AS AQA Statistics: Case Study Practical Exercises | AS AQA 统计：案例分析实战演练

📚 AS AQA Statistics: Case Study Practical Exercises | AS AQA 统计：案例分析实战演练

Real-world data analysis brings statistical theory to life. In this article, we walk through a complete case study based on the AQA AS Statistics syllabus, applying core techniques – from sampling design to hypothesis testing using the binomial distribution – in a single, coherent investigation. The scenario involves a school canteen satisfaction survey, giving you a practical feel for how statisticians draw conclusions from sample data.

真实数据分析让统计理论鲜活起来。本文基于 AQA AS 统计课程，通过一个完整的案例实战，将抽样设计、二项分布假设检验等核心技术串联应用。场景围绕某学校食堂满意度调查展开，帮助你切身感受统计学家如何从样本数据中得出结论。

1. Case Background | 案例背景

Greenwood Academy claims that 80% of its students are satisfied with the new lunch menu. A team of student journalists suspects the true satisfaction rate may actually be lower. They decide to gather evidence by conducting a small-scale survey and performing a statistical test. The population of interest is all students who regularly eat school lunches, counted in the hundreds. From this population, a sample of 20 students is selected at random and asked whether they are satisfied with the menu (Yes/No). In the sample, 14 students answer ‘Yes’.

格林伍德学院声称 80% 的学生对新午餐菜单感到满意。学生记者团队怀疑真实满意率可能更低。他们决定通过小规模调查收集证据并进行统计检验。所关心的总体是所有经常在校用餐的学生，数量达数百人。从该总体中随机抽取 20 名学生作为样本，询问他们对菜单是否满意（是/否）。样本中有 14 名学生回答“是”。

2. Data Collection and Sampling Methods | 数据收集与抽样方法

Good statistical practice starts with careful data collection. The student journalists considered several sampling methods:

良好的统计实践始于严谨的数据收集。学生记者考虑了多种抽样方法：

Simple random sampling: Every student has an equal chance of being selected, e.g. using a random number generator on the school register. This minimises selection bias.
简单随机抽样：每名学生被抽中的机会均等，例如利用随机数生成器从全校名单中抽取。该方法可最大限度减少选择偏差。
Systematic sampling: Choosing every k-th student from an ordered list. Quick to implement but may introduce bias if there is an underlying pattern.
系统抽样：从有序名单中每隔 k 名学生抽选一名。实施快捷，但如果名单存在周期性模式则可能引入偏差。
Stratified sampling: Dividing the population into subgroups (e.g. year groups) and sampling proportionally from each. This ensures representation but is more complex.
分层抽样：将总体按年级等分成亚组，并从每个亚组中按比例抽样。这能确保代表性，但操作更复杂。

For this investigation, a simple random sample of 20 students was drawn using a random number generator. The responses were recorded as 1 for ‘satisfied’ and 0 for ‘not satisfied’. No non-responses occurred.

在本调查中，使用随机数生成器抽取了 20 名学生的简单随机样本。回答记录为 1（满意）和 0（不满意），且没有无应答出现。

3. Descriptive Statistics: Exploring the Data | 描述性统计：数据探索

Before conducting a formal test, we summarise the sample. The number of satisfied students is X = 14 out of n = 20. The sample proportion is:

在进行正式检验之前，我们先对样本进行汇总。满意的学生人数为 X = 14，样本容量 n = 20。样本比例为：

p̂ = 14 ÷ 20 = 0.7 (70%)

This is noticeably lower than the claimed 80%. A bar chart or simple table (Table 1) can help visualise the distribution of responses.

该数值明显低于声称的 80%。可以使用条形图或简单表格（表 1）直观展示回答的分布情况。

Response / 回答	Frequency / 频数
Satisfied / 满意	14
Not satisfied / 不满意	6

While 70% looks different from 80%, we need to determine whether this difference could simply be due to sampling variability.

