Hypothesis Testing for the Mean of a Poisson Distribution | 泊松分布均值的假设检验方法

📚 Hypothesis Testing for the Mean of a Poisson Distribution | 泊松分布均值的假设检验方法

In A-Level Further Mathematics (Edexcel), hypothesis testing for the mean of a Poisson distribution is a core topic in statistical inference. It allows us to make decisions about an unknown population mean λ (lambda) based on sample data, using the elegant properties of the Poisson model.

在爱德思（Edexcel）进阶数学中，对泊松分布均值进行假设检验是统计推断的核心内容。它使我们能够利用泊松模型的优美性质，基于样本数据对未知的总体均值 λ（lambda）做出决策。

1. The Poisson Distribution and Its Parameter λ | 泊松分布及其参数 λ

The Poisson distribution models the number of events occurring in a fixed interval of time or space, when events happen independently at a constant average rate. If a random variable X follows a Poisson distribution with mean λ, we write X ~ Po(λ). The probability mass function is given by P(X = x) = e⁻λ λˣ / x!, for x = 0, 1, 2, …

泊松分布用于模拟在固定时间或空间区间内事件发生的次数，要求事件独立发生且平均速率恒定。若随机变量 X 服从均值为 λ 的泊松分布，记作 X ~ Po(λ)。其概率质量函数为 P(X = x) = e⁻λ λˣ / x!，其中 x = 0, 1, 2, …

An important property of the Poisson distribution is that both the mean and the variance are equal to λ. This single parameter completely determines the distribution, making hypothesis testing particularly elegant.

泊松分布的一个重要性质是均值和方差都等于 λ。这个单一参数完全决定了分布的形状，使得假设检验格外优雅简洁。

Typical examples include: the number of accidents at a junction per week, the number of customers arriving at a shop per hour, or the number of defects in a metre of fabric.

典型例子包括：某个路口每周的事故次数、每小时到达商店的顾客人数，或者每米布料上的瑕疵数量。

2. Setting Up Hypotheses | 建立假设

As with all hypothesis tests, we begin by stating a null hypothesis and an alternative hypothesis. The null hypothesis, denoted H₀, always takes the form of an equality: H₀ : λ = λ₀, where λ₀ is a known value specified in the question.

与所有假设检验一样，我们首先要陈述原假设和备择假设。原假设记为 H₀，总是一个等式：H₀ : λ = λ₀，其中 λ₀ 是题目中给出的已知值。

The alternative hypothesis, denoted H₁, can take one of three forms depending on the context of the problem:

备择假设记为 H₁，根据问题的实际情况可以取三种形式之一：

H₁ : λ ≠ λ₀ (two-tailed test) – used when we are testing for a change in either direction.

H₁ : λ ≠ λ₀（双尾检验）——用于检验两个方向上的变化。
H₁ : λ > λ₀ (one-tailed test) – used when we suspect the mean has increased.

H₁ : λ > λ₀（单尾检验）——用于怀疑均值增大的情况。
H₁ : λ < λ₀ (one-tailed test) – used when we suspect the mean has decreased.

H₁ : λ < λ₀（单尾检验）——用于怀疑均值减小的情况。

The choice between a one-tailed and two-tailed test must be made before looking at the data, based on the wording of the question. Words like ‘increased’, ‘decreased’, ‘more than’, ‘fewer than’ suggest one-tailed tests, while ‘changed’, ‘different from’, ‘not equal to’ suggest a two-tailed test.

选择单尾还是双尾检验必须在查看数据之前做出，依据是题目的措辞。像“增加”“减少”“多于”“少于”这样的词暗示单尾检验，而“改变”“不同于”“不等于”则暗示双尾检验。

3. The Test Statistic and Its Distribution | 检验统计量及其分布

In a Poisson hypothesis test, we collect a single observation x from the random variable X ~ Po(λ). This observed count x is our test statistic. Under the null hypothesis, we assume that X ~ Po(λ₀), and we use this assumption to calculate probabilities.

