📚 Hypothesis Testing for the Mean of a Normal Distribution with Unknown Variance | 方差未知时正态分布均值的假设检验
When conducting a hypothesis test for a population mean, we often face the realistic situation where the population variance σ² is unknown. In such cases, the z-test is no longer appropriate, and we must rely on the t-test, which uses the sample standard deviation s as an estimate of σ. This article covers the Edexcel A-Level Mathematics approach to performing a one-sample t-test for the mean of a normal distribution when the variance is unknown.
在对总体均值进行假设检验时，我们经常会遇到总体方差 σ² 未知的实际情形。此时 z 检验不再适用，必须转而使用 t 检验，用样本标准差 s 来估计 σ。本文介绍 Edexcel A-Level 数学中方差未知时对正态分布均值进行单样本 t 检验的方法。
1. When the Variance is Unknown | 当方差未知时
In many real-world scenarios, the population standard deviation σ is not given, and the sample size is small. Instead of the standard normal distribution, we use Student's t-distribution. This adjustment accounts for the extra uncertainty introduced by estimating σ with the sample standard deviation s. The t-test is more conservative, producing wider confidence intervals and larger p-values than the z-test for the same data, which reduces the chance of incorrectly rejecting the null hypothesis.
在许多实际情形中，总体标准差 σ 并未给出，且样本容量较小。此时我们使用学生 t 分布来代替标准正态分布。这一调整考虑了用样本标准差 s 估计 σ 所引入的额外不确定性。t 检验更为保守，对于相同的数据，其置信区间更宽、p 值更大，从而降低了错误拒绝原假设的风险。
2. Setting up Hypotheses | 建立假设
The null hypothesis H₀ always assumes that the population mean μ equals a specified value μ₀. The alternative hypothesis H₁ can be two-tailed (μ ≠ μ₀) or one-tailed (μ < μ₀ or μ > μ₀). The choice depends on the wording of the problem. For example, “test whether the mean has changed” suggests a two-tailed test, while “test whether the mean has increased” implies a right-tailed test. It is crucial to state the hypotheses in terms of population parameters, not sample statistics.
原假设 H₀ 总是假定总体均值 μ 等于某个指定值 μ₀。备择假设 H₁ 可以是双尾的 (μ ≠ μ₀) 或单尾的 (μ < μ₀ 或 μ > μ₀)。选择取决于题目表述。例如，“检验均值是否发生了变化” 提示双尾检验，而 “检验均值是否增加” 则意味着右侧单尾检验。务必用总体参数而非样本统计量来表述假设。
3. The t-distribution and Degrees of Freedom | t 分布与自由度
The test statistic follows a t-distribution with ν = n – n, especially for small degrees of freedom. As ν increases, the t-distribution approaches the standard normal distribution. In practice, when n > 30, the difference becomes negligible, but the t-test remains the theoretically correct procedure whenever σ is unknown.
检验统计量服从自由度为 ν = n – 1 的 t 分布，其中 n 为样本容量。t 分布对称且呈钟形，但尾部比正态分布更厚，尤其是在自由度较小时。随着 ν 增大，t 分布趋近于标准正态分布。在实践中，当 n > 30 时，差异变得可以忽略，但只要 σ 未知，t 检验在理论上始终是正确的做法。
4. Test Statistic for One-sample t-test | 单样本 t 检验的检验统计量
The test statistic t is calculated as:
检验统计量 t 的计算公式为：
t = (x̄ – μ₀) / (s / √n)
where x̄ is the sample mean, μ₀ is the hypothesised population mean, s is the sample standard deviation, and n is the sample size. This statistic measures how many standard errors the sample mean is away from μ₀. A large absolute value of t indicates a discrepancy between the data and the null hypothesis.
其中 x̄ 为样本均值，μ₀ 为假设的总体均值，s 为样本标准差，n 为样本容量。该统计量衡量样本均值与 μ₀ 相差多少个标准误。|t| 值很大表明数据与原假设之间存在较大差异。
5. Critical Values and Rejection Regions | 临界值与拒绝域
Using the significance level α (commonly 5% or 1%) and the degrees of freedom ν, we find the critical value(s) from t-tables. For a two-tailed test at significance level α, we reject H₀ if |t| > t(α/2, ν). For a one-tailed test, we compare t with t(α, ν) (or –t(α, ν)) depending on the direction. The rejection region is the set of values of the test statistic that lead to the rejection of H₀.
利用显著性水平 α（通常为 5% 或 1%）和自由度 ν，我们从 t 值表中查找临界值。对于显著性水平为 α 的双尾检验，若 |t| > t(α/2, ν)，则拒绝 H₀。对于单尾检验，根据方向将 t 与 t(α, ν)（或 –t(α, ν)）进行比较。拒绝域是导致拒绝 H₀ 的检验统计量取值的集合。
If the test statistic falls in the rejection region, we reject the null hypothesis; otherwise, we do not reject it. Remember that 'not rejecting' is not the same as 'accepting' H₀.
如果检验统计量落入拒绝域，我们拒绝原假设；否则，不拒绝原假设。请记住，“不拒绝” 并不等同于 “接受” H₀。
6. p-value Method | p 值方法
An alternative approach is to compute the p-value – than, the observed value, assuming H₀ is true. For a two-tailed test, p-value = 2 × P(T > |t| d provides more information, as it indicates how strong the evidence against H₀ really is.
