相关笔记

Slide 1 — Statistical inference about the difference of population means（第1页——总体均值差的统计推断）

Knowledge Points (知识点)

1. Inference for the difference of two population means（两个总体均值差的统计推断）
2. Case 1: population standard deviations σ₁ and σ₂ known（情形一：已知总体标准差 σ₁、σ₂）
3. Case 2: population standard deviations σ₁ and σ₂ unknown（情形二：未知总体标准差 σ₁、σ₂）

Explanation（解释）

• We study how to use two independent samples to make inferences about the difference in population means ${\mu }_1-{\mu }_2$.
• There are two main scenarios:
  • When both population standard deviations ${\sigma }_1$ and ${\sigma }_2$ are known, we use the z-distribution.
  • When at least one of ${\sigma }_1,{\sigma }_2$ is unknown, we estimate them using sample standard deviations and use t-based methods.
• The goal is to construct confidence intervals and perform hypothesis tests about ${\mu }_1-{\mu }_2$.

📖 点击查看中文解释

• 本章讨论如何利用两个独立样本来推断两个总体均值之差 ${\mu }_1-{\mu }_2$。
• 主要分为两种情形：
  • 当总体标准差 ${\sigma }_1$ 与 ${\sigma }_2$ 已知 时，采用 z 分布 进行推断；
  • 当 ${\sigma }_1$ 或 ${\sigma }_2$ 至少有一个 未知 时，用样本标准差估计，并使用 t 分布 方法。
• 我们的目标是对 ${\mu }_1-{\mu }_2$ 构造置信区间并进行假设检验。

Example（例子）

• Suppose we compare average spending of Group 1 (students who mainly shop online) and Group 2 (students who mainly shop offline).
• We want to know whether the population means differ, that is, whether ${\mu }_1-{\mu }_2\ne 0$.
• Depending on whether ${\sigma }_1,{\sigma }_2$ are known or unknown, we choose different formulas and distributions, but the target parameter is always ${\mu }_1-{\mu }_2$.

📖 点击查看中文解释

• 例如比较两组学生的平均消费水平：第一组主要线上消费（总体 1），第二组主要线下消费（总体 2）。
• 我们关心总体均值是否不同，即 ${\mu }_1-{\mu }_2\ne 0$。
• 根据 ${\sigma }_1,{\sigma }_2$ 是否已知，选择不同的公式和分布，但关注的参数始终是 ${\mu }_1-{\mu }_2$。

Extension（拓展）

• This framework applies to many business problems: comparing two products, two branches, or two marketing strategies.
• Later sections will also discuss assumptions such as independence, normality, and large-sample approximations that justify using z or t procedures.

📖 点击查看中文解释

• 该框架可用于大量商业问题：比较两种产品、两家门店或两种营销策略的平均效果。
• 后续内容还会讨论使用 z 或 t 方法所需的前提假设，如样本独立性、总体近似正态以及大样本近似等。

Summary（小结）

• This chapter introduces two-sample inference for the difference in means, with separate methods for known and unknown population standard deviations.

📖 点击查看中文解释

• 本章介绍针对总体均值差的两样本推断，并根据总体标准差已知或未知采用不同的方法。

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Slide 2 — Notation for two-population mean comparison（第2页——两个总体均值比较的符号约定）

Knowledge Points (知识点)

1. Population parameters ${\mu }_1,{\mu }_2,{\sigma }_1,{\sigma }_2$（总体参数：均值与标准差）
2. Sample statistics ${\bar{X}}_1,{\bar{X}}_2,n_1,n_2$（样本统计量：均值与样本量）
3. Difference of population means ${\mu }_1-{\mu }_2$ and difference of sample means ${\bar{X}}_1-{\bar{X}}_2$（总体均值差与样本均值差）

Explanation（解释）

• Population 1 has mean ${\mu }_1$ and standard deviation ${\sigma }_1$; Population 2 has mean ${\mu }_2$ and standard deviation ${\sigma }_2$.
• We draw independent samples:
  • Sample 1: size $n_1$, sample mean ${\bar{X}}_1$ from Population 1.
  • Sample 2: size $n_2$, sample mean ${\bar{X}}_2$ from Population 2.
• The parameter of interest is the difference in population means ${\mu }_1-{\mu }_2$.
• The point estimator of ${\mu }_1-{\mu }_2$ is the difference in sample means ${\bar{X}}_1-{\bar{X}}_2$.