在泊松假设检验中，我们从随机变量 X ~ Po(λ) 中收集一个观测值 x。这个观测计数 x 就是我们的检验统计量。在原假设下，我们假设 X ~ Po(λ₀)，并利用这个假设来计算概率。

Sometimes the observation consists of a total over several intervals or a combined count. In such cases, we use the additive property of the Poisson distribution: if X₁ ~ Po(λ₁) and X₂ ~ Po(λ₂) are independent, then X₁ + X₂ ~ Po(λ₁ + λ₂). This allows us to rescale λ appropriately.

有时观测值是多个区间上的总数或合并计数。在这种情况下，我们使用泊松分布的可加性：若 X₁ ~ Po(λ₁) 与 X₂ ~ Po(λ₂) 独立，则 X₁ + X₂ ~ Po(λ₁ + λ₂)。这使我们能够适当地重新调整 λ 的值。

If the observation is the total over n intervals, then under H₀: Total ~ Po(nλ₀).

如果观测值是 n 个区间的总和，则在 H₀ 下：总和服从 Po(nλ₀)。

For example, if X ~ Po(λ) represents the number of calls per minute and we observe over 5 minutes, the total number of calls follows a Poisson distribution with mean 5λ. This rescaling is essential for obtaining correct probabilities.

例如，如果 X ~ Po(λ) 表示每分钟的电话数量，而我们观测了 5 分钟，那么电话总数服从均值为 5λ 的泊松分布。这种重新调整对于获得正确的概率至关重要。

4. One-Tailed Tests | 单尾检验

For a one-tailed test with H₁ : λ > λ₀, we calculate the probability P(X ≥ x) under H₀. This is the probability of observing a value as large as, or larger than, the actual observed value. If this probability is less than the significance level α, we reject H₀.

对于 H₁ : λ > λ₀ 的单尾检验，我们在 H₀ 下计算 P(X ≥ x)。这是观察到与实际观测值一样大或更大的值的概率。如果这个概率小于显著性水平 α，我们就拒绝 H₀。

For a one-tailed test with H₁ : λ < λ₀, we calculate P(X ≤ x) under H₀. If this probability is less than α, we reject H₀.

对于 H₁ : λ < λ₀ 的单尾检验，我们在 H₀ 下计算 P(X ≤ x)。如果这个概率小于 α，我们就拒绝 H₀。

When using Edexcel formula booklets, you will often need to use cumulative Poisson probability tables. These tables typically give P(X ≤ r) for selected values of λ. For the right tail, you must use the complement rule: P(X ≥ x) = 1 − P(X ≤ x − 1).

使用爱德思公式手册时，你通常需要查泊松累积概率表。这些表通常给出选定 λ 值下的 P(X ≤ r)。对于右尾，必须使用补集法则：P(X ≥ x) = 1 − P(X ≤ x − 1)。

Worked Example (Right-tailed): A council believes the mean number of potholes per kilometre on a road is 2. After heavy rain, they suspect the number has increased. In a 1 km sample, 6 potholes are found. Test at the 5% significance level.

示例（右尾检验）：市政委员会认为某条路每公里坑洼的平均数为 2。大雨过后，他们怀疑数量增加了。在 1 公里的样本中发现了 6 个坑洼。在 5% 显著性水平下进行检验。

H₀ : λ = 2, H₁ : λ > 2. Under H₀, X ~ Po(2). We compute P(X ≥ 6) = 1 − P(X ≤ 5). From Poisson tables with λ = 2, P(X ≤ 5) = 0.9834. Therefore P(X ≥ 6) = 1 − 0.9834 = 0.0166.

H₀ : λ = 2，H₁ : λ > 2。在 H₀ 下，X ~ Po(2)。我们计算 P(X ≥ 6) = 1 − P(X ≤ 5)。查 λ = 2 的泊松表，P(X ≤ 5) = 0.9834。因此 P(X ≥ 6) = 1 − 0.9834 = 0.0166。

Since 0.0166 < 0.05, we reject H₀. There is sufficient evidence at the 5% level that the mean number of potholes has increased.