另一种方法是计算 p 值——在 H₀ 为真的条件下，得到比观测值更极端或同样极端的检验统计量的概率。对于双尾检验，p 值 = 2 × P(T > |t|)；对于单尾检验，为 P(T > t) 或 P(T < t)。若 p 值小于 α，则拒绝 H₀。p 值方法提供了更多信息，因为它能表明反对 H₀ 的证据有多强。
In A-Level exams, p-values are often obtained from a calculator's t-distribution functions. Most calculators will report the p-value directly when you perform a t-test.
在 A-Level 考试中，p 值通常可通过计算器的 t 分布功能求得。大多数计算器在执行 t 检验时会直接给出 p 值。
7. Assumptions for the Test | 检验的假设条件
The one-sample t-test requires that the sample is drawn from a normally distributed population. If the population is not normal, the test is still robust for moderate departures from normality provided the sample size is sufficiently large (central limit theorem). However, for small samples, the normality assumption is critical. The data must also be independent observations. In exam questions, you may be asked to state these assumptions or comment on their validity based on given information.
单样本 t 检验要求样本来自正态分布的总体。如果总体非正态，但只要样本容量足够大（中心极限定理），该检验对中等程度的非正态性仍具有稳健性。然而，对于小样本，正态性假设至关重要。数据还必须是独立观测值。在考试题目中，可能要求你陈述这些假设，或根据给定信息评论其合理性。
8. Worked Example: Two-tailed Test | 例题：双尾检验
A machine fills bags of sugar labelled as 1 kg. A random sample of 10 bags has a mean mass of 1.005 kg and a standard deviation of 0.012 kg. Test, at the 5% significance level, whether the machine is over- or under-filling.
一台机器包装标示为 1 kg 的糖袋。随机抽取 10 袋，样本均值为 1.005 kg，标准差为 0.012 kg。在 5% 的显著性水平下检验该机器是否存在过量或不足填充。
Solution:
解：
H₀: μ = 1, H₁: μ ≠ 1 (two-tailed).
H₀: μ = 1, H₁: μ ≠ 1 (双尾)。
x̄ = 1.005, s = 0.012, n = 10, μ₀ = 1.
Test statistic: t = (1.005 – 1) / (0.012/√10) = 0.005 / (0.012/3.1623) ≈ 0.005 / 0.003795 = 1.317.
检验统计量：t = (1.005 – 1) / (0.012/√10) = 0.005 / (0.012/3.1623) ≈ 0.005 / 0.003795 = 1.317。
Degrees of freedom ν = 9. From t-table, two-tailed critical value at 5% is t(0.025, 9) = 2.262.
自由度 ν = 9。查 t 值表，5% 双尾临界值 t(0.025, 9) = 2.262。
Since |1.317| < 2.262, we do not reject H₀. There is insufficient evidence to suggest the mean fill is different from 1 kg.
由于 |1.317| < 2.262，我们不拒绝 H₀。没有足够证据表明平均填充量与 1 kg 有差异。
9. Worked Example: One-tailed Test | 例题：单尾检验
A teacher claims that her students' average score is above 70. A sample of 20 students yields a mean of 74.2 and a standard deviation of 8.1. Test the claim at the 1% significance level.
一位老师声称她学生的平均成绩高于 70。抽取 20 名学生，样本均值为 74.2，标准差为 8.1。在 1% 显著性水平下检验该声称。
H₀: μ = 70, H₁: μ > 70 (right-tailed).
H₀: μ = 70，H₁: μ > 70（右侧单尾）。
t = (74.2 – 70) / (8.1/√20) = 4.2 / (8.1/4.4721) = 4.2 / 1.811 = 2.319.
ν = 19. One-tailed critical value t(0.01, 19) ≈ 2.539 (from table).
ν = 19。单尾临界值 t(0.01, 19) ≈ 2.539（查表）。
Since 2.319 < 2.539, we do not reject H₀ at the 1% level. The evidence is not strong enough to support the teacher's claim at this strict significance level. (If we had used 5%, critical value would be 1.729 and we would reject H₀; this shows the importance of α.)
由于 2.319 < 2.539，在 1% 水平下我们不拒绝 H₀。证据不够有力支持老师的声称在这种严格的显著性水平下。（若采用 5%，临界值为 1.729，则拒绝 H₀；这显示了 α 的重要性。）
10. Using a Calculator | 使用计算器
Edexcel A-Level expects students to use scientific calculators with statistical functions. You can directly compute the t-statistic and p-value from sample data. Common models (Casio fx-991EX, TI-84) have a 't-test' function where you input the sample mean, sample standard deviation, sample size, and μ₀. The output provides t, p-value, and sometimes the confidence interval. Remember to choose the correct alternative hypothesis option.
Edexcel A-Level 要求学生使用具有统计功能的科学计算器。可以直接从样本数据计算 t 统计量和 p 值。常见机型（Casio fx-991EX、TI-84）具有 “t 检验” 功能，输入样本均值、样本标准差、样本容量和 μ₀。输出将提供 t、p 值，有时还有置信区间。记得选择正确的备择假设选项。
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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