📖 点击查看中文解释

• 总体 1 的均值为 ${\mu }_1$，标准差为 ${\sigma }_1$；总体 2 的均值为 ${\mu }_2$，标准差为 ${\sigma }_2$。
• 我们各自从两个总体中抽取独立样本：
  • 样本 1：容量 $n_1$，样本均值 ${\bar{X}}_1$，来自总体 1；
  • 样本 2：容量 $n_2$，样本均值 ${\bar{X}}_2$，来自总体 2。
• 我们关注的总体参数是总体均值差 ${\mu }_1-{\mu }_2$。
• 其点估计量是样本均值差 ${\bar{X}}_1-{\bar{X}}_2$。

Example（例子）

• Population 1: monthly online spending of all business students; Population 2: monthly online spending of all non-business students.
• We sample $n_1=40$ business students and $n_2=35$ non-business students and compute their sample means ${\bar{X}}_1,{\bar{X}}_2$.
• We will use ${\bar{X}}_1-{\bar{X}}_2$ to estimate ${\mu }_1-{\mu }_2$.

📖 点击查看中文解释

• 总体 1：所有商科学生的每月线上消费；总体 2：所有非商科学生的每月线上消费。
• 抽取 $n_1=40$ 个商科学生和 $n_2=35$ 个非商科学生，计算各自的样本均值 ${\bar{X}}_1,{\bar{X}}_2$。
• 用样本均值差 ${\bar{X}}_1-{\bar{X}}_2$ 来估计总体均值差 ${\mu }_1-{\mu }_2$。

Extension（拓展）

• The notation extends naturally to paired-sample designs, but here we focus on independent samples.
• Clear notation helps when we later derive the sampling distribution of ${\bar{X}}_1-{\bar{X}}_2$ and construct confidence intervals.

📖 点击查看中文解释

• 这些符号也可以扩展到配对样本的情形，但本节主要讨论独立样本。
• 统一的符号约定有助于后续推导 ${\bar{X}}_1-{\bar{X}}_2$ 的抽样分布并构造置信区间。

Summary（小结）

• We distinguish clearly between population parameters and sample statistics, and recognize ${\bar{X}}_1-{\bar{X}}_2$ as the point estimator of ${\mu }_1-{\mu }_2$.

📖 点击查看中文解释

• 本页明确区分了总体参数与样本统计量，并认识到样本均值差 ${\bar{X}}_1-{\bar{X}}_2$ 是总体均值差 ${\mu }_1-{\mu }_2$ 的点估计量。

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Slide 3 — Distribution of the difference of sample means（第3页——样本均值差的分布）

Knowledge Points (知识点)

1. Expected value of ${\bar{X}}_1-{\bar{X}}_2$（样本均值差的期望）
2. Standard deviation / standard error of ${\bar{X}}_1-{\bar{X}}_2$ when ${\sigma }_1,{\sigma }_2$ are known（已知总体标准差时的标准差/标准误）
3. Role of sample sizes $n_1,n_2$（样本量对变异性的影响）

Explanation（解释）

• Under the usual assumptions (independent samples, each from a population with mean ${\mu }_i$ and variance ${\sigma }_i^2$), the expected value of the difference in sample means is

$$E({\bar{X}}_1-{\bar{X}}_2)={\mu }_1-{\mu }_2.$$

• When the population standard deviations are known, the standard deviation (also called standard error) of ${\bar{X}}_1-{\bar{X}}_2$ is

$${\sigma }_{{\bar{X}}_1-{\bar{X}}_2}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}.$$

• Larger sample sizes $n_1,n_2$ make the standard error smaller, leading to more precise estimates.

📖 点击查看中文解释

• 在常见假设下（两个样本相互独立，各自来自均值为 ${\mu }_i$、方差为 ${\sigma }_i^2$ 的总体），样本均值差的期望值为

$$E({\bar{X}}_1-{\bar{X}}_2)={\mu }_1-{\mu }_2.$$

• 当总体标准差已知时，样本均值差 ${\bar{X}}_1-{\bar{X}}_2$ 的标准差/标准误为

$${\sigma }_{{\bar{X}}_1-{\bar{X}}_2}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}.$$

• 样本量 $n_1,n_2$ 越大，标准误越小，估计就越精确。

Example（例子）

• Let ${\sigma }_1=10,{\sigma }_2=12,n_1=50,n_2=60$.
• Then

$${\sigma }_{{\bar{X}}_1-{\bar{X}}_2}=\sqrt{\frac{10^2}{50}+\frac{12^2}{60}}=\sqrt{\frac{100}{50}+\frac{144}{60}}=\sqrt{2+2.4}=\sqrt{4.4}\approx \mathrm{2.10.}$$

• This tells us the typical sampling variation of the difference in sample means.

📖 点击查看中文解释

• 设 ${\sigma }_1=10,{\sigma }_2=12,n_1=50,n_2=60$。
• 则

$${\sigma }_{{\bar{X}}_1-{\bar{X}}_2}=\sqrt{\frac{10^2}{50}+\frac{12^2}{60}}=\sqrt{\frac{100}{50}+\frac{144}{60}}=\sqrt{2+2.4}=\sqrt{4.4}\approx \mathrm{2.10.}$$

• 该值表示样本均值差在重复抽样中的典型波动大小。

Extension（拓展）

• When both populations are normal or when sample sizes are large, the distribution of ${\bar{X}}_1-{\bar{X}}_2$ is approximately normal with mean ${\mu }_1-{\mu }_2$ and standard deviation ${\sigma }_{{\bar{X}}_1-{\bar{X}}_2}$.
• This normality justifies using z-based confidence intervals and hypothesis tests in the “σ known” case.

