📚 Hypothesis test for the mean of a normal distribution with unknown variance | 正态分布均值未知方差的假设检验
In statistical hypothesis testing, we often need to make inferences about a population mean μ. When the population is normally distributed and the population variance σ² is known, a z-test is used. However, in many practical situations σ² is unknown, and only the sample variance s² is available. In such cases, the t-test provides a more accurate method for testing the population mean, especially for small samples.
在统计假设检验中，我们经常需要对总体均值 μ 进行推断。当总体服从正态分布且总体方差 σ² 已知时，我们使用 z 检验。然而，在许多实际情况下 σ² 未知，只有样本方差 s² 可用。此时，t 检验为检验总体均值提供了更准确的方法，尤其是在小样本情况下。
1. Background of Hypothesis Testing | 假设检验的背景
Hypothesis testing is a formal procedure for deciding whether a claim about a population parameter is supported by sample evidence. We first state two competing hypotheses: the null hypothesis H₀, which represents the status quo or no effect, and the alternative hypothesis H₁, which reflects the claim we are testing. A test statistic is then calculated from the sample data, and we compare it with a critical value to make a decision.
假设检验是一种正式的统计程序，用于判断关于总体参数的声称是否得到样本证据的支持。我们首先提出两个对立的假设：原假设 H₀，表示现状或无效应；备择假设 H₁，反映我们正在检验的声称。然后根据样本数据计算检验统计量，并与临界值进行比较以做出决策。
The decision is based on a significance level α, typically 0.05 or 0.01, which defines the probability of rejecting H₀ when it is actually true. If the test statistic falls in the rejection region, we reject H₀ in favour of H₁; otherwise, we do not reject H₀.
决策基于显著性水平 α，通常取 0.05 或 0.01，它定义了当 H₀ 实际为真时拒绝它的概率。如果检验统计量落入拒绝域，我们就拒绝 H₀ 并接受 H₁；否则，我们无法拒绝 H₀。
2. Why Unknown Variance Requires a Different Approach | 为什么未知方差需要不同方法
When σ² is known, the sample mean x̄ follows a normal distribution and the test statistic z = (x̄ - μ₀)/(σ/√n) is standard normal. But when σ² is unknown, we replace σ with the sample standard deviation s. This substitution introduces extra uncertainty because s is itself a random variable that varies from sample to sample.
当 σ² 已知时，样本均值 x̄ 服从正态分布，检验统计量 z = (x̄ - μ₀)/(σ/√n) 为标准正态分布。但 σ² 未知时，我们用样本标准差 s 代替 σ。这种替代引入了额外的不确定性，因为 s 本身就是一个随机变量，会随样本的不同而变化。
Using s instead of σ changes the sampling distribution. The resulting test statistic no longer follows a standard normal distribution, especially when the sample size n is small. Instead, it follows a Student's t-distribution with n - 1 degrees of freedom. The t-distribution has heavier tails than the normal distribution, which compensates for the additional variability in estimating σ.
用 s 代替 σ 会改变抽样分布。由此得到的检验统计量不再服从标准正态分布，尤其是当样本容量 n 较小时。它服从具有 n - 1 个自由度的学生 t 分布。t 分布的尾部比正态分布更厚，从而补偿了估计 σ 时带来的额外变异性。
3. The t-Distribution | t 分布
Student' of freedom, denoted ν (nu). For a sample of size n, the degrees of freedom are ν = n - 1.
学生 t 分布是一种连续概率分布，它对称且呈钟形，类似于标准正态分布，但尾部更厚。其具体形状取决于自由度，记为 ν（nu）。对于大小为 n 的样本，自由度为 ν = n - 1。
wo distributions are nearly identical, which is why z-tests are often used for large samples even when σ is unknown. However, for small samples, using the t-distribution is essential to avoid misleading conclusions.
随着 ν 增大，t 分布趋近于标准正态分布。当 n 较大时（如 n ≥ 30），两者几乎相同，这就是为什么在大样本下即使 σ 未知也常使用 z 检验。但在小样本情况下，使用 t 分布对于避免误导性结论至关重要。
hese critical values are available in t-tables published by Edexcel and other exam boards.
我们用 tₕ 表示临界值，其中下标 h 表示自由度。例如，t₉₀.₀₅ 表示当 ν = 9 时上尾面积为 0.05 的临界值。这些临界值可在 Edexcel 和其他考试局发布的 t 分布表中查到。
4. Setting Up Hypotheses | 建立假设
Before conducting a t-test, we must clearly define H₀ and H₁. The null hypothesis always specifies a specific value for the population mean, denoted μ₀. The alternative hypothesis can be one-sided or two-sided depending on the research question.
在进行 t 检验之前，我们必须明确 H₀ 和 H₁。原假设总是为总体均值指定一个特定值，记为 μ₀。备择假设可以是单尾或双尾，取决于研究问题。
: μ < 800. This is a left-tailed test. Conversely, if we want to test whether the mean has increased, we use H₁: μ > μ₀ (right-tailed test).
