定量方法（Quantitative Methods）— 抽样与估计 · 第 6 课（复习实战）
一、知识回顾：抽样与估计五课速览
| 课次 | 主题 | 核心内容 |
|---|---|---|
| L125 | 抽样方法 | 简单随机抽样、分层抽样、系统抽样 |
| L126 | 中心极限定理 | CLT 的三个条件与结论 |
| L127 | 标准误 | SE 的定义、与标准差的区别 |
| L128 | 点估计 vs 区间估计 | 无偏性、有效性、一致性 |
| L129 | 置信区间构建 | 四种场景的 CI 公式与计算 |
二、核心公式速查卡
2.1 标准误公式
| 场景 | 标准误 SE |
|---|---|
| 样本均值的 SE（σ 已知） | σ / √n |
| 样本均值的 SE（σ 未知） | s / √n |
| 样本比例的 SE | √[p̂(1-p̂) / n] |
2.2 置信区间公式
| 场景 | 公式 | 条件 |
|---|---|---|
| 均值 CI · z 法 | x̄ ± z(α/2) × σ/√n | σ 已知 或 n ≥ 30 |
| 均值 CI · t 法 | x̄ ± t(α/2, n-1) × s/√n | σ 未知 且 n < 30 |
| 比例 CI | p̂ ± z(α/2) × √[p̂(1-p̂)/n] | np̂ ≥ 10 且 n(1-p̂) ≥ 10 |
2.3 关键临界值
| 置信水平 | z(α/2) | t(0.025, df=10) | t(0.025, df=20) |
|---|---|---|---|
| 90% | 1.645 | 1.812 | 1.725 |
| 95% | 1.96 | 2.228 | 2.086 |
| 99% | 2.576 | 3.169 | 2.845 |
2.4 中心极限定理（CLT）
当样本量 n ≥ 30，无论总体分布是什么形状，样本均值的抽样分布 近似正态，均值为 μ，标准误为 σ/√n。
三、综合练习题
📝 Block 1：抽样方法（L125）
Q1. 某基金公司要将 500 只股票按市值大小分成 5 层，每层随机抽取 10 只构成样本。这属于什么抽样方法？
A. 简单随机抽样 B. 系统抽样 C. 分层抽样 D. 整群抽样
Q2. 关于分层抽样的优点，以下哪个说法最准确？
A. 比简单随机抽样更容易实施 B. 保证每个个体被抽到的概率完全相等 C. 能降低抽样误差，尤其在层内同质性强时 D. 不需要抽样框
Q3. 以下哪种情况最适合使用系统抽样？
A. 总体有明显周期模式时 B. 总体名单按某种随机顺序排列时 C. 需要保证各层都有代表时 D. 总体规模很小
📝 Block 2：中心极限定理（L126）
Q4. 某研究收集了 64 个样本，总体服从右偏分布，均值为 50，标准差为 16。根据 CLT，样本均值的抽样分布近似：
A. 右偏分布，均值 = 50 B. 正态分布，均值 = 50，标准差 = 2 C. 正态分布，均值 = 50，标准差 = 16 D. t 分布，自由度 = 63
Q5. CLT 的关键结论中，样本均值的标准差（即标准误）等于：
A. σ B. σ / n C. σ / √n D. s / n
Q6. 如果样本量 n = 9，总体为正态分布，则样本均值的抽样分布是：
A. 近似正态（由 CLT 保证） B. 精确正态（因为总体正态） C. t 分布（必须用 t） D. 无法确定
📝 Block 3：标准误（L127）
Q7. 以下哪项最能区分"标准差"与"标准误"？
A. 标准差描述总体，标准误描述样本 B. 标准差描述个体数据的离散程度，标准误描述估计量的精度 C. 标准差总是大于标准误 D. 两者是同一概念的不同名称
Q8. 当样本量 n 从 25 增加到 100 时，样本均值的标准误会如何变化？
A. 减为原来的 1/4 B. 减为原来的 1/2 C. 增加 2 倍 D. 增加 4 倍
Q9. 某比例估计 p̂ = 0.40，n = 200，标准误是多少？
A. 0.0245 B. 0.0346 C. 0.0490 D. 0.0693
📝 Block 4：点估计 vs 区间估计（L128）
Q10. 一个好的点估计量应满足三个属性：无偏性、有效性和一致性。以下描述正确的是：
A. 无偏性意味着估计量 = 真值 B. 有效性意味着在所有无偏估计量中方差最小 C. 一致性意味着标准误始终为 0 D. 无偏性意味着每次估计结果都相同
Q11. 点估计的主要局限是什么？
A. 计算过于复杂 B. 只能用于大样本 C. 无法反映估计的不确定性 D. 总是有偏的
Q12. 以下哪个是区间估计相对点估计的优势？
A. 计算更快 B. 不受样本量影响 C. 提供了估计精度的信息 D. 总是比点估计更准确
