📚 A-Level Further Mathematics: Hypothesis Testing for the Mean with Unknown Variance | A-Level 进阶数学：方差未知时均值的假设检验
In A-Level Further Mathematics, hypothesis tests for a population mean are usually introduced with the simplifying assumption that the population variance is known. In real statistical work, however, the population variance is almost never known. When the variance is unknown, we must estimate it from the sample, and the test statistic no longer follows a Normal distribution. Instead, it follows a Student’s t-distribution.
在 A-Level 进阶数学中，均值假设检验通常先假设总体方差已知，以便简化计算。但在实际统计工作中，总体方差几乎总是未知的。当方差未知时，我们必须用样本估计方差，此时检验统计量不再服从正态分布，而是服从 Student 的 t 分布。
1. The Problem with Unknown Variance | 方差未知带来的问题
When the population variance σ² is known, the test statistic for testing a mean is z = (x̄ – μ₀)/(σ/√n), which follows a standard Normal distribution. We can then compare z with critical values such as 1.645 or 1.960. The Normal distribution works because σ is a fixed, known constant.
当总体方差 σ² 已知时，检验均值的统计量为 z = (x̄ – μ₀)/(σ/√n)，它服从标准正态分布。由于 σ 是已知的固定常数，我们可以直接将 z 与 1.645、1.960 等临界值进行比较，正态分布可以直接使用。
If σ² is unknown, we cannot calculate this z statistic. The natural idea is to replace σ with the sample standard deviation s. However, s itself is a random variable: it changes from sample to sample. Using s in place of σ introduces extra uncertainty, so the resulting statistic has a distribution with heavier tails than the Normal distribution.
当 σ² 未知时，我们无法计算上述 z 统计量。自然的想法是用样本标准差 s 代替 σ。但 s 本身是随机变量，会随着样本改变而改变。用 s 代替 σ 会引入额外的不确定性，因此所得统计量的分布比正态分布有更厚的尾部。
For a sample of size n, the sample variance is defined as s² = Σ(xᵢ – x̄)²/(n – 1). The denominator n – 1 is the number of degrees of freedom associated with the estimate. This is not just a technical detail; it directly affects the shape of the sampling distribution we use for testing.
对于容量为 n 的样本，样本方差定义为 s² = Σ(xᵢ – x̄)²/(n – 1)。其中 n – 1 是该估计量所对应的自由度。这不仅是技术细节，它直接决定检验时所使用的抽样分布的形状。
2. Student’s t-Distribution and Degrees of Freedom | Student 的 t 分布与自由度
The statistic T = (x̄ – μ₀)/(s/√n) follows a Student’s t-distribution with n – 1 degrees of freedom, provided that the population is Normally distributed. The t-distribution is symmetric and bell-shaped, like the standard Normal distribution, but it has more probability in the tails.
统计量 T = (x̄ – μ₀)/(s/√n) 服从自由度为 n – 1 的 Student 的 t 分布，前提是总体服从正态分布。t 分布是对称的钟形分布，与标准正态分布相似，但其尾部概率更大。
As n increases, s becomes a more reliable estimate of σ, and the t-distribution approaches the standard Normal distribution. For small sample sizes, particularly n < 30, the difference is noticeable and cannot be ignored.
随着 n 增大，s 对 σ 的估计越来越可靠，t 分布也越来越接近标准正态分布。当样本量较小，尤其是 n < 30 时，二者之间的差别非常明显，不可忽略。
T = (x̄ – μ₀)/(s/√n) ~ t(n – 1)
The only extra parameter in this distribution is the degrees of freedom. In a one-sample t-test we write degrees of freedom as ν = n – 1. The notation t(ν) means the t-distribution with ν degrees of freedom.
该分布唯一的额外参数是自由度。在单样本 t 检验中，自由度记为 ν = n – 1。记号 t(ν) 表示自由度为 ν 的 t 分布。
3. Hypotheses and the Test Statistic | 原假设、备择假设与检验统计量
As with any hypothesis test, we begin by stating a null hypothesis and an alternative hypothesis. For a test about a population mean μ, the null hypothesis is usually H₀: μ = μ₀, where μ₀ is a claimed value. The alternative hypothesis is one of:
与所有假设检验一样，我们首先写出原假设和备择假设。对于总体均值 μ 的检验，原假设通常为 H₀: μ = μ₀，其中 μ₀ 是被声称的数值。备择假设为以下三种之一：
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H₁: μ ≠ μ₀ for a two-tailed test;
H₁: μ ≠ μ₀，用于双尾检验；
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H₁: μ > μ₀ for a one-tailed upper-tail test;
H₁: μ > μ₀，用于单尾右侧检验；
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H₁: μ < μ₀ for a one-tailed lower-tail test.