虽然 70% 看起来与 80% 不同，但我们需要判断这一差异是否可能仅仅源于抽样波动。

4. Setting Up a Probability Model: The Binomial Distribution | 建立概率模型：二项分布

The response variable is binary (satisfied/not satisfied) and the sample was drawn independently from a large population. If the true proportion of satisfied students is p, then the number X of satisfied students in a sample of size 20 follows a binomial distribution:

响应变量是二值的（满意/不满意），且样本是从一个较大总体中独立抽取的。如果真正满意的学生比例为 p，那么容量为 20 的样本中满意学生数 X 服从二项分布：

X ~ B(20, p)

Under the academy’s claim, p = 0.8, so X ~ B(20, 0.8). The binomial probability function is given by:

在学院声称的条件下，p = 0.8，因此 X ~ B(20, 0.8)。二项概率函数为：

P(X = k) = ²⁰Cₖ × 0.8ᵏ × 0.2²⁰⁻ᵏ

We will use this model to assess how likely a result as extreme as 14 (or fewer) satisfied students is, assuming the claim is true.

我们将利用该模型评估，在声称属实的前提下，得到 14 个（或更少）满意学生这样极端的结果的可能性有多大。

5. Stating Hypotheses | 陈述假设

The student journalists suspect the satisfaction rate is lower than 80%. Their suspicions form the alternative hypothesis. The null hypothesis represents the status quo – the academy’s claim.

学生记者怀疑满意率低于 80%。他们的怀疑构成备择假设。原假设代表现状，即学院的声称。

H₀: p = 0.8
H₁: p < 0.8

This is a one-tailed (lower-tail) test because the alternative hypothesis specifies a direction. A significance level of 5% (α = 0.05) is chosen before looking at the data.

这是一个单尾（左尾）检验，因为备择假设指明了方向。在查看数据之前，选定了 5% 的显著性水平（α = 0.05）。

6. Test Statistic and Significance Level | 检验统计量与显著性水平

The test statistic is the observed number of satisfied students, X = 14. Under H₀, X ~ B(20, 0.8). The significance level α = 0.05 determines the threshold for rejecting H₀. We will calculate the probability of obtaining a result at least as extreme as 14 in the direction of H₁.

检验统计量为观察到的满意学生数 X = 14。在 H₀ 下，X ~ B(20, 0.8)。显著性水平 α = 0.05 决定了拒绝 H₀ 的门槛。我们将计算在 H₁ 方向上得到至少与 14 同样极端的结果的概率。

Two approaches are common in AQA AS Statistics: the p-value method and the critical region method. Both lead to the same conclusion.

AQA AS 统计中常见两种方法：p 值法和临界区域法。两者得出相同的结论。

7. Calculating p-value and Critical Region | 计算 p 值与临界区域

p-value approach: The p-value is P(X ≤ 14 | p = 0.8). Using cumulative binomial tables or calculator:

p 值法：p 值 = P(X ≤ 14 | p = 0.8)。利用二项累积概率表或计算器可得：

p-value = P(X ≤ 14) = 0.1958 (4 d.p.)

Since 0.1958 > 0.05, the p-value is greater than the significance level.

由于 0.1958 > 0.05，p 值大于显著性水平。

Critical region method: The critical region for a lower-tail test at α = 0.05 is the set of values of X such that P(X ≤ c) ≤ 0.05. From tables:

临界区域法：在 α = 0.05 的左尾检验中，临界区域是满足 P(X ≤ c) ≤ 0.05 的 X 的值的集合。查表可得：

P(X ≤ 12) = 0.0321 ≤ 0.05
P(X ≤ 13) = 0.0875 > 0.05

Thus the critical region is X ≤ 12. The observed value X = 14 does not fall in the critical region.

因此临界区域为 X ≤ 12。观察值 X = 14 未落入临界区域。

8. Making a Decision and Conclusion | 做出决策与结论

Because the p-value (0.1958) is not less than 0.05, and X = 14 is not in the critical region, we fail to reject H₀. There is insufficient evidence at the 5% significance level to support the journalists’ suspicion that the true satisfaction rate is below 80%.

因为 p 值 (0.1958) 不小于 0.05，且 X = 14 不在临界区域内，我们无法拒绝 H₀。在 5% 的显著性水平下，没有足够证据支持记者们关于真实满意率低于 80% 的怀疑。

This does not prove that p = 0.8 is true, only that the sample data are not inconsistent with it.