📖 点击查看中文解释

• 当两个总体为正态分布，或样本量足够大时，${\bar{X}}_1-{\bar{X}}_2$ 近似服从均值为 ${\mu }_1-{\mu }_2$、标准差为 ${\sigma }_{{\bar{X}}_1-{\bar{X}}_2}$ 的正态分布。
• 这种正态性为“已知 σ” 情形下使用 z 置信区间与 z 检验提供了理论依据。

Summary（小结）

• The difference in sample means is an unbiased estimator of the difference in population means, and its standard error depends on both population variances and sample sizes.

📖 点击查看中文解释

• 样本均值差是总体均值差的无偏估计量，其标准误由两总体的方差和样本量共同决定。

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Slide 4 — Confidence interval for ${\mu }_1-{\mu }_2$ when σ₁, σ₂ are known（第4页——已知 σ₁、σ₂ 时的均值差置信区间）

Knowledge Points (知识点)

1. Point estimate ${\bar{X}}_1-{\bar{X}}_2$（总体均值差的点估计）
2. $100(1-\alpha )\%$ confidence interval formula with known σ₁, σ₂（已知总体标准差时的置信区间公式）
3. Meaning of significance level $\alpha$ in a two-tailed interval（双侧置信区间中的显著性水平）

Explanation（解释）

• The point estimate of the difference in population means is

$${\hat{\mu }}_1-{\hat{\mu }}_2={\bar{X}}_1-{\bar{X}}_2.$$

• When ${\sigma }_1,{\sigma }_2$ are known and the sampling distribution of ${\bar{X}}_1-{\bar{X}}_2$ is normal, a $100(1-\alpha )\%$ confidence interval for ${\mu }_1-{\mu }_2$ is

$$({\bar{X}}_1-{\bar{X}}_2)\pm z_{\alpha /2}\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}.$$

• Here $z_{\alpha /2}$ is the critical value from the standard normal distribution such that the two tails together have area $\alpha$.

📖 点击查看中文解释

• 总体均值差的点估计为

$${\hat{\mu }}_1-{\hat{\mu }}_2={\bar{X}}_1-{\bar{X}}_2.$$

• 当 ${\sigma }_1,{\sigma }_2$ 已知且 ${\bar{X}}_1-{\bar{X}}_2$ 近似正态时，$100(1-\alpha )\%$ 的置信区间为

$$({\bar{X}}_1-{\bar{X}}_2)\pm z_{\alpha /2}\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}.$$

• 其中 $z_{\alpha /2}$ 是标准正态分布的临界值，使得两侧尾部的总面积为 $\alpha$。

Example（例子）

• Suppose ${\bar{X}}_1=120,{\bar{X}}_2=110,{\sigma }_1=15,{\sigma }_2=20,n_1=50,n_2=60$.
• For a 95% confidence interval, $\alpha =0.05$ and $z_{\alpha /2}=z_{0.025}\approx 1.96$.
• The standard error is

$$\sqrt{\frac{15^2}{50}+\frac{20^2}{60}}=\sqrt{\frac{225}{50}+\frac{400}{60}}=\sqrt{4.5+6.67}\approx \sqrt{11.17}\approx \mathrm{3.34.}$$

• The interval is

$$(120-110)\pm 1.96\times 3.34=10\pm 6.54=(3.46,16.54).$$

• We are 95% confident that ${\mu }_1-{\mu }_2$ lies between 3.46 and 16.54.

📖 点击查看中文解释

• 设 ${\bar{X}}_1=120,{\bar{X}}_2=110,{\sigma }_1=15,{\sigma }_2=20,n_1=50,n_2=60$。
• 对于 95% 置信区间，$\alpha =0.05$，$z_{\alpha /2}=z_{0.025}\approx 1.96$。
• 标准误为

$$\sqrt{\frac{15^2}{50}+\frac{20^2}{60}}\approx \mathrm{3.34.}$$

• 置信区间为

$$10\pm 1.96\times 3.34=(3.46,16.54).$$

• 我们有 95% 的把握认为，总体均值差 ${\mu }_1-{\mu }_2$ 介于 3.46 与 16.54 之间。

Extension（拓展）

• The same formula can be adapted for one-sided intervals by using $z_{\alpha }$ instead of $z_{\alpha /2}$.
• The structure parallels the one-sample z-interval, but now the standard error combines the variability from both populations.

📖 点击查看中文解释

• 若构造单侧置信区间，只需将临界值改为 $z_{\alpha }$ 而非 $z_{\alpha /2}$。
• 该公式的结构与单总体 z 置信区间类似，只是标准误中合并了两个总体的变异性。

Summary（小结）

• With known population standard deviations, the confidence interval for ${\mu }_1-{\mu }_2$ is built from the point estimate ${\bar{X}}_1-{\bar{X}}_2$ plus/minus a z-critical value times the combined standard error.