H₁: μ < μ₀，用于单尾左侧检验。
After collecting a random sample of size n, we calculate the sample mean x̄ and the sample standard deviation s. The test statistic is:
在收集到容量为 n 的随机样本后，我们计算样本均值 x̄ 和样本标准差 s。检验统计量为：
t = (x̄ – μ₀)/(s/√n)
This t value measures how many standard errors the sample mean is away from the claimed mean μ₀. If the null hypothesis is true, t should usually be close to zero. A t value far away from zero provides evidence against H₀.
这个 t 值衡量样本均值与声称均值 μ₀ 相差多少个标准误。如果原假设为真，t 通常应接近 0。如果 t 值远远偏离 0，则提供了拒绝 H₀ 的证据。
4. Critical Values, Significance Level and Rejection Regions | 临界值、显著性水平与拒绝域
The significance level α is the probability of rejecting H₀ when it is actually true. Common levels in A-Level Further Mathematics are α = 0.05, α = 0.01 and α = 0.10. We compare our test statistic with a critical value from the t-table.
显著性水平 α 是原假设实际为真时却拒绝原假设的概率。A-Level 进阶数学中常用的显著性水平为 α = 0.05、α = 0.01 和 α = 0.10。我们需要将检验统计量与 t 分布表中的临界值进行比较。
For a one-tailed test with significance level α, the critical value is the value c such that P(T > c) = α for an upper-tail test, or P(T < -c) = α for a lower-tail test. For a two-tailed test, each tail contains α/2 probability.
对于单尾检验，显著性水平为 α 时，临界值 c 满足：右侧检验中 P(T > c) = α，左侧检验中 P(T < -c) = α。对于双尾检验，每个尾部各包含 α/2 的概率。
In a t-table, the degrees of freedom appear in the left column. The column headings are often labelled as the upper-tail probability p. For a two-tailed test at α = 0.05, we must read the column for p = 0.025, because each tail has probability 0.025.
在 t 分布表中，左侧列通常为自由度，列标题为右侧尾部概率 p。对于 α = 0.05 的双尾检验，我们必须读取 p = 0.025 那一列，因为每个尾部概率为 0.025。
| Degrees of freedom ν | p = 0.10 | p = 0.05 | p = 0.025 | p = 0.01 |
| 9 | 1.383 | 1.833 | 2.262 | 2.821 |
| 15 | 1.341 | 1.753 | 2.131 | 2.602 |
| 24 | 1.318 | 1.711 | 2.064 | 2.492 |
We reject H₀ if the observed test statistic lies in the critical region. In a one-tailed upper-tail test, we reject if t > critical value. In a one-tailed lower-tail test, we reject if t < -critical value. In a two-tailed test, we reject if |t| > critical value.
若观测到的检验统计量落在拒绝域内，则拒绝 H₀。在右侧单尾检验中，拒绝条件为 t > 临界值；在左侧单尾检验中，拒绝条件为 t < -临界值；在双尾检验中，拒绝条件为 |t| > 临界值。
5. Worked Example 1: One-Tailed Test | 例题一：单尾检验
A manufacturer claims that a new type of battery has a mean lifetime of 50 hours. A sample of 10 batteries is selected at random. The sample mean is 49.7 hours and the sample standard deviation is 0.8 hours. Assume the battery lifetimes are Normally distributed. Test at the 5% significance level whether the mean lifetime is actually less than 50 hours.
某制造商声称一种新型电池的平均寿命为 50 小时。现随机抽取 10 节电池，测得样本均值为 49.7 小时，样本标准差为 0.8 小时。假设电池寿命服从正态分布。在 5% 显著性水平下检验平均寿命是否真的小于 50 小时。
Step 1: State the hypotheses.
第一步：写出假设。
H₀: μ = 50, H₁: μ < 50
Step 2: Calculate the test statistic. We have n = 10, x̄ = 49.7, s = 0.8 and μ₀ = 50.
