If not, it is approximately normal when both n₁ ≥ 30 and n₂ ≥ 30, by the central limit theorem.
The two samples must be independent of each other (not paired). Within each sample, when sampling without replacement, check the 10% condition: n₁ ≤ 10% of population 1 and n₂ ≤ 10% of population 2.
To find a probability, compute z = (value − (μ₁ − μ₂)) ÷ σ(x̄₁ − x̄₂), then use the normal distribution. On a calculator, normalcdf with the mean and SD of the difference does the same thing.
Everything you need to describe the sampling distribution of x̄₁ − x̄₂.
| Feature | Formula or rule | Condition needed | Example (σ₁ = 6, n₁ = 36, σ₂ = 8, n₂ = 16) |
|---|---|---|---|
| Mean | μ₁ − μ₂ | None; always true | If μ₁ = 50 and μ₂ = 42, the mean is 8 |
| Standard deviation | √(σ₁²/n₁ + σ₂²/n₂) | Independent samples; 10% condition for each sample | √(36/36 + 64/16) = √5 ≈ 2.24 |
| Shape: normal | Exactly normal | Both populations normal | Heights, if each population is normal |
| Shape: approximately normal | Central limit theorem | n₁ ≥ 30 and n₂ ≥ 30 | Here n₂ = 16, so we would need population 2 to be normal |
| Probability | z = (value − (μ₁ − μ₂)) ÷ SD | Shape is (approximately) normal | P(x̄₁ − x̄₂ > 10) → z = (10 − 8) ÷ 2.24 ≈ 0.89, area ≈ 0.186 |
AP-style multiple choice. Pick an answer to see the explanation.
The vocabulary for topic 4.6. Read the list, flip through flashcards, or quiz yourself. For every term in the unit, see the Unit 4 terminology guide.