定量方法（Quantitative Methods）— 抽样与估计 · 第 5 课
一、引言
前情回顾（L128）：我们学过了置信区间的概念、基本公式和解释方式。 置信区间 = 点估计 ± 临界值 × 标准误
本节课，我们把这个框架落地到四种具体场景，逐一掌握每种场景的公式、适用条件和计算细节。
二、场景一：总体均值的置信区间（σ 已知 / 大样本）
2.1 公式
x̄ ± z(α/2) × σ/√n
2.2 适用条件
| 条件 | 说明 |
|---|---|
| 总体标准差 σ 已知 | 罕见但考试爱考 |
| 或 n ≥ 30（大样本） | 中心极限定理兜底，可用 s 替代 σ |
2.3 案例
某银行信用卡部调查了 100 名客户的月均消费额，样本均值 x̄ = ¥4,200，历史数据显示总体标准差 σ = ¥800。构造 95% 置信区间。
SE = 800/√100 = 80
95% CI = 4200 ± 1.96 × 80 = 4200 ± 156.8 = [4043.2, 4356.8]
📌 解读： 我们有 95% 的把握认为，全体客户月均消费额在 ¥4,043 到 ¥4,357 之间。
2.4 关键考点：置信水平 ↔ z 值对照
| 置信水平 | α | z(α/2) |
|---|---|---|
| 90% | 0.10 | 1.645 |
| 95% | 0.05 | 1.96 |
| 99% | 0.01 | 2.576 |
三、场景二：总体均值的置信区间（σ 未知 + 小样本）
3.1 公式
x̄ ± t(α/2, df) × s/√n, df = n - 1
3.2 t 分布的关键特征
| 特征 | z 分布（标准正态） | t 分布 |
|---|---|---|
| 形状 | 固定，精确的钟形 | 钟形，但尾部更厚（fat tail） |
| 依赖参数？ | 无 | 依赖自由度 df |
| df → ∞ | — | 趋近于 z 分布 |
| 同置信水平下的临界值 | 较小 | 更大（更保守） |
💡 直观理解： σ 未知时，我们用 s 估计 σ，相当于又多了一层不确定性。t 分布的厚尾就是对"连 σ 都不知道"的惩罚。
3.3 常用 t 临界值速查表
| df | t(0.05)（90%） | t(0.025)（95%） | t(0.005)（99%） |
|---|---|---|---|
| 5 | 2.015 | 2.571 | 4.032 |
| 10 | 1.812 | 2.228 | 3.169 |
| 15 | 1.753 | 2.131 | 2.947 |
| 20 | 1.725 | 2.086 | 2.845 |
| 30 | 1.697 | 2.042 | 2.750 |
| ∞ | 1.645 | 1.96 | 2.576 |
🔥 记忆技巧：df 越小，t 和 z 差距越大；df ≥ 30，t ≈ z（近似）
3.4 案例：小样本下的 CI
分析师随机选取了 16 只新能源股票，计算其过去一年收益率的均值为 12%，样本标准差 s = 8%。请构造 95% 置信区间。
df = 16 - 1 = 15, t(0.025, 15) = 2.131
SE = 8%/√16 = 2%
95% CI = 12% ± 2.131 × 2% = 12% ± 4.262% = [7.738%, 16.262%]
⚠️ 如果用 z 值 1.96 近似计算，CI = [8.08%, 15.92%] —— 区间偏窄，不够保守！
四、场景三：总体比例的置信区间
4.1 公式
p̂ ± z(α/2) × √(p̂(1-p̂)/n)
其中 p̂ = 样本比例（成功数/样本量）
4.2 适用条件
| 条件 | 说明 |
|---|---|
| n·p̂ ≥ 10 | 成功的期望数足够 |
| n·(1-p̂) ≥ 10 | 失败的期望数足够 |
| 用 z 临界值 | 比例问题始终用 z（不用 t） |
⚠️ 如果上述条件不满足（即 n·p̂ < 10 或 n·(1-p̂) < 10），需要用精确方法（如二项分布），但一级不考。
4.3 案例：市场调研
某电商平台随机调查了 400 名用户，其中 260 人对新界面表示满意。构造 90% 置信区间。
p̂ = 260/400 = 0.65
SE = √(0.65 × 0.35 / 400) = √(0.00056875) ≈ 0.02385
