📚 Sampling & Estimation in Statistics | 统计方法：抽样与估计
Statistics is the science of collecting, summarising and drawing conclusions from data. In an exam context, few topics are as consistently tested as sampling and estimation, because they link the real-world practicalities of data collection with the theoretical machinery of probability. This article will walk you through every key concept, formula and trap you need to master.
统计学是收集、概括并从数据中得出结论的科学。在考试中，几乎没有哪个专题像抽样与估计这样稳定地被考查，因为它将数据收集的现实操作与概率的理论工具紧密联系在一起。本文将带你梳理每一个关键概念、公式和易错点，帮助你在考场上游刃有余。
1. Populations and Samples | 总体与样本
A population is the entire set of individuals or objects of interest in a study. A parameter is a numerical summary of a population, such as the population mean μ or the population proportion p. Because we rarely have time or resources to measure the whole population, we take a sample: a subset of the population.
总体是研究中所关心的全部个体或对象的集合。参数是总体的数值概括，例如总体均值 μ 或总体比例 p。由于我们很少有时间或资源去测量整个总体，因此我们会抽取样本：总体中的一个子集。
A sample statistic, such as the sample mean x̄ or the sample proportion p̂, is a numerical summary of the sample. Crucially, we use these statistics to estimate the unknown population parameters. The quality of that estimation depends entirely on how the sample is obtained.
样本统计量，例如样本均值 x̄ 或样本比例 p̂，是对样本的数值概括。关键的是，我们使用这些统计量来估计未知的总体参数。估计质量的好坏完全取决于样本是如何获取的。
| Population 总体 | Sample 样本 |
| Entire group 全部个体 | Subset selected 选取的子集 |
| Parameter (μ, p) 参数 | Statistic (x̄, p̂) 统计量 |
| Usually unknown 通常未知 | Computed from data 由数据计算 |
2. Sampling Methods | 抽样方法
The choice of sampling method determines whether your inferences are valid. The gold standard is a method that gives every member of the population a known, non-zero chance of being selected. There are four principal methods you must know.
抽样方法的选择决定了你的推断是否有效。黄金标准是让总体中每个成员都有已知且非零的概率被选中。你必须掌握以下四种主要方法。
Simple Random Sampling (SRS) – every possible sample of size n has an equal chance of being selected. This is typically done using random numbers, or drawing names from a hat. It removes selection bias but requires a complete list of the population, called a sampling frame.
简単随机抽样（SRS） – 每个大小为 n 的样本被选中的概率都相等。通常使用随机数表或抽签方式完成。它消除了选择偏差，但需要一份完整的总体名单，即抽样框。
Stratified Sampling – the population is split into strata (subgroups) such as age groups or gender. The number taken from each stratum is proportional to its size in the population, then SRS is used within each stratum. This guarantees representation of every subgroup and often reduces sampling variability.
分层抽样 – 将总体分成若干层（子组），例如年龄组或性别组。每层抽取的数量与该层在总体中的大小成比例，然后在每层内部采用简单随机抽样。这保证了每个子组都有代表，通常还能减小抽样变异性。
For example, if a school has 60% girls and 40% boys and we want a stratified sample of 50 students, we need 30 girls and 20 boys.
例如，一所学校有 60% 的女生和 40% 的男生，若想抽取 50 名学生的分层样本，则需要抽取 30 名女生和 20 名男生。
Systematic Sampling – the population is ordered, then every kᵗʰ member is selected after a random starting point. This is simple and fast, but can be biased if there is a periodic pattern in the list.
系统抽样 – 先将总体排序，然后在随机起点之后每第 k 个成员被选中。这种方法简单快捷，但如果名单中存在周期性模式，则可能产生偏差。
Cluster Sampling – the population is divided into clusters (often geographic areas). A random sample of clusters is chosen, and then all members within chosen clusters are surveyed. It is practical for large populations but can be less precise if clusters are heterogeneous.
整群抽样 – 将总体划分为若干群（通常是地理区域）。随机选取若干群，然后对选中群中的所有成员进行调查。它适用于大规模总体，但如果群内部差异很大，精度可能较低。
There is also Quota Sampling (non-random): interviewers choose people to meet quotas for subgroup sizes. It is cheap and quick but not based on probability, so it risks interviewer bias and cannot produce reliable confidence intervals.
