📚 A-Level Further Mathematics: Hypothesis Test for Difference of Means | A-Level 进阶数学：均值之差的假设检验
In advanced statistics, comparing two population means is a fundamental skill. This article explains the hypothesis test for the difference between two means, covering assumptions, test statistics, critical values, p-values, and worked examples in full line with the Edexcel Further Mathematics specification.
在进阶统计学中，比较两个总体均值是基本且重要的技能。本文围绕均值之差的假设检验展开，系统讲解检验假设、检验统计量、临界值、p 值以及完整例题，完全贴合 Edexcel 进阶数学大纲。
1. When to Use the Difference of Means Test | 何时使用均值之差检验
The difference of means test is used when we want to compare the means μ₁ and μ₂ of two independent populations. We have a random sample of size n₁ from Population 1 and an independent random sample of size n₂ from Population 2. This test allows us to determine whether the observed difference between the sample means x̄₁ and x̄₂ is statistically significant.
均值之差检验用于比较两个独立总体的均值 μ₁ 和 μ₂。我们从总体 1 中抽取容量为 n₁ 的随机样本，并从总体 2 中抽取另一个独立的随机样本。该检验帮助我们判断样本均值 x̄₁ 与 x̄₂ 之间的观测差异是否具有统计学意义。
This test is particularly useful in A-Level Further Mathematics when comparing two treatments, products, or populations where the underlying data are normally distributed or the sample sizes are large enough to apply the Central Limit Theorem.
在 A-Level 进阶数学中，当比较两种处理、两种产品或两个总体，且数据服从正态分布或样本量足够大而适用中心极限定理时，该检验尤为常用。
2. Setting Up Hypotheses | 建立假设
The null hypothesis is usually that there is no difference between the population means: H₀: μ₁ – μ₂ = 0 (or μ₁ = μ₂). The alternative hypothesis can be two-tailed: H₁: μ₁ ≠ μ₂; or one-tailed: H₁: μ₁ < μ₂ or H₁: μ₁ > μ₂. The direction of the alternative hypothesis determines the type of test.
原假设通常设为两个总体均值无差异：H₀: μ₁ – μ₂ = 0（即 μ₁ = μ₂）。备择假设可以是双尾的：H₁: μ₁ ≠ μ₂；也可以是单尾的：H₁: μ₁ < μ₂ 或 H₁: μ₁ > μ₂。备择假设的方向决定了检验的类型。
It is important to decide whether the test is one-tailed or two-tailed before collecting data. This decision should be based on the research question, not on the observed sample means.
在收集数据之前就要确定检验是单尾还是双尾。这一决定应基于研究问题本身，而不能依据已观测到的样本均值。
3. Assumptions and Conditions | 假设与条件
For the test to be valid, we need independent random samples from the two populations. If the population variances σ₁² and σ₂² are known, we can use a normal distribution for the test statistic. If they are unknown, the sample variances s₁² and s₂² are used, and the normal approximation is acceptable when both sample sizes are large (n₁, n₂ ≥ 30) by the Central Limit Theorem.
为了使检验有效，我们需要来自两个总体的独立随机样本。如果总体方差 σ₁² 和 σ₂² 已知，可以直接使用正态分布构造检验统计量。如果方差未知，则使用样本方差 s₁² 和 s₂²；只要两个样本量均较大（n₁, n₂ ≥ 30），根据中心极限定理，正态近似仍然成立。
The populations should ideally be normally distributed, especially for small samples. In many exam problems the sample sizes are large enough, so the normal distribution is a safe approximation.