📖 点击查看中文解释

• 当总体标准差已知时，均值差的置信区间由“样本均值差 ± z 临界值 × 合并标准误”构成，是两总体均值比较的基本工具。

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Slide 5 — Confidence interval for μ₁ − μ₂（第5页——均值差置信区间）

Knowledge Points (知识点)

1. Point estimate ${\bar{x}}_1-{\bar{x}}_2$ for ${\mu }_1-{\mu }_2$（总体均值差的点估计）
2. Confidence interval formula with known ${\sigma }_1,{\sigma }_2$（已知总体标准差时的置信区间公式）
3. Significance level $\alpha$ and two-tailed intervals（显著性水平与双侧区间）

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Explanation（解释）

• The point estimate of the difference in population means is

$${\hat{\mu }}_1-{\hat{\mu }}_2={\bar{x}}_1-{\bar{x}}_2.$$

• When population standard deviations ${\sigma }_1,{\sigma }_2$ are known and samples are independent, a $100(1-\alpha )\%$ confidence interval for ${\mu }_1-{\mu }_2$ is

$$({\bar{x}}_1-{\bar{x}}_2)\pm z_{\alpha /2}\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}.$$

• Here $z_{\alpha /2}$ is the critical value from the standard normal distribution such that each tail has probability $\alpha /2$.
• $\alpha$ is the significance level for a two-tailed interval: it is the total probability outside the interval.

📖 点击查看中文解释

• 总体均值差 ${\mu }_1-{\mu }_2$ 的点估计为

$${\hat{\mu }}_1-{\hat{\mu }}_2={\bar{x}}_1-{\bar{x}}_2.$$

• 当总体标准差 ${\sigma }_1,{\sigma }_2$ 已知且样本相互独立时，$100(1-\alpha )\%$ 的置信区间为

$$({\bar{x}}_1-{\bar{x}}_2)\pm z_{\alpha /2}\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}.$$

• 其中 $z_{\alpha /2}$ 是标准正态分布的临界值，使得每个尾部的概率为 $\alpha /2$。
• $\alpha$ 是双侧置信区间的显著性水平，表示区间之外的总概率。

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Example（例子）

• For a 95% confidence interval, we set $\alpha =0.05$.
• The corresponding critical value is $z_{\alpha /2}=z_{0.025}\approx 1.96$.
• Any confidence interval using 95% confidence will have the form

$$({\bar{x}}_1-{\bar{x}}_2)\pm 1.96\times \text{(standard error)}.$$

📖 点击查看中文解释

• 若需要 95% 置信区间，则取 $\alpha =0.05$。
• 此时临界值为 $z_{\alpha /2}=z_{0.025}\approx 1.96$。
• 任意 95% 置信区间都可以写成

$$({\bar{x}}_1-{\bar{x}}_2)\pm 1.96\times \text{标准误}.$$

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Extension（拓展）

• If the hypothesized difference (often $0$) lies outside the confidence interval, we will later reject the null hypothesis in a two-tailed test.
• Thus confidence intervals and hypothesis tests are closely connected.

📖 点击查看中文解释

• 若某个假设的均值差（通常为 $0$）落在置信区间之外，那么在后面进行的双侧假设检验中，我们会拒绝原假设。
• 因此，置信区间与假设检验之间有紧密联系。

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Summary（小结）

• With known ${\sigma }_1,{\sigma }_2$, the confidence interval for ${\mu }_1-{\mu }_2$ is constructed by taking the point estimate ${\bar{x}}_1-{\bar{x}}_2$ plus/minus a z-critical value times the combined standard error.

📖 点击查看中文解释

• 当总体标准差已知时，均值差的置信区间由“样本均值差 ± z 临界值 × 合并标准误”构成，是比较两个总体均值的基本工具。

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Slide 6 — Example setup: ABC company vs competitor（第6页——示例设定：ABC 公司与竞争对手）

Knowledge Points (知识点)

1. Comparing two population means using sample information（用样本信息比较两个总体均值）
2. Identifying $n_1,n_2,{\bar{x}}_1,{\bar{x}}_2,{\sigma }_1,{\sigma }_2$ from a table（从表格中识别参数）
3. Significance level $\alpha =0.05$ in practice（实际问题中的显著性水平）

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Explanation（解释）

• The ABC company compares the average life of its own product (Sample 1) with that of a competitor (Sample 2).
• Both samples are independent, and population standard deviations are assumed known.