📝 Block 5：置信区间构建（L129）
Q13. 某研究测量了 36 个家庭的月支出，样本均值 = ¥8,500，样本标准差 = ¥1,500。求总体均值的 95% 置信区间：
A. [¥8,010, ¥8,990] B. [¥8,200, ¥8,800] C. [¥7,960, ¥9,040] D. [¥8,110, ¥8,890]
Q14. 某调查随机抽取 400 人，其中 240 人表示满意。满意率的 95% 置信区间约为：
A. [0.55, 0.65] B. [0.56, 0.64] C. [0.53, 0.67] D. [0.50, 0.70]
Q15. 关于置信区间的解读，正确的是：
A. 总体均值有 95% 的概率落在该区间内 B. 该区间有 95% 的概率包含总体均值 C. 如果重复抽样 100 次构造 100 个 CI，其中约 95 个包含总体均值 D. 该区间包含了 95% 的样本数据
Q16. 样本量增加时，置信区间会：
A. 变宽 B. 变窄 C. 不变 D. 取决于置信水平
Q17. 以下哪个因素会使置信区间变宽？
A. 增加样本量 B. 降低置信水平 C. 增加置信水平（如从 95% → 99%） D. 减小标准差
Q18. 一个 90% 的置信区间比相同数据的 95% 置信区间：
A. 更宽 B. 更窄 C. 一样宽 D. 无法比较
📝 Block 6：综合应用题
Q19. 某基金经理分析了过去 5 年（60 个月）的月度收益率，月均收益率 = 0.8%，月收益率标准差 = 4.5%。请构造月均收益率的 95% 置信区间：
A. [-0.34%, 1.94%] B. [0.08%, 1.52%] C. [-0.10%, 1.70%] D. [0.40%, 1.20%]
Q20. 接上题。如果该基金经理将分析窗口扩大到 15 年（180 个月），月均收益率和标准差不变，新的 95% CI 宽度大约变为原来的：
A. 1/3 B. 1/√3 ≈ 0.577 C. 1/2 D. 2/3
Q21. 某公司质检员从一批电子产品中抽取 10 个进行寿命测试，得到样本均值 4,200 小时，样本标准差 380 小时。已知行业标准为 4,000 小时，在 95% 置信水平下，总体均值是否显著高于行业标准？
A. 是，因为 4,200 > 4,000 B. 是，因为 95% CI 下限 > 4,000 C. 否，因为 95% CI 包含 4,000 D. 不能确定，因为使用 t 分布临界值后 CI 包含 4,000
Q22. 一项研究将 500 个被调查者随机分成两组，第一组（n₁ = 250）对某政策的支持率为 52%，第二组（n₂ = 250）的支持率为 48%。在 95% 置信水平下，两组支持率差是否显著？
提示：两独立样本比例差的 SE = √[p̂₁(1-p̂₁)/n₁ + p̂₂(1-p̂₂)/n₂]
A. 不显著，95% CI 包含 0 B. 显著，95% CI 不包含 0 C. 显著，因为 52% > 48% D. 需要更大的样本才能判断
四、答案与解析
Block 1：抽样方法
Q1. 答案：C（分层抽样）
按市值分 5 层 → 每层独立抽取 → 经典分层抽样（Stratified Random Sampling）
Q2. 答案：C
分层抽样的核心优势：层内同质性强时，层内方差小，抽样误差低。
Q3. 答案：B
系统抽样适合总体名单随机排列时，每隔 k 个抽取一个。若总体有周期模式，系统抽样会产生偏差。
Block 2：中心极限定理
Q4. 答案：B
n = 64 ≥ 30，CLT 适用。抽样分布 ≈ 正态，均值 = μ = 50，SE = σ/√n = 16/8 = 2。
Q5. 答案：C
标准误 = σ/√n，这是 CLT 的核心结论之一。
Q6. 答案：B
总体为正态 + 任何样本量 → 样本均值的抽样分布是精确正态（不需要 CLT 近似）。但若 σ 未知，构造 CI 时使用 t 分布。
Block 3：标准误
Q7. 答案：B
标准差 (s or σ) = 个体数据点围绕均值的离散程度；标准误 (SE) = 估计量（如 x̄）的精度。
Q8. 答案：B
SE = σ/√n，n 从 25→100，√100/√25 = 10/5 = 2，SE 减半。
Q9. 答案：B
SE = √[0.40 × 0.60 / 200] = √(0.24/200) = √0.0012 = 0.0346
Block 4：点估计 vs 区间估计
Q10. 答案：B
无偏性 = E(估计量) = 真值；有效性 = 方差最小；一致性 = n 增大时估计量趋近真值。
Q11. 答案：C
点估计给出单一数值，无法反映「估计有多靠谱」。区间估计补上了这个缺口。
Q12. 答案：C