由于 0.0166 < 0.05，我们拒绝 H₀。有充分证据在 5% 水平上表明坑洼的平均数量增加了。

5. Two-Tailed Tests | 双尾检验

For a two-tailed test with H₁ : λ ≠ λ₀, we split the significance level α equally between the two tails. At the 5% significance level, we allocate 2.5% to the lower tail and 2.5% to the upper tail. The critical region consists of the extreme values in both tails.

对于 H₁ : λ ≠ λ₀ 的双尾检验，我们将显著性水平 α 平均分配到两个尾端。在 5% 显著性水平下，下尾分配 2.5%，上尾分配 2.5%。临界区域由两个尾端的极端值组成。

To determine the critical region in the lower tail, we find the largest value c₁ such that P(X ≤ c₁) ≤ α/2. To determine the critical region in the upper tail, we find the smallest value c₂ such that P(X ≥ c₂) ≤ α/2.

要确定下尾的临界区域，我们找到最大的 c₁ 使得 P(X ≤ c₁) ≤ α/2。要确定上尾的临界区域，我们找到最小的 c₂ 使得 P(X ≥ c₂) ≤ α/2。

Worked Example (Two-tailed): A shop records a mean of 3 customers per 10-minute interval. After a promotional campaign, the manager wants to know if this mean has changed. In a 10-minute interval, 8 customers arrive. Test at the 5% significance level.

示例（双尾检验）：一家商店记录每 10 分钟平均有 3 位顾客。促销活动后，经理想知道这个均值是否发生了变化。在一个 10 分钟的时间段内，有 8 位顾客到达。在 5% 显著性水平下进行检验。

H₀ : λ = 3, H₁ : λ ≠ 3. Under H₀, X ~ Po(3). At the 5% level, each tail has probability 0.025.

H₀ : λ = 3，H₁ : λ ≠ 3。在 H₀ 下，X ~ Po(3)。在 5% 水平下，每个尾端概率为 0.025。

Lower tail: P(X ≤ 0) = 0.0498 > 0.025, so we look for the largest c₁ with P(X ≤ c₁) ≤ 0.025. P(X ≤ 0) = 0.0498, which is already too large, so the lower critical region contains no values at all.

下尾：P(X ≤ 0) = 0.0498 > 0.025，所以我们寻找最大的 c₁ 使得 P(X ≤ c₁) ≤ 0.025。P(X ≤ 0) = 0.0498，已经过大，因此下尾临界区域不包含任何值。

Upper tail: we need P(X ≥ c₂) ≤ 0.025. Compute P(X ≥ 7) = 1 − P(X ≤ 6). From tables, P(X ≤ 6) = 0.9665, so P(X ≥ 7) = 0.0335 > 0.025. Next, P(X ≥ 8) = 1 − P(X ≤ 7) = 1 − 0.9881 = 0.0119 < 0.025. Therefore the upper critical region is X ≥ 8.

上尾：我们需要 P(X ≥ c₂) ≤ 0.025。计算 P(X ≥ 7) = 1 − P(X ≤ 6)。查表得 P(X ≤ 6) = 0.9665，所以 P(X ≥ 7) = 0.0335 > 0.025。接着，P(X ≥ 8) = 1 − P(X ≤ 7) = 1 − 0.9881 = 0.0119 < 0.025。因此上尾临界区域为 X ≥ 8。

Since 8 lies in the critical region, we reject H₀. There is sufficient evidence at the 5% level that the mean number of customers has changed.

由于 8 落在临界区域内，我们拒绝 H₀。有充分证据在 5% 水平上表明顾客的平均数量发生了变化。

6. Critical Regions and p-Values | 临界区域与 p 值

There are two equivalent methods for conducting a hypothesis test: the critical region method and the p-value method. In the critical region method, we pre-determine the set of values that lead to rejection of H₀. In the p-value method, we calculate the probability of obtaining a result at least as extreme as the observed one, and compare this p-value directly with α.

进行假设检验有两种等价方法：临界区域法和 p 值法。临界区域法预先确定导致拒绝 H₀ 的取值集合。p 值法则计算得到至少与观测值一样极端的结果的概率，并将这个 p 值直接与 α 比较。

Critical region method steps:

临界区域法步骤：

State H₀ and H₁ clearly.