还有一种配额抽样（非随机）：调查者按子组的配额来挑选受访者。它成本低、速度快，但并非基于概率，因此存在调查者偏差的风险，且无法产生可靠的置信区间。
3. Parameters vs Statistics | 参数与统计量
A parameter is a fixed but unknown number describing the population. A statistic is a number computed from a sample and is known once the sample is drawn. The value of a statistic varies from sample to sample; this variability is captured by the sampling distribution.
参数是描述总体的一个固定但未知的数字。统计量是从样本中计算出来的数字，一旦样本被抽取，其值就确定了。统计量的值因样本而异；这种变异性由抽样分布来刻画。
The sampling distribution of a statistic is the distribution of all possible values of that statistic across all possible samples of a given size. This concept is the heart of statistical inference.
统计量的抽样分布，是该统计量在给定大小的所有可能样本中所有可能取值的分布。这一概念是统计推断的核心。
For the sample mean x̄ from a population with mean μ and variance σ²:
对于来自均值为 μ、方差为 σ² 的总体的样本均值 x̄：
E(x̄) = μ, Var(x̄) = σ² / n, SD(x̄) = σ / √n
Notice that the standard deviation of x̄ decreases as n increases. A larger sample produces a more precise estimate of μ. This formula assumes independent observations.
注意，x̄ 的标准差随 n 的增大而减小。更大的样本能产生对 μ 更精确的估计。该公式假设观测值相互独立。
4. Point Estimation | 点估计
A point estimate is a single value used to estimate an unknown population parameter. For example, we use the sample mean x̄ to estimate the population mean μ, and the sample proportion p̂ to estimate the population proportion p.
点估计是用于估计未知总体参数的单个数值。例如，我们用样本均值 x̄ 来估计总体均值 μ，用样本比例 p̂ 来估计总体比例 p。
An estimator is unbiased if its expected value equals the population parameter: E(x̄) = μ and E(p̂) = p. This is a key property tested in exams. The sample variance s² is an unbiased estimator of the population variance σ² when the denominator is n − 1, not n.
如果估计量的期望值等于总体参数，则称该估计量是无偏的：E(x̄) = μ，E(p̂) = p。这是考试中的一个关键性质。当分母为 n − 1 而非 n 时，样本方差 s² 是总体方差 σ² 的无偏估计量。
s² = Σ(xᵢ − x̄)² / (n − 1)
The formula with n − 1 is called Bessel’s correction. It corrects the bias that arises because x̄ is itself estimated from the data, making deviations slightly smaller than they are relative to the true mean.
使用 n − 1 的公式称为贝塞尔校正。它修正了因 x̄ 本身由数据估计而产生的偏差，因为相对于真实均值，实际偏差会略微偏小。
The notion of a good estimator also involves precision: a smaller standard error implies a more reliable estimate. We usually prefer estimators that are both unbiased and have minimal variance.
一个好的估计量还涉及精度：标准误越小意味着估计越可靠。我们通常倾向于既无偏又方差最小的估计量。
5. The Central Limit Theorem | 中心极限定理
The Central Limit Theorem (CLT) is perhaps the most important theorem in statistics. It states that for a random sample of size n taken from a population with mean μ and variance σ², the distribution of the sample mean x̄ is approximately normal when n is large:
中心极限定理（CLT）或许是统计学中最重要的定理。它指出，来自均值为 μ、方差为 σ² 的总体的随机样本，当 n 足够大时，样本均值 x̄ 的分布近似正态：
x̄ ~ N(μ, σ² / n) approximately, for large n
For proportions, if we have a sample of n independent Bernoulli trials with success probability p, then the sample proportion p̂ is approximately normal for sufficiently large n:
对于比例，如果我们有 n 次独立的伯努利试验，成功概率为 p，那么当 n 足够大时，样本比例 p̂ 近似服从正态分布：
p̂ ~ N(p, p(1 − p) / n) approximately
What counts as “large n”? A common rule of thumb is n ≥ 30 for means, but for proportions we require both np ≥ 5 and n(1 − p) ≥ 5 to ensure the normal approximation is valid. The beauty of the CLT is that it holds regardless of the shape of the original population distribution.