第二步：计算检验统计量。已知 n = 10，x̄ = 49.7，s = 0.8，μ₀ = 50。
t = (49.7 – 50)/(0.8/√10) = -0.3/0.25298 = -1.186
Step 3: Find the critical value. For a one-tailed test at α = 0.05 with ν = 9, the critical value is 1.833. Since this is a lower-tail test, the critical region is t < -1.833.
第三步：查找临界值。对于 α = 0.05 的单尾检验，自由度 ν = 9，临界值为 1.833。由于这是左侧检验，拒绝域为 t < -1.833。
Step 4: Compare and conclude. The observed t = -1.186 is greater than -1.833, so it does not lie in the critical region. We do not reject H₀. There is insufficient evidence to conclude that the mean lifetime is less than 50 hours.
第四步：比较并下结论。观测值 t = -1.186 大于 -1.833，因此不落在拒绝域内。我们不能拒绝 H₀。没有足够证据表明平均寿命小于 50 小时。
6. Worked Example 2: Two-Tailed Test | 例题二：双尾检验
A coffee machine is designed to fill cups with a mean volume of 120 ml. A random sample of 16 cups is taken. The sample mean volume is 118.5 ml and the sample standard deviation is 3.2 ml. Assuming the fill volumes are Normally distributed, test at the 5% significance level whether the mean volume differs from 120 ml.
某咖啡机设计的目标是每杯平均填充 120 ml。现随机抽取 16 杯，测得样本均值为 118.5 ml，样本标准差为 3.2 ml。假设填充体积服从正态分布，在 5% 显著性水平下检验平均体积是否与 120 ml 不同。
Step 1: State the hypotheses.
第一步：写出假设。
H₀: μ = 120, H₁: μ ≠ 120
Step 2: Calculate the test statistic. Here n = 16, x̄ = 118.5, s = 3.2 and μ₀ = 120.
第二步：计算检验统计量。这里 n = 16，x̄ = 118.5，s = 3.2，μ₀ = 120。
t = (118.5 – 120)/(3.2/√16) = -1.5/0.8 = -1.875
Step 3: Find the critical value. For a two-tailed test at α = 0.05, each tail has probability 0.025. With ν = 15, the critical value is 2.131. Therefore we reject H₀ if |t| > 2.131.
第三步：查找临界值。对于 α = 0.05 的双尾检验，每个尾部概率为 0.025。自由度 ν = 15 时临界值为 2.131。因此当 |t| > 2.131 时拒绝 H₀。
Step 4: Compare and conclude. Since |t| = 1.875 is less than 2.131, we do not reject H₀. There is not enough evidence to say that the mean volume differs from 120 ml.
第四步：比较并下结论。因为 |t| = 1.875 小于 2.131，我们不能拒绝 H₀。没有足够证据表明平均体积不等于 120 ml。
7. The p-Value Approach | p 值方法
Instead of comparing the test statistic with a critical value, we can calculate the p-value. The p-value is the probability, assuming H₀ is true, of observing a test statistic as extreme as, or more extreme than, the one actually observed. In a one-tailed test, the p-value is the area in one tail. In a two-tailed test, the p-value is the sum of the areas in both tails.
除了将检验统计量与临界值比较，我们还可以计算 p 值。p 值是在原假设为真的前提下，观察到与当前统计量同样极端或更极端的检验统计量的概率。单尾检验中，p 值是一个尾部的面积；双尾检验中，p 值是两侧尾部面积之和。
A large p-value means the observed data are consistent with H₀. A small p-value, usually less than or equal to α, means the data are unlikely under H₀ and we reject H₀. The p-value and the critical-value method always give the same conclusion.
p 值较大说明观测数据与原假设一致。p 值较小，通常小于或等于 α，说明在原假设下出现该数据较为罕见，因此拒绝 H₀。p 值法与临界值法得出的结论总是一致的。
In examinations, you may be asked to estimate a p-value using a t-table. For example, in the battery question we found t = -1.186 with 9 degrees of freedom. Since 1.186 is less than 1.383, the one-tailed p-value is greater than 0.10. Because 0.10 > 0.05, we do not reject H₀.