📖 点击查看中文解释

• ABC 公司想比较自己产品（样本 1）与竞争对手产品（样本 2）的平均寿命。
• 两个样本相互独立，并且假设总体标准差已知。

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Example（例子）

Data table（数据表）

	Sample 1 (ABC)	Sample 2 (Competitor)
Sample size $n$	120 units	80 units
Sample mean $\bar{x}$	275 min	258 min
Standard deviation $\sigma$	15 min	20 min

• Significance level: $\alpha =0.05$.

📖 点击查看中文解释

• 样本 1（ABC 产品）：
  • 样本量 $n_1=120$，样本均值 ${\bar{x}}_1=275$ 分钟，标准差 ${\sigma }_1=15$ 分钟；
• 样本 2（竞争对手）：
  • 样本量 $n_2=80$，样本均值 ${\bar{x}}_2=258$ 分钟，标准差 ${\sigma }_2=20$ 分钟；
• 显著性水平：$\alpha =0.05$。

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Extension（拓展）

• The question “Is there any difference?” corresponds to testing whether

$${\mu }_1-{\mu }_2=0$$

or estimating a confidence interval to see if $0$ is included.

📖 点击查看中文解释

• “是否有差异？”这一问题在统计上对应于检验

$${\mu }_1-{\mu }_2=0$$

或者构造置信区间，看 $0$ 是否落在区间之内。

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Summary（小结）

• This example provides realistic sample data and a chosen significance level, allowing us to compute a confidence interval for ${\mu }_1-{\mu }_2$ and judge whether ABC’s product differs from its competitor’s.

📖 点击查看中文解释

• 本例给出了具体的样本数据和显著性水平，为我们计算 ${\mu }_1-{\mu }_2$ 的置信区间、判断 ABC 产品是否优于竞争对手提供了基础。

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Slide 7 — Example calculation: confidence interval and conclusion（第7页——示例计算：置信区间与结论）

Knowledge Points (知识点)

1. Computing the standard error of ${\bar{x}}_1-{\bar{x}}_2$（计算样本均值差的标准误）
2. Finding the confidence interval numerically（数值上求出置信区间）
3. Using the interval to judge significance（利用置信区间判断显著性）

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Explanation（解释）

• From the ABC example, the point estimate is

$${\bar{x}}_1-{\bar{x}}_2=275-258=17\text{ min}.$$

• The standard error of ${\bar{x}}_1-{\bar{x}}_2$ is

$$\text{SE}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}=\sqrt{\frac{15^2}{120}+\frac{20^2}{80}}\approx \mathrm{2.62.}$$

• With $\alpha =0.05$, the critical value is $z_{0.025}=1.96$.
• The confidence interval is

$$17\pm 1.96\times 2.62=17\pm 5.14=(11.86,22.14).$$

📖 点击查看中文解释

• 对于 ABC 例子，均值差的点估计为

$${\bar{x}}_1-{\bar{x}}_2=275-258=17\text{ 分钟}.$$

• 样本均值差的标准误为

$$\text{SE}=\sqrt{\frac{15^2}{120}+\frac{20^2}{80}}\approx \mathrm{2.62.}$$

• 显著性水平 $\alpha =0.05$ 时，临界值 $z_{0.025}=1.96$。
• 置信区间为

$$17\pm 1.96\times 2.62=17\pm 5.14=(11.86,22.14).$$

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Example（例子）：Interpretation（区间解释）

• Because the entire interval $(11.86,22.14)$ is above 0, we can say:
  • At the 5% significance level, there is significant evidence that ABC’s product has a larger mean life than its competitor’s.
  • The estimated difference in mean life is between about 12 and 22 minutes.

📖 点击查看中文解释

• 因为区间 $(11.86,22.14)$ 整体大于 0，所以在 5% 的显著性水平下：
  • 有显著证据表明 ABC 产品的平均寿命高于竞争对手；
  • 平均寿命的差异估计在约 12 到 22 分钟之间。

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Extension（拓展）

• If 0 had been contained in the interval, we would conclude that the data are consistent with no difference in mean lifetimes.
• This logic is equivalent to performing a two-tailed hypothesis test with null hypothesis ${\mu }_1-{\mu }_2=0$.

📖 点击查看中文解释

• 如果 0 落在置信区间内，我们会认为数据与“平均寿命无差异”的假设一致。
• 这种判断与对原假设 ${\mu }_1-{\mu }_2=0$ 进行双侧假设检验的结论是等价的。

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Summary（小结）

• For the ABC example, the 95% confidence interval shows a positive difference far from 0, indicating that ABC’s product performs significantly better than the competitor’s in terms of mean life.