区间估计给出一个范围，直接传达了估计的精度（宽度越窄越精确）。
Block 5：置信区间构建
Q13. 答案：A
n = 36（大样本），用 z 法： SE = 1500/√36 = 1500/6 = 250 95% CI = 8500 ± 1.96 × 250 = 8500 ± 490 = [8010, 8990]
Q14. 答案：A
p̂ = 240/400 = 0.60 SE = √[0.60 × 0.40 / 400] = √(0.24/400) = √0.0006 = 0.0245 95% CI = 0.60 ± 1.96 × 0.0245 = 0.60 ± 0.048 = [0.552, 0.648] ≈ [0.55, 0.65]
Q15. 答案：C
最严谨的解读是频率学派视角：如果无限重复实验，95% 的 CI 会包含 μ。不能简单说「这个区间有 95% 概率包含 μ」（因为 μ 是常数，不是随机变量）。
Q16. 答案：B
CI 宽度 ∝ 1/√n，n ↑ → CI ↓（变窄）。
Q17. 答案：C
置信水平 ↑ → 临界值 ↑ → CI 变宽。A 和 D 会使 CI 变窄，B 也会使 CI 变窄。
Q18. 答案：B
90% 的 z(0.05) = 1.645 < 95% 的 z(0.025) = 1.96，所以 90% CI 更窄。你要求的把握越大，区间就越宽。
Block 6：综合应用
Q19. 答案：A
n = 60（大样本），用 z 法： SE = 4.5%/√60 = 4.5%/7.746 = 0.581% 95% CI = 0.8% ± 1.96 × 0.581% = 0.8% ± 1.139% = [-0.339%, 1.939%] 注意：区间包含 0 → 月均收益率可能为 0，不能断言显著为正。
Q20. 答案：B
CI 宽度 ∝ SE = σ/√n 新 SE / 旧 SE = √60/√180 = 1/√3 ≈ 0.577 时间扩大 3 倍，精度提升 √3 倍（不是 3 倍！）。
Q21. 答案：D
n = 10（小样本，σ 未知）→ 必须用 t 分布！ SE = 380/√10 = 120.2 t(0.025, df=9) = 2.262 95% CI = 4200 ± 2.262 × 120.2 = 4200 ± 271.8 = [3928, 4472] 区间包含 4000 → 在 95% 置信水平下，不能断言总体均值显著高于 4000。 🔥 易错点：小样本忘记从 z 切换到 t 分布！
Q22. 答案：A
p̂₁ = 0.52, p̂₂ = 0.48 SE_diff = √[0.52×0.48/250 + 0.48×0.52/250] = √[2 × 0.2496/250] = √0.001997 = 0.0447 95% CI_diff = (0.52-0.48) ± 1.96 × 0.0447 = 0.04 ± 0.0876 = [-0.0476, 0.1276] CI 包含 0 → 4% 的差异在统计上不显著。
五、错题诊断表
| 题号 | 对应课次 | 易错原因 | 建议 |
|---|---|---|---|
| Q6 | L126 | 混淆「总体正态→精确正态」和「CLT→近似正态」 | 记住：总体正态时不需要 CLT，但 CI 构造仍需看 σ 是否已知 |
| Q7 | L127 | 标准误和标准差混为一谈 | SE 是估计量的标准差，不是原始数据的标准差 |
| Q13 | L129 | 小样本用了 z 而不是 t | 口诀：小样本+σ未知=t；大样本或σ已知=z |
| Q15 | L128-L129 | CI 的概率解读常出错 | 记住频率学派的标准答案 C |
| Q19 | L129 | 忘记 CI 可以跨 0，跨 0 = 不显著 | 实战中 CI 是否跨 0 是快速判断显著性的方法 |
| Q21 | L126+L129 | 看到 n=10 还条件反射用 z=1.96 | 小样本用 t 分布，临界值更大，CI 更宽 |
| Q22 | L129 | 两样本比例差的 SE 公式不熟 | 独立样本 SE 是各自 SE² 之和再开方 |
六、本模块要点总结
🎯 抽样与估计模块（L125-L130）完成！以下是必须带走的 5 个要点：
| # | 要点 | 一句话总结 |
|---|---|---|
| 1 | CLT | n≥30 时样本均值近似正态，无论总体什么形状 |
| 2 | SE | 衡量估计量精度，等于 σ/√n，与 n 的平方根成反比 |
| 3 | z vs t | σ 已知或用大样本→z；σ 未知且小样本→t |
| 4 | CI 解读 | CI 说的是「方法的可靠性」，不是「某个特定区间的概率」 |
| 5 | CI 宽度 | 宽度 ∝ 临界值 × σ/√n，受 n、置信水平、数据变异性三方影响 |
