定量方法（Quantitative Methods）— 假设检验 · 第 6 课（收官）
一、六课知识全景图
| 课次 | 主题 | 核心能力 |
|---|---|---|
| L131 | 假设检验框架 | H0/Ha 设定、检验统计量、决策规则 |
| L132 | p 值与显著性水平 | p < alpha → 拒绝 H0；p 的本质 |
| L133 | Type I & Type II 错误 | alpha(冤枉) vs beta(遗漏)，Power = 1-beta |
| L134 | 单尾 vs 双尾 | Ha 方向决定尾巴，双尾 p = 单尾 p x 2 |
| L135 | z-test 与 t-test | sigma 已知→Z，sigma 未知→T（核心区分） |
| L136 | 综合练习 + 周测 | 30 题全面检验，查漏补缺 |
二、核心公式速查表
2.1 检验统计量
| 场景 | 公式 | 自由度 |
|---|---|---|
| Z 检验（sigma 已知） | z = (x-bar - mu0) / (sigma/sqrt(n)) | — |
| T 检验（sigma 未知） | t = (x-bar - mu0) / (s/sqrt(n)) | df = n-1 |
2.2 关键临界值（alpha = 0.05）
| 双尾 | 左尾 | 右尾 | |
|---|---|---|---|
| Z | +/-1.96 | -1.645 | +1.645 |
| t(df=10) | +/-2.228 | -1.812 | +1.812 |
| t(df=20) | +/-2.086 | -1.725 | +1.725 |
| t(df=30) | +/-2.042 | -1.697 | +1.697 |
2.3 决策规则
| 方法 | 拒绝 H0 条件 |
|---|---|
| 临界值法 | |
| p 值法 | p < alpha |
| 置信区间法 | mu0 不在置信区间内（双尾 Ha: mu /= mu0） |
2.4 四类错误与 Power
| H0 为真 | H0 为假 | |
|---|---|---|
| 不拒绝 H0 | OK 正确 | X Type II (beta) |
| 拒绝 H0 | X Type I (alpha) | OK Power (1-beta) |
三、综合练习（30 题）
Part 1：概念基础（Q1–Q10）
Q1. H0 代表： A. 研究者想要证明的主张 B. 默认的/现状的陈述，包含等号 C. 通过样本数据直接计算的结论 D. 总是包含"不等于"的关系式
Q2. 关于 p 值，正确的是： A. p 值 = H0 为真的概率 B. p 值越大，越应拒绝 H0 C. p 值 = H0 为真时，观察到当前或更极端结果的概率 D. p 值恒等于 alpha
Q3. alpha = 0.05 的含义： A. 95% 把握 H0 为真 B. 5% 概率 H0 为真 C. H0 为真时，5% 概率错误拒绝 H0 D. H0 为假时，5% 概率错误接受 H0
Q4. Type I 和 Type II 错误： A. Type I=H0假但接受；Type II=H0真但拒绝 B. Type I=H0真但拒绝；Type II=H0假但接受 C. 同一检验中两者可同时降低 D. alpha 增大则 beta 一定增大
Q5. Power 是： A. 1 - alpha B. 1 - beta C. 正确接受 H0 的概率 D. Type I 错误概率
Q6. 单尾 vs 双尾： A. 双尾 p = 单尾 p x 0.5 B. Ha: mu /= mu0 对应单尾 C. 同一数据双尾比单尾更易拒绝 H0 D. 问题为「是否更好」→ 用单尾
Q7. 合理的 H0 是： A. H0: x-bar = 100 B. H0: mu > 100 C. H0: mu <= 100 D. H0: mu /= 100
Q8. z=1.80, alpha=0.05, 双尾： A. 拒绝 H0，1.80 > 1.645 B. 不拒绝 H0，|1.80| < 1.96 C. 拒绝 H0，p < 0.05 D. 需知道 sigma
Q9. n 增大的影响，错误的是： A. Type I 概率 alpha 自动减小 B. 标准误减小 C. Power 增大 D. t 临界值趋近 Z
Q10. p=0.03, alpha=0.05： A. 拒绝 H0，可能犯 Type I B. 拒绝 H0，可能犯 Type II C. 不拒绝，p 接近 0.05 D. 信息不足
Part 2：Z 检验与 T 检验（Q11–Q18）
Q11. sigma=5, n=36, x-bar=28, H0: mu=30, Ha: mu /= 30。z 值： A. -0.40 B. -1.20 C. -2.40 D. -4.80
Q12. 接 Q11, alpha=0.05 双尾，结论： A. 拒绝 H0，|z|>1.96 B. 不拒绝，|z|<1.96 C. 拒绝，p<0.05 D. 需查 t 表
Q13. 必须用 T 检验的是： A. n=15, sigma 已知 B. n=100, sigma 已知 C. n=25, sigma 未知，给了 s D. n=200, sigma 已知
Q14. 组合 16 月月均收益 1.2%, s=3.6%。检验 mu=0（双尾）。t 值： A. 0.33 B. 0.53 C. 1.33 D. 2.33
Q15. 接 Q14，df=15, t(0.025,15)=2.131，结论： A. 拒绝 H0，有显著正收益 B. 拒绝 H0，收益为负 C. 不拒绝，无充分证据表明收益 /= 0 D. 接受 H0，证明收益恰好为 0
Q16. t 分布的正确描述： A. 方差恒为 1 B. df 减小 → 尾部变薄 C. 比正态分布有更肥的尾部 D. df=30 时等价于正态分布
