定量方法(Quantitative Methods)— 假设检验 · 第 5 课
一、前情回顾:检验框架已搭建完成
| 课次 | 内容 | 关键收获 |
|---|---|---|
| L131 | 假设检验基本框架 | H₀/Hₐ 设定、检验统计量、决策规则 |
| L132 | p 值与显著性水平 | p < α → 拒绝 H₀;p 是"极端程度" |
| L133 | Type I & Type II 错误 | α(冤枉) vs β(遗漏),Power = 1-β |
| L134 | 单尾 vs 双尾 | Hₐ 方向决定尾巴,双尾 p = 单尾 p × 2 |
今天回答核心问题:具体怎么检验一个均值是否等于(大于/小于)某个值?
二、均值检验的两大工具
2.1 直觉
想检验「全港基金平均年化收益是不是 5%」,你需要:
- 抽一个样本,计算样本均值 x̄
- 评估 x̄ 和 5% 的差距有多大
- 判断这个差距是「随机波动」还是「真实差异」
这个判断过程,就需要检验统计量。对均值来说,有两个选择:z 和 t。
2.2 什么时候用哪个?
总体标准差 σ 是否已知?
├─ 已知 → Z 检验
└─ 未知 → T 检验(用样本标准差 s 代替 σ)
| Z 检验 | T 检验 | |
|---|---|---|
| σ 已知? | ✅ 已知 | ❌ 未知,用 s 估计 |
| 分布 | 正态分布 N(0,1) | t 分布(df = n-1) |
| 临界值 | Zα 固定(1.645, 1.96…) | tα,n-1 取决于自由度 |
| 适用场合 | 学术题、已知总体参数 | 实际金融分析(99% 情况) |
| CFA 一级 | 概念层面常见 | 计算题主力 |
📌 CFA 一级实战规则:σ 已知 → Z;σ 未知 → T。 大部分金融场景 σ 未知 → T 检验为主。
三、Z 检验(Z-Test)
3.1 适用条件
- 总体服从正态分布,或样本量足够大(n ≥ 30,CLT)
- 总体标准差 σ 已知
- 数据独立随机抽样
3.2 检验统计量
$$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$$
其中: - x̄ = 样本均值 - μ₀ = H₀ 中假设的总体均值 - σ = 已知的总体标准差 - n = 样本容量 - SE = σ/√n(标准误)
3.3 决策规则速查
| Hₐ | 拒绝 H₀ 的条件 |
|---|---|
| Hₐ: μ ≠ μ₀(双尾) | |
| Hₐ: μ > μ₀(右尾) | z > Zα |
| Hₐ: μ < μ₀(左尾) | z < −Zα |
3.4 案例:面粉重量检验
某面粉厂包装规格为 1000g,σ = 15g(经验已知)。质检随机抽取 50 袋: x̄ = 995g。α = 0.05。检验是否缺斤少两。
H₀: μ ≥ 1000(足秤)
Hₐ: μ < 1000(缺秤)→ 左尾检验
n = 50,σ = 15,x̄ = 995
z = (995 - 1000) / (15/√50)
= -5 / 2.121
= -2.357
α = 0.05,左尾 Zα = -1.645
-2.357 < -1.645 → 拒绝 H₀ ✅
结论:有显著证据表明机器装填不足 1000g。建议校准。
🎯 Z 检验的特点:如果 σ 已知,z 统计量服从精确正态分布,临界值固定。
四、T 检验(T-Test)
4.1 为什么需要 T 检验?
