Difference of Two Means
Introduced| English |
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| difference of two sample means/ˈdɪfrəns ɒv tuː ˈsæmpl miːnz/ |
Comparing two means
- To compare two population means, use the difference of two sample means 两个样本均值之差 $\bar{x}_1 - \bar{x}_2$.
- Its sampling distribution is centered at the true difference: $\mu_1 - \mu_2$.
- This statistic drives every two-sample $t$-procedure in Unit 7.
- It tells us whether two groups' averages genuinely differ.
Combining the two spreads
- The standard deviation combines both groups' variability:
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$$\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$$
- Again the variances add — each group contributes its own $\frac{\sigma^2}{n}$.
- The $+$ holds even though we're taking a difference of means.
Conditions for both samples
- Normality: each $\bar{x}$ must be normal — a normal population or a large enough $n$ (CLT) for each group.
- Independence: the two samples are independent of each other.
- The 10% condition for each if sampling without replacement.
- Verify these separately for group $1$ and group $2$.
Using the normal model
- With conditions met, $\bar{x}_1 - \bar{x}_2$ is approximately normal.
- Standardize: (observed difference $-$ true difference) $\div$ the combined SD.
- Read the probability from the standard normal.
- This foundation becomes the two-sample $t$-test and interval later.
Add the variances, even for a difference of means — $\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}$, then square-root; never subtract, and never add the SDs directly. And each group needs its own normality justification (normal population or large $n$) — the CLT must be satisfied per group, not for the combined data.
Group $1$: $\mu_1 = 50$, $\sigma_1 = 6$, $n_1 = 9$. Group $2$: $\mu_2 = 45$, $\sigma_2 = 8$, $n_2 = 16$.
- Center: $\mu_1 - \mu_2 = 50 - 45 = 5$.
- SD: $\sqrt{\dfrac{36}{9} + \dfrac{64}{16}} = \sqrt{4 + 4} = \sqrt{8} \approx 2.83$.
- Standardize any observed difference against this center and SD.
The difference of two sample means $\bar{x}_1 - \bar{x}_2$ has mean $\mu_1 - \mu_2$ and SD $\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$ (variances add). Model it as normal when the normality (per group) and independence conditions hold — the basis for two-sample inference about means.
Distribution of x-bar-1 − x-bar-2
Centered at μ1 − μ2; the two variances add under the root.
μ1 = 50, μ2 = 45. Find the center of x-bar-1 − x-bar-2.
μ1 − μ2 = 50 − 45 = 5.
σ1=6,n1=9 and σ2=8,n2=16. Find SD = √(36/9 + 64/16). Round to two decimals.
√(4 + 4) = √8 ≈ 2.83.
The variance of a difference of two independent sample means is the sum of the two variances.
Variances add even for a difference: σ1²/n1 + σ2²/n2.
For a normal model of x-bar-1 − x-bar-2, normality must be justified...
Each group needs a normal population or a large enough n.
You may add the two standard deviations directly to get the SD of the difference.
Add the variances, then take the square root — not the SDs.