Q17. n=20, x-bar=45, s=8, H0: mu=50, Ha: mu<50。方法： A. Z 检验 B. T 检验, df=19, 左尾 C. T 检验, df=20, 左尾 D. Z 检验查正态表
Q18. 判断：「n>30 时无论 sigma 是否已知都用 Z 检验」： A. 对，CLT 保证 B. 错，sigma 未知始终用 T C. 对，大样本 s≈sigma D. 错，n>50 才可以用 Z
Part 3：综合推理（Q19–Q24）
Q19. alpha 从 0.05 降到 0.01 的影响： I. Type I 概率降 II. Type II 概率升 III. Power 降 IV. 拒绝域扩大 A. I,II B. II,IV C. I,II,III D. 全部
Q20. 双尾检验 p=0.08, alpha=0.05： A. 8% 概率 H0 为真 B. alpha=0.10 下会拒绝 H0 C. H0 一定为真 D. Type II 错误不严重
Q21. p > alpha 时： A. 证明 H0 为真 B. 接受 H0 C. 不拒绝 H0 D. 数据有误
Q22. Type II 错误最可能大幅降低的是： A. alpha 从 0.05 降到 0.01 B. n 增加 4 倍 C. 单尾替代双尾 D. sigma 增大
Q23. p=0.001 最恰当的解释： A. H0 极大概率为真 B. 如果 H0 为真，极端结果概率仅 0.1% C. alpha 应设为 0.001 D. Type II 概率是 0.001
Q24. 置信区间与假设检验（双尾, alpha=0.05）： A. mu0 在 95%CI 内 → 拒绝 B. mu0 在 95%CI 外 → 不拒绝 C. mu0 在 95%CI 外 → 拒绝 D. 两者互不相关
Part 4：实战应用题（Q25–Q30）
Q25. 验证新策略产生正 Alpha。正确假设： A. H0: mu<=0, Ha: mu>0（右尾） B. H0: mu>=0, Ha: mu<0（左尾） C. H0: mu=0, Ha: mu/=0（双尾） D. H0: mu=0, Ha: mu<0（左尾）
Q26. 标称 500ml, sigma=3, n=25, x-bar=498, Ha: mu<500。结论： A. z=-3.33, 拒绝 B. z=-1.67, 拒绝 C. t=-3.33, 拒绝 D. z=-0.67, 不拒绝
Q27. 基金 49 周周均 0.30%, s=1.05%。检验 mu=0 vs mu>0。t 值： A. 0.30 B. 1.00 C. 2.00 D. 3.00
Q28. 接 Q27, df=48, t(0.05,48)≈1.677： A. 拒绝 H0，显著正收益 B. 不拒绝 H0 C. 接受 H0，证明无收益 D. t=2.00>1.677，拒绝 H0
Q29. 研究表明有效（alpha=0.05），实际无效。这是： A. Type I B. Type II C. 正确 D. Power 不足
Q30. p 值说法错误的是： A. p=0.03 意味着 H0 真时仅 3% 概率看到如此极端结果 B. p<0.05 自动意味结果有实际重要性 C. p 越小反对 H0 证据越强 D. p 值依赖样本量
四、答案与详解
Part 1：概念基础
| 题号 | 答案 | 详解 |
|---|---|---|
| Q1 | B | H0 是默认状态/现状，必须包含等号（=, <=, >=）。A 是 Ha（备择假设/研究主张）。 |
| Q2 | C | p 值 =「H0 为真时观察到当前或更极端结果的概率」。A 是经典误解（p 值 != H0 为真概率）。 |
| Q3 | C | alpha = P(Type I) = P(拒绝 H0 |
| Q4 | B | Type I = H0 真但拒绝（冤枉好人）；Type II = H0 假但接受（漏网之鱼）。alpha 降则 beta 升，反向关系。 |
| Q5 | B | Power = 1 - beta = P(H0 假时正确拒绝 H0)。不是 1-alpha。 |
| Q6 | D | A 反了（双尾 p=单尾 p×2）；B 错（mu /= mu0 是双尾）；C 错（双尾更难拒绝，临界值更大）。 |
| Q7 | C | H0 必须包含等号。A：x-bar 是样本统计量不能放 H0；B：>不含等号；D：/= 不含等号。 |
| Q8 | B | 双尾 alpha=0.05→临界值 ±1.96。 |
| Q9 | A | n 增大不影响 alpha（alpha 是预设的！）。SE 减小 OK，Power 增大 OK，t→Z OK。 |
| Q10 | A | p=0.03<0.05→拒绝 H0。若 H0 实际为真，则犯了 Type I 错误（拒绝了真的 H0）。 |
Part 2：Z 检验与 T 检验
| 题号 | 答案 | 详解 |
|---|---|---|
| Q11 | C | SE=5/6=0.8333。z=(28-30)/0.8333=-2.40。 |
| Q12 | A | |
| Q13 | C | sigma 未知→必须用 T 检验。A/B/D：sigma 已知用 Z。n=25, df=24。 |
| Q14 | C | SE=3.6%/4=0.9%。t=1.2%/0.9%=1.33。 |
| Q15 | C | |
| Q16 | C | t 分布比正态有更肥的尾部（fat tails）。A 错：方差=df/(df-2)；B 错：df 小则尾部肥；D 错：df→∞时才趋近正态。 |
| Q17 | B | sigma 未知→T 检验。df=n-1=19。Ha: mu<50 是左尾。 |
| Q18 | B | CFA 标准：sigma 未知时始终用 T 检验（无论 n 多大）。n>30 不是转 Z 的条件。 |
