1. Sampling Distribution of the Difference of Means
Suppose we have two independent populations:
- Population 1: mean μ1, variance σ12, sample size n1
- Population 2: mean μ2, variance σ22, sample size n2
Sample means: Xˉ1 and Xˉ2
We are interested in the difference of sample means:
D=Xˉ1−Xˉ2
Properties
-
Mean of D:
E(D)=E(Xˉ1−Xˉ2)=μ1−μ2
-
Variance of D (assuming independence):
Var(D)=Var(Xˉ1)+Var(Xˉ2)=n1σ12+n2σ22
-
Standard Error (SE) of Difference:
SE(D)=n1σ12+n2σ22
-
Distribution Shape:
- If populations are normal → D is exactly normal
- If sample sizes are large → D is approximately normal (Central Limit Theorem)
Example:
- Population 1 mean = 50, σ1=10, n1=25
- Population 2 mean = 45, σ2=8, n2=36
SE(D)=25102+3682=4+1.78=5.78≈2.41
- Difference in means = (50 - 45 = 5), SE = 2.41
2. Sampling Distribution of the Difference of Proportions
Suppose we have two independent populations with proportions of success:
- Population 1: p1, sample size n1
- Population 2: p2, sample size n2
Sample proportions: p^1 and p^2
We are interested in the difference of sample proportions:
Dp=p^1−p^2
Properties
-
Mean of Dp:
E(Dp)=p1−p2
-
Variance of Dp (assuming independence):
Var(Dp)=n1p1(1−p1)+n2p2(1−p2)
-
Standard Error (SE) of Difference:
SE(Dp)=n1p1(1−p1)+n2p2(1−p2)
-
Distribution Shape:
- For large n1 and n2 → approximately normal
Example:
- Population 1 proportion p1=0.6, n1=100
- Population 2 proportion p2=0.4, n2=120
SE(Dp)=1000.6∗0.4+1200.4∗0.6=0.0024+0.002=0.0044≈0.066
- Difference in proportions = (0.6 - 0.4 = 0.2), SE ≈ 0.066
3. Key Points to Remember
| Statistic |
Mean |
Standard Error (SE) |
Distribution |
| Difference of Means (Xˉ1−Xˉ2) |
μ1−μ2 |
n1σ12+n2σ22 |
Normal (exact or approx.) |
| Difference of Proportions (p^1−p^2) |
p1−p2 |
n1p1(1−p1)+n2p2(1−p2) |
Approximately Normal (large n) |
Important Notes:
- Populations should be independent.
- For small sample sizes, consider using t-distribution for means.
- For proportions, normal approximation works if n1p1,n1(1−p1),n2p2,n2(1−p2)≥5.