对于单尾检验，备择假设是 μ > μ₀ 或 μ < μ₀。例如，如果制造商声称电池平均寿命至少为 800 小时，我们可能检验 H₀: μ = 800 对 H₁: μ < 800。这是左尾检验。相反，如果我们想检验均值是否增加，则使用 H₁: μ > μ₀（右尾检验）。
For a two-tailed test, the alternative is μ ≠ μ₀. This is used when we only want to detect a difference in either direction. The significance level α is split equally between the two tails: α/2 in each tail.
对于双尾检验，备择假设为 μ ≠ μ₀。当我们只想检测任意方向的差异时使用。显著性水平 α 被平均分配到两个尾部：每尾 α/2。
ains the inequality (<, >, or ≠). The hypotheses must be set before looking at the data to ensure objectivity.
重要的是，H₀ 总是包含等号（=），而 H₁ 包含不等号（<、> 或 ≠）。为了避免主观性，必须在查看数据之前设定假设。
5. The Test Statistic | 检验统计量
When the population variance σ² is unknown, the test statistic for testing a hypothesis about the population mean μ is calculated as:
当总体方差 σ² 未知时，检验总体均值 μ 的检验统计量计算如下：
t = (x̄ − μ₀) / (s / √n)
where x̄ is the sample mean, μ₀ is the hypothesized population mean under H₀, s is the sample standard deviation, and n is the sample size. The statistic t follows a t-distribution with n − 1 degrees of freedom.
其中 x̄ 是样本均值，μ₀ 是 H₀ 中假设的总体均值，s 是样本标准差，n 是样本容量。统计量 t 遵循自由度为 n − 1 的 t 分布。
Notice that the formula has the same structure as the z-test statistic, but with s replacing σ. The denominator s/√n is the standard error of the sample mean when σ is unknown. Because s varies from sample to sample, the distribution of t has heavier tails than the standard normal distribution.
注意，这个公式与 z 检验统计量结构相同，只是用 s 代替了 σ。分母 s/√n 是 σ 未知时样本均值的标准误。由于 s 因样本而异，t 的分布尾部比标准正态分布更厚。
When calculating s, use the sample standard deviation with n − 1 in the denominator:
计算 s 时，应使用分母为 n − 1 的样本标准差：
s = √[ Σ(xᵢ − x̄)² / (n − 1) ]
This is an unbiased estimator of the population standard deviation σ.
这是总体标准差 σ 的无偏估计量。
6. Degrees of Freedom and t-Table | 自由度与 t 分布表
ent the number of independent pieces of information used to estimate the sample variance. In the t-table, the rows correspond to different degrees of freedom, and the columns correspond to the right-tail probabilities.
单样本 t 检验的自由度为 ν = n − 1。它们代表用于估计样本方差的独立信息数量。在 t 分布表中，行对应不同的自由度，列对应右尾概率。
= 0.05 and ν = 10, the critical value is t₁₀,₀.₀₅ ≈ 1.812. This means P(t > 1.812) = 0.05.
对于显著性水平为 α 的单尾检验，我们查找标记为 α 的列。例如，α = 0.05 且 ν = 10 时，临界值为 t₁₀,₀.₀₅ ≈ 1.812。这意味着 P(t > 1.812) = 0.05。
α/2 = 0.025, giving t₁₀,₀.₀₂₅ ≈ 2.228. The rejection regions are t < −2.228 and t > 2.228.
对于双尾检验，我们将显著性水平拆分：右尾概率为 α/2。α = 0.05 且 ν = 10 时，使用 α/2 = 0.025 列，得到 t₁₀,₀.₀₂₅ ≈ 2.228。拒绝域为 t < −2.228 和 t > 2.228。
When the degrees of freedom are large (ν > 30), the t-table entries become close to the corresponding z-values. In the limit, the t-distribution is exactly the standard normal distribution.
当自由度较大（ν > 30）时，t 分布表中的值接近相应的 z 值。极限情况下，t 分布就是标准正态分布。
7. Carrying Out the Test: Steps | 进行检验的步骤
The t-test for the mean of a normal distribution with unknown variance can be summarised in the following steps:
未知方差的正态分布均值 t 检验可归纳为以下步骤：
Step 1: State the hypotheses. Write down H₀ and H₁ in terms of μ₀ the significance level α.
步骤 1：建立假设。写出 H₀ 和 H₁，明确 μ₀ 和显著性水平 α。
Step 2: Calculate the test statistic. Compute the sample mean x̄, sample standard deviation s, and then t = (x̄ − μ₀)/(s/√n).
步骤 2：计算检验统计量。计算样本均值 x̄、样本标准差 s，然后 t = (x̄ − μ₀)/(s/√n)。
Step 3: Determine the critical value. se . Use the t-table with ν = n − 1 to find the critical value(s) for the given α and tail type.