理想情况下，总体应遵从正态分布，尤其在样本量较小时。在多数考试题目中，样本量足够大，因此正态近似是可靠的。
4. The Test Statistic and Its Distribution | 检验统计量及其分布
Under H₀, the difference of the sample means follows a normal distribution with mean 0 and variance σ₁²/n₁ + σ₂²/n₂. When population variances are known, the test statistic is calculated by standardising:
在原假设 H₀ 下，样本均值之差近似服从均值为 0、方差为 σ₁²/n₁ + σ₂²/n₂ 的正态分布。当总体方差已知时，检验统计量通过标准化过程得到：
Z = (x̄₁ – x̄₂) / √(σ₁²/n₁ + σ₂²/n₂)
This statistic follows a standard normal distribution N(0,1), so we compare it with the appropriate critical value from the normal distribution table.
该统计量服从标准正态分布 N(0,1)，因此我们可以将其与正态分布表中的临界值进行比较。
When the population variances are unknown, we replace σ₁² and σ₂² by the sample variances s₁² and s₂². The statistic is still called Z and is approximately standard normal for large samples.
当总体方差未知时，我们用样本方差 s₁² 和 s₂² 代替 σ₁² 和 σ₂²。该统计量仍记为 Z，并在大样本下近似服从标准正态分布。
5. Critical Values and Rejection Regions | 临界值与拒绝域
For a two-tailed test at significance level α, the critical values are ±zα/2. Reject H₀ if Z ≤ -zα/2 or Z ≥ zα/2.
对于显著性水平 α 的双尾检验，临界值为 ±zα/2。若 Z ≤ -zα/2 或 Z ≥ zα/2，则拒绝 H₀。
For a one-tailed test with H₁: μ₁ > μ₂, reject H₀ if Z ≥ zα. For H₁: μ₁ < μ₂, reject H₀ if Z ≤ -zα.
对于备择假设 H₁: μ₁ > μ₂ 的单尾检验，若 Z ≥ zα，拒绝 H₀；对于 H₁: μ₁ < μ₂，若 Z ≤ -zα，拒绝 H₀。
| 显著性水平 α | 0.10 | 0.05 | 0.01 |
| 单尾临界值 zα | 1.282 | 1.645 | 2.326 |
| 双尾临界值 zα/2 | 1.645 | 1.960 | 2.576 |
These critical values are used frequently in Edexcel Further Statistics papers. You may be given them in a formula booklet, but it is safer to memorise the common ones.
这些临界值在 Edexcel 进阶统计试卷中频繁出现。考试中可能会在公式册中给出，但最稳妥的做法是记住常用值。
6. The p-Value Approach | p 值方法
Alternatively, we can compute the p-value, which is the probability of obtaining a test statistic at least as extreme as the one observed, assuming H₀ is true.
另一种方法是在假定 H₀ 成立的前提下计算 p 值，即观测到与当前检验统计量同样极端（或更极端）结果的概率。
For a two-tailed test, the p-value is 2 × P(Z ≥ |z|). For a one-tailed test where the alternative is μ₁ > μ₂, the p-value is P(Z ≥ z); for μ₁ < μ₂, it is P(Z ≤ z).
对于双尾检验，p 值 = 2 × P(Z ≥ |z|)。对于备择假设为 μ₁ > μ₂ 的单尾检验，p 值 = P(Z ≥ z)；对于 μ₁ < μ₂，p 值 = P(Z ≤ z)。
We reject H₀ if the p-value is less than or equal to the significance level α. This method gives the same conclusion as the critical-value method, but is often more convenient when using statistical tables or calculators.
当 p 值 ≤ α 时拒绝 H₀。p 值法与临界值法会给出相同的结论，但在使用统计表或计算器时往往更加便捷。
7. Worked Example 1: Two-Tailed Test (Known Variances) | 例题 1：双尾检验（方差已知）
A manufacturer produces two types of battery. The lifetimes of Type A are normally distributed with σ₁ = 5.0 hours, and Type B with σ₂ = 4.0 hours. A sample of n₁ = 50 Type A batteries has mean x̄₁ = 105 hours, and n₂ = 40 Type B batteries has mean x̄₂ = 102 hours. Test at the 5% significance level whether the mean lifetimes differ significantly.