90% CI = 0.65 ± 1.645 × 0.02385 = 0.65 ± 0.0392 = [0.6108, 0.6892]
📌 解读： 我们有 90% 的把握认为，整体用户满意度在 61.1% 到 68.9% 之间。
五、场景四：两总体均值之差的置信区间
5.1 为什么需要它？
点估计只能告诉你 x̄₁ - x̄₂ = 3.5，但 3.5 是否"明显不同于 0"？区间估计告诉我们答案。
5.2 两种子场景
子场景 A：独立样本，两总体方差已知（或大样本）
(x̄₁ - x̄₂) ± z(α/2) × √(σ₁²/n₁ + σ₂²/n₂)
子场景 B：独立样本，两总体方差未知但假设相等（pooled）
(x̄₁ - x̄₂) ± t(α/2, n₁+n₂-2) × sp × √(1/n₁ + 1/n₂)
其中 sp² = ((n₁-1)s₁² + (n₂-1)s₂²) / (n₁ + n₂ - 2)（pooled variance）
5.3 案例：新药 vs 安慰剂
临床试验：新药组 50 人，平均血压下降 x̄₁ = 12 mmHg，s₁ = 5 mmHg 安慰剂组 50 人，平均下降 x̄₂ = 4 mmHg，s₂ = 4 mmHg 构造 95% 置信区间（假设方差相等）。
sp² = (49×25 + 49×16) / 98 = 2009/98 ≈ 20.5, sp ≈ 4.527
SE = 4.527 × √(1/50 + 1/50) = 4.527 × 0.2 = 0.9054
df = 50 + 50 - 2 = 98, t(0.025, 98) ≈ 1.984
95% CI = (12 - 4) ± 1.984 × 0.9054 = 8 ± 1.796 = [6.20, 9.80]
🟢 解读： 区间全部 > 0，说明在 95% 置信水平下，新药的降压效果确实优于安慰剂，差异约为 6.2 ~ 9.8 mmHg。
六、影响置信区间宽度的因素（全场景总结）
CI 宽度 ∝ 临界值 × 波动性/√n
| 因素 | 变化 | 宽度 | 直观理解 |
|---|---|---|---|
| 样本量 n | ↑ | ↓ | 更多信息 → 更精确 |
| 置信水平 (1-α) | ↑ | ↑ | 更高把握 → 更保守 |
| 总体波动 σ | ↑ | ↑ | 数据本身噪声大 |
| z → t（小样本） | — | ↑ | t 尾部厚，更保守 |
🔥 考试重点： 给定一组条件变化，判断 CI 变宽还是变窄。
七、常见陷阱与考试套路
7.1 选错临界值
| 情况 | 用 z 还是 t？ |
|---|---|
| σ 已知 | z |
| σ 未知，n ≥ 30 | z（近似） |
| σ 未知，n < 30 | t（df = n-1） |
| 比例问题 | 始终用 z |
| 方差/标准差 | 用 χ²（一级了解即可） |
7.2 混淆 SE 的公式
| 场景 | SE 公式 |
|---|---|
| 单个均值 | σ/√n |
| 单个比例 | √(p̂(1-p̂)/n) |
| 两均值之差（独立） | √(σ₁²/n₁ + σ₂²/n₂) |
7.3 置信区间的判断技巧
如果 95% CI 包含零（或比例的 0.5 等参考值）→ 该参数与参考值之间无统计学显著差异。
如果 95% CI 全部为正 → 参数显著 大于 参考值。
如果 95% CI 全部为负 → 参数显著 小于 参考值。
八、实战综合案例
案例：基金经理的选股能力评估
某基金经理声称其选股能产生超额收益（alpha）。你收集了其 25 个季度的 alpha 数据： - 均值 x̄ = 0.8%/季度 - 标准差 s = 1.5%/季度
问题 1： 构造 95% 置信区间，判断 alpha 是否显著大于零？
问题 2： 如果 n = 100（其他不变），结论会变吗？
解答 1：
df = 24, t(0.025, 24) ≈ 2.064
SE = 1.5%/√25 = 0.3%
95% CI = 0.8% ± 2.064 × 0.3% = 0.8% ± 0.619% = [0.181%, 1.419%]
🟢 CI 全部为正，alpha 显著 > 0。基金经理的选股能力有统计证据。
解答 2（快速版）：
SE = 1.5%/√100 = 0.15%
95% CI ≈ 0.8% ± 1.96 × 0.15% = [0.506%, 1.094%]
CI 更窄，结论更稳健。这就是为什么机构投资者重视长业绩记录（大 n = 小 CI = 强结论）。
九、本节要点总结