步骤 3：确定临界值。使用 t 分布表，根据 ν = n − 1 查找给定 α 和单/双尾的临界值。
Step 4: Compare and conclude. If the absolute value of the test statistic exceeds the critical value, reject H₀; otherwise, do not reject H₀. State the conclusion in the context of the problem.
步骤 4：比较并得出结论。如果检验统计量的绝对值超过临界值，则拒绝 H₀；否则不拒绝 H₀。并结合问题背景陈述结论。
Always use clear notation and show your substitution of values. In an exam, full marks require every step to be visible.
始终使用清晰的符号并展示代入过程。在考试中，要拿到满分必须展示每一步。
8. Example 1: One-Tailed t-Test | 示例 1：单尾 t 检验
A company claims that the average lifetime of its light bulbs is at least 800 hours. A random sample of 10 bulbs has a mean lifetime of 790 hours and a standard deviation of 20 hours. Test the claim at the 5% significance level, assuming the lifetimes are normally distributed.
某公司声称其灯泡的平均寿命至少为 800 小时。随机抽取 10 只灯泡，样本平均寿命为 790 小时，标准差为 20 小时。在 5% 显著性水平下检验该宣称，假设寿命服从正态分布。
Solution: We test H₀: μ = 800 against H₁: μ < 800 (left-tailed test). Here n = 10, x̄ = 790, s = 20, μ₀ = 800.
解答：我们检验 H₀: μ = 800 对 H₁: μ < 800（左尾检验）。这里 n = 10，x̄ = 790，s = 20，μ₀ = 800。
Step 2: Compute the test statistic:
步骤 2：计算检验统计量：
t = (790 − 800) / (20 / √10) = −10 / (20 / 3.1623) = −10 / 6.3246 ≈ −1.581
is t₉,₀.₀₅ = 1.833. Since this is a left-tailed test, the rejection region is t < −1.833.
步骤 3：自由度 ν = 10 − 1 = 9。对于 α = 0.05 的单尾检验，临界值为 t₉,₀.₀₅ = 1.833。由于这是左尾检验，拒绝域为 t < −1.833。
Step 4: Because −1.581 > −1.833, the test statistic is not in the rejection region. Therefore, we do not reject H₀. There is insufficient evidence at the 5% level to conclude that the average lifetime is less than 800 hours.
步骤 4：因为 −1.581 > −1.833，检验统计量不在拒绝域中。因此，我们不拒绝 H₀。在 5% 显著性水平下，没有足够证据表明平均寿命小于 800 小时。
Note: The claim was "at least 800 hours," so failing to reject H₀ means the sample evidence is consistent with the claim. It does not prove the claim is true, but it does not contradict it.
注意：宣称是“至少 800 小时”，因此未能拒绝 H₀ 意味着样本证据与宣称一致。这并不能证明宣称正确，但也不与其矛盾。
9. Example 2: Two-Tailed t-Test | 示例 2：双尾 t 检验
A food manufacturer states that the average weight of a bag of chips is 500 g. To verify this, a random sample of 25 bags is taken, giving a mean weight of 502 g and a standard deviation of 6 g. Is there evidence at the 10% significance level to suggest the average weight differs from 500 g?
一家食品制造商声称每袋薯片的平均重量为 500 g。为验证这一说法，随机抽取 25 袋，平均重量为 502 g，标准差为 6 g。在 10% 显著性水平下，是否有证据表明平均重量与 500 g 有差异？
Solution: We test H₀: μ = 500 against H₁: μ ≠ 500 (two-tailed). Here n = 25, x̄ = 502, s = 6, μ₀ = 500.
解答：我们检验 H₀: μ = 500 对 H₁: μ ≠ 500（双尾）。这里 n = 25，x̄ = 502，s = 6，μ₀ = 500。
Compute the test statistic:
计算检验统计量：
t = (502 − 500) / (6 / √25) = 2 / (6 / 5) = 2 / 1.2 ≈ 1.667
l. From the t-table, t₂₄,₀.₀₅ ≈ 1.711. The rejection regions are t < −1.711 and t > 1.711.
自由度 ν = 25 − 1 = 24。对于 α = 0.10 的双尾检验，右尾使用 α/2 = 0.05。从 t 分布表查得 t₂₄,₀.₀₅ ≈ 1.711。拒绝域为 t < −1.711 和 t > 1.711。
Since 1.667 < 1.711, the test statistic is not in the rejection region. Therefore, we do not reject H₀. There is insufficient evidence at the 10% level to conclude that the average bag weight differs from 500 g.
由于 1.667 < 1.711，检验统计量不在拒绝域中。因此，我们不拒绝 H₀。在 10% 显著性水平下，没有足够证据表明袋装薯片的平均重量与 500 g 有差异。
This example illustrates that a two-tailed test requires a larger absolute critical value, making it harder to reject H₀ for a given level of significance.
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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