现实世界中 σ 几乎总是未知的。
当我们用 s(样本标准差)代替 σ 时: - 分子 x̄ − μ₀ 仍近似正态 - 但分母多了一层不确定性(s 本身在波动) - → 结果:检验统计量不再是正态分布,而是 t 分布
Z 统计量: (x̄ − μ₀) / (σ/√n) → 正态分布
T 统计量: (x̄ − μ₀) / (s/√n) → t 分布
↑
这个 s 引入了额外不确定性
4.2 t 分布的特征
t 分布 vs 正态
正态(尖峰)
╱ ╲ t(df=2) —— 肥尾
╱ ╲ t(df=5) —— 中等
╱ ╲ t(df=30) —— 趋近正态
────┴────────────────┴────
| 特征 | 说明 |
|---|---|
| 形状 | 对称、钟形,中心为 0 |
| 肥尾 | 比正态分布尾部更厚(不确定性更大) |
| 自由度 df | df = n − 1 控制肥度 |
| n↑ → | t 趋近正态(n > 120 时几乎等同) |
| df = 1 → | 柯西分布(最肥) |
📌 同 α 下,t 临界值 > Z 临界值。因为 s 代替 σ 带来的额外不确定性要求更大的「安全边际」。
4.3 检验统计量
$$t_{n-1} = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$$
其中: - s = 样本标准差 = $\sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}$ - df = n − 1
4.4 决策规则
| Hₐ | 拒绝 H₀ 的条件 |
|---|---|
| Hₐ: μ ≠ μ₀(双尾) | |
| Hₐ: μ > μ₀(右尾) | t > tα, n−1 |
| Hₐ: μ < μ₀(左尾) | t < −tα, n−1 |
p < α → 拒绝 H₀(与 Z 检验相同)
五、实战案例:基金 Alpha 检验
5.1 案例数据
某基金 36 个月月均超额收益 0.65%,s = 2.80%。 H₀: μ = 0(无 Alpha),Hₐ: μ ≠ 0(双尾),α = 0.05
n = 36,x̄ = 0.65%,s = 2.80%
SE = 2.80% / √36 = 0.4667%
t = (0.65% - 0) / 0.4667% = 1.393
df = 35
t₀.₀₂₅,₃₅ ≈ 2.030(查表或计算器)
|t| = 1.393 < 2.030 → 不拒绝 H₀ ❌
结论:月均超额 0.65%,但波动太大,统计上不足以证明存在 Alpha。
5.2 直观理解
月均 0.65%,表面看不错
但 36 个月标准差 2.80%,波动剧烈
→ 这 0.65% 可能只是运气好的 36 个月
→ 需要更长时间或更大 Alpha 才能证明
六、Z vs T 对比速查
| 维度 | Z 检验 | T 检验 |
|---|---|---|
| 分布 | 标准正态 N(0,1) | t(df) 分布 |
| σ 已知? | ✅ 是 | ❌ 否(用 s) |
| 临界值 | 固定(1.645, 1.96,D) | 随 df 变化 |
| 尾部 | 标准 | 更肥(n 小更明显) |
| 大样本 n→∞ | → | → 趋近 Z 检验 |
| CFA 实战 | 概念题为主 | 计算题主力 |
| p 值计算 | 查 Z 表 | 查 t 表(按 df) |
七、关键临界值对比(α = 0.05)
| df | t₀.₀₂₅(双尾) | Z₀.₀₂₅ | t₀.₀₅(单尾) | Z₀.₀₅ |
|---|---|---|---|---|
| 1 | 12.706 | 1.96 | 6.314 | 1.645 |
| 5 | 2.571 | 1.96 | 2.015 | 1.645 |
| 10 | 2.228 | 1.96 | 1.812 | 1.645 |
| 20 | 2.086 | 1.96 | 1.725 | 1.645 |
| 30 | 2.042 | 1.96 | 1.697 | 1.645 |
| 60 | 2.000 | 1.96 | 1.671 | 1.645 |
| 120 | 1.980 | 1.96 | 1.658 | 1.645 |
| ∞ | 1.960 | 1.96 | 1.645 | 1.645 |
📌 n 越小(df 越小)→ t 临界值越大 → 拒绝 H₀ 的门槛越高 → 对「样本量惩罚」
八、样本量的三重影响
| 方向 | 效果 |
|---|---|
| SE ↓ | n↑ → SE = s/√n ↓ → 检验统计量 ↑ → 更易拒绝 H₀ |
| 临界值 ↓ | n↑ → df↑ → t 临界值 ↓ → 更易拒绝 H₀ |
| Power ↑ | n↑ → 分布更集中 → 更容易检测到真实效应 |
🎯 增大样本量是提升检验能力的「万能药」——同时降低 SE、降低临界值、提升 Power。
九、CFA 经典陷阱
陷阱 1:什么时候用 Z 检验?