明确写出 H₀ 和 H₁。
Find the critical region as described above.

按上述方法找出临界区域。
State whether the observed value falls inside or outside the critical region.

说明观测值是否落在临界区域内。
Write a conclusion in the context of the problem.

结合题目背景写出结论。

p-value method steps:

p 值法步骤：

State H₀ and H₁.

写出 H₀ 和 H₁。
Calculate the probability of observing a value as extreme as the data (the p-value). For a right-tailed test, p-value = P(X ≥ x). For a left-tailed test, p-value = P(X ≤ x). For a two-tailed test at significance level α, we compare the one-tail probability with α/2 or double the one-tail p-value and compare with α.

计算得到与数据一样极端的概率（p 值）。右尾检验中，p 值 = P(X ≥ x)。左尾检验中，p 值 = P(X ≤ x)。双尾检验在显著性水平 α 下，将单尾概率与 α/2 比较，或者将单尾 p 值加倍后与 α 比较。
If p-value < α (or p-value < α/2 for two-tailed using one-tail probability), reject H₀.

如果 p 值 < α（双尾检验使用单尾概率时比较 α/2），则拒绝 H₀。

Both methods always lead to the same conclusion. In Edexcel exams, you may use either, but you must present your reasoning clearly.

两种方法总是得出相同的结论。在爱德思考场中，你可以使用任何一种，但必须清晰地展示你的推理过程。

7. Using Poisson Tables Effectively | 高效使用泊松表

The Edexcel formula booklet provides cumulative Poisson probability tables P(X ≤ r) for various values of λ. These tables are essential for performing hypothesis tests efficiently. When λ is not in the table, use the closest value or interpolate if instructed.

爱德思公式手册提供了不同 λ 值下的泊松累积概率表 P(X ≤ r)。这些表对于高效完成假设检验至关重要。当 λ 不在表中时，使用最接近的值，或者根据指示进行插值。

Here is a sample extract from a Poisson cumulative table for λ = 2:

以下是 λ = 2 的泊松累积概率表示例摘录：

x	0	1	2	3	4	5	6	7
P(X ≤ x)	0.1353	0.4060	0.6767	0.8571	0.9473	0.9834	0.9955	0.9989

For a right-tailed probability such as P(X ≥ 6), remember: P(X ≥ 6) = 1 − P(X ≤ 5) = 1 − 0.9834 = 0.0166.

对于右尾概率如 P(X ≥ 6)，记住：P(X ≥ 6) = 1 − P(X ≤ 5) = 1 − 0.9834 = 0.0166。

In addition to tables, your calculator may have a Poisson distribution function. You can use this to compute exact probabilities quickly. However, in written exams, you should still show the relevant probability statements and the comparison with the significance level.

除了查表，你的计算器也可能带有泊松分布功能。你可以用它快速计算精确概率。然而，在笔试中，你仍然需要写出相关的概率表达式以及与显著性水平的比较过程。

8. Worked Example with Aggregated Data | 汇总数据的示例

When data are collected over several intervals, we must aggregate both the count and the rate. This is a common examination trap.

当数据跨越多个区间收集时，我们必须同时汇总计数和速率。这是一个常见的考试陷阱。

Example: Historically, the mean number of defects in a roll of cloth is 0.5 per metre. The manufacturer introduces a new production process and suspects the defect rate has decreased. Over 4 metres of cloth, only 1 defect is found. Test at the 1% significance level.

示例：历史上，一卷布料每米平均缺陷数为 0.5。制造商引入新的生产工艺后怀疑缺陷率下降了。在 4 米布料中，仅发现 1 个缺陷。在 1% 显著性水平下进行检验。

Let X be the number of defects in 4 metres. Under H₀, the mean for 4 metres is 4 × 0.5 = 2. Therefore X ~ Po(2). H₀ : λ = 0.5 per metre (equivalently, mean over 4 m = 2); H₁ : λ < 0.5 per metre.

