Sampling Distribution of x-bar · 样本均值的抽样分布
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| sample mean/ˈsæmpl miːn/ | 样本均值 | yàng běn jūn zhí |
| standard error/ˈstændəd ˈerə/ | 标准误 | biāo zhǔn wù |
The distribution of x-bar
- A sample mean 样本均值 $\bar{x}$ estimates the true population mean $\mu$.
- Its sampling distribution is centered at $\mu$: $\mu_{\bar{x}} = \mu$ (so $\bar{x}$ is unbiased).
- Each sample gives a different $\bar{x}$; the spread shrinks with sample size.
- This is the workhorse statistic for inference about means.
x-bar 的分布
- 样本均值 $\bar{x}$ 估计真实的总体均值 $\mu$。
- 它的抽样分布以 $\mu$ 为中心：$\mu_{\bar{x}} = \mu$（所以 $\bar{x}$ 是无偏的）。
- 每个样本给出不同的 $\bar{x}$；分散随样本量缩小。
- 这是关于均值的推断的主力统计量。
The standard deviation
- The standard deviation of $\bar{x}$ is:
-
$$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$$
- The population SD divided by $\sqrt{n}$ — often called the standard error.
- Quadruple $n$ and the spread only halves (because of the square root).
标准差
- $\bar{x}$ 的标准差是：
-
$$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$$
- 总体标准差除以 $\sqrt{n}$——常称为标准误。
- $n$ 变为四倍，分散却只减半（因为那个平方根）。
Normality via the CLT
- Is $\bar{x}$ normal? Two ways to justify it:
- If the population is normal, $\bar{x}$ is exactly normal for any $n$.
- Otherwise, the Central Limit Theorem gives approximate normality when $n \ge 30$.
- Always state which justification your problem relies on.
借助 CLT 得到正态性
- $\bar{x}$ 正态吗？有两种论证方式：
- 如果总体正态，那么对任何 $n$，$\bar{x}$ 都恰好正态。
- 否则，当 $n \ge 30$ 时，中心极限定理给出近似正态。
- 总要说明你的题目依赖哪一种论证。
Conditions and probabilities
- Check the 10% condition ($n \le 0.10N$) when sampling without replacement.
- With normality justified, standardize: $z = \dfrac{\bar{x} - \mu}{\sigma/\sqrt{n}}$.
- Then find probabilities for $\bar{x}$ from the standard normal.
- This answers "how likely is a sample mean this far from $\mu$?"
条件与概率
- 不放回抽样时检查 10% 条件（$n \le 0.10N$）。
- 正态性有依据后，标准化：$z = \dfrac{\bar{x} - \mu}{\sigma/\sqrt{n}}$。
- 然后从标准正态求 $\bar{x}$ 的概率。
- 这回答“一个离 $\mu$ 这么远的样本均值有多可能？”
Use $\sigma/\sqrt{n}$, not $\sigma$, for the spread of $\bar{x}$ — the sample mean varies far less than individual data values. A frequent error is dividing by $n$ instead of $\sqrt{n}$. And justify normality explicitly: either a normal population or the CLT with a large $n$ — don't assume $\bar{x}$ is normal for a small sample from a skewed population.
$\bar{x}$ 的分散用 $\sigma/\sqrt{n}$，而不是 $\sigma$——样本均值的变动远小于个别数据值。一个常见错误是除以 $n$ 而不是 $\sqrt{n}$。并且要明确论证正态性：要么总体正态，要么用大 $n$ 的 CLT——不要对来自偏斜总体的小样本假定 $\bar{x}$ 正态。
Population mean $\mu = 100$, SD $\sigma = 20$; take $n = 25$.
- Center: $\mu_{\bar{x}} = 100$. SD: $\sigma_{\bar{x}} = \dfrac{20}{\sqrt{25}} = \dfrac{20}{5} = 4$.
- $P(\bar{x} > 108)$: $z = \dfrac{108 - 100}{4} = 2$ → about $2.5\%$.
- So a sample mean above $108$ is fairly unlikely.
总体均值 $\mu = 100$、标准差 $\sigma = 20$；取 $n = 25$。
- 中心：$\mu_{\bar{x}} = 100$。标准差：$\sigma_{\bar{x}} = \dfrac{20}{\sqrt{25}} = \dfrac{20}{5} = 4$。
- $P(\bar{x} > 108)$：$z = \dfrac{108 - 100}{4} = 2$ → 约 $2.5\%$。
- 所以样本均值高于 $108$ 相当不太可能。
The sample mean $\bar{x}$ has $\mu_{\bar{x}} = \mu$ and $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$. Justify a normal model either by a normal population or by the Central Limit Theorem ($n\ge 30$); check the 10% condition when sampling without replacement, then find probabilities via $z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$.
样本均值 $\bar{x}$ 有 $\mu_{\bar{x}} = \mu$ 和 $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$。用总体正态或中心极限定理（$n\ge 30$）论证正态模型；不放回抽样时检查 10% 条件，再用 $z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ 求概率。
The sampling distribution of x-bar · x-bar 的抽样分布
Centered at μ, spread σ/√n (the standard error). · 以 μ 为中心，分散 σ/√n（标准误）。
Population σ = 20, n = 25. Find the SD of x-bar, σ/√n. · 总体 σ = 20，n = 25。求 x-bar 的标准差 σ/√n。
20/√25 = 20/5 = 4. · 20/√25 = 20/5 = 4。
With μ=100 and SD of x-bar = 4, find the z-score of x-bar = 108. · μ=100 且 x-bar 的标准差 = 4，求 x-bar = 108 的 z 分数。
z = (108 − 100)/4 = 2. · z = (108 − 100)/4 = 2。
The standard deviation of the sample mean is... · 样本均值的标准差是……
Divide σ by √n — not by n. · 用 σ 除以 √n——而不是除以 n。
If the population is already normal, x-bar is normal for any sample size n. · 如果总体本就是正态的，那么对任何样本量 n，x-bar 都正态。
A normal population gives an exactly normal x-bar at any n. · 正态总体在任何 n 下给出恰好正态的 x-bar。
To make the sampling distribution of x-bar have half the spread, you should... · 要让 x-bar 抽样分布的分散减半，你应该……
Spread ∝ 1/√n, so 4× n halves the SD. · 分散 ∝ 1/√n，所以 n 变 4 倍使标准差减半。