❌ 样本量 > 30 → 自动用 Z ✅ 只有 σ 已知 才用 Z!n > 30 但 σ 未知 → 仍用 T
Z vs T 的区分标准是 σ 是否已知,不是 n 多大。
陷阱 2:用 s 做 Z 检验的临界值
❌ 如果题目给了 s 而不是 σ,但你强行用 Z 临界值 ✅ s → 必须用 T,临界值查 t 表(按 n-1)
陷阱 3:自由度选错
❌ df = n(总把 n 当自由度) ✅ 单样本均值检验:df = n − 1
陷阱 4:n>30 认为 t 与正态无区别
❌ n=35 → 心里直接当 Z 临界值用 ✅ n=35 时 t₀.₀₂₅,₃₄ ≈ 2.032 > 1.96,仍有差异! 接近 n=120 时才基本等同。
陷阱 5:p 值判定
❌ t 统计量算出来直接查 Z 表 ✅ t 统计量必须查 t 表(按 df),否则 p 值不准确
十、完整检验步骤模板
进行一个均值假设检验,标准五步:
第 1 步:设定假设
H₀: μ = μ₀(或 ≤, ≥)
Hₐ: μ ≠ μ₀(或 >, <)
第 2 步:选择检验统计量
├─ σ 已知 → z = (x̄ - μ₀) / (σ/√n)
└─ σ 未知 → t = (x̄ - μ₀) / (s/√n), df = n-1
第 3 步:确定显著性水平与临界值
α = 0.05 → 查 Z 表或 t 表
第 4 步:计算并比较
计算统计量 → 与临界值比较 / 查 p 值
第 5 步:做出决策并解释
「在 α = 0.05 水平下,[拒绝 / 不拒绝] H₀。
有/无充分证据表明...」
📝 课堂练习
Part A:选择检验方法
Q1. 已知 σ = 8,n = 25,检验 μ = 100。应用:
A. Z 检验 B. T 检验 C. 卡方检验 D. F 检验
Q2. σ 未知,n = 50,x̄ = 3.2,s = 1.5。检验 μ = 3。应用:
A. Z 检验,查正态表 B. T 检验,df = 49 C. T 检验,df = 50 D. Z 检验,因为 n > 30
Part B:假设设定与决策
Q3. 某分析师想检验基金 Alpha 是否为正。H₀ 应设为:
A. H₀: μ = 0 B. H₀: μ ≤ 0 C. H₀: μ ≥ 0 D. H₀: μ ≠ 0
Q4. n = 16,x̄ = 21.3,s = 3.8,H₀: μ = 20,Hₐ: μ ≠ 20,α = 0.05。t 值约为:
A. 0.34 B. 1.37 C. 2.06 D. 2.74
Part C:综合判断
Q5. n = 25,x̄ = 48,s = 10,H₀: μ = 50,Hₐ: μ < 50,α = 0.05。以下正确的是:
A. Z 检验,z = -1.0 B. T 检验,df = 24 C. T 检验,df = 25 D. Z 检验,因为 σ 未知
Q6. 关于 t 分布,正确的是:
A. t 分布总是比正态分布更尖 B. 随 df 增大,t 分布趋近正态分布 C. t 检验在小样本中不能用 D. n > 30 时 t 临界值等于正态临界值
Q7. 同一组数据,哪些因素会让 t 统计量增大?
I. 增大 n II. 增大 s III. 增大 |x̄ − μ₀| IV. 增大 α
A. I 和 III B. II 和 IV C. I, II, III D. 全部
Q8. σ 已知 vs σ 未知,同一组数据下哪个更容易拒绝 H₀?