📖 点击查看中文解释

• 在 ABC 示例中，95% 置信区间完全为正且远离 0，说明 ABC 产品在平均寿命上显著优于竞争对手。

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Slide 8 — Hypothesis tests about μ₁ − μ₂ with known σ₁, σ₂（第8页——已知 σ₁、σ₂ 时的均值差假设检验）

Knowledge Points (知识点)

1. Null and alternative hypotheses for comparing two means（比较两个均值的原假设与备择假设）
2. Three types of tests: left-tailed, right-tailed, two-tailed（三种检验形式：左尾、右尾、双尾）
3. z test statistic for ${\mu }_1-{\mu }_2$ when σ’s are known（已知总体标准差时的 z 检验统计量）

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Explanation（解释）

Hypotheses（假设形式）

• Left-tailed test (testing if population 1 mean is smaller):

$$H_0:{\mu }_1-{\mu }_2\ge D_0,H_a:{\mu }_1-{\mu }_2<D_0.$$

• Right-tailed test (testing if population 1 mean is larger):

$$H_0:{\mu }_1-{\mu }_2\le D_0,H_a:{\mu }_1-{\mu }_2>D_0.$$

• Two-tailed test (testing if there is any difference):

$$H_0:{\mu }_1-{\mu }_2=D_0,H_a:{\mu }_1-{\mu }_2\ne D_0.$$

• Test statistic (known ${\sigma }_1,{\sigma }_2$):

$$z=\frac{({\bar{x}}_1-{\bar{x}}_2)-D_0}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}.$$

📖 点击查看中文解释

• 左尾检验（检验总体 1 均值是否更小）：

$$H_0:{\mu }_1-{\mu }_2\ge D_0,H_a:{\mu }_1-{\mu }_2<D_0.$$

• 右尾检验（检验总体 1 均值是否更大）：

$$H_0:{\mu }_1-{\mu }_2\le D_0,H_a:{\mu }_1-{\mu }_2>D_0.$$

• 双尾检验（检验是否存在差异）：

$$H_0:{\mu }_1-{\mu }_2=D_0,H_a:{\mu }_1-{\mu }_2\ne D_0.$$

• 当 ${\sigma }_1,{\sigma }_2$ 已知时，检验统计量为

$$z=\frac{({\bar{x}}_1-{\bar{x}}_2)-D_0}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}.$$

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Example（例子）

• For the ABC company, to test whether there is any difference, we set $D_0=0$ and use the two-tailed hypotheses:

$$H_0:{\mu }_1-{\mu }_2=0,H_a:{\mu }_1-{\mu }_2\ne 0.$$

• The test statistic becomes

$$z=\frac{({\bar{x}}_1-{\bar{x}}_2)-0}{\text{SE}}=\frac{17}{2.62}\approx 6.49,$$

which is far into the rejection region for $\alpha =0.05$.

📖 点击查看中文解释

• 对 ABC 公司例子，若要检验“是否存在差异”，取 $D_0=0$，建立双尾假设：

$$H_0:{\mu }_1-{\mu }_2=0,H_a:{\mu }_1-{\mu }_2\ne 0.$$

• 检验统计量为

$$z=\frac{17}{2.62}\approx 6.49,$$

远大于 $\alpha =0.05$ 时的临界值 $1.96$，因此拒绝原假设。

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Extension（拓展）

• Decision rules:
  • Left-tailed: reject $H_0$ if $z<-z_{\alpha }$.
  • Right-tailed: reject $H_0$ if $z>z_{\alpha }$.
  • Two-tailed: reject $H_0$ if $|z|>z_{\alpha /2}$.
• p-value methods lead to equivalent conclusions.

📖 点击查看中文解释

• 决策规则：
  • 左尾检验：若 $z<-z_{\alpha }$，则拒绝 $H_0$；
  • 右尾检验：若 $z>z_{\alpha }$，则拒绝 $H_0$；
  • 双尾检验：若 $|z|>z_{\alpha /2}$，则拒绝 $H_0$。
• 使用 p 值方法也会得到等价的结论。

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Summary（小结）

• Hypothesis tests about ${\mu }_1-{\mu }_2$ specify a null value $D_0$, choose the appropriate tail form, and use the z statistic based on the combined standard error when ${\sigma }_1,{\sigma }_2$ are known.

📖 点击查看中文解释

• 针对 ${\mu }_1-{\mu }_2$ 的假设检验，需要给定假设差值 $D_0$，确定是左尾、右尾还是双尾检验，并在已知标准差时使用基于合并标准误的 z 统计量进行决策。

Slide 9 — One-sided test: ABC product vs competitor (p-value)（第9页——单侧检验：ABC 产品与竞争对手（p 值法））

Knowledge Points (知识点)

1. Right-tailed test for difference of means（均值差的右尾检验）
2. p-value vs. significance level $\alpha$（p 值与显著性水平的比较）
3. Interpretation: “significantly higher” vs “not higher”（“显著更高”的解释）

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Explanation（解释）

• Question: Is ABC’s mean product life higher than the competitor’s?
• We use a right-tailed test for the difference of two population means.