什么算”足够大”？对于均值，常用经验法则是 n ≥ 30；但对于比例，我们要求 np ≥ 5 且 n(1 − p) ≥ 5，以确保正态近似有效。中心极限定理的优美之处在于，无论原始总体分布的形状如何，它都成立。
In practice, we usually use the sample standard deviation s to estimate σ, so the standard error is estimated as s/√n. This adds an extra layer of uncertainty, which is why the t-distribution is used for small samples with unknown σ.
在实际中，我们通常用样本标准差 s 来估计 σ，因此标准误估计为 s/√n。这增加了一层额外的不确定性，这就是为什么当 σ 未知且样本较小时使用 t 分布。
6. Confidence Intervals for Means | 均值的置信区间
A confidence interval gives a range of plausible values for an unknown parameter, together with a level of confidence, usually 95%. For a population mean, when σ is known and n is large (or data are normal), the 95% confidence interval is:
置信区间为未知参数提供一个可行的取值范围，并伴随一个置信水平，通常为 95%。对于总体均值，当 σ 已知且 n 较大（或数据正态）时，95% 置信区间为：
x̄ ± 1.96 × σ / √n
The value 1.96 is the critical value z* such that P(−1.96 < Z < 1.96) = 0.95, where Z is the standard normal distribution. For 90% the critical value is 1.645, and for 99% it is 2.576. You must memorise these three.
1.96 是临界值 z*，满足 P(−1.96 < Z < 1.96) = 0.95，其中 Z 是标准正态分布。90% 的临界值为 1.645，99% 的临界值为 2.576。这三个值必须牢记。
If σ is unknown, we estimate it with s and use the t-distribution with n − 1 degrees of freedom, provided the data come from a normal distribution:
如果 σ 未知，我们使用 s 来估计它，并在数据来自正态分布的前提下，使用自由度为 n − 1 的 t 分布：
x̄ ± t*(n − 1) × s / √n
When n is large, t*(n − 1) approaches z*, so many exam boards allow the normal approximation for n ≥ 30 even when σ is unknown.
当 n 较大时，t*(n − 1) 趋近于 z*，因此许多考试局允许当 n ≥ 30 时，即使 σ 未知也使用正态近似。
Interpreting the interval: if we repeated the sampling procedure many times and constructed a 95% confidence interval each time, approximately 95% of those intervals would contain the true population mean. It does not mean there is a 95% probability that μ lies in this particular interval.
对区间的解释：如果我们重复抽样过程多次，并每次构造一个 95% 置信区间，那么大约 95% 的这些区间会包含真实的总体均值。这并不意味着 μ 落在这个特定区间内有 95% 的概率。
7. Confidence Intervals for Proportions | 比例的置信区间
For a categorical variable, we estimate the population proportion using p̂ = x/n, where x is the number of successes in n trials. The 95% confidence interval for a proportion is:
对于分类变量，我们用 p̂ = x/n 来估计总体比例，其中 x 是 n 次试验中的成功次数。比例的 95% 置信区间为：
p̂ ± 1.96 × √(p̂(1 − p̂) / n)
This interval is valid when np̂ ≥ 5 and n(1 − p̂) ≥ 5. If the sample proportion is near 0 or 1, the interval can extend beyond 0 or 1, which is logically impossible; in such cases, many examiners recommend using a continuity correction or an alternative method.
该区间在 np̂ ≥ 5 且 n(1 − p̂) ≥ 5 时有效。如果样本比例接近 0 或 1，区间可能超出 0 或 1 的范围，这在逻辑上是不可能的；在这种情况下，许多出题者建议使用连续性校正或替代方法。
As an example, suppose 120 of 400 surveyed students study mathematics beyond A-level. Then p̂ = 120/400 = 0.30. The standard error is √(0.30 × 0.70 / 400) = √0.000525 ≈ 0.0229. The 95% confidence interval is:
举个例子，假设调查的 400 名学生中有 120 名在 A-level 之后继续学习数学。则 p̂ = 120/400 = 0.30。标准误为 √(0.30 × 0.70 / 400) = √0.000525 ≈ 0.0229。95% 置信区间为：
0.30 ± 1.96 × 0.0229 = 0.30 ± 0.0449 = (0.255, 0.345)
We can conclude with 95% confidence that the true proportion is between 25.5% and 34.5%. Notice that this interval is symmetric about p̂.