设 X 为 4 米布料中的缺陷数。在 H₀ 下，4 米的均值为 4 × 0.5 = 2。因此 X ~ Po(2)。H₀ : λ = 0.5 每米（等价地，4 米上的均值为 2）；H₁ : λ < 0.5 每米。

We calculate P(X ≤ 1) with λ = 2. From the table, P(X ≤ 1) = 0.4060. Since 0.4060 is not less than 0.01, we do not reject H₀.

我们计算 λ = 2 时的 P(X ≤ 1)。查表得 P(X ≤ 1) = 0.4060。由于 0.4060 不小于 0.01，我们不拒绝 H₀。

There is insufficient evidence at the 1% level to conclude that the defect rate has decreased. The observation of 1 defect in 4 metres is quite consistent with a mean of 2 defects per 4 metres.

在 1% 水平上没有足够证据表明缺陷率下降了。4 米中观察到 1 个缺陷与每 4 米平均 2 个缺陷是相当一致的。

9. Worked Example with p-Value Method | 使用 p 值法的示例

Let us revisit an earlier example but solve it using the p-value method to see the comparison.

让我们重新审视之前的一个示例，但使用 p 值法来求解，以观察比较过程。

Example: A hospital records an average of 4 emergency admissions per hour. New staffing protocols are introduced, and the hospital director believes the average has increased. In the next hour, 9 admissions occur. Test at the 5% significance level.

示例：一家医院记录平均每小时有 4 例急诊入院。新的人员配置方案实施后，院长认为平均数量增加了。在接下来的一小时内，发生了 9 例入院。在 5% 显著性水平下进行检验。

Step 1 – Hypotheses: H₀ : λ = 4, H₁ : λ > 4.

步骤 1 – 假设：H₀ : λ = 4，H₁ : λ > 4。

Step 2 – Test statistic: We observe x = 9.

步骤 2 – 检验统计量：我们观测到 x = 9。

Step 3 – p-value: Under H₀, X ~ Po(4). p-value = P(X ≥ 9) = 1 − P(X ≤ 8). From Poisson tables with λ = 4, P(X ≤ 8) = 0.9786. Thus p-value = 1 − 0.9786 = 0.0214.

步骤 3 – p 值：在 H₀ 下，X ~ Po(4)。p 值 = P(X ≥ 9) = 1 − P(X ≤ 8)。查 λ = 4 的泊松表，P(X ≤ 8) = 0.9786。因此 p 值 = 1 − 0.9786 = 0.0214。

Step 4 – Decision: Since 0.0214 < 0.05, we reject H₀.

步骤 4 – 决策：由于 0.0214 < 0.05，我们拒绝 H₀。

Step 5 – Conclusion: There is sufficient evidence at the 5% level that the mean number of emergency admissions per hour has increased.

步骤 5 – 结论：有充分证据在 5% 水平上表明每小时急诊入院的平均数量增加了。

10. Finding Critical Regions Explicitly | 显式求临界区域

Sometimes a question asks you to “find the critical region” without giving an observed value. In this case, you must determine the full set of values of X that would cause you to reject H₀.

有时题目不给出观测值，而是要求你“求出临界区域”。在这种情况下，你必须确定会导致拒绝 H₀ 的 X 的完整取值集合。

Example: Let X ~ Po(λ). Find the critical region for a test of H₀ : λ = 1.5 against H₁ : λ > 1.5 using the 5% significance level.

示例：设 X ~ Po(λ)。对 H₀ : λ = 1.5 对 H₁ : λ > 1.5 的检验，在 5% 显著性水平下求临界区域。

We need the smallest value c such that P(X ≥ c) ≤ 0.05. Equivalently, P(X ≤ c − 1) ≥ 0.95. Using λ = 1.5 tables:

我们需要最小的 c 使得 P(X ≥ c) ≤ 0.05。等价地，P(X ≤ c − 1) ≥ 0.95。使用 λ = 1.5 的表：

x	0	1	2	3	4	5	6
P(X ≤ x)	0.2231	0.5578	0.8088	0.9344	0.9814	0.9955	0.9991

We see that P(X ≤ 4) = 0.9814 ≥ 0.95, but P(X ≤ 3) = 0.9344 < 0.95. Therefore c − 1 = 4, so c = 5. The critical region is X ≥ 5.

