Mathfolis

Sampling Distribution for a Difference of Means

Unit 5 · Sampling Distributions

What AP Stats asks here

The two-sample difference of means inherits variability from both samples. The mean of xˉ1−xˉ2\bar{x}_1 - \bar{x}_2 is μ1−μ2\mu_1 - \mu_2 and its SE combines σ12/n1\sigma_1^2/n_1 and σ22/n2\sigma_2^2/n_2 under the square root. The trap mirrors the proportions version: variances add even when subtracting because Var(−Y)=(-Y) = Var(Y)(Y).

Sampling distribution of xˉ1−xˉ2\bar{x}_1 - \bar{x}_2

Center
μxˉ1−xˉ2=μ1−μ2\mu_{\bar{x}_1 - \bar{x}_2} = \mu_1 - \mu_2
Standard error
σxˉ1−xˉ2=σ12n1+σ22n2\sigma_{\bar{x}_1 - \bar{x}_2} = \sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}

Conditions

Independent random samples
from each population\text{from each population}
Normality (each sample)
population normal OR ni≥30\text{population normal OR } n_i \ge 30
10% condition (each)
ni<0.10Nin_i < 0.10 N_i
AP Tip: Estimating a difference is noisier than estimating either mean alone. To target a fixed SE for the difference, each group typically needs roughly the same effort as one-sample work — variances accumulate.
Caution: Both samples must clear the normality condition. If one sample has a skewed population and small n, the difference's distribution is not approximately normal.
Type 1

Describe the sampling distribution

Verify the conditions and compute σ12/n1+σ22/n2\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2} for the SE.

Example 1
Population 1: μ1=170\mu_1 = 170 cm, σ1=8\sigma_1 = 8 cm; Population 2: μ2=165\mu_2 = 165 cm, σ2=7\sigma_2 = 7 cm. Independent SRSs n1=40n_1 = 40, n2=50n_2 = 50. Which best describes the sampling distribution of xˉ1−xˉ2\bar{x}_1 - \bar{x}_2? (A) N(5,1.606)N(5, 1.606) — variances add even when subtracting. (B) N(5,1.0)N(5, 1.0) — only the larger SD matters. (C) N(5,0.30)N(5, 0.30) (D) Cannot apply normal because n1<30n_1 < 30.

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Type 2

Variance still adds on subtraction

Var(−Y)=(-Y) = Var(Y)(Y), so variances of independent variables add regardless of sign.

Example 2
Which justifies why variances add when computing xˉ1−xˉ2\bar{x}_1 - \bar{x}_2 instead of subtracting? (A) Subtracting two random variables is identical to adding the negative; for independent variables, variance is unchanged by sign, so Var(X − Y) = Var(X) + Var(Y). (B) Variances subtract — the textbook formula is wrong. (C) Variances cancel when sample sizes are equal. (D) The CLT is required for variances to add.

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Type 3

Compute probabilities

Standardize xˉ1−xˉ2\bar{x}_1 - \bar{x}_2 with the combined SE and look up the z-table.

Example 3
For xˉ1−xˉ2∼N(5,1.606)\bar{x}_1 - \bar{x}_2 \sim N(5, 1.606), what is P(xˉ1−xˉ2>7)P(\bar{x}_1 - \bar{x}_2 > 7)? (A) ≈ 0.11 (B) ≈ 0.30 (C) ≈ 0.50 (D) ≈ 0.90

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Sampling Distribution for a Difference of Means | AP Statistics — Mathfolis