考试中，你可能需要利用 t 分布表估计 p 值。例如，在电池问题中我们得到 t = -1.186，自由度为 9。由于 1.186 小于 1.383，因此单尾 p 值大于 0.10。因为 0.10 > 0.05，所以不拒绝 H₀。
8. Conditions and Assumptions | 使用条件与假设
The t-test for the mean is only valid when certain conditions are met. First, the sample must be a random sample, so that the observations are independent of one another. Second, the population must be approximately Normally distributed, especially when the sample size is small. For larger samples, the Central Limit Theorem makes the test more robust to non-Normality.
均值 t 检验只有在满足一定条件时才有效。首先，样本必须是随机样本，以保证观测值相互独立。其次，总体应近似服从正态分布，尤其是当样本量较小时。对于较大样本，中心极限定理使检验对非正态性更加稳健。
Third, the population variance is unknown, but the sample standard deviation s must be calculated correctly using n – 1 in the denominator. Using the biased estimator with n in the denominator is a common error that leads to wrong conclusions.
第三，总体方差未知，但样本标准差 s 必须用 n – 1 作为分母正确计算。使用 n 作分母的有偏估计是常见错误，会导致错误结论。
Finally, when n is very small, the t-test relies heavily on the Normality assumption. Outliers can also distort both x̄ and s, so the data should be inspected carefully before applying the test.
最后，当 n 很小时，t 检验对正态性假设的依赖性很强。异常值会同时影响 x̄ 和 s，因此在应用检验前应仔细检查数据。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
One common mistake is using the Normal distribution when the variance is unknown. A student may calculate s and then proceed as if the statistic were z. This is only acceptable when n is large enough for the t-distribution to be almost identical to the Normal distribution, but in an exam you should always use t when σ is unknown.
常见错误之一是在方差未知时使用正态分布。有的学生计算出 s 后仍将其统计量当作 z。只有当 n 很大，使 t 分布几乎与正态分布相同时才可以这样处理；但在考试中，只要 σ 未知，就应该使用 t 分布。
Another mistake is using the wrong degrees of freedom. For a one-sample t-test, degrees of freedom are n – 1, not n. Using the wrong row in the table produces the wrong critical value.
另一个错误是使用错误的自由度。对于单样本 t 检验，自由度是 n – 1，而不是 n。在表中查错行会得到错误的临界值。
Students also sometimes confuse one-tailed and two-tailed critical columns. For a two-tailed test at α = 0.05, always look up the column p = 0.025. For a one-tailed test at α = 0.05, look up p = 0.05.
学生也常混淆单尾和双尾检验的临界值列。对于 α = 0.05 的双尾检验，应查找 p = 0.025 的列；对于 α = 0.05 的单尾检验，应查找 p = 0.05 的列。
Finally, do not say ‘accept H₀’. We only say ‘do not reject H₀’. A failure to reject H₀ does not prove that μ equals μ₀; it only means the evidence against H₀ is not strong enough.
最后，不要使用”接受 H₀”的说法。我们只能说”不拒绝 H₀”。未能拒绝 H₀ 并不证明 μ 就等于 μ₀，只说明反对 H₀ 的证据还不够强。
10. Summary | 总结
When the population variance is unknown, the one-sample t-test is the correct procedure for testing a population mean. The test statistic is t = (x̄ – μ₀)/(s/√n), and it follows a t-distribution with n – 1 degrees of freedom under H₀, provided the sample is random and the population is approximately Normal.
当总体方差未知时，单样本 t 检验是检验总体均值的正确方法。检验统计量为 t = (x̄ – μ₀)/(s/√n)，在原假设成立条件下服从自由度为 n – 1 的 t 分布，前提是样本随机且总体近似正态。
To carry out the test, state H₀ and H₁, calculate t, determine the critical value from a t-table, and decide whether the observed t lies in the rejection region. Alternatively, estimate the p-value and compare it with α.
进行检验时，先写出 H₀ 和 H₁，计算 t 值，通过 t 分布表确定临界值，然后判断观测 t 值是否落在拒绝域内。也可以估计 p 值并与 α 比较。
The t-test is a fundamental tool in A-Level Further Mathematics and in real-world statistics. Understanding why it is needed, how it differs from the z-test, and how to apply it correctly will help you succeed in exams and in future statistical work.
t 检验是 A-Level 进阶数学以及实际统计中的基本工具。理解为什么需要 t 检验、它与 z 检验有何不同，以及如何正确应用，将帮助你在考试和未来的统计工作中取得成功。
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