| # | 要点 |
|---|---|
| 1 | 均值 CI（σ 已知）：x̄ ± z(α/2) · σ/√n |
| 2 | 均值 CI（σ 未知小样本）：x̄ ± t(α/2, n-1) · s/√n |
| 3 | 比例 CI：p̂ ± z(α/2) · √(p̂(1-p̂)/n)，条件：n·p̂, n·(1-p̂) ≥ 10 |
| 4 | 两均值差 CI：用 pooled SE 或两个独立 SE 加总 |
| 5 | σ 已知 → z；σ 未知大样本 → z；σ 未知小样本 → t |
| 6 | 比例问题永远用 z |
| 7 | CI 不跨零 → 显著；跨零 → 不显著 |
| 8 | 缩小 CI 最有效的方法：增大 n（唯一可控方式） |
十、测试题
Q1： 抽样 36 个数据，样本均值 = 50，总体标准差 σ = 12。总体均值的 90% 置信区间为：
A. [46.71, 53.29] B. [46.08, 53.92] C. [47.00, 53.00]
Q2： 分析师用 n = 16 的样本构造 t 区间，样本均值 = 30，s = 8。自由度为：
A. 16 B. 15 C. 取决于是否知道 σ
Q3： 某民调调查 500 人，300 人支持某政策。支持率的 95% CI 为：
A. [0.556, 0.644] B. [0.557, 0.643] C. [0.560, 0.640]
Q4： 在 σ 未知、n = 10 时，用 z 值代替 t 值构造 CI，会导致：
A. CI 比正确的更宽 B. CI 比正确的更窄 C. CI 不变
Q5： A 组（n₁ = 40, x̄₁ = 82）、B 组（n₂ = 40, x̄₂ = 78），两组方差均已知且相等 σ² = 36。μ₁ - μ₂ 的 95% CI 为：
A. [1.37, 6.63] B. [0.56, 7.44] C. [1.04, 6.96]
Q6（判断）： 对于比例估计，如果 n·p̂ = 8（< 10），我们应当用 t 分布代替 z 分布来构造 CI。
答案与解析
Q1：A SE = 12/√36 = 2 90% CI = 50 ± 1.645 × 2 = 50 ± 3.29 = [46.71, 53.29]
Q2：B t 分布的自由度 = n - 1 = 16 - 1 = 15，与是否知道 σ 无关。
Q3：B p̂ = 300/500 = 0.6 SE = √(0.6 × 0.4 / 500) = √0.00048 ≈ 0.0219 95% CI = 0.6 ± 1.96 × 0.0219 = 0.6 ± 0.0429 = [0.5571, 0.6429] 最接近的是 [0.557, 0.643]。
Q4：B t 值（t(0.025,9) ≈ 2.262）> z 值（1.96），所以用 z 会使 CI 比正确值更窄。这在考试中是一个常用陷阱：小样本忘了用 t，CI 偏窄，显得更精确，实则是虚假精确。
Q5：A SE = √(36/40 + 36/40) = √1.8 ≈ 1.3416 95% CI = (82 - 78) ± 1.96 × 1.3416 = 4 ± 2.6295 = [1.37, 6.63]
Q6：错误 比例问题始终用 z 分布。当 n·p̂ < 10 或 n·(1-p̂) < 10 时，应使用精确方法（如基于二项分布），而不是切换为 t 分布。t 分布不适用于比例估计。
Q6：错误 比例问题始终用 z 分布。当 n·p̂ < 10 或 n·(1-p̂) < 10 时，应使用精确方法（如基于二项分布），而不是切换为 t 分布。t 分布不适用于比例估计。
计。
Quantitative Methods — Sampling and Estimation · Lesson 5
I. Introduction
Recap (L128): We learned the concept, basic formula, and interpretation of confidence intervals. CI = Point Estimate ± Critical Value × Standard Error
In this lesson, we apply this framework to four specific scenarios, mastering the formulas, conditions, and calculation details for each.