Step 1: Hypotheses

$$H_0:{\mu }_1-{\mu }_2\le 0\text{vs.}{H}_a:{\mu }_1-{\mu }_2>0$$

• ${\mu }_1$: mean life of ABC’s product
• ${\mu }_2$: mean life of competitor’s product

Step 2: Significance level

$$\alpha =0.01$$

Step 3: Test statistic and p-value

$$z=\frac{({\bar{x}}_1-{\bar{x}}_2)-0}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}=\frac{(275-258)-0}{\sqrt{\frac{15^2}{120}+\frac{20^2}{80}}}\approx 6.49$$

• For $z=6.49$, the p-value < 0.001 (very close to 0).

Decision rule (right-tailed):
If $\text{p-value}<\alpha$, reject $H_0$.

Since $\text{p-value}<0.001<0.01$, we reject $H_0$ and support $H_a$.

📖 点击查看中文解释

• 问题：ABC 产品的平均寿命是否高于竞争对手？
• 使用右尾检验比较两个总体均值。
  步骤 1：提出假设

$$H_0:{\mu }_1-{\mu }_2\le 0,H_a:{\mu }_1-{\mu }_2>0$$

• 其中 ${\mu }_1$ 为 ABC 产品的总体平均寿命，${\mu }_2$ 为竞争对手产品的总体平均寿命。
  步骤 2：显著性水平

$$\alpha =0.01$$

步骤 3：计算检验统计量与 p 值

$$z=\frac{(275-258)-0}{\sqrt{\frac{15^2}{120}+\frac{20^2}{80}}}\approx 6.49$$

• 对应的 p 值 < 0.001，远小于 0.01。
• 决策规则（右尾）：若 $\text{p-value}<\alpha$，则拒绝 $H_0$。
• 因为 $\text{p-value}<0.001<0.01$，所以拒绝 $H_0$，支持 $H_a$。

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Example（例子）结论

• We conclude that ABC’s mean life is significantly higher than the competitor’s at the 1% significance level.

📖 点击查看中文解释

• 结论：在 1% 的显著性水平下，ABC 产品的平均寿命显著高于竞争对手产品。

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Extension（拓展）

• The p-value approach does not require computing the critical value.
• It directly tells how extreme the observed $z$ is under $H_0$.

📖 点击查看中文解释

• p 值方法不需要先求临界值，而是直接衡量在原假设成立时观测到当前 $z$ 值“有多极端”。
• p 值越小，反对 $H_0$ 的证据越强。

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Summary（小结）

• For a one-sided right-tailed test, if the computed p-value is smaller than $\alpha$, we conclude that population 1’s mean is significantly larger than population 2’s mean.

📖 点击查看中文解释

• 在单侧右尾检验中，若 p 值小于显著性水平 $\alpha$，则说明总体 1 的均值显著大于总体 2 的均值。

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Slide 10 — One-sided test: ABC vs competitor (critical value)（第10页——单侧检验：ABC 与竞争对手（临界值法））

Knowledge Points (知识点)

1. z-critical value $z_{\alpha }$ for right-tailed test（右尾检验的临界值 $z_{\alpha }$）
2. Compare test statistic $z$ with $z_{\alpha }$（用 $z$ 与 $z_{\alpha }$ 比较做决策）
3. Connection to p-value approach（与 p 值法的一致性）

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Explanation（解释）

Same hypotheses and data as Slide 9:

$$H_0:{\mu }_1-{\mu }_2\le 0,H_a:{\mu }_1-{\mu }_2>0,\alpha =0.01$$

• Test statistic is still $z\approx 6.49$.

Critical value for right-tailed test

$$z_{\alpha }=z_{0.01}\approx 2.33$$

Decision rule (right-tailed):
Reject $H_0$ if $z>z_{\alpha }$.

Since

$$z\approx 6.49>2.33=z_{\alpha },$$

we reject $H_0$ and conclude that ABC’s mean is significantly higher.

📖 点击查看中文解释

• 与第 9 页相同的假设与数据：

$$H_0:{\mu }_1-{\mu }_2\le 0,H_a:{\mu }_1-{\mu }_2>0,\alpha =0.01$$

• 检验统计量仍为 $z\approx 6.49$。
  右尾检验的临界值

$$z_{\alpha }=z_{0.01}\approx 2.33$$

• 决策规则：若 $z>z_{\alpha }$，则拒绝 $H_0$。
• 因为

$$z\approx 6.49>2.33=z_{\alpha },$$

所以拒绝 $H_0$，认为 ABC 产品的平均寿命显著更高。

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Example（例子）比较两种方法

• p-value approach (Slide 9): compare p-value with $\alpha$.
• Critical value approach (this slide): compare $z$ with $z_{\alpha }$.
• Both give the same conclusion.

📖 点击查看中文解释

• p 值法（第 9 页）：比较 p 值与 $\alpha$。
• 临界值法（本页）：比较 $z$ 与 $z_{\alpha }$。
• 两种方法得到的结论完全一致。

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Extension（拓展）

• For a left-tailed test, we would use $z_{1-\alpha }=-z_{\alpha }$.
• For a two-tailed test, we use $\pm z_{\alpha /2}$ as critical values.