七、自检清单
- [ ] 能区分简单随机、分层、系统三种抽样方法并判断适用场景
- [ ] 能准确说出 CLT 的三个结论（正态性、均值、标准误）
- [ ] 能在 3 秒内判断该用 z 还是 t
- [ ] 能用公式 x̄ ± z × σ/√n 和 x̄ ± t × s/√n 完成 CI 计算
- [ ] 能正确解读 CI（频率学派版本）
- [ ] 能判断 CI 是否跨 0（快速显著性判断）
- [ ] 能分析样本量变化对 CI 宽度的影响
📈 下一模块预告：L131-L136 假设检验 — 从「估计」到「决策」，学会用 p 值和显著性水平做出统计判断！
Quantitative Methods — Sampling & Estimation · Lesson 6 (Review & Practice)
I. Knowledge Recap: Five-Lesson Overview
| Lesson | Topic | Core Content |
|---|---|---|
| L125 | Sampling Methods | Simple random, stratified, systematic sampling |
| L126 | Central Limit Theorem | Three conditions and conclusions of CLT |
| L127 | Standard Error | Definition of SE, difference from standard deviation |
| L128 | Point vs Interval Estimation | Unbiasedness, efficiency, consistency |
| L129 | Confidence Interval Construction | CI formulas and calculations for four scenarios |
II. Formula Cheat Sheet
2.1 Standard Error Formulas
| Scenario | Standard Error (SE) |
|---|---|
| SE of sample mean (σ known) | σ / √n |
| SE of sample mean (σ unknown) | s / √n |
| SE of sample proportion | √[p̂(1-p̂) / n] |
2.2 Confidence Interval Formulas
| Scenario | Formula | Condition |
|---|---|---|
| Mean CI · z-method | x̄ ± z(α/2) × σ/√n | σ known or n ≥ 30 |
| Mean CI · t-method | x̄ ± t(α/2, n-1) × s/√n | σ unknown and n < 30 |
| Proportion CI | p̂ ± z(α/2) × √[p̂(1-p̂)/n] | np̂ ≥ 10 and n(1-p̂) ≥ 10 |
2.3 Key Critical Values
| Confidence Level | z(α/2) | t(0.025, df=10) | t(0.025, df=20) |
|---|---|---|---|
| 90% | 1.645 | 1.812 | 1.725 |
| 95% | 1.96 | 2.228 | 2.086 |
| 99% | 2.576 | 3.169 | 2.845 |
2.4 Central Limit Theorem (CLT)
When sample size n ≥ 30, regardless of the population distribution shape, the sampling distribution of the sample mean is approximately normal, with mean = μ and standard error = σ/√n.
III. Comprehensive Practice Questions
📝 Block 1: Sampling Methods (L125)
Q1. A fund company divides 500 stocks into 5 strata by market capitalization and randomly selects 10 stocks from each stratum. What sampling method is this?