II. Scenario 1: CI for Population Mean (σ Known / Large Sample)
2.1 Formula
x̄ ± z(α/2) × σ/√n
2.2 Conditions
| Condition | Note |
|---|---|
| Population SD σ known | Rare in practice, common on exams |
| Or n ≥ 30 (large sample) | CLT kicks in, use s as σ proxy |
2.3 Example
A bank surveys 100 customers on monthly spending. Sample mean x̄ = ¥4,200, historical σ = ¥800. Construct a 95% CI.
SE = 800/√100 = 80
95% CI = 4200 ± 1.96 × 80 = 4200 ± 156.8 = [4043.2, 4356.8]
📌 Interpretation: We are 95% confident the population mean monthly spending is between ¥4,043 and ¥4,357.
2.4 Key: Confidence Level ↔ z-Value
| Confidence Level | α | z(α/2) |
|---|---|---|
| 90% | 0.10 | 1.645 |
| 95% | 0.05 | 1.96 |
| 99% | 0.01 | 2.576 |
III. Scenario 2: CI for Population Mean (σ Unknown + Small Sample)
3.1 Formula
x̄ ± t(α/2, df) × s/√n, df = n - 1
3.2 Key Features of t-Distribution
| Feature | z (Standard Normal) | t |
|---|---|---|
| Shape | Fixed, exact bell | Bell-shaped, thicker tails |
| Depends on parameter? | No | Depends on df |
| df → ∞ | — | Approaches z |
| Critical value for same CL | Smaller | Larger (more conservative) |
💡 Intuition: When σ is unknown, using s to estimate σ adds another layer of uncertainty. The fat tails of the t-distribution "penalize" us for not knowing σ.
3.3 Quick Reference: t Critical Values
| df | t(0.05) (90%) | t(0.025) (95%) | t(0.005) (99%) |
|---|---|---|---|
| 5 | 2.015 | 2.571 | 4.032 |
| 10 | 1.812 | 2.228 | 3.169 |
| 15 | 1.753 | 2.131 | 2.947 |
| 20 | 1.725 | 2.086 | 2.845 |
| 30 | 1.697 | 2.042 | 2.750 |
| ∞ | 1.645 | 1.96 | 2.576 |
🔥 Memory tip: Smaller df → larger gap between t and z; df ≥ 30, t ≈ z.
3.4 Example: Small-Sample CI
An analyst randomly selects 16 clean energy stocks. Their past-year mean return is 12%, sample SD s = 8%. Construct a 95% CI.
df = 16 - 1 = 15, t(0.025, 15) = 2.131
SE = 8%/√16 = 2%
95% CI = 12% ± 2.131 × 2% = 12% ± 4.262% = [7.738%, 16.262%]
⚠️ Using z = 1.96 would give CI = [8.08%, 15.92%] — narrower, less conservative!
IV. Scenario 3: CI for Population Proportion
4.1 Formula
p̂ ± z(α/2) × √(p̂(1-p̂)/n)
where p̂ = sample proportion (successes / sample size)
4.2 Conditions
| Condition | Note |
|---|---|
| n·p̂ ≥ 10 | Sufficient expected successes |
| n·(1-p̂) ≥ 10 | Sufficient expected failures |
| Use z critical values | Proportions ALWAYS use z (never t) |
⚠️ If conditions are not met (n·p̂ < 10 or n·(1-p̂) < 10), exact methods (binomial) are required — not tested at Level I.