A. σ 已知(Z 检验) B. σ 未知(T 检验) C. 取决于 n D. 完全一样
📊 答案与解析
| 题号 | 答案 | 解析 |
|---|---|---|
| Q1 | A | σ 已知 → Z 检验。n=25 < 30 但不影响选择(选 Z/T 看 σ 不是看 n)。 |
| Q2 | B | σ 未知 → T 检验。df = n − 1 = 49。❌ 常见错误 D:不能因为 n > 30 就用 Z。 |
| Q3 | B | 「Alpha 为正」是研究主张 → Hₐ: μ > 0。对应 H₀: μ ≤ 0。这是右尾检验的标准 H₀ 写法。 |
| Q4 | B | SE = 3.8/√16 = 0.95。t = (21.3−20)/0.95 = 1.368 ≈ 1.37。 |
| Q5 | B | σ 未知 → T 检验。df = 25−1 = 24。t = (48−50)/(10/5) = −1.0。t = −1.0 > −1.711(t₀.₀₅,₂₄)→ 不拒绝 H₀。 |
| Q6 | B | A 错:t 更肥尾更扁。C 错:小样本正需要 t 检验。D 错:n > 30 不完全等同。B 正确。 |
| Q7 | A | 公式:t = (x̄ − μ₀)/(s/√n)。n↑ → SE↓ → t↑ ✅。s↑ → SE↑ → t↓ ❌。 |
| Q8 | A | σ 已知用 Z,临界值 1.96;σ 未知用 T,临界值 > 1.96(df 越小越大)。同数据下 Z 更容易拒绝 H₀。⚠️ 但现实中 σ 几乎未知,这更像「理想 vs 现实」的比较。 |
📌 本课核心记忆卡
| 概念 | 一句话 |
|---|---|
| Z vs T 选择 | σ 已知 → Z;σ 未知 → T(跟 n 无关!) |
| Z 统计量 | z = (x̄ − μ₀)/(σ/√n),分布 N(0,1) |
| T 统计量 | t = (x̄ − μ₀)/(s/√n),分布 t(df),df = n−1 |
| t 分布特征 | 对称、肥尾,df↑ → 趋近正态 |
| 临界值 | 同 α 下 t 临界值 > Z 临界值(T 更保守) |
| n 的三重作用 | SE↓ / 临界值↓ / Power↑ |
| CFA 陷阱 | n > 30 不是用 Z 的理由!关键看 σ 是否已知 |
| 标准误 | SE = σ/√n 或 s/√n,是检验统计量的分母 |
🔑 σ 已知 = Z;σ 未知 = T。这个判定规则,CFA 一级计算题第一步必问。
🧠 扩展思考:实际金融中为什么绝大多数用 T?
金融场景的 σ(总体标准差):
- 股票收益率总体标准差?不知道,只能推
- 基金行业平均 alpha 波动?无官方数据
- 房贷利率离散度?历史在变,总体值不存在
→ 你永远在用 s 估计 σ
→ 所以 T 检验是 CFA 一级计算题的绝对主力
→ Z 检验更多出现在概念题中:"如果 σ = 10..."
💡 理解 Z vs T 的差异,本质上理解「已知 vs 估计」带来的额外不确定性。
下节课预告:L136 — 均值差异检验(成对与独立样本),学习比较两个总体的均值。
Quantitative Methods — Hypothesis Testing · Lesson 5
I. Recap: The Testing Framework Is Built
| Lesson | Topic | Key Takeaway |
|---|---|---|
| L131 | Hypothesis Testing Framework | H₀/Hₐ setup, test statistic, decision rule |
| L132 | p-Value & Significance Level | p < α → Reject H₀; p measures "extremeness" |
| L133 | Type I & Type II Errors | α (false positive) vs β (false negative), Power = 1−β |
| L134 | One-Tailed vs Two-Tailed | Hₐ direction determines tails; two-tailed p = one-tailed p × 2 |
Today's core question: How exactly do we test whether a mean equals (is greater/less than) some value?
II. Two Tools for Testing Means
2.1 The Intuition
To test whether "the average annual return of Hong Kong funds is 5%":
- Draw a sample, compute the sample mean x̄
- Assess how far x̄ is from 5%
- Determine if the gap is "random noise" or "real difference"
This judgment requires a test statistic. For means, we have two choices: z and t.
2.2 When to Use Which?
Population σ known?
├─ Yes → Z-Test
└─ No → T-Test (use sample s instead of σ)
| Z-Test | T-Test | |
|---|---|---|
| σ known? | ✅ Yes | ❌ No, use s |
| Distribution | Standard Normal N(0,1) | t-distribution (df = n−1) |
| Critical value | Fixed Zα (1.645, 1.96…) | tα,n−1 varies by df |
| Usage | Textbook problems, known σ | Real-world financial analysis (99%) |
| CFA Level I | Concept-level questions | Calculation questions — primary |
📌 CFA Level I golden rule: σ known → Z; σ unknown → T. Most financial scenarios have unknown σ → T-test dominates.