A. Simple random sampling B. Systematic sampling C. Stratified sampling D. Cluster sampling
Q2. Regarding the advantages of stratified sampling, which statement is the most accurate?
A. Easier to implement than simple random sampling B. Ensures every individual has an exactly equal probability of selection C. Reduces sampling error, especially when within-stratum homogeneity is high D. Does not require a sampling frame
Q3. In which situation is systematic sampling most appropriate?
A. When the population has a clear periodic pattern B. When the population list is in some random order C. When representation from each stratum is required D. When the population size is very small
📝 Block 2: Central Limit Theorem (L126)
Q4. A study collects 64 observations from a right-skewed population with mean = 50 and standard deviation = 16. According to the CLT, the sampling distribution of the sample mean is approximately:
A. Right-skewed, mean = 50 B. Normal, mean = 50, standard deviation = 2 C. Normal, mean = 50, standard deviation = 16 D. t-distribution, df = 63
Q5. According to the CLT, the standard deviation of the sample mean (i.e., the standard error) equals:
A. σ B. σ / n C. σ / √n D. s / n
Q6. If n = 9 and the population is normally distributed, the sampling distribution of the sample mean is:
A. Approximately normal (guaranteed by CLT) B. Exactly normal (because the population is normal) C. t-distribution (must use t) D. Cannot be determined
📝 Block 3: Standard Error (L127)
Q7. Which best distinguishes "standard deviation" from "standard error"?
A. Standard deviation describes the population; standard error describes the sample B. Standard deviation describes dispersion of individual data points; standard error describes precision of an estimator C. Standard deviation is always larger than standard error D. They are different names for the same concept
Q8. When sample size increases from 25 to 100, the standard error of the sample mean will:
A. Decrease to 1/4 of its original value B. Decrease to 1/2 of its original value C. Increase by a factor of 2 D. Increase by a factor of 4
Q9. Given p̂ = 0.40 and n = 200, what is the standard error?
A. 0.0245 B. 0.0346 C. 0.0490 D. 0.0693
📝 Block 4: Point Estimate vs Interval Estimate (L128)
Q10. A good point estimator should satisfy three properties: unbiasedness, efficiency, and consistency. Which statement is correct?
A. Unbiasedness means the estimator equals the true value B. Efficiency means the estimator has the smallest variance among all unbiased estimators C. Consistency means the standard error is always zero D. Unbiasedness means every estimate produces the same result
Q11. What is the main limitation of a point estimate?
A. Too complex to compute B. Can only be used with large samples C. Cannot reflect the uncertainty of the estimate D. Is always biased
Q12. Which of the following is an advantage of interval estimation over point estimation?
A. Faster to compute B. Not affected by sample size C. Provides information about estimation precision D. Always more accurate than point estimates
📝 Block 5: Confidence Interval Construction (L129)
Q13. A study measures monthly household expenditures for 36 families. Sample mean = ¥8,500, sample standard deviation = ¥1,500. What is the 95% CI for the population mean?
A. [¥8,010, ¥8,990] B. [¥8,200, ¥8,800] C. [¥7,960, ¥9,040] D. [¥8,110, ¥8,890]
Q14. A survey randomly samples 400 people, of whom 240 express satisfaction. The 95% CI for the satisfaction rate is approximately:
A. [0.55, 0.65] B. [0.56, 0.64] C. [0.53, 0.67] D. [0.50, 0.70]
Q15. Regarding the interpretation of a confidence interval, which is correct?
A. There is a 95% probability that the population mean falls within this interval B. This interval has a 95% probability of containing the population mean C. If we repeatedly sample 100 times and construct 100 CIs, approximately 95 of them will contain the population mean D. This interval contains 95% of the sample data
Q16. When sample size increases, the confidence interval will:
A. Become wider B. Become narrower C. Remain unchanged D. Depend on the confidence level
Q17. Which factor will make a confidence interval wider?
