📚 Finding Type I and Type II Errors Using the Normal Distribution | 利用正态分布求第I类和第II类错误
In A-Level Edexcel Statistics, hypothesis testing is a core topic where we make decisions about population parameters based on sample data. Understanding and calculating Type I and Type II errors under the normal distribution is essential for evaluating the reliability of these decisions. This article provides a structured approach to finding both error types, building from the fundamentals of normal hypothesis tests to worked examples involving means and proportions, all while connecting theory to typical Edexcel exam questions.
在 A-Level Edexcel 统计学中，假设检验是一个核心主题，我们基于样本数据对总体参数做出决策。理解并计算正态分布下的第 I 类和第 II 类错误对于评估这些决策的可靠性至关重要。本文提供了一个结构化的方法来求解这两种错误类型，从正态假设检验的基础知识构建到涉及均值和比例的应用实例，同时将理论与典型的 Edexcel 考试题目联系起来。
1. Recap of Hypothesis Testing for a Normal Mean | 正态均值假设检验回顾
population mean μ when the population variance σ² is known uses the test statistic Z = (X̄ − μ₀) / (σ/√n), where X̄ is the sample mean, μ₀ is the hypothesised mean, σ is the population standard deviation, and n is the sample size. The Z-statistic follows a standard normal distribution N(0,1) under the null hypothesis H₀.
当总体方差 σ² 已知时，对总体均值 μ 的假设检验使用检验统计量 Z = (X̄ − μ₀) / (σ/√n)，其中 X̄ 是样本均值，μ₀ 是假设的均值，σ 是总体标准差，n 是样本容量。在原假设 H₀ 下，Z 统计量服从标准正态分布 N(0,1)。
We define a significance level α (commonly 5% or 1% l value(s) are found using the inverse normal distribution, for example zₐ = 1.645 for a one-tailed test at α = 0.05.
我们定义一个显著性水平 α（通常为 5% 或 1%），并确定 Z 的拒绝域。如果计算出的 Z 落入该区域，我们拒绝 H₀ 而支持备择假设 H₁。临界值使用逆正态分布求得，例如在 α = 0.05 的单尾检验中 zₐ = 1.645。
2. What Are Type I and Type II Errors? | 什么是第I类和第II类错误？
A Type I error occurs when we reject a true null hypothesis. Its probability is exactly the significance level α, provided the test is designed correctly. A Type II error occurs when we fail to reject a false null hypothesis; its probability is denoted by β. The power of a test is 1 − β, which is the probability of correctly rejecting a false H₀.
第 I 类错误发生在拒绝了一个真实的原假设时。如果检验设计正确，其概率恰好是显著性水平 α。第 II 类错误发生在未能拒绝一个错误的原假设时；其概率用 β 表示。检验的功效是 1 − β，即正确拒绝错误 H₀ 的概率。
H₀ and under a specific true alternative value μ = μ₁. The calculation involves standardising and finding tail areas.
在正态分布的背景下，两种错误的概率都可以通过考虑检验统计量在 H₀ 下以及在某个特定真实备择值 μ = μ₁ 下的分布来计算。计算涉及标准化并求出尾部面积。
3. Significance Level and the Probability of a Type I Error | 显著性水平与第I类错误概率
For a correctly specified test, P(Type I error) = α. For example, if we use a 5% ion is needed for the Type I error probability itself, but you may be asked to confirm it or to find α for a given critical value.
对于正确设定的检验，P(第 I 类错误) = α。例如，如果我们使用 5% 显著性水平，单尾临界值 z = 1.645，那么当 μ = μ₀ 时 Z > 1.645 的概率恰好是 0.05。第 I 类错误概率本身不需要额外计算，但你可能会被要求确认它，或给定临界值求 α。
However, in practical situations where the actual distributional assumptions are slightly off or the test is approximate, the true Type I error rate may differ slightly, but Edexcel questions assume the normal model holds perfectly.
然而，在实际情况下当实际分布假设略有偏离或检验是近似的时候，真实的第 I 类错误率可能会略有不同，但 Edexcel 题目假定正态模型完全成立。
4. Finding Type I Error Using the Normal Distribution | 利用正态分布求第I类错误
Suppose a test statistic T ~ N(μ, σ²/√n) under H₀. We define the critical region as values of X̄ greater than some threshold c. The Type I error probability is P(X̄ > c | μ = μ₀). This is found by standardising: P(Z > (c − μ₀)/(σ/√n)). The resulting value is α.