📖 点击查看中文解释

• 左尾检验中，临界值为 $z_{1-\alpha }=-z_{\alpha }$。
• 双尾检验中，临界值为 $\pm z_{\alpha /2}$。

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Summary（小结）

• In the critical-value approach, if the standardized test statistic lies in the rejection region (beyond the critical value), we reject $H_0$; otherwise, we fail to reject $H_0$.

📖 点击查看中文解释

• 在临界值法中，若标准化统计量落入拒绝域（超过临界值），就拒绝 $H_0$；否则就“不拒绝” $H_0$。

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Slide 11 — Practice: TOEFL scores of two universities（第11页——练习：两所大学托福成绩比较）

Knowledge Points (知识点)

1. Two-sample z test for difference in mean scores（两总体均值差的 z 检验）
2. Setting up hypotheses for “significant difference”（“是否有显著差异”的假设设定）
3. Interpreting test results in context（在情境中解释统计结论）

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Explanation（解释）

We compare TOEFL scores of Newland University and ABC University.

Data table（数据表）

Group	Score ( $\bar{x}$ )	Standard deviation ( $\sigma$ )	Sample size ( $n$ )
Newland University	103	15	50
ABC University	96	10	50

• Significance level: $\alpha =0.05$.

Step 1: Hypotheses

“Is there significant difference?” → two-tailed test

$$H_0:{\mu }_1-{\mu }_2=0\text{vs.}{H}_a:{\mu }_1-{\mu }_2\ne 0$$

where

• ${\mu }_1$: mean TOEFL score of Newland students,
• ${\mu }_2$: mean TOEFL score of ABC students.

Step 2: Test statistic

Point estimate:

$${\bar{x}}_1-{\bar{x}}_2=103-96=7$$

Standard error:

$$\text{SE}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}=\sqrt{\frac{15^2}{50}+\frac{10^2}{50}}=\sqrt{\frac{225}{50}+\frac{100}{50}}=\sqrt{6.5}\approx 2.55$$

z-score:

$$z=\frac{({\bar{x}}_1-{\bar{x}}_2)-0}{\text{SE}}=\frac{7}{2.55}\approx 2.75$$

Step 3: Decision

• For a two-tailed test with $\alpha =0.05$, critical values:

$$\pm z_{0.025}=\pm \mathrm{1.96.}$$

• Since $|z|\approx 2.75>1.96$, we reject $H_0$.

📖 点击查看中文解释

• 比较 Newland University 与 ABC University 学生的托福成绩。
• 数据见上表，显著性水平 $\alpha =0.05$。
  步骤 1：假设（是否有差异 → 双尾检验）

$$H_0:{\mu }_1-{\mu }_2=0,H_a:{\mu }_1-{\mu }_2\ne 0$$

• ${\mu }_1$：Newland 学生托福平均分；${\mu }_2$：ABC 学生托福平均分。
  步骤 2：检验统计量

$${\bar{x}}_1-{\bar{x}}_2=103-96=7$$

$$\text{SE}=\sqrt{\frac{15^2}{50}+\frac{10^2}{50}}=\sqrt{6.5}\approx 2.55$$

$$z=\frac{7}{2.55}\approx 2.75$$

步骤 3：决策

• 双尾检验、$\alpha =0.05$ 时临界值为 $\pm 1.96$。
• 因为 $|z|\approx 2.75>1.96$，所以拒绝 $H_0$。

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Example（例子）结论

• There is a significant difference in mean TOEFL scores between Newland and ABC at the 5% level.
• Since ${\bar{x}}_1=103>96={\bar{x}}_2$, Newland students have higher average TOEFL scores.

📖 点击查看中文解释

• 在 5% 的显著性水平下，两所大学的托福平均分存在显著差异。
• 且 Newland 的样本均值更高，说明 Newland 学生的托福成绩平均更高。

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Extension（拓展）

• We could also construct a 95% confidence interval for ${\mu }_1-{\mu }_2$:

$$({\bar{x}}_1-{\bar{x}}_2)\pm z_{0.025}\times \text{SE}=7\pm 1.96\times 2.55\approx 7\pm 5.00=(2.0,12.0)$$

• Because 0 is not in this interval, the confidence-interval approach gives the same conclusion as the hypothesis test.

📖 点击查看中文解释

• 亦可构造 95% 置信区间：

$$7\pm 1.96\times 2.55\approx (2.0,12.0)$$

• 由于区间不包含 0，与假设检验得到的“有显著差异”结论一致。

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Summary（小结）

• For the TOEFL example, we used a two-sample z test (and an equivalent confidence interval) to show that Newland University’s mean TOEFL score is significantly higher than ABC University’s at $\alpha =0.05$.

📖 点击查看中文解释

• 托福示例通过两样本 z 检验（以及等价的置信区间）表明，在显著性水平 $\alpha =0.05$ 下，Newland University 学生的平均托福分数显著高于 ABC University。