某制造商生产两种电池。A 型电池寿命服从正态分布，σ₁ = 5.0 小时；B 型电池 σ₂ = 4.0 小时。抽取 n₁ = 50 个 A 型电池，样本均值 x̄₁ = 105 小时；抽取 n₂ = 40 个 B 型电池，样本均值 x̄₂ = 102 小时。在 5% 显著性水平下检验两种电池的平均寿命是否存在显著差异。
Step 1: State hypotheses. H₀: μ₁ – μ₂ = 0, H₁: μ₁ – μ₂ ≠ 0.
步骤 1：写出假设。 H₀: μ₁ – μ₂ = 0，H₁: μ₁ – μ₂ ≠ 0。
Step 2: Calculate the test statistic.
步骤 2：计算检验统计量。
Z = (105 – 102) / √(5²/50 + 4²/40) = 3 / √(0.5 + 0.4) = 3/√0.9 = 3.162
Step 3: Compare with the critical value. For α = 0.05 and a two-tailed test, the critical value is z0.025 = 1.960. Since 3.162 > 1.960, we reject H₀.
步骤 3：与临界值比较。 对于 α = 0.05 的双尾检验，临界值 z0.025 = 1.960。因为 3.162 > 1.960，所以我们拒绝 H₀。
Step 4: Conclusion. There is sufficient evidence at the 5% level that the mean lifetimes of the two battery types differ significantly.
步骤 4：结论。 在 5% 显著性水平下有充分证据表明两种电池的平均寿命存在显著差异。
8. Worked Example 2: One-Tailed Test (Large Samples) | 例题 2：单尾检验（大样本）
A school compares two teaching methods. In Group 1 (n₁ = 100 students), the mean score is x̄₁ = 72.5 with sample variance s₁² = 12.0. In Group 2 (n₂ = 90 students), the mean score is x̄₂ = 70.0 with sample variance s₂² = 15.0. Is there evidence at the 1% level that Method 1 gives a higher mean score than Method 2?
某学校比较两种教学方法。第 1 组 n₁ = 100 名学生，平均分 x̄₁ = 72.5，样本方差 s₁² = 12.0；第 2 组 n₂ = 90 名学生，平均分 x̄₂ = 70.0，样本方差 s₂² = 15.0。在 1% 显著性水平下，是否有证据表明方法 1 的平均分高于方法 2？
Step 1: State hypotheses. H₀: μ₁ – μ₂ = 0, H₁: μ₁ – μ₂ > 0.
步骤 1：写出假设。 H₀: μ₁ – μ₂ = 0，H₁: μ₁ – μ₂ > 0。
Step 2: Calculate the test statistic. Using s₁² and s₂² because the population variances are unknown,
步骤 2：计算检验统计量。 由于总体方差未知，使用样本方差 s₁² 和 s₂²：
Z = (72.5 – 70.0) / √(12.0/100 + 15.0/90) = 2.5 / √(0.12 + 0.16667) = 2.5 / √0.28667 = 4.669
Step 3: Compare with the critical value. For α = 0.01 and a one-tailed test, z0.01 = 2.326. Since 4.669 > 2.326, we reject H₀.
步骤 3：与临界值比较。 对于 α = 0.01 的单尾检验，z0.01 = 2.326。因为 4.669 > 2.326，所以我们拒绝 H₀。
Step 4: Conclusion. There is strong evidence at the 1% level that Method 1 produces a higher mean score than Method 2.
步骤 4：结论。 在 1% 显著性水平下有充分的证据表明，方法 1 的平均分高于方法 2。
9. Common Mistakes and Pitfalls | 常见错误与陷阱
Students often confuse one-tailed and two-tailed hypotheses. Remember that a two-tailed test uses ±zα/2, while a one-tailed test uses only zα in the direction suggested by H₁.
学生经常混淆单尾与双尾假设。请记住：双尾检验使用 ±zα/2
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