A. Increasing sample size B. Decreasing confidence level C. Increasing confidence level (e.g., from 95% to 99%) D. Decreasing standard deviation
Q18. A 90% confidence interval, compared to a 95% CI for the same data, is:
A. Wider B. Narrower C. The same width D. Cannot be compared
📝 Block 6: Integrated Application
Q19. A portfolio manager analyzes monthly returns over the past 5 years (60 months). Average monthly return = 0.8%, monthly return standard deviation = 4.5%. Construct the 95% CI for the average monthly return:
A. [-0.34%, 1.94%] B. [0.08%, 1.52%] C. [-0.10%, 1.70%] D. [0.40%, 1.20%]
Q20. Continuing from above. If the manager expands the analysis window to 15 years (180 months), with mean and standard deviation unchanged, the new 95% CI width will be approximately:
A. 1/3 of the original B. 1/√3 ≈ 0.577 of the original C. 1/2 of the original D. 2/3 of the original
Q21. A quality inspector tests 10 electronic products for lifespan, obtaining sample mean = 4,200 hours and sample standard deviation = 380 hours. The industry standard is 4,000 hours. At the 95% confidence level, is the population mean significantly higher than the industry standard?
A. Yes, because 4,200 > 4,000 B. Yes, because the 95% CI lower bound > 4,000 C. No, because the 95% CI contains 4,000 D. Cannot be determined; using the t-distribution critical value, the CI contains 4,000
Q22. A study randomly divides 500 respondents into two groups. Group 1 (n₁ = 250) has a 52% approval rate for a policy; Group 2 (n₂ = 250) has 48% approval. At the 95% confidence level, is the difference in approval rates significant?
Hint: SE of the difference between two independent sample proportions = √[p̂₁(1-p̂₁)/n₁ + p̂₂(1-p̂₂)/n₂]
A. Not significant; the 95% CI contains 0 B. Significant; the 95% CI does not contain 0 C. Significant, because 52% > 48% D. A larger sample is needed to judge
IV. Answers & Explanations
Block 1: Sampling Methods
Q1. Answer: C (Stratified sampling)
Divided into 5 strata by market cap → independent sampling within each stratum → classic stratified random sampling.
Q2. Answer: C
The core advantage of stratified sampling: when within-stratum homogeneity is high, within-stratum variance is small, yielding lower sampling error.
Q3. Answer: B
Systematic sampling is suitable when the population list is randomly ordered, selecting every k-th unit. If the population has a periodic pattern, systematic sampling can introduce bias.
Block 2: Central Limit Theorem
Q4. Answer: B
n = 64 ≥ 30, CLT applies. Sampling distribution ≈ normal, mean = μ = 50, SE = σ/√n = 16/8 = 2.
Q5. Answer: C
Standard error = σ/√n — this is one of the core conclusions of the CLT.
Q6. Answer: B
Population is normal + any sample size → the sampling distribution of the sample mean is exactly normal (CLT approximation is not needed). However, if σ is unknown, constructing a CI requires the t-distribution.
Block 3: Standard Error
Q7. Answer: B
Standard deviation (s or σ) = dispersion of individual data points around the mean; Standard error (SE) = precision of an estimator (e.g., x̄).
Q8. Answer: B
SE = σ/√n. n from 25→100, √100/√25 = 10/5 = 2, so SE is halved.
Q9. Answer: B
SE = √[0.40 × 0.60 / 200] = √(0.24/200) = √0.0012 = 0.0346
Block 4: Point Estimate vs Interval Estimate
Q10. Answer: B
Unbiasedness = E(estimator) = true value; Efficiency = minimum variance; Consistency = estimator converges to true value as n increases.
Q11. Answer: C
A point estimate provides a single number and cannot reflect "how reliable the estimate is." Interval estimation fills that gap.
Q12. Answer: C
An interval estimate provides a range, directly conveying the precision of estimation (narrower interval = greater precision).
Block 5: Confidence Interval Construction
Q13. Answer: A
n = 36 (large sample), use z-method: SE = 1500/√36 = 1500/6 = 250 95% CI = 8500 ± 1.96 × 250 = 8500 ± 490 = [8010, 8990]
Q14. Answer: A
p̂ = 240/400 = 0.60 SE = √[0.60 × 0.40 / 400] = √(0.24/400) = √0.0006 = 0.0245 95% CI = 0.60 ± 1.96 × 0.0245 = 0.60 ± 0.048 = [0.552, 0.648] ≈ [0.55, 0.65]
Q15. Answer: C
The most rigorous interpretation follows the frequentist perspective: if we repeat the experiment infinitely many times, 95% of the CIs will contain μ. We cannot simply say "this interval has a 95% probability of containing μ" (because μ is a constant, not a random variable).