假设检验统计量 T ~ N(μ, σ²/√n) 在 H₀ 下。我们定义拒绝域为 X̄ 大于某个阈值 c 的值。第 I 类错误概率为 P(X̄ > c | μ = μ₀)。通过标准化求得：P(Z > (c − μ₀)/(σ/√n))。所得值即为 α。
In Edexcel problems, you might be given a critical value in terms of the original measurement units and asked to find the significance level. Always convert to the standard normal scale.
在 Edexcel 题目中，你可能得到用原始测量单位表示的临界值，并被要求求显著性水平。始终转换到标准正态尺度。
an weight is 500 g with σ = 4 g. A sample of 25 packets is taken. The critical region is X̄ > 501.2 g. Find α.
示例：一台机器分装糖包。在 H₀ 下，平均重量为 500 克，σ = 4 克。抽取 25 包的样本。拒绝域为 X̄ > 501.2 克。求 α。
z = (501.2 − 500) / (4/√25) = 1.2 / 0.8 = 1.5
P(Type I error) = P(Z > 1.5) = 1 − Φ(1.5) ≈ 0.0668.
P(第 I 类错误) = P(Z > 1.5) = 1 − Φ(1.5) ≈ 0.0668.
5. Type II Error and the Power of a Test | 第II类错误与检验的功效
c value for the true population parameter under H₁, say μ = μ₁. The probability β is the area under the distribution of X̄ when μ = μ₁ that falls within the acceptance region of the test.
当 H₀ 为假而检验统计量没有落入拒绝域时，发生第 II 类错误。要计算 β = P(第 II 类错误)，我们需要一个备择假设下真实总体参数的具体值，比如 μ = μ₁。β 是当 μ = μ₁ 时 X̄ 的分布落在检验接受域内的面积。
The power of the test is then 1 − β, which indicates the test's ability to detect a difference of a certain size. Edexcel questions often ask for the probability of a Type II error given a specific alternative mean.
检验的功效是 1 − β，它表明检验检测某一特定差异大小的能力。Edexcel 题目经常要求给定一个特定备择均值，计算第 II 类错误的概率。
6. General Method for Calculating Type II Error under Normality | 正态分布下计算第II类错误的一般方法
Step 1: Identify the critical value(s) for the test in terms of the sample mean X̄. For a one-tailed test with H₁: μ > μ₀, the critical value is c = μ₀ + zₐ × (σ/√n). For a two-tailed test, two critical values c₁ and c₂ are used.
第 1 步：确定检验统计量 X̄ 的临界值。对于 H₁: μ > μ₀ 的单尾检验，临界值为 c = μ₀ + zₐ × (σ/√n)。对于双尾检验，使用两个临界值 c₁ 和 c₂。
Step 2: Assume the true mean is μ₁ (where μ₁ satisfies H₁). Then X̄ ~ N(μ₁, σ²/n). Calculate the probability that X̄ falls in the acceptance region: β = P( X̄ < c | μ = μ₁ ) for a right-tailed test, or P( X̄ > c | μ = μ₁ ) for a left-tailed test. For two-tailed, β = P( c₁ < X̄ < c₂ | μ = μ₁ ).
第 2 步：假设真实均值为 μ₁（μ₁ 满足 H₁）。那么 X̄ ~ N(μ₁, σ²/n)。计算 X̄ 落在接受域内的概率：对于右尾检验，β = P( X̄ < c | μ = μ₁ )；对于左尾检验，β = P( X̄ > c | μ = μ₁ )。对于双尾检验，β = P( c₁ < X̄ < c₂ | μ = μ₁ )。
Step 3: Standardise X̄ using μ₁ and compute the probability from the standard normal table. The result is the Type II error probability.
第 3 步：使用 μ₁ 对 X̄ 标准化，并从标准正态表中计算概率。结果即为第 II 类错误概率。
It is crucial to use the correct mean when standardising: use μ₀ for Type I error, and μ₁ for Type II error. Mixing these up is a common mistake.
标准化时使用正确的均值至关重要：计算第 I 类错误用 μ₀，计算第 II 类错误用 μ₁。混淆两者是常见错误。
7. Worked Example: One-Tailed Test for a Mean | 实例分析：均值的单尾检验
g. A consumer group suspects the true mean loss is less than 5 kg and takes a sample of 36 members. They test at the 5% significance level. H₀: μ = 5, H₁: μ < 5. Find the critical value for X̄, and then calculate the Type II error probability if the true mean loss is 4.5 kg.
Edexcel 风格题目：一家健身房声称会员在其计划中平均减重 5 公斤。已知减重的标准差为 2.1 公斤。一个消费者组织怀疑真实平均减重少于 5 公斤，并抽取了 36 名会员的样本。他们在 5% 显著性水平下检验。H₀: μ = 5，H₁: μ < 5。求 X̄ 的临界值，然后计算如果真实平均减重为 4.5 公斤时第 II 类错误的概率。
e need the z-value such that P(Z < z) = 0.05. That is z = −1.645. The critical value for X̄: c = μ₀ + z × (σ/√n) = 5 + (−1.645) × (2.1/6) = 5 − 1.645 × 0.35 = 5 − 0.57575 = 4.42425 (approx 4.424).