4.3 Example: Market Survey
An e-commerce platform surveys 400 users; 260 are satisfied with a new UI. Construct a 90% CI.
p̂ = 260/400 = 0.65
SE = √(0.65 × 0.35 / 400) = √(0.00056875) ≈ 0.02385
90% CI = 0.65 ± 1.645 × 0.02385 = 0.65 ± 0.0392 = [0.6108, 0.6892]
📌 Interpretation: We are 90% confident the true satisfaction rate is between 61.1% and 68.9%.
V. Scenario 4: CI for Difference Between Two Population Means
5.1 Why Do We Need This?
A point estimate tells you x̄₁ - x̄₂ = 3.5, but is 3.5 "meaningfully different from 0"? The interval estimate answers this.
5.2 Two Sub-Scenarios
Sub-scenario A: Independent samples, variances known (or large samples)
(x̄₁ - x̄₂) ± z(α/2) × √(σ₁²/n₁ + σ₂²/n₂)
Sub-scenario B: Independent samples, variances unknown but assumed equal (pooled)
(x̄₁ - x̄₂) ± t(α/2, n₁+n₂-2) × sp × √(1/n₁ + 1/n₂)
where sp² = ((n₁-1)s₁² + (n₂-1)s₂²) / (n₁ + n₂ - 2) (pooled variance)
5.3 Example: New Drug vs. Placebo
Clinical trial: Drug group n₁ = 50, mean BP reduction x̄₁ = 12 mmHg, s₁ = 5 mmHg Placebo group n₂ = 50, mean reduction x̄₂ = 4 mmHg, s₂ = 4 mmHg Construct 95% CI (assume equal variances).
sp² = (49×25 + 49×16) / 98 = 2009/98 ≈ 20.5, sp ≈ 4.527
SE = 4.527 × √(1/50 + 1/50) = 4.527 × 0.2 = 0.9054
df = 50 + 50 - 2 = 98, t(0.025, 98) ≈ 1.984
95% CI = (12 - 4) ± 1.984 × 0.9054 = 8 ± 1.796 = [6.20, 9.80]
🟢 Interpretation: CI is entirely > 0, meaning at 95% confidence, the drug truly outperforms placebo, with a reduction difference of ~6.2 to 9.8 mmHg.
VI. Factors Affecting CI Width (All Scenarios)
CI Width ∝ Critical Value × Variability/√n
| Factor | Change | Width | Intuition |
|---|---|---|---|
| Sample size n | ↑ | ↓ | More information → more precise |
| Confidence level (1-α) | ↑ | ↑ | Higher certainty → wider range |
| Population variability σ | ↑ | ↑ | Noisier data |
| z → t (small n) | — | ↑ | t has fatter tails, more conservative |
🔥 Exam focus: Given a change in conditions, determine whether CI widens or narrows.
VII. Common Pitfalls & Exam Tricks
7.1 Choosing the Wrong Critical Value
| Situation | z or t? |
|---|---|
| σ known | z |
| σ unknown, n ≥ 30 | z (approximate) |
| σ unknown, n < 30 | t (df = n-1) |
| Proportions | Always z |
| Variance/SD | Use χ² (Level I awareness only) |
7.2 Confusing SE Formulas
| Scenario | SE Formula |
|---|---|
| Single mean | σ/√n |
| Single proportion | √(p̂(1-p̂)/n) |
| Difference of two means (independent) | √(σ₁²/n₁ + σ₂²/n₂) |
7.3 CI Judgment Shortcut
If 95% CI includes zero → parameter is NOT significantly different from the reference value.
If 95% CI is entirely positive → parameter is significantly greater than the reference.
If 95% CI is entirely negative → parameter is significantly less than the reference.
VIII. Hands-On Case Study
Case: Evaluating a Fund Manager's Stock-Picking Skill
A fund manager claims to generate alpha. You collect 25 quarters of alpha data: - Mean x̄ = 0.8%/quarter - SD s = 1.5%/quarter
Q1: Construct a 95% CI. Is alpha significantly > 0?