Q16. Answer: B
CI width ∝ 1/√n, so as n ↑, CI ↓ (narrower).
Q17. Answer: C
Confidence level ↑ → critical value ↑ → CI wider. A and D would narrow the CI; B would also narrow it.
Q18. Answer: B
90% z(0.05) = 1.645 < 95% z(0.025) = 1.96, so the 90% CI is narrower. The more confidence you demand, the wider the interval.
Block 6: Integrated Application
Q19. Answer: A
n = 60 (large sample), use z-method: SE = 4.5%/√60 = 4.5%/7.746 = 0.581% 95% CI = 0.8% ± 1.96 × 0.581% = 0.8% ± 1.139% = [-0.339%, 1.939%] Note: the interval includes 0 → the average monthly return may be 0; cannot assert it is significantly positive.
Q20. Answer: B
CI width ∝ SE = σ/√n New SE / Old SE = √60/√180 = 1/√3 ≈ 0.577 Time expanded 3×, precision improved √3× (not 3×!).
Q21. Answer: D
n = 10 (small sample, σ unknown) → must use the t-distribution! SE = 380/√10 = 120.2 t(0.025, df=9) = 2.262 95% CI = 4200 ± 2.262 × 120.2 = 4200 ± 271.8 = [3928, 4472] The interval contains 4000 → at the 95% confidence level, cannot assert the population mean is significantly higher than 4000. 🔥 Common trap: forgetting to switch from z to t for small samples!
Q22. Answer: A
p̂₁ = 0.52, p̂₂ = 0.48 SE_diff = √[0.52×0.48/250 + 0.48×0.52/250] = √[2 × 0.2496/250] = √0.001997 = 0.0447 95% CI_diff = (0.52-0.48) ± 1.96 × 0.0447 = 0.04 ± 0.0876 = [-0.0476, 0.1276] CI contains 0 → the 4% difference is not statistically significant.
V. Common Error Diagnosis Table
| Q# | Lesson | Common Mistake | Recommendation |
|---|---|---|---|
| Q6 | L126 | Confusing "population normal → exactly normal" vs "CLT → approximately normal" | Remember: when population is normal, CLT is not needed, but CI construction still depends on whether σ is known |
| Q7 | L127 | Confusing standard error with standard deviation | SE is the standard deviation of an estimator, not of the raw data |
| Q13 | L129 | Using z instead of t for small samples | Rule of thumb: small sample + σ unknown = t; large sample or σ known = z |
| Q15 | L128-L129 | Incorrect probabilistic interpretation of CI | Memorize the frequentist answer: option C |
| Q19 | L129 | Forgetting that CI can cross 0; crossing 0 = not significant | In practice, whether CI crosses 0 is a quick significance test |
| Q21 | L126+L129 | Reflexively using z=1.96 for n=10 | Small sample → use t-distribution, larger critical value, wider CI |
| Q22 | L129 | Unfamiliar with difference-of-proportions SE formula | Independent sample SE = sqrt of sum of individual SE² |
VI. Key Module Takeaways
🎯 Sampling & Estimation module (L125-L130) complete! Here are the 5 must-remember takeaways:
| # | Takeaway | One-Liner |
|---|---|---|
| 1 | CLT | n≥30 → sample mean ≈ normal, regardless of population shape |
| 2 | SE | Measures estimator precision, equals σ/√n, inversely proportional to √n |
| 3 | z vs t | σ known or large sample → z; σ unknown and small sample → t |
| 4 | CI Interpretation | CI describes the reliability of the method, not the probability of a specific interval |
| 5 | CI Width | Width ∝ critical value × σ/√n, affected by n, confidence level, and data variability |
VII. Self-Check Checklist
- [ ] Can distinguish simple random, stratified, and systematic sampling and identify appropriate scenarios
- [ ] Can accurately state the three conclusions of CLT (normality, mean, standard error)
- [ ] Can decide within 3 seconds whether to use z or t
- [ ] Can compute CIs using x̄ ± z × σ/√n and x̄ ± t × s/√n
- [ ] Can correctly interpret a CI (frequentist version)
- [ ] Can determine whether a CI crosses 0 (quick significance check)
- [ ] Can analyze the impact of sample size changes on CI width
📈 Next Module Preview: L131-L136 Hypothesis Testing — From estimation to decision-making: learn to make statistical judgments using p-values and significance levels!