首先求临界值。由于是左尾检验，我们需要 P(Z < z) = 0.05 的 z 值，即 z = −1.645。X̄ 的临界值：c = μ₀ + z × (σ/√n) = 5 + (−1.645) × (2.1/6) = 5 − 1.645 × 0.35 = 5 − 0.57575 = 4.42425（约 4.424）。
The acceptance region is X̄ > 4.424. Now, if μ = 4.5, X̄ ~ N(4.5, 2.1²/36) = N(4.5, 0.35²). The Type II error β is P(X̄ > 4.424 | − (1 − Φ(0.217)) = Φ(0.217) ≈ 0.586. The test has low power (≈0.414) for this alternative.
接受域为 X̄ > 4.424。现在，如果 μ = 4.5，X̄ ~ N(4.5, 2.1²/36) = N(4.5, 0.35²)。第 II 类错误 β = P(X̄ > 4.424 | μ = 4.5)。标准化：z = (4.424 − 4.5) / 0.35 = −0.076 / 0.35 ≈ −0.217。则 β = P(Z > −0.217) = 1 − Φ(−0.217) = 1 − (1 − Φ(0.217)) = Φ(0.217) ≈ 0.586。对于这个备择假设，检验的功效较低（≈0.414）。
8. Worked Example: Two-Tailed Test for a Mean | 实例分析：均值的双尾检验
on is 0.2 mm. A sample of 40 bolts is taken to test H₀: μ = 10 against H₁: μ ≠ 10 at the 5% ce region limits and then the probability of a Type II error if the true mean is 10.05 mm.
一个制造商声明螺栓直径为 10 毫米。标准差为 0.2 毫米。抽取 40 个螺栓的样本，在 5% 显著性水平下检验 H₀: μ = 10 对 H₁: μ ≠ 10。求接受域界限，然后计算如果真实均值为 10.05 毫米时第 II 类错误的概率。
For a two-tailed test at 5%, the critical z-values are ±1.96. The critical values for X̄: c₁ = 10 − 1.96 × (0.2/√40 162 ≈ 10 − 0.0620 = 9.9380; c₂ = 10 + 0.0620 = 10.0620. The acceptance region is 9.938 < X̄ < 10.062.
对于 5% 的双尾检验，临界 z 值为 ±1.96。X̄ 的临界值：c₁ = 10 − 1.96 × (0.2/√40) = 10 − 1.96 × 0.03162 ≈ 10 − 0.0620 = 9.9380；c₂ = 10 + 0.0620 = 10.0620。接受域为 9.938 < X̄ < 10.062。
Now assume μ = 10.05. Then X̄ ~ N(10.05, 0.2²/40) = N(10.05, 0.03162²). The Type II error β = P(9.938 < X̄ < 10.062 | n β = Φ(0.379) − Φ(−3.54). Φ(−3.54) is virtually 0, so β ≈ Φ(0.379) ≈ 0.648. The power is about 0.352.
现在假设 μ = 10.05。那么 X̄ ~ N(10.05, 0.2²/40) = N(10.05, 0.03162²)。第 II 类错误 β = P(9.938 < X̄ < 10.062 | μ = 10.05)。对两端标准化：z₁ = (9.938 − 10.05) / 0.03162 = −0.112 / 0.03162 ≈ −3.54；z₂ = (10.062 − 10.05) / 0.03162 = 0.012 / 0.03162 ≈ 0.379。则 β = Φ(0.379) − Φ(−3.54)。Φ(−3.54) 几乎为 0，因此 β ≈ Φ(0.379) ≈ 0.648。功效约为 0.352。
9. Factors Affecting Type I and Type II Errors | 影响第I类和第II类错误的因素
The Type I error probability is entirely determined by the chosen significance level α. You can reduce α to be more conservative, but this will increase β for a given alternative. Thus, α and β are inversely related for fixed sample size and effect size.
第 I 类错误概率完全由所选的显著性水平 α 决定。你可以降低 α 以更加保守，但这会导致对于给定备择假设的 β 增大。因此，在样本容量和效应量固定的情况下，α 和 β 是反向相关的。
Increasing sample size n shrinks the standard error σ/√n, making the sampling distribution of X̄ narrower. This reduces both α and β for a given critical value, or more practically, it allows smaller α and higher power simultaneously. Edexcel exam questions might ask you to explain how n affects errors.
