所属模块：
M1 · 数学与统计基础 (Mathematics & Statistics Fundamentals)| 专题分类：假设检验 (Hypothesis Testing)| 难度等级：Medium
一、核心一句话结论 (One-Sentence Summary)
样本量与效应量平方成反比；功效分析反解所需 n。
Sample size scales inversely with the square of Minimum Detectable Effect ($N propto 1/delta^2$); halving the effect size requires quadrupling the sample size.
二、核心考点要义 (Key Insights)
- 📌 A/B 上线前必做功效分析，避免’欠功效实验’
- 📌 方差可用历史数据或 CUPED 降低
English Insights:
– Standard sample size formula: $N = frac{2(z_{1-alpha/2} + z_{1-beta})^2 sigma^2}{delta^2}$ for two-sample equal-allocation tests.
– Square-root law: Standard error shrinks at rate $O(1/sqrt{N})$, compelling exponential sample requirements for tiny metrics.
– Key inputs: Significance level $alpha$ (typically 0.05), Power $1-beta$ (typically 0.80), Baseline variance $sigma^2$, and MDE $delta$.
三、核心数学原理与机理推导 (Mathematical Principles & Derivation)
$$n propto frac{(z_{1-alpha/2}+z_{1-beta})^2,sigma^2}{Delta^2}$$
推导（两样本均值比较）：检验统计量 Z=(X̄A−X̄B)/√(2σ²/n)，在 H₀ 下 ~N(0,1)，在 H₁（真实差异 Δ）下 ~N(Δ/√(2σ²/n), 1)。拒绝域为 |Z|>z)²/Δ²。核心结论：n 与效应量 Δ 的平方成反比——效应减半，样本量需增至 4 倍；n 与方差 σ² 成正比（方差翻倍，样本量翻倍）。这两个关系决定了实验设计的基本约束：小效应实验极其昂贵。}，要求 H₁ 下落入拒绝域的概率 ≥ 1−β，即 (Δ/√(2σ²/n)) − z{1−α/2} ≥ z_{1−β}，解出 n ≥ 2σ²(z_{1−α/2}+z_{1−β
📖 查看英文严格数学推导 (English Mathematical Derivation)
For two-sample z-test with $N$ users per variant, the difference in sample means $Delta = bar{X}_B – bar{X}_A$ has variance $text{Var}(Delta) = frac{sigma_A^2}{N} + frac{sigma_B^2}{N} approx frac{2sigma^2}{N}$. Under $H_0$, critical threshold is $C = z_{1-alpha/2} sqrt{frac{2sigma^2}{N}}$. Under $H_1: Delta = delta$, statistical power requires $P(Delta > C mid H_1) = 1 – beta$, which means $C = delta – z_{1-beta} sqrt{frac{2sigma^2}{N}}$. Equating the two expressions for $C$: $z_{1-alpha/2} sqrt{frac{2sigma^2}{N}} = delta – z_{1-beta} sqrt{frac{2sigma^2}{N}} implies (z_{1-alpha/2} + z_{1-beta}) sqrt{frac{2sigma^2}{N}} = delta$. Squaring and solving for $N$: $N = frac{2(z_{1-alpha/2} + z_{1-beta})^2 sigma^2}{delta^2}$. For $alpha=0.05, 1-beta=0.80$, $(1.96 + 0.84)^2 = 7.84$, giving rule-of-thumb $N approx frac{16sigma^2}{delta^2}$.
四、工业级落地权衡与工程考量 (Industrial Trade-offs)
实践要点：① MDE 必须先定——MDE（最小可检测效应）应由业务价值决定（如’至少提升 0.5% 转化率才值得上线’），而非由样本量倒推；反过来做（先定样本量再问能检出什么）会导致’实验能检出什么就上什么’的荒谬决策。② 降方差等于免费加样本——由 n∝σ²，方差降低 50% 等价于样本量翻倍；CUPED 用实验前协变量常能降低 30–50% 方差，是工业界最高杠杆的优化。③ 分流比例的影响——若处理组只占 p 比例，有效样本量下降，最优通常是 50/50（方差在等分时最小），但当处理有风险或成本高时可用不均衡分流。④ 指标类型不同公式不同——比率型指标（CTR）用 p(1−p) 代替 σ²，计数型需考虑过度离散，重尾指标应改用 bootstrap 或稳健估计。
⚙️ 查看英文落地权衡分析 (English Systems & Trade-offs)
If traffic is constrained and MDE cannot be achieved within acceptable timeframes, applied scientists have three options: (1) Variance reduction (CUPED): Reduces effective $sigma^2$ by factor $(1-rho^2)$, cutting required $N$ by up to 50%. (2) Transform metrics: Switch from noisy continuous revenue to conversion rate or binarized flags with lower coefficient of variation. (3) Multi-week run durations to capture weekly seasonality.
五、常见面试避坑陷阱 (Common Pitfalls & Traps)
- ⚠️ 先定样本量再倒推能检出什么（应由业务价值定 MDE）
- ⚠️ 忽视方差降低（CUPED）对样本量的等效替代作用
English Pitfalls:
– Setting MDE based on wishful thinking rather than business ROI thresholds.
– Ignoring clustering in sample size calculations when randomization unit differs from analysis unit (e.g. User-level treatment with page-level metrics).
六、高频深度面试追问与预测 (Follow-Up Questions)
- 最小可检测效应（MDE）怎么定？
- How does sample size calculation change for binary proportion metrics ($p(1-p)$ variance)?
- 为什么要先定 MDE 再算样本量？
- Why must sample size be inflated by the Design Effect (DEFF) in cluster-randomized experiments?
七、知识图谱对齐 (Knowledge Graph Anchor)
- 🔗 关联底层卡片：
数理统计假说检验、P 值、I/II 类错误与统计功效(Hypothesis Testing, P-Values, Power & Type I/II Error) - 🗺️ 知识图谱模块：
数理基础思维导图
🔬 算法科学家与机器学习深度考察全量题库 (Science Depth)
本题收录于 TalentMe 算法科学家深度考察真题库 (Science Depth)。全库共 856 道硬核考点，深度覆盖数学统计、经典ML、深度学习、Transformer、大语言模型、多模态、推荐系统与 MLOps。支持 Jev 面经智能匹配、一键离线单文件 HTML 手册导出并直连 Obsidian 本地记忆。