我们看到 P(X ≤ 4) = 0.9814 ≥ 0.95，但 P(X ≤ 3) = 0.9344 < 0.95。因此 c − 1 = 4，所以 c = 5。临界区域为 X ≥ 5。

To check: P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − 0.9814 = 0.0186 < 0.05, while P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − 0.9344 = 0.0656 > 0.05. Hence 5 is indeed the boundary of the critical region.

验证：P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − 0.9814 = 0.0186 < 0.05，而 P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − 0.9344 = 0.0656 > 0.05。因此 5 确实是临界区域的边界。

11. Common Mistakes and Pitfalls | 常见错误与陷阱

Students frequently make the following errors in Poisson hypothesis tests. Being aware of them will help you avoid losing marks.

学生在泊松假设检验中经常犯以下错误。了解这些错误有助于你避免丢分。

Forgetting to rescale λ when the observation period is different from the given rate period. Always check the units.

忘记重新调整 λ：当观测时间段与给定速率的时间段不同时。始终检查单位。
Using the wrong tail for the alternative hypothesis. H₁ : λ > λ₀ requires the right tail; H₁ : λ < λ₀ requires the left tail.

用错尾端：H₁ : λ > λ₀ 需要用右尾；H₁ : λ < λ₀ 需要用左尾。
Forgetting to use α/2 for two-tailed tests when finding critical values. The total significance level must be split equally.

双尾检验忘记使用 α/2：求临界值时必须将总显著性水平平分。
Writing the conclusion without context. Always state what the decision means in the context of the original problem: ‘there is evidence that the mean number of … has increased’, not just ‘reject H₀’.

结论没有结合背景：始终说明决策在原问题中的含义：“有证据表明……的平均数量增加了”，而不仅仅是“拒绝 H₀”。
Confusing ≥ with > when using the complement rule. Recall P(X ≥ c) = 1 − P(X ≤ c − 1), not 1 − P(X ≤ c).

混淆 ≥ 与 >：使用补集法则时注意 P(X ≥ c) = 1 − P(X ≤ c − 1)，而不是 1 − P(X ≤ c)。

It is also important to remember that the Poisson distribution is discrete. Hypothesis tests with discrete distributions are often conservative, meaning the actual significance level may be slightly less than the nominal α. Examiners accept this inherent feature.

还要记住泊松分布是离散的。使用离散分布进行假设检验通常是保守的，即实际显著性水平可能略小于名义上的 α。考官接受这一固有特性。

12. Summary of the Procedure | 程序总结

The complete procedure for conducting a Poisson mean hypothesis test can be summarised as follows:

泊松均值假设检验的完整程序总结如下：

Define the random variable and the context. Write X ~ Po(λ) and identify the relevant time or space interval.

定义随机变量和背景。写出 X ~ Po(λ) 并确定相关的时间或空间区间。
State H₀ : λ = λ₀ and the appropriate H₁ (based on whether the test is one- or two-tailed).

写出 H₀ : λ = λ₀ 和合适的 H₁（基于检验是单尾还是双尾）。
Identify the distribution of the test statistic under H₀, rescaling λ if the observation covers multiple intervals.

确定检验统计量在 H₀ 下的分布，如果观测覆盖多个区间则重新调整 λ。
Compute the probability of the observed result (or find the critical region) at the given significance level.

计算观测结果的概率（或求出临界区域），使用给定的显著性水平。
Compare the p-value with α (or check whether the observed value lies in the critical region).

将 p 值与 α 比较（或检查观测值是否落在临界区域内）。
Make a decision and write a conclusion in the context of the question.

做出决策，并结合题目背景写出结论。

Mastering this procedure will enable you to tackle any Poisson hypothesis testing question in the Edexcel Further Mathematics exam with confidence. The key is consistent practice with both table-based calculations and calculator-based checks.

掌握这一程序将使你能够自信地应对爱德思进阶数学考试中的任何泊松假设检验问题。关键在于持之以恒的练习，既要用查表计算，也要用计算器进行验证。

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Hypothesis Testing for the Mean of a Poisson Distribution | 泊松分布均值的假设检验方法