增大样本容量 n 会缩小标准误 σ/√n，使 X̄ 的抽样分布更窄。对于给定的临界值，这同时降低了 α 和 β，或更实际地说，它允许更小的 α 和更高的功效。Edexcel 考试题可能会要求你解释 n 如何影响错误。
The size of the true difference (μ₁ − μ₀) also matters. A larger effect size separates the two distributions further, decreasing β for the same α. This is why a test may have low power for subtle effects.
真实差异的大小（μ₁ − μ₀）也很重要。更大的效应量会使两个分布拉开更大的距离，在相同的 α 下降低 β。这就是为什么检验对于细微的效应可能功效较低。
10. Visualising Errors with Normal Distributions | 用正态分布可视化错误
Drawing two overlapping bell curves—one centred at μ₀ (under H₀) and one at μ₁ (under H₁)—is extremely helpful. The critical value vertical line separates the acceptance and rejection regions. The area under the H₀ curve in the rejection region is α; the area under the H₁ curve in the acceptance region is β. Power is the rest of the area under H₁.
绘制两个重叠的钟形曲线——一个以 μ₀ 为中心（H₀ 下），另一个以 μ₁ 为中心（H₁ 下）——非常有帮助。临界值的垂直线分隔接受域和拒绝域。拒绝域中 H₀ 曲线下的面积为 α；接受域中 H₁ 曲线下的面积为 β。功效是 H₁ 曲线下其余部分的面积。
This visual approach clarifies why shifting the critical value to reduce α increases β, and why a larger sample size makes both curves taller and narrower, reducing overlap.
这种可视化方法阐明了为何移动临界值以降低 α 会增加 β，以及为何较大的样本量会使两条曲线都更高更窄，减少重叠。
11. Common Mistakes and Exam Tips | 常见错误与应试技巧
Confusing which mean (μ₀ or μ₁) to use when standardising is a typical error. Always ask: for Type I, assume H₀ true; for Type II, assume H₁ true. Also, remember that a Type II error can only be calculated for a specific alternative value, not for a vague 'H₁ is true'.
混淆在标准化时使用哪个均值（μ₀ 还是 μ₁）是典型错误。始终要问：对于第 I 类错误，假定 H₀ 为真；对于第 II 类错误，假定 H₁ 为真。另外记住，第 II 类错误只能针对一个具体的备择值来计算，不能针对模糊的“H₁ 为真”。
In two-tailed tests, ensure you use both acceptance boundaries when computing β. Check whether the question assumes the population variance is known or estimated; Edexcel usually gives σ known when dealing with normal errors, but if σ is estimated, the t-distribution applies, though error calculations with t are beyond the usual scope.
在双尾检验中，确保在计算 β 时使用两个接受域边界。检查题目是假设总体方差已知还是估计的；处理正态错误时 Edexcel 通常给出已知的 σ，但如果 σ 是估计的，则适用 t 分布，尽管使用 t 进行错误计算通常超出范围。
= β, Power = 1 − β. Always draw a diagram if possible; it will help you avoid sign errors.
使用清晰的符号：写出 P(第 I 类错误) = α，P(第 II 类错误) = β，功效 = 1 − β。如果可能，总是画一个图表；它有助于避免符号错误。
12. Summary and Key Takeaways | 总结与关键要点
- Type I error (α): Rejecting H₀ when it is true. Its probability is the significance level, easily found from critical values and the null distribution.
- 第 I 类错误 (α)：当 H₀ 为真时拒绝 H₀。其概率为显著性水平，容易通过临界值和原假设分布求得。
- Type II error (β): Failing to reject H₀ when it is false. Requires a specific alternative mean μ₁ and is computed as the probability that the sample mean falls in the acceptance region under that alternative distribution.
- 第 II 类错误 (β)：当 H₀ 为假时未能拒绝 H₀。需要一个特定的备择均值 μ₁，并计算在该备择分布下样本均值落入接受域的概率。
- Power = 1 − β, the probability of detecting a true effect.
- 功效 = 1 − β，即检测到真实效应的概率。
- Inversely related α and β; increasing n decreases both errors simultaneously.
- α 与 β 反向相关；增大 n 可同时降低两种错误。
- Standardisation is key: use μ₀ for α, μ₁ for β.
- 标准化是关键：求 α 用 μ₀，求 β 用 μ₁。
Mastering these calculations not only secures marks on direct error questions but also deepens your understanding of the logic behind hypothesis testing—an invaluable asset for the A-Level Edexcel exam.
掌握这些计算不仅能在直接的错误问题中得分，还能加深你对假设检验逻辑的理解——这对 A-Level Edexcel 考试来说是宝贵的财富。
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