这并不证明 p = 0.8 为真，只表明样本数据与之并无明显矛盾。

9. Assessing Assumptions | 评估假设检验的前提条件

For the binomial test to be valid, several assumptions must be checked:

为使二项检验有效，必须检查若干假设条件：

Independence: Students were selected randomly and independently. The population is large enough that the sample proportion does not affect subsequent selections.
独立性：学生是随机且独立抽取的。总体足够大，样本比例不会影响后续抽取。
Fixed number of trials: The sample size n = 20 was fixed in advance.
固定试验次数：样本容量 n = 20 是事先固定的。
Constant probability: The probability of a student being satisfied is assumed constant throughout the sampling process.
恒定概率：假设在整个抽样过程中，每位学生满意的概率保持不变。
Binary outcomes: Each response is simply ‘satisfied’ or ‘not satisfied’.
二值结果：每个回答只有“满意”或“不满意”。

If the sample had been taken without replacement from a very small year group, a correction might be needed, but here the population is large relative to n.

如果是从一个极小的年级中无放回抽样，则可能需要校正，但此处相对 n 而言总体很大。

10. Communicating the Results | 沟通结果

A formal statistical report would summarise the investigation: “From a random sample of 20 students, 14 expressed satisfaction. A binomial hypothesis test gave a p-value of 0.1958. At the 5% significance level, there is insufficient evidence to reject the academy’s claim of 80% satisfaction.”

一份正式的统计报告会这样总结调查：“从 20 名学生的随机样本中，14 人表示满意。二项假设检验得出的 p 值为 0.1958。在 5% 显著性水平下，没有足够证据拒绝学院关于 80% 满意的声称。”

It is important to convey both the conclusion and the uncertainty, perhaps including a confidence interval for the true proportion in a more advanced analysis.

传达结论及其不确定性很重要，在更深入的分析中或许可以加入真实比例的置信区间。

11. Sensitivity Analysis: Changing the Significance Level | 敏感性分析：改变显著性水平

What if the significance level had been 10% (α = 0.10)? The critical value would then be c = 13 because P(X ≤ 13) = 0.0875 ≤ 0.10, making the critical region X ≤ 13. Since 14 > 13, the conclusion would still be to not reject H₀. This reinforces that the evidence is not particularly compelling.

假如显著性水平设为 10% (α = 0.10) 会怎样？此时临界值为 c = 13，因为 P(X ≤ 13) = 0.0875 ≤ 0.10，临界区域为 X ≤ 13。由于 14 > 13，结论仍然是无法拒绝 H₀。这进一步说明证据并不十分有力。

If the journalists wished to be more confident in detecting a real drop, they would need a larger sample size to increase the power of the test.

如果记者们希望更有把握地检测出真实下降，他们需要更大的样本量以提高检验的功效。

12. Summary and Exam Tips | 总结与考试技巧

This case study demonstrates a complete cycle of a statistical investigation for the AS AQA Statistics exam: question formulation, data collection, descriptive summary, probability modelling, hypothesis statements, test execution, and conclusion in context. Key points to remember:

本案例展示了 AQA AS 统计考试中一个完整的统计调查循环：提出问题、收集数据、描述汇总、概率建模、假设陈述、执行检验并结合情境得出结论。需要记住的关键点：

Always define the population, sample, and variables clearly.
始终清晰定义总体、样本与变量。
State the hypotheses using parameters (p) and specify the direction (one-tail or two-tail).
使用参数 (p) 陈述假设，并指明方向（单尾或双尾）。
Use correct binomial notation and show full working for the p-value or critical region.
使用正确的二项记号，并完整展示 p 值或临界区域的计算过程。
Link the conclusion back to the context: never just say ‘reject H₀’; explain what it means for the real-world question.
将结论联系回情境：不要只说“拒绝 H₀”，要解释它对实际问题意味着什么。
Check assumptions and comment on possible limitations.
检查假设条件，并评论可能的局限性。

Practising such structured case studies will strengthen your ability to tackle both routine exercises and the more open-ended questions on the AS paper.

练习这类结构化的案例分析，将增强你应对 AS 试卷中常规习题和更具开放性问题时的能力。

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AS AQA Statistics: Case Study Practical Exercises | AS AQA 统计：案例分析实战演练