Part 3：综合推理
| 题号 | 答案 | 详解 |
|---|---|---|
| Q19 | C | alpha↓→Type I↓（I对），beta↑（II对），Power=1-beta↓（III对），拒绝域缩小（IV错）。 |
| Q20 | B | p=0.08<0.10→alpha=0.10 下拒绝。A 是经典谬误。C 错：不能说一定为真。 |
| Q21 | C | 假设检验结论只能是「拒绝 H0」或「不拒绝 H0」，不能说「接受 H0」或「证明 H0」。 |
| Q22 | B | n 增大→标准误↓→检验更灵敏→beta 大幅降低。A 反而增 beta。C 有影响但不如 B 大。 |
| Q23 | B | p=0.001 表示「若 H0 为真，如此极端结果概率仅 0.1%」，证据很强地反对 H0。 |
| Q24 | C | mu0 在 95%CI 外↔在 alpha=0.05 下拒绝 H0（双尾等价关系）。 |
Part 4：实战应用
| 题号 | 答案 | 详解 |
|---|---|---|
| Q25 | A | 「正 Alpha」→ 想证明 mu>0 → Ha: mu>0 → H0: mu<=0，右尾检验。 |
| Q26 | A | SE=3/5=0.6。z=(498-500)/0.6=-2/0.6=-3.33。左尾临界值-1.645，-3.33<-1.645→拒绝。 |
| Q27 | C | SE=1.05%/7=0.15%。t=0.30%/0.15%=2.00。 |
| Q28 | D | t=2.00>t(0.05,48)≈1.677→拒绝 H0，有显著正收益。 |
| Q29 | A | 结论「有效」但实际「无效」→拒绝了真的 H0（H0: 无效）→ Type I 错误。 |
| Q30 | B | p<0.05 只说明统计显著性，不等于实际重要性（经济显著性/效应大小）。A/C/D 均正确。 |
五、本周知识点总结
假设检验六步法
- 设定 H0 和 Ha（H0 必含等号）
- 确定检验统计量（sigma 已知→Z，未知→T）
- 选择显著性水平 alpha
- 确定决策规则（临界值法 / p 值法）
- 收集数据计算统计量
- 做出决策（拒绝 H0 或不拒绝 H0）
CFA 一级核心考点
- p 值本质：H0 为真时极端结果的概率，不是 H0 为真的概率
- Z vs T：sigma 已知→Z，sigma 未知→T（不分 n 大小）
- 错误类型：Type I = 冤枉好人(alpha)，Type II = 漏网之鱼(beta)
- Power：真的拒绝假的 = 1-beta
- 结论措辞：只说「拒绝」或「不拒绝」，不说「接受」或「证明」
- 单/双尾：双尾更难拒绝（临界值更大），双尾 p = 单尾 p x 2
恭喜完成假设检验章节！🎉 下周进入下一主题。
Quantitative Methods — Hypothesis Testing · Lesson 6 (Finale)
I. Six-Lesson Knowledge Map
| Lesson | Topic | Core Competency |
|---|---|---|
| L131 | Hypothesis Testing Framework | H0/Ha setup, test statistic, decision rule |
| L132 | p-value & Significance Level | p < alpha → Reject H0; essence of p-value |
| L133 | Type I & Type II Errors | alpha (false positive) vs beta (false negative), Power = 1-beta |
| L134 | One-tailed vs Two-tailed | Ha direction determines tail; two-tailed p = one-tailed p x 2 |
| L135 | z-test & t-test | sigma known → Z; sigma unknown → T (key distinction) |
| L136 | Comprehensive Review + Weekly Quiz | 30 questions to test full understanding |
II. Key Formula Reference Sheet
2.1 Test Statistics
| Scenario | Formula | df |
|---|---|---|
| Z-test (sigma known) | z = (x-bar - mu0) / (sigma/sqrt(n)) | — |
| T-test (sigma unknown) | t = (x-bar - mu0) / (s/sqrt(n)) | df = n-1 |
2.2 Key Critical Values (alpha = 0.05)
| Two-tailed | Left-tailed | Right-tailed | |
|---|---|---|---|
| Z | +/-1.96 | -1.645 | +1.645 |
| t(df=10) | +/-2.228 | -1.812 | +1.812 |
| t(df=20) | +/-2.086 | -1.725 | +1.725 |
| t(df=30) | +/-2.042 | -1.697 | +1.697 |
2.3 Decision Rules
| Method | Reject H0 When |
|---|---|
| Critical Value | |
| p-value | p < alpha |
