R7 练习: 估计与推断
考纲范围
- Compare and contrast simple random, stratified random, cluster, convenience, and judgmental sampling and their implications for sampling error in an investment problem.
- Explain the central limit theorem and its importance for the distribution and standard error of the sample mean.
- Describe the use of resampling (bootstrap, jackknife) to estimate the sampling distribution of a statistic.
Q1.
The stratified random sampling means to:
A. divide the population into subpopulations and draw data evenly from them.
B. divide the population into subpopulations and draw data in sizes proportional to each subpopulation.
C. select samples randomly from the population.
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答案：B
解析：分层随机抽样（stratified random sampling）的定义。
选项 判断 解析 A ✗ 均匀抽取不是分层随机抽样的特征，分层抽样按比例抽取 B ✓ 分层随机抽样将总体分为子群体，然后按各子群体的比例从中抽样 C ✗ 这是简单随机抽样的描述
Q2.
An analyst is studying the net income (NI) of technology stocks globally. She uses all the data in the internal database that she has established when researching dozens of representative stocks in the technology sector. This sampling method is an example of:
A. stratified random sampling.
B. judgmental sampling.
C. convenience sampling.
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答案：C
解析：分析师使用自己已有的内部数据库中的数据，而非通过随机方式从总体中抽取。
选项 判断 解析 A ✗ 分层随机抽样需要将总体分层后随机抽取 B ✗ 判断抽样是基于研究者的专业判断选择样本 C ✓ 便利抽样是使用手边方便获取的数据，如已有的内部数据库
Q3.
Jiang, a risk control manager, is checking last year’s transactions of the overseas derivative trading department through sampling. Based on his expertise and profession, Jiang checks mostly the over-the-counter (OTC) derivative instruments. Which of the following statements best describes Jiang’s sampling method? His method:
A. may result in a non-representative sample.
B. can cover each transaction with equal probability.
C. is time-efficient and cost-effective.
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答案：A
解析：Jiang基于专业判断主要检查OTC衍生品，这是判断抽样（judgmental sampling）。
选项 判断 解析 A ✓ 判断抽样可能导致样本不具代表性，因为它依赖于研究者的主观判断，可能遗漏某些重要交易类型 B ✗ 只有简单随机抽样才能以相等概率覆盖每个交易 C ✗ 虽然判断抽样可能高效，但这不是对其方法最恰当的描述
Q4.
An analyst calculates the bond returns of 125 firms in the US market. If the sample mean is 7.2% while the mean of the entire US bond returns is 7.9%, the difference between 7.9% and 7.2% is best described as:
A. the standard error of sampling distribution.
B. the data snooping bias.
C. the sampling error.
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答案：C
解析：样本统计量与总体参数之间的差异。
选项 判断 解析 A ✗ 标准误差是样本均值分布的标准差，不是样本均值与总体均值的差异 B ✗ 数据窥探偏差是反复测试直到找到显著结果的偏差 C ✓ 抽样误差（sampling error）= 样本统计量 - 总体参数 = 7.2% - 7.9% = -0.7%（或其绝对值0.7%）
Q5.
Which of the following statements about the sampling methods is incorrect?
A. Compared with simple random sampling, stratified random sampling ensures that the sample includes the subpopulations of interest.
B. In systematic sampling, the sample can be drawn by selecting every kth observation until reaching the desired size.
C. In contrast to other probability sampling methods, keeping the same sample size, cluster sampling usually yields higher accuracy.
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答案：C
解析：各种抽样方法的特点比较。
选项 判断 解析 A ✓ 正确，分层随机抽样确保样本包含所有感兴趣的子群体 B ✓ 正确，系统抽样每隔k个观察值抽取一个 C ✗ 不正确。相同样本量下，聚类抽样通常精度较低（不是更高），因为同一聚类内的个体往往相似
Q6.
