Difference of Two Means · 两个均值之差
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| difference of two sample means/ˈdɪfrəns ɒv tuː ˈsæmpl miːnz/ | 两个样本均值之差 | liǎng gè yàng běn jūn zhí zhī chà |
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.
比较两个均值
- 要比较两个总体均值，用两个样本均值之差 $\bar{x}_1 - \bar{x}_2$。
- 它的抽样分布以真实差为中心：$\mu_1 - \mu_2$。
- 这个统计量驱动第 7 单元里每一个两样本 $t$ 程序。
- 它告诉我们两个组的平均是否真的不同。
Combining the two spreads
- The standard deviation combines both groups' variability:
-
$$\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$$
- Again the variances add — each group contributes its own $\frac{\sigma^2}{n}$.
- The $+$ holds even though we're taking a difference of means.
合并两个分散
- 标准差把两个组的变异性合并起来：
-
$$\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$$
- 又是方差相加——每个组贡献它自己的 $\frac{\sigma^2}{n}$。
- 即便我们取的是均值之差，那个 $+$ 仍然成立。
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 $2$.
两个样本的条件
- **正态性：**每个 $\bar{x}$ 都必须正态——每个组要么总体正态、要么 $n$ 足够大（CLT）。
- **独立性：**两个样本彼此独立。
- 不放回抽样时每个组还要满足 10% 条件。
- 对组 $1$ 和组 $2$ 分别验证这些。
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.
使用正态模型
- 条件满足时，$\bar{x}_1 - \bar{x}_2$ 近似正态。
- 标准化：（观察到的差 $-$ 真实的差）$\div$ 合并的标准差。
- 从标准正态读出概率。
- 这个基础后面成为两样本 $t$ 检验和区间。
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.
相加方差，即使是均值之差——$\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}$，再开平方根；绝不相减，也绝不直接相加标准差。并且每个组都需要它自己的正态性论证（总体正态或大 $n$）——CLT 必须逐组满足，而非对合并数据。
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.
组 $1$：$\mu_1 = 50$，$\sigma_1 = 6$，$n_1 = 9$。组 $2$：$\mu_2 = 45$，$\sigma_2 = 8$，$n_2 = 16$。
- 中心：$\mu_1 - \mu_2 = 50 - 45 = 5$。
- 标准差：$\sqrt{\dfrac{36}{9} + \dfrac{64}{16}} = \sqrt{4 + 4} = \sqrt{8} \approx 2.83$。
- 用这个中心和标准差去标准化任何观察到的差。
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.
两个样本均值之差 $\bar{x}_1 - \bar{x}_2$ 的均值为 $\mu_1 - \mu_2$，标准差为 $\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$（方差相加）。当正态性（逐组）和独立性条件成立时，把它建模为正态——这是关于均值的两样本推断的基础。
Distribution of x-bar-1 − x-bar-2 · x-bar-1 − x-bar-2 的分布
Centered at μ1 − μ2; the two variances add under the root. · 以 μ1 − μ2 为中心；两个方差在根号下相加。
μ1 = 50, μ2 = 45. Find the center of x-bar-1 − x-bar-2. · μ1 = 50，μ2 = 45。求 x-bar-1 − x-bar-2 的中心。
μ1 − μ2 = 50 − 45 = 5. · μ1 − μ2 = 50 − 45 = 5。
σ1=6,n1=9 and σ2=8,n2=16. Find SD = √(36/9 + 64/16). Round to two decimals. · σ1=6,n1=9 且 σ2=8,n2=16。求标准差 = √(36/9 + 64/16)。保留两位小数。
√(4 + 4) = √8 ≈ 2.83. · √(4 + 4) = √8 ≈ 2.83。
The variance of a difference of two independent sample means is the sum of the two variances. · 两个独立样本均值之差的方差是两个方差之和。
Variances add even for a difference: σ1²/n1 + σ2²/n2. · 即使是差，方差也相加：σ1²/n1 + σ2²/n2。
For a normal model of x-bar-1 − x-bar-2, normality must be justified... · 对 x-bar-1 − x-bar-2 的正态模型，正态性必须……被论证。
Each group needs a normal population or a large enough n. · 每个组都需要正态总体或足够大的 n。
You may add the two standard deviations directly to get the SD of the difference. · 你可以直接把两个标准差相加得到差的标准差。
Add the variances, then take the square root — not the SDs. · 先加方差再开平方根——而不是加标准差。