| Confidence Interval | mu0 not in CI (two-tailed Ha: mu /= mu0) |
2.4 Errors & Power
| H0 True | H0 False | |
|---|---|---|
| Do Not Reject H0 | OK Correct | X Type II (beta) |
| Reject H0 | X Type I (alpha) | OK Power (1-beta) |
III. Comprehensive Practice (30 Questions)
Part 1: Conceptual Foundations (Q1-Q10)
Q1. In hypothesis testing, H0 represents: A. The claim the researcher wants to prove B. The default/status quo statement, containing equality C. A conclusion directly computed from sample data D. A relationship always containing "not equal to"
Q2. Regarding the p-value, which is correct? A. p-value = probability that H0 is true B. The larger the p-value, the more we should reject H0 C. p-value = probability of observing current or more extreme results, given H0 is true D. p-value is always equal to alpha
Q3. alpha = 0.05 means: A. 95% confidence that H0 is true B. 5% probability that H0 is true C. When H0 is true, 5% probability of incorrectly rejecting H0 D. When H0 is false, 5% probability of incorrectly accepting H0
Q4. Type I and Type II errors: A. Type I = H0 false but accepted; Type II = H0 true but rejected B. Type I = H0 true but rejected; Type II = H0 false but accepted C. Both can be reduced simultaneously in the same test D. Increasing alpha always increases beta
Q5. Power is defined as: A. 1 - alpha B. 1 - beta C. Probability of correctly accepting H0 D. Probability of Type I error
Q6. One-tailed vs two-tailed tests: A. Two-tailed p-value = one-tailed p-value x 0.5 B. Ha: mu /= mu0 corresponds to a one-tailed test C. Two-tailed tests are easier to reject H0 than one-tailed D. If the research question is "is it better?", use a one-tailed test
Q7. Which is a valid H0 specification? A. H0: x-bar = 100 B. H0: mu > 100 C. H0: mu <= 100 D. H0: mu /= 100
Q8. z = 1.80, alpha = 0.05, two-tailed. Conclusion: A. Reject H0, since 1.80 > 1.645 B. Do not reject H0, since |1.80| < 1.96 C. Reject H0, since p < 0.05 D. Need to know sigma to decide
Q9. Increasing sample size n — which effect is INCORRECT? A. Type I error probability alpha automatically decreases B. Standard error decreases C. Power increases D. t critical values converge to Z critical values
Q10. p = 0.03, alpha = 0.05. Decision: A. Reject H0, may commit Type I error B. Reject H0, may commit Type II error C. Do not reject, p is close to 0.05 D. Insufficient information
Part 2: Z-test & T-test (Q11-Q18)
Q11. sigma = 5, n = 36, x-bar = 28, H0: mu = 30, Ha: mu /= 30. z-value: A. -0.40 B. -1.20 C. -2.40 D. -4.80