Which of the following statements about the central limit theorem is least accurate? To conclude the sample mean is normally distributed:
A. the population should follow a normal distribution.
B. the population must have a finite variance.
C. the sample size has to be large enough, generally larger than 30.
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答案：A
解析：中心极限定理的条件。
选项 判断 解析 A ✗ 不准确。中心极限定理的要点恰恰是：即使总体不服从正态分布，只要样本量足够大，样本均值也近似服从正态分布 B ✓ 正确，总体必须有有限方差 C ✓ 正确，样本量需要足够大，一般大于30
Q7.
The sample mean follows a normal distribution with a standard error of 1.5, and the population variance is 144. The sample size is closest to:
A. 8.
B. 64.
C. 96.
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答案：B
解析：由标准误差和总体方差反推样本量。
计算过程： $1.5 = \frac{12}{\sqrt{n}}\sqrt{n} = \frac{12}{1.5} = 8n = 64$$
选项 判断 解析 A ✗ 8是sqrt(n)的值，不是n B ✓ n = (12/1.5)² = 8² = 64 C ✗ 计算错误
Q8.
With regard to the central limit theorem, which of the following statements is most likely correct?
A. If the population distribution is lognormal, the sample mean will follow a normal distribution with any sample size.
B. The standard error of the sample mean may be greater than the standard deviation of the population.
C. Even if the population distribution is unknown, the central limit theorem can be used to construct the confidence interval for the population mean when the sample size is large.
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答案：C
解析：中心极限定理的应用。
选项 判断 解析 A ✗ 只有总体本身为正态分布时，任何样本量的样本均值都服从正态分布。对数正态分布需要大样本才能使样本均值近似正态 B ✗ 标准误差 = σ/√n，当n≥1时，SE ≤ σ，因此标准误差不会大于总体标准差 C ✓ 正确，中心极限定理允许在总体分布未知时，利用大样本构建总体均值的置信区间
Q9.
Which of the following statements regarding the distinction between jackknife resampling and bootstrap resampling is most likely accurate?
A. Jackknife resampling is done with replacement, while bootstrap resampling is not.
B. Jackknife resampling usually requires that the number of repetitions equals the sample size, while bootstrap resampling does not.
C. Jackknife resampling is particularly useful when the analytical formula such as z-statistic or t-statistic is not available, while bootstrap resampling is not.
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答案：B
解析：刀切法（jackknife）和自助法（bootstrap）的区别。
选项 判断 解析 A ✗ 恰好相反：bootstrap是有放回抽样，jackknife是无放回的（每次去掉一个观察值） B ✓ Jackknife通常需要n次重复（样本量大小），每次去掉一个观察值；bootstrap的重复次数不受限制 C ✗ Bootstrap和jackknife都适用于没有解析公式的情况
Q10.
Compared with jackknife resampling, bootstrap resampling:
A. uses an analytical formula such as a z-statistic or t-statistic.
B. requires that each resample is of the same size as the original sample.
C. usually requires n repetitions for a sample of size n.
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答案：B
解析：Bootstrap重抽样的特征。
选项 判断 解析 A ✗ Bootstrap不使用解析公式，而是通过模拟获得结果 B ✓ Bootstrap的每次重抽样大小与原始样本相同（有放回） C ✗ 这是jackknife的特征（n次重复），bootstrap的重复次数通常远大于n
Q11.
Resampling (bootstrap, jackknife) is widely used to estimate the sampling distribution of a statistic. Which of the following is (are) true about resampling?
I. Bootstrap method mimics the random sampling process by regarding the randomly selected sample as if it were the population. II. Jackknife resampling usually requires n repetitions for a sample of size n. III. Bootstrap usually produces similar results, whereas Jackknife gives different results for every run.
A. I only.
B. I and II only.
C. I, II, and III.
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答案：B
解析：重抽样方法的特征。
选项 判断 解析 I ✓ Bootstrap将随机样本视为总体，从中有放回地重抽样 II ✓ Jackknife每次去掉一个观察值，对n个样本需要n次重复 III ✗ 恰好相反：Jackknife是确定性的（每次结果相同），Bootstrap由于随机抽样可能每次结果不同 因此I和II正确，答案为B。