我们可以以 95% 的置信度得出结论：真实比例在 25.5% 到 34.5% 之间。注意该区间关于 p̂ 对称。
8. Margin of Error and Sample Size | 误差界限与样本量
The margin of error is the half-width of a confidence interval. For a mean it is z* × σ/√n, and for a proportion it is z* × √(p̂(1 − p̂)/n). It measures the maximum expected difference between the sample estimate and the true population parameter.
误差界限是置信区间宽度的一半。对于均值，它为 z* × σ/√n；对于比例，它为 z* × √(p̂(1 − p̂)/n)。它衡量样本估计值与真实总体参数之间的最大预期差异。
To find the required sample size for estimating a mean with a specified margin of error E, set E = z* × σ/√n and solve:
为求在指定误差界限 E 下估计均值所需的样本量，令 E = z* × σ/√n 并求解：
n = (z* × σ / E)²
For proportions, the formula is n = (z* / E)² × p̂(1 − p̂). A conservative approach is to use p̂ = 0.5, which maximizes p̂(1 − p̂) = 0.25, giving the largest safe sample size:
对于比例，公式为 n = (z* / E)² × p̂(1 − p̂)。保守的做法是取 p̂ = 0.5，它使 p̂(1 − p̂) = 0.25 达到最大，从而给出最安全的较大样本量：
n = (z*)² / (4E²)
For example, to be 95% confident of a margin of error no larger than 0.03 for a proportion, use n = (1.96)² / (4 × 0.03²) = 3.8416 / 0.0036 ≈ 1067.1, so we round up to 1068. Always round the sample size up, never down.
例如，要以 95% 的置信度使比例的误差界限不超过 0.03，使用 n = (1.96)² / (4 × 0.03²) = 3.8416 / 0.0036 ≈ 1067.1，因此向上取整到 1068。样本量永远要向上取整，不能向下取整。
Note that for the sample size formula involving σ, if σ is unknown we may use a pilot sample estimate or values from similar studies. A smaller margin of error requires a much larger sample size because n scales with the square of the ratio.
注意，在涉及 σ 的样本量公式中，如果 σ 未知，我们可以使用预调查的估计值或类似研究中的数值。更小的误差界限需要大得多的样本量，因为 n 与比值的平方成正比。
9. Bias vs Precision | 偏差与精度
Bias is a systematic error that pushes estimates away from the true parameter. Precision is the degree of scatter of estimates around their own average. A sampling method can be unbiased but imprecise, or biased but precise — though of course we want neither bias nor excessive variability.
偏差是把估计值推离真实参数的系统性错误。精度是估计值围绕其自身均值的离散程度。一种抽样方法可能无偏但不精确，或者有偏但精确——当然我们既不想要偏差，也不想要过大的变异性。
Imagine a target: accurate and precise means all arrows cluster on the bullseye. If the arrows cluster tightly but in the lower-left corner, the method is precise but biased. If the arrows are scattered all over but centered on the bullseye, it is accurate (unbiased) but imprecise.
想象一个靶子：准确且精确意味着所有箭都密集分布在靶心。如果箭紧密聚集但都在左下角，这种方法精确但有偏。如果箭四处散落但以靶心为中心，则它准确（无偏）但不精确。
Increasing the sample size reduces the standard error and therefore improves precision. However, a large sample cannot fix a biased sampling method: you simply get a very precise estimate of the wrong value. This is why random sampling matters so much.
增加样本量可以减小标准误，从而提高精度。然而，大的样本无法修复有偏的抽样方法：你只会得到一个非常精确却错误的值。这就是随机抽样如此重要的原因。
10. Exam Traps and Tips | 常见考点与易错点
The most common exam errors relate to the following pitfalls. First, forgetting to divide the population variance by n when finding the standard error. Second, using n instead of n − 1 in the sample variance formula. Third, misquoting critical values: 1.96 for 95%, 1.645 for 90%, 2.576 for 99%.
最常见的考试错误集中在以下几点。第一，求标准误时忘记用总体方差除以 n。第二，在样本方差公式中使用 n 而非 n − 1。第三，记错临界值：95% 为 1.96，90% 为 1.645，99% 为 2.576。
Another frequent issue is interpreting confidence intervals incorrectly. Saying “there is a 95% probability that μ is in this interval” is wrong for a frequentist setting; the correct phrasing is about the long-run proportion of intervals that contain μ. This distinction is tested directly.