Q12. From Q11, alpha = 0.05 two-tailed, conclusion: A. Reject H0, |z| > 1.96 B. Do not reject, |z| < 1.96 C. Reject, p < 0.05 D. Need t-table
Q13. Which MUST use a T-test instead of Z-test? A. n = 15, sigma known B. n = 100, sigma known C. n = 25, sigma unknown, s given D. n = 200, sigma known
Q14. Portfolio: 16 months, mean return 1.2%, s = 3.6%. Test mu = 0 (two-tailed). t-value: A. 0.33 B. 0.53 C. 1.33 D. 2.33
Q15. From Q14, df = 15, t(0.025,15) = 2.131, conclusion: A. Reject H0, significant positive return B. Reject H0, return is negative C. Do not reject, insufficient evidence return /= 0 D. Accept H0, proves return is exactly 0
Q16. Correct description of t-distribution: A. Variance is always 1 B. Lower df → thinner tails C. Fatter tails than normal distribution D. df = 30 makes it equivalent to normal distribution
Q17. n = 20, x-bar = 45, s = 8, H0: mu = 50, Ha: mu < 50. Method: A. Z-test B. T-test, df = 19, left-tailed C. T-test, df = 20, left-tailed D. Z-test with normal table
Q18. Judge: "When n > 30, always use Z-test regardless of sigma." A. True, CLT guarantees it B. False, always use T-test when sigma is unknown C. True, s approximates sigma in large samples D. False, only when n > 50
Part 3: Integrated Reasoning (Q19-Q24)
Q19. Reducing alpha from 0.05 to 0.01 results in: I. Type I error probability decreases II. Type II error probability increases III. Power decreases IV. Rejection region expands A. I, II B. II, IV C. I, II, III D. All
Q20. Two-tailed test, p = 0.08, alpha = 0.05. Correct statement: A. 8% probability H0 is true B. Would reject H0 at alpha = 0.10 C. H0 must be true D. Type II error is not serious
Q21. When p > alpha, the correct statement is: A. Proves H0 is true B. Accept H0 C. Do not reject H0 D. Data is erroneous
Q22. Which most reduces Type II error (beta)? A. Reduce alpha from 0.05 to 0.01 B. Increase sample size n by 4x C. Switch from two-tailed to one-tailed D. Increase sigma
Q23. p = 0.001. Best interpretation: A. H0 is very likely true B. If H0 is true, probability of such extreme results is only 0.1% C. Alpha should be set to 0.001 D. Type II error probability is 0.001
Q24. CI and hypothesis testing (two-tailed, alpha = 0.05): A. mu0 in 95% CI → Reject H0 B. mu0 outside 95% CI → Do not reject C. mu0 outside 95% CI → Reject H0 D. CI and hypothesis testing are unrelated
Part 4: Applied Problems (Q25-Q30)
Q25. Analyst wants to verify: new strategy produces positive Alpha (vs 0). Correct setup: A. H0: mu <= 0, Ha: mu > 0 (right-tailed) B. H0: mu >= 0, Ha: mu < 0 (left-tailed) C. H0: mu = 0, Ha: mu /= 0 (two-tailed) D. H0: mu = 0, Ha: mu < 0 (left-tailed)