III. Z-Test
3.1 When Applicable
- Population normally distributed, or large sample (n ≥ 30, CLT)
- Population standard deviation σ is known
- Independent random sampling
3.2 Test Statistic
$$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$$
Where: - x̄ = sample mean - μ₀ = hypothesized population mean under H₀ - σ = known population standard deviation - n = sample size - SE = σ/√n (standard error)
3.3 Decision Rules at a Glance
| Hₐ | Reject H₀ if |
|---|---|
| Hₐ: μ ≠ μ₀ (two-tailed) | |
| Hₐ: μ > μ₀ (right-tailed) | z > Zα |
| Hₐ: μ < μ₀ (left-tailed) | z < −Zα |
3.4 Example: Flour Weight Inspection
A flour factory packages at 1000g standard, σ = 15g (known from experience). QC randomly samples 50 bags: x̄ = 995g. α = 0.05. Test for underweight.
H₀: μ ≥ 1000 (meets standard)
Hₐ: μ < 1000 (underweight) → left-tailed test
n = 50, σ = 15, x̄ = 995
z = (995 − 1000) / (15/√50)
= −5 / 2.121
= −2.357
α = 0.05, left-tailed Zα = −1.645
−2.357 < −1.645 → Reject H₀ ✅
Conclusion: Significant evidence the machine fills below 1000g. Calibration recommended.
🎯 Z-test feature: if σ is known, the z statistic follows an exact normal distribution with fixed critical values.
IV. T-Test
4.1 Why Do We Need the T-Test?
In reality, σ is almost never known.
When we substitute s (sample standard deviation) for σ: - The numerator x̄ − μ₀ is still approximately normal - But the denominator gains an extra layer of uncertainty (s itself varies) - → Result: the test statistic follows a t-distribution, not a normal distribution
Z statistic: (x̄ − μ₀) / (σ/√n) → Normal distribution
T statistic: (x̄ − μ₀) / (s/√n) → t-distribution
↑
This s introduces extra uncertainty
4.2 Characteristics of the t-Distribution
t-Distribution vs Normal
Normal (higher peak, thinner tails)
╱ ╲ t(df=2) — fattest tails
╱ ╲ t(df=5) — moderate
╱ ╲ t(df=30) — approaching normal
────┴────────────────┴────
| Feature | Description |
|---|---|
| Shape | Symmetric, bell-shaped, centered at 0 |
| Fat tails | Thicker tails than normal (more uncertainty) |
| Degrees of freedom df | df = n − 1 controls tail thickness |
| n↑ → | t approaches normal (virtually identical at n > 120) |
| df = 1 → | Cauchy distribution (fattest) |
📌 At the same α, t critical value > Z critical value. The extra uncertainty from using s in place of σ requires a larger "safety margin."
4.3 Test Statistic
$$t_{n-1} = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$$
Where: - s = sample standard deviation = $\sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}$ - df = n − 1
4.4 Decision Rules
| Hₐ | Reject H₀ if |
|---|---|
| Hₐ: μ ≠ μ₀ (two-tailed) | |
| Hₐ: μ > μ₀ (right-tailed) | t > tα, n−1 |
| Hₐ: μ < μ₀ (left-tailed) | t < −tα, n−1 |
p < α → Reject H₀ (same as Z-test)
V. Case Study: Fund Alpha Test
5.1 Data
A fund has 36 months of monthly excess returns averaging 0.65%, with s = 2.80%. H₀: μ = 0 (no alpha), Hₐ: μ ≠ 0 (two-tailed), α = 0.05
n = 36, x̄ = 0.65%, s = 2.80%
SE = 2.80% / √36 = 0.4667%
t = (0.65% − 0) / 0.4667% = 1.393
df = 35
t₀.₀₂₅,₃₅ ≈ 2.030 (table or calculator)
|t| = 1.393 < 2.030 → Fail to reject H₀ ❌
Conclusion: Monthly excess of 0.65%, but volatility is too high — statistically insufficient to prove alpha exists.
