抽样分布

A statistic 统计量 (like a sample mean 样本均值 $\bar{x}$ or sample proportion 样本比例 $\hat{p}$) is computed from a sample and varies from sample to sample – this is sampling variability 抽样变异性. A parameter 参数 ($\mu$ or $p$) is the fixed truth about the population 总体. The sampling distribution 抽样分布 is the distribution of a statistic over all possible samples of a given size – it is the bridge from one sample to inference 推断.

For large enough samples, many sampling distributions are approximately normal 正态的. That lets us describe a statistic by a center 中心 (its mean), a spread 离散程度 (its standard error 标准误), and a normal shape – and then compute how likely a given sample result is.

The Central Limit Theorem 中心极限定理 (CLT): for independent 独立 observations from the same population with finite, positive variance 方差, the sampling distribution of a sample mean becomes approximately normal as $n$ increases. A strongly skewed 偏斜 population may need a much larger sample than $30$; there is no universal sample-size cutoff. The standard error decreases as $n$ increases.

A statistic is unbiased 无偏 if the mean of its sampling distribution equals the parameter – it is correct on average. Bias 偏差 is about the center being off; variability 变异性 is about the spread. A good estimator 估计量 is both unbiased (centered right) and low-variability (precise 精确的); larger samples reduce variability but do not fix bias from bad sampling.

For a sample proportion $\hat{p}$ from an SRS 简单随机样本: the mean is $p$ (unbiased), and the standard deviation 标准差 is

This spread is the standard error: the standard deviation of the statistic's sampling distribution. When $p$ is unknown, replace it by $\hat p$ to estimate that standard error, as in the later inference units. It is approximately normal when $np\ge 10$ and $n(1-p)\ge 10$ (the Large Counts condition 大计数条件), and the $10\%$ condition ($n\le 0.10N$) keeps the observations near-independent.

Worked example. Suppose $40\%$ of voters favor a measure ($p=0.4$) and you sample $n=100$. The standard error is $\sigma_{\hat p}=\sqrt{\dfrac{0.4(0.6)}{100}}=0.049$. The chance a sample gives $\hat{p}>0.5$ is $z=\dfrac{0.5-0.4}{0.049}=2.04$, so $P(\hat p>0.5)\approx0.02$ – a majority in the sample would be surprising.

Common mistakes. Checking “$n\ge30$” for a proportion: that rule belongs to means; for $\hat p$ check $np\ge10$ and $n(1-p)\ge10$ (here $40$ and $60$). Treating the $10\%$ condition as a normality check: it keeps the observations close to independent, so the formula for $\sigma_{\hat p}$ is valid.

For $\hat{p}_1-\hat{p}_2$ from two independent samples: the mean is $p_1-p_2$, and because the samples are independent the variances add 方差相加:

It is approximately normal when the Large Counts condition holds in both samples.

Worked example. In town 1, $60\%$ of adults cycle to work; in town 2, $50\%$. Independent random samples of $100$ adults are taken from each town. How likely is it that the town 1 sample shows a lower proportion than town 2?

The mean of $\hat p_1-\hat p_2$ is $0.6-0.5=0.10$, and

Large Counts: $60$, $40$, $50$ and $50$ are all at least $10$, so use a normal model. $z=\dfrac{0-0.10}{0.07}\approx-1.43$, so $P(\hat p_1-\hat p_2<0)\approx 0.077$. About $8\%$ of such pairs of samples would put town 1 behind, even though town 1 really is ahead.

For a sample mean $\bar{x}$ from an SRS: the mean is $\mu$ (unbiased), and the standard deviation is

Its shape is normal if the population is normal, or approximately normal for large $n$ by the CLT. Note the spread shrinks like $\sqrt{n}$ – quadrupling the sample halves the standard error.

Worked example. A population has $\mu=70$ and $\sigma=12$. For samples of $n=36$, the sampling distribution of $\bar{x}$ is centered at $70$ with standard error $\dfrac{12}{\sqrt{36}}=2$. The chance a sample mean exceeds $73$ is $z=\dfrac{73-70}{2}=1.5$, so $P(\bar x>73)\approx0.067$.

Common mistakes. Using $\sigma=12$ instead of $\sigma_{\bar x}=2$ for a question about a mean of $36$ values. For one individual, $z=(73-70)/12=0.25$ and $P(X>73)\approx0.40$: a single value above $73$ is common, a sample mean above $73$ is not. Saying the CLT makes the population (or the sample) normal: it is the sampling distribution of $\bar x$ that becomes approximately normal.

For $\bar{x}_1-\bar{x}_2$ from two independent samples: the mean is $\mu_1-\mu_2$, and (independent, so variances add)

This is the foundation for two-sample inference in the next units.

Worked example. Machine 1 fills bottles with $\mu_1=500\ \text{mL}$ and $\sigma_1=4\ \text{mL}$; machine 2 with $\mu_2=498\ \text{mL}$ and $\sigma_2=3\ \text{mL}$. Both fill amounts are normal. Independent random samples of $16$ bottles from machine 1 and $9$ from machine 2 are taken. The mean of $\bar x_1-\bar x_2$ is $500-498=2\ \text{mL}$, and

Both populations are normal, so the difference is normal even for these small samples. The machine 2 sample has the larger mean when $\bar x_1-\bar x_2<0$:

The variances add even though the means subtract: $1+1$ under the square root, never $1-1$.

Answer each before reading the answers below.

1. A population proportion is $p=0.30$ and samples of $n=50$ are taken. Give the mean and standard deviation of $\hat p$, and check the Large Counts condition.
2. A population has $\mu=100$ and $\sigma=15$. For samples of $n=25$, find $P(\bar x>106)$.
3. By what factor must the sample size grow to halve the standard deviation of $\bar x$?
4. A population of incomes is strongly right-skewed. Is $\bar x$ approximately normal for samples of $n=5$? For $n=50$?
5. A student says: "A bigger sample removes bias." Correct the student.

1. Mean $0.30$; standard deviation $\sqrt{0.30(0.70)/50}=0.065$. Large Counts: $np=15$ and $n(1-p)=35$, both at least 10.
2. $\sigma_{\bar x}=15/\sqrt{25}=3$, so $z=\dfrac{106-100}{3}=2.0$ and $P(\bar x>106)\approx0.023$.
3. Four times as large, because $\sigma_{\bar x}=\sigma/\sqrt n$.
4. Not for $n=5$: with a skewed population, a small sample's mean is still skewed. For $n=50$, probably yes: $n\ge 30$ is the usual AP guideline, so by the Central Limit Theorem $\bar x$ is approximately normal. An extremely skewed population (a few huge incomes) could still need a larger sample.
5. A larger sample reduces variability (the spread of the sampling distribution), not bias (where its center sits). A biased method such as a voluntary-response 自愿回应 poll stays biased, however large.

• A sampling distribution describes a statistic over repeated samples. It is centered on the true parameter when the statistic is unbiased.
• The Central Limit Theorem: for a large enough sample the sample mean is approximately normal, even if the population is not.
• Larger samples give less variability (a smaller standard error).
• Check the conditions (random, independent/10%, large enough) before using a normal model.
• Keep straight what varies — the statistic — versus the fixed parameter.