Q26. QC: nominal 500ml, sigma = 3. n = 25, x-bar = 498, Ha: mu < 500. Conclusion: A. z = -3.33, Reject B. z = -1.67, Reject C. t = -3.33, Reject D. z = -0.67, Do not reject
Q27. Fund: 49 weeks, mean 0.30%, s = 1.05%. Test mu = 0 vs mu > 0. t-value: A. 0.30 B. 1.00 C. 2.00 D. 3.00
Q28. From Q27, df = 48, t(0.05,48) ≈ 1.677: A. Reject H0, significant positive return B. Do not reject H0 C. Accept H0, proves no return D. t = 2.00 > 1.677, Reject H0
Q29. Study concludes "drug works" (alpha = 0.05), it actually doesn't. This is: A. Type I error B. Type II error C. Correct D. Insufficient Power
Q30. Which statement about p-value is INCORRECT? A. p = 0.03 means only 3% chance of such extreme results if H0 is true B. p < 0.05 automatically means the result has practical importance C. Smaller p-value means stronger evidence against H0 D. p-value depends on sample size
IV. Answers & Explanations
Part 1: Conceptual Foundations
| # | Ans | Explanation |
|---|---|---|
| Q1 | B | H0 is the default/status quo, must contain equality (=, <=, >=). A describes Ha (alternative/research claim). |
| Q2 | C | p-value = P(observed or more extreme |
| Q3 | C | alpha = P(Type I) = P(Reject H0 |
| Q4 | B | Type I = H0 true but rejected (false positive); Type II = H0 false but accepted (false negative). C wrong: alpha down → beta up (inverse). |
| Q5 | B | Power = 1 - beta = P(correctly reject H0 when H0 is false). Not 1 - alpha. |
| Q6 | D | A reversed (two-tailed p = one-tailed p x 2); B: mu /= mu0 is two-tailed; C: two-tailed is harder to reject (larger critical value). |
| Q7 | C | H0 must contain equality. A: x-bar is a sample statistic; B: > has no equality; D: /= has no equality. |
| Q8 | B | Two-tailed alpha = 0.05 → critical value +/-1.96. |
| Q9 | A | n does NOT affect alpha (alpha is preset!). SE down OK, Power up OK, t→Z convergence OK. |
| Q10 | A | p = 0.03 < 0.05 → Reject H0. If H0 is actually true, we committed Type I error. |
Part 2: Z-test & T-test
| # | Ans | Explanation |
|---|---|---|
| Q11 | C | SE = 5/6 = 0.8333. z = (28-30)/0.8333 = -2.40. |
| Q12 | A | |
| Q13 | C | sigma unknown → must use T-test. A/B/D: sigma known, use Z. n = 25, df = 24. |
| Q14 | C | SE = 3.6%/4 = 0.9%. t = 1.2%/0.9% = 1.33. |
| Q15 | C | |
| Q16 | C | t-distribution has fatter tails than normal. A: variance = df/(df-2); B: lower df → fatter tails; D: only as df→∞. |
| Q17 | B | sigma unknown → T-test. df = n-1 = 19. Ha: mu < 50 is left-tailed. |