另一个常见问题是错误地解读置信区间。在频率学派框架下，说”μ 落入该区间的概率是 95%”是错误的；正确的表述是关于长期来看包含 μ 的区间所占的比例。这一区别会被直接考查。
A further trap is forgetting to check conditions before applying the normal approximation. For proportions, you must verify np̂ ≥ 5 and n(1 − p̂) ≥ 5. For means, if the population is heavily skewed, a small sample will not satisfy the CLT, so you should use the t-distribution only if data are reasonably normal.
还有一个陷阱是应用正态近似前忘记检验条件。对于比例，你必须验证 np̂ ≥ 5 且 n(1 − p̂) ≥ 5。对于均值，如果总体严重偏斜，小样本不满足中心极限定理，此时只有在数据近似正态时才能使用 t 分布。
When computing sample size, always round up to the next integer, even if the formula gives 1067.02. Reversing the roles of p̂ and q̂ in the margin of error formula is also a common slip. Writing q̂ = 1 − p̂ and checking p̂ ± 1.96√(p̂q̂/n) helps avoid mistakes.
计算样本量时，即使公式给出 1067.02，也要向上取整到下一个整数。在误差界限公式中互换 p̂ 和 q̂ 的角色也是常见的失误。写出 q̂ = 1 − p̂，并检查 p̂ ± 1.96√(p̂q̂/n)，有助于避免错误。
Finally, in systematic sampling, remember the sampling interval k = N/n, and the starting point is chosen at random from the first k units, not from the whole list. In stratified sampling, the allocation in each stratum must be proportional: nᵢ = n × (Nᵢ / N).
最后，在系统抽样中，抽样间隔 k = N/n 必须记住，起点从前 k 个单元中随机选择，而不是从整个名单中随机选择。在分层抽样中，每层的分配量必须成比例：nᵢ = n × (Nᵢ / N)。
11. Worked Summary Example | 综合示例
Let’s combine everything into a full solution. A factory produces bolts. A random sample of 400 bolts has mean length x̄ = 5.02 cm and standard deviation s = 0.10 cm. Find a 95% confidence interval for the population mean length, and estimate the sample size needed for a margin of error of ±0.01 cm.
让我们将全部内容整合到一个完整的解答中。一家工厂生产螺栓。随机抽取 400 个螺栓，样本均值 x̄ = 5.02 cm，样本标准差 s = 0.10 cm。求总体均值的 95% 置信区间，并估计误差界限为 ±0.01 cm 时所需的样本量。
Since n = 400 is large, we can use z* = 1.96. The standard error is s/√n = 0.10/√400 = 0.10/20 = 0.005. The confidence interval is:
因为 n = 400 较大，我们使用 z* = 1.96。标准误为 s/√n = 0.10/√400 = 0.10/20 = 0.005。置信区间为：
5.02 ± 1.96 × 0.005 = 5.02 ± 0.0098 = (5.0102, 5.0298)
We are 95% confident that the true mean length is between 5.0102 cm and 5.0298 cm. For the required sample size, use n = (z* × s / E)², treating s = 0.10 as a pilot estimate of σ:
我们有 95% 的信心认为真实均值长度在 5.0102 cm 和 5.0298 cm 之间。对于所需样本量，使用 n = (z* × s / E)²，将 s = 0.10 视为 σ 的预估计值：
n = (1.96 × 0.10 / 0.01)² = (19.6)² = 384.16 → round up to 385
Thus a sample of at least 385 bolts is needed to achieve the desired precision.
因此，至少需要 385 个螺栓的样本才能达到所需的精度。
Mastering sampling and estimation means understanding the bridge between data and inference. Always check the sampling method, verify the conditions, use the correct standard error, and interpret intervals with confidence — but not overconfidence. With the tools above, you are well equipped for any statistics paper.
掌握抽样与估计，意味着理解数据与推断之间的桥梁。始终检查抽样方法，验证适用条件，使用正确的标准误，并以有信心的姿态解释区间——但不要过度自信。借助以上工具，你已经为任何统计学试卷做好了充分准备。
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