5.2 Intuitive Explanation
0.65% monthly average — looks decent on the surface
But 2.80% std dev over 36 months — extreme volatility
→ This 0.65% could just be a lucky 36-month run
→ Need a longer track record or larger alpha to prove it
VI. Z vs T Quick Comparison
| Dimension | Z-Test | T-Test |
|---|---|---|
| Distribution | Standard Normal N(0,1) | t(df) distribution |
| σ known? | ✅ Yes | ❌ No (use s) |
| Critical value | Fixed (1.645, 1.96, etc.) | Varies with df |
| Tails | Standard | Fatter (more so when n is small) |
| Large sample n→∞ | → | → Converges to Z-test |
| CFA practical | Conceptual questions | Calculation questions — primary |
| p-value lookup | Z-table | t-table (by df) |
VII. Key Critical Value Comparison (α = 0.05)
| df | t₀.₀₂₅ (two-tailed) | Z₀.₀₂₅ | t₀.₀₅ (one-tailed) | Z₀.₀₅ |
|---|---|---|---|---|
| 1 | 12.706 | 1.96 | 6.314 | 1.645 |
| 5 | 2.571 | 1.96 | 2.015 | 1.645 |
| 10 | 2.228 | 1.96 | 1.812 | 1.645 |
| 20 | 2.086 | 1.96 | 1.725 | 1.645 |
| 30 | 2.042 | 1.96 | 1.697 | 1.645 |
| 60 | 2.000 | 1.96 | 1.671 | 1.645 |
| 120 | 1.980 | 1.96 | 1.658 | 1.645 |
| ∞ | 1.960 | 1.96 | 1.645 | 1.645 |
📌 Smaller n (smaller df) → larger t critical value → higher bar to reject H₀ → "sample size penalty"
VIII. The Triple Effect of Sample Size
| Direction | Effect |
|---|---|
| SE ↓ | n↑ → SE = s/√n ↓ → test statistic ↑ → easier to reject H₀ |
| Critical value ↓ | n↑ → df↑ → t critical value ↓ → easier to reject H₀ |
| Power ↑ | n↑ → distribution more concentrated → easier to detect true effects |
🎯 Increasing sample size is the "universal remedy" for boosting test power — simultaneously reduces SE, lowers critical values, and increases power.
IX. Classic CFA Traps
Trap 1: When to Use Z-Test?
❌ Sample size > 30 → automatically use Z ✅ Only use Z when σ is known! n > 30 but σ unknown → still use T
The Z vs T criterion is whether σ is known, NOT how large n is.
Trap 2: Using s with Z Critical Values
❌ If the question gives s instead of σ, but you apply Z critical values anyway ✅ s → must use T, look up critical values from the t-table (by n−1)
Trap 3: Wrong Degrees of Freedom
❌ df = n (always treating n as df) ✅ One-sample mean test: df = n − 1
Trap 4: Thinking n > 30 Means t = Normal
❌ n = 35 → mentally default to Z critical values ✅ At n = 35, t₀.₀₂₅,₃₄ ≈ 2.032 > 1.96 — still a difference! Only at n ≈ 120+ do they become virtually identical.
Trap 5: p-Value Determination
❌ Compute the t statistic, then look it up in the Z-table ✅ t statistics must be looked up in the t-table (by df), otherwise p-values are incorrect
X. Complete Hypothesis Testing Template
Standard five-step procedure for a mean hypothesis test:
Step 1: State the hypotheses
H₀: μ = μ₀ (or ≤, ≥)
Hₐ: μ ≠ μ₀ (or >, <)
Step 2: Select the test statistic
├─ σ known → z = (x̄ − μ₀) / (σ/√n)
└─ σ unknown → t = (x̄ − μ₀) / (s/√n), df = n−1
Step 3: Determine significance level and critical value
α = 0.05 → look up Z-table or t-table
Step 4: Compute and compare
Calculate statistic → compare with critical value / find p-value
Step 5: Make decision and interpret
"At α = 0.05, [reject / fail to reject] H₀.
There [is / is not] sufficient evidence that..."
📝 Practice Questions
Part A: Choose the Test Method
Q1. Given σ = 8, n = 25, test μ = 100. Use:
A. Z-test B. T-test C. Chi-square test D. F-test
Q2. σ unknown, n = 50, x̄ = 3.2, s = 1.5. Test μ = 3. Use:
A. Z-test, use normal table B. T-test, df = 49 C. T-test, df = 50 D. Z-test, because n > 30
Part B: Hypothesis Setup & Decision
Q3. An analyst wants to test whether fund alpha is positive. H₀ should be:
A. H₀: μ = 0 B. H₀: μ ≤ 0 C. H₀: μ ≥ 0 D. H₀: μ ≠ 0
Q4. n = 16, x̄ = 21.3, s = 3.8, H₀: μ = 20, Hₐ: μ ≠ 20, α = 0.05. The t-value is approximately:
A. 0.34 B. 1.37 C. 2.06 D. 2.74
Part C: Comprehensive Judgment
Q5. n = 25, x̄ = 48, s = 10, H₀: μ = 50, Hₐ: μ < 50, α = 0.05. Which is correct?