| Q18 | B | CFA standard: when sigma is unknown, always use T-test regardless of n. n > 30 is NOT the switch criterion. |
Part 3: Integrated Reasoning
| # | Ans | Explanation |
|---|---|---|
| Q19 | C | alpha down → Type I down (I OK), beta up (II OK), Power = 1-beta down (III OK), rejection region shrinks (IV wrong). |
| Q20 | B | p = 0.08 < 0.10 → would reject at alpha = 0.10. A is classic fallacy. C wrong: cannot say "must be true." |
| Q21 | C | Hypothesis test conclusions are only: "Reject H0" or "Do not reject H0." Never "accept H0" or "prove H0." |
| Q22 | B | Increasing n → SE down → test more sensitive → beta greatly reduced. A increases beta. C has some effect but B is largest. |
| Q23 | B | p = 0.001 = "If H0 is true, only 0.1% chance of results this extreme" — very strong evidence against H0. |
| Q24 | C | mu0 outside 95% CI ↔ Reject H0 at alpha = 0.05 (two-tailed equivalence). |
Part 4: Applied Problems
| # | Ans | Explanation |
|---|---|---|
| Q25 | A | "Positive Alpha" → want to prove mu > 0 → Ha: mu > 0 → H0: mu <= 0, right-tailed. |
| Q26 | A | SE = 3/5 = 0.6. z = (498-500)/0.6 = -2/0.6 = -3.33. Left-tail critical = -1.645. -3.33 < -1.645 → Reject. |
| Q27 | C | SE = 1.05%/7 = 0.15%. t = 0.30%/0.15% = 2.00. |
| Q28 | D | t = 2.00 > t(0.05,48) ≈ 1.677 → Reject H0, significant positive return. |
| Q29 | A | Conclude "works" but actually doesn't → rejected a true H0 (H0: no effect) → Type I error. |
| Q30 | B | p < 0.05 only means statistical significance, NOT practical/economic importance. A, C, D are all correct. |
V. Weekly Summary: Key CFA Level I Takeaways
Hypothesis Testing 6-Step Framework
- State H0 and Ha (H0 must contain equality)
- Identify test statistic (sigma known → Z, unknown → T)
- Choose significance level alpha
- Determine decision rule (critical value / p-value)
- Collect data and compute statistic
- Make decision (Reject H0 or Do Not Reject H0)
CFA Level I Core Exam Points
- p-value essence: Probability of extreme results given H0 true; NOT the probability H0 is true
- Z vs T: sigma known → Z; sigma unknown → T (regardless of n size)
- Error types: Type I = false positive (alpha); Type II = false negative (beta)
- Power: Correctly rejecting a false H0 = 1 - beta
- Conclusion wording: Only "Reject" or "Do Not Reject"; never "Accept" or "Prove"
- One vs Two-tailed: Two-tailed harder to reject (larger critical value); two-tailed p = one-tailed p x 2
Congratulations on completing the Hypothesis Testing module! 🎉 Next topic coming next week.