Q2: If n = 100 (everything else unchanged), does the conclusion change?
Solution 1:
df = 24, t(0.025, 24) ≈ 2.064
SE = 1.5%/√25 = 0.3%
95% CI = 0.8% ± 2.064 × 0.3% = 0.8% ± 0.619% = [0.181%, 1.419%]
🟢 CI entirely positive → alpha is significantly > 0. Statistical evidence supports the manager's stock-picking skill.
Solution 2 (quick version):
SE = 1.5%/√100 = 0.15%
95% CI ≈ 0.8% ± 1.96 × 0.15% = [0.506%, 1.094%]
CI is narrower, conclusion is more robust. This is why institutional investors value long track records (large n = narrow CI = strong conclusion).
IX. Key Takeaways
| # | Takeaway |
|---|---|
| 1 | CI for mean (σ known): x̄ ± z(α/2) · σ/√n |
| 2 | CI for mean (σ unknown, small n): x̄ ± t(α/2, n-1) · s/√n |
| 3 | CI for proportion: p̂ ± z(α/2) · √(p̂(1-p̂)/n); conditions: n·p̂, n·(1-p̂) ≥ 10 |
| 4 | CI for difference of means: use pooled SE or sum of independent SEs |
| 5 | σ known → z; σ unknown + large n → z; σ unknown + small n → t |
| 6 | Proportions ALWAYS use z |
| 7 | CI does not cross zero → significant; crosses zero → not significant |
| 8 | Best way to narrow CI: increase n (the only controllable factor) |
X. Practice Questions
Q1: Sample of 36 observations, sample mean = 50, population σ = 12. The 90% CI for the population mean is:
A. [46.71, 53.29] B. [46.08, 53.92] C. [47.00, 53.00]
Q2: An analyst uses n = 16 to construct a t-interval, sample mean = 30, s = 8. Degrees of freedom:
A. 16 B. 15 C. Depends on whether σ is known
Q3: A poll surveys 500 people; 300 support a policy. The 95% CI for the support rate is:
A. [0.556, 0.644] B. [0.557, 0.643] C. [0.560, 0.640]
Q4: When σ is unknown and n = 10, using z instead of t to construct the CI will result in:
A. CI wider than the correct one B. CI narrower than the correct one C. CI unchanged
Q5: Group A (n₁ = 40, x̄₁ = 82), Group B (n₂ = 40, x̄₂ = 78), both variances known and equal σ² = 36. The 95% CI for μ₁ - μ₂ is:
A. [1.37, 6.63] B. [0.56, 7.44] C. [1.04, 6.96]
Q6 (True/False): For proportion estimation, if n·p̂ = 8 (< 10), we should use the t-distribution instead of z to construct the CI.
Answers & Explanations
Q1: A SE = 12/√36 = 2 90% CI = 50 ± 1.645 × 2 = 50 ± 3.29 = [46.71, 53.29]
Q2: B t-distribution df = n - 1 = 16 - 1 = 15, regardless of whether σ is known.
Q3: B p̂ = 300/500 = 0.6 SE = √(0.6 × 0.4 / 500) = √0.00048 ≈ 0.0219 95% CI = 0.6 ± 1.96 × 0.0219 = 0.6 ± 0.0429 = [0.5571, 0.6429] Closest match: [0.557, 0.643].
Q4: B t(0.025, 9) ≈ 2.262 > z = 1.96, so using z produces a CI narrower than the correct one. Classic exam trap: forgetting to use t for small samples makes the CI falsely precise.
Q5: A SE = √(36/40 + 36/40) = √1.8 ≈ 1.3416 95% CI = (82 - 78) ± 1.96 × 1.3416 = 4 ± 2.6295 = [1.37, 6.63]
Q6: False Proportions ALWAYS use z. When n·p̂ < 10 or n·(1-p̂) < 10, use exact methods (e.g., binomial-based), NOT the t-distribution. t is never appropriate for proportion estimation.