A. Z-test, z = −1.0 B. T-test, df = 24 C. T-test, df = 25 D. Z-test, because σ is unknown
Q6. Regarding the t-distribution, which is correct?
A. The t-distribution always has a higher peak than the normal distribution B. As df increases, the t-distribution approaches the normal distribution C. The t-test cannot be used with small samples D. When n > 30, the t critical value equals the normal critical value
Q7. For the same dataset, which factors will increase the t statistic?
I. Increase n II. Increase s III. Increase |x̄ − μ₀| IV. Increase α
A. I and III B. II and IV C. I, II, III D. All of the above
Q8. σ known vs σ unknown — with the same data, which makes it easier to reject H₀?
A. σ known (Z-test) B. σ unknown (T-test) C. Depends on n D. Exactly the same
📊 Answers & Explanations
| # | Ans | Explanation |
|---|---|---|
| Q1 | A | σ known → Z-test. n = 25 < 30 does not affect the choice (Z vs T depends on σ, not n). |
| Q2 | B | σ unknown → T-test. df = n − 1 = 49. ❌ Common error D: n > 30 does not justify using Z. |
| Q3 | B | "Alpha is positive" is the research claim → Hₐ: μ > 0. The corresponding H₀: μ ≤ 0. This is the standard H₀ form for a right-tailed test. |
| Q4 | B | SE = 3.8/√16 = 0.95. t = (21.3−20)/0.95 = 1.368 ≈ 1.37. |
| Q5 | B | σ unknown → T-test. df = 25−1 = 24. t = (48−50)/(10/5) = −1.0. t = −1.0 > −1.711 (t₀.₀₅,₂₄) → fail to reject H₀. |
| Q6 | B | A is wrong: t has fatter tails and is flatter. C is wrong: small samples are exactly when t-tests are needed. D is wrong: n > 30 does not mean full equivalence. B is correct. |
| Q7 | A | Formula: t = (x̄ − μ₀)/(s/√n). n↑ → SE↓ → t↑ ✅. s↑ → SE↑ → t↓ ❌. |
| Q8 | A | σ known → Z, critical value 1.96; σ unknown → T, critical value > 1.96 (larger when df is smaller). With the same data, Z makes it easier to reject H₀. ⚠️ But in reality σ is almost never known — this is more of an "ideal vs reality" comparison. |
📌 Key Memory Card
| Concept | One-Liner |
|---|---|
| Z vs T selection | σ known → Z; σ unknown → T (nothing to do with n!) |
| Z statistic | z = (x̄ − μ₀)/(σ/√n), distributed N(0,1) |
| T statistic | t = (x̄ − μ₀)/(s/√n), distributed t(df), df =n−1 |
| T-distribution features | Symmetric, fat-tailed, df↑ → approaches normal |
| Critical value | At same α, t critical value > Z critical value (T is more conservative) |
| Triple role of n | SE↓ / Critical value↓ / Power↑ |
| CFA trap | n > 30 is NOT a reason to use Z! The key is whether σ is known |
| Standard error | SE = σ/√n or s/√n — the denominator of the test statistic |
🔑 σ known = Z; σ unknown = T. This decision rule is the CFA Level I calculation question's mandatory first step.
🧠 Extended Thinking: Why T-Tests Dominate in Real Finance?
σ (population standard deviation) in finance:
- True std dev of stock returns? Unknown — we can only infer it
- Industry-average fund alpha volatility? No official data
- Mortgage rate dispersion? History changes, the population value doesn't exist
→ You are always using s to estimate σ
→ So the T-test is the absolute workhorse of CFA Level I calculation questions
→ The Z-test primarily appears in conceptual questions: "If σ = 10..."
💡 Understanding the Z vs T difference is fundamentally about understanding the extra uncertainty that comes from "known vs estimated."
Next up: L136 — Tests for Mean Differences (Paired and Independent Samples), learning to compare means of two populations.