📚 Case Study Practice for A-Level AQA Statistics | A-Level AQA 统计：案例分析实战演练
Statistical case studies form the heart of A-Level AQA Statistics examinations – they require you not only to recall formulas but to structure investigations, choose appropriate models, interpret outputs, and communicate findings in a real-world context. This article walks you through three complete case studies, covering hypothesis testing for means and proportions, correlation analysis, and the critical thinking needed to excel in exam-style questions. Each section pairs English explanation with its Chinese equivalent, followed by step-by-step solutions, so you can build confidence in applying theory to data.
统计案例分析是 AQA A-Level 统计学考试的核心 —— 它不仅要求你记住公式，更要求你构建调查框架、选择合适的模型、解读输出结果，并在真实情境中清晰地表达结论。本文将通过三个完整的案例，涵盖均值和比例的假设检验、相关与回归分析，以及应对考试所需的关键思维。每一小节先给出英文讲解，再提供对应的中文说明，并配合分步解答，帮助你建立将理论应用于数据的信心。
1. Introduction to Statistical Case Studies | 统计案例分析简介
In AQA Statistics, a case study typically presents a real or realistic scenario accompanied by a dataset or summary statistics. Your task is to frame a statistical question, decide on the appropriate test (z-test, t-test, binomial, chi-squared, or correlation), check necessary assumptions, carry out calculations, and write a conclusion that addresses the original problem. The examiner expects clear notation, correct use of tables, interpretation of p-values or critical regions, and a contextualised final statement that refers back to the scenario.
在 AQA 统计学中，案例分析通常会给出一个真实或贴近现实的场景，并附上数据集或汇总统计量。你的任务是提出统计问题、选择合适的检验方法（z 检验、t 检验、二项检验、卡方检验或相关分析）、验证必要假设、进行计算，并撰写出能够回应原始问题的结论。阅卷官希望看到清晰的符号、正确查表、对 p 值或拒绝域的解释，以及一个联系回场景的、有背景意义的最终陈述。
2. Case Study 1: Testing a Manufacturer’s Claim about Battery Lifetime | 案例1：检验制造商关于电池寿命的声称
A manufacturer claims that its AA batteries have a mean lifetime of 25.0 hours. A consumer group suspects the true mean is lower. They test a random sample of 30 batteries and record the lifetimes (in hours). The sample mean is 24.3 hours and the sample standard deviation is 1.8 hours. Assume that battery lifetimes are normally distributed. Using a 5% significance level, determine whether there is sufficient evidence to reject the manufacturer’s claim.
一家制造商声称其 AA 电池的平均寿命为 25.0 小时。一个消费者组织怀疑真实均值低于该声称值。他们随机抽取了 30 节电池并记录了寿命（小时）。样本均值为 24.3 小时，样本标准差为 1.8 小时。假设电池寿命服从正态分布。在 5% 的显著性水平下，判断是否有充分证据拒绝制造商的声称。
3. Data Summary and Preliminary Checks | 数据汇总与初步检查
We have n = 30, sample mean x̄ = 24.3, sample standard deviation s = 1.8. The population standard deviation σ is unknown, so we use a one-sample t-test. The assumption of normality is reasonable because the population is stated to be normal, and the sample size is moderate. We also assume the sample is random and independent.
我们有 n = 30，样本均值 x̄ = 24.3，样本标准差 s = 1.8。总体标准差 σ 未知，因此我们使用单样本 t 检验。由于题目已说明总体服从正态分布，且样本量为中等规模，正态性假设合理。同时我们假设样本是随机且独立的。
4. Formulating Hypotheses and Choosing Significance Level | 建立假设与选择显著性水平
A one-tailed test is appropriate because the consumer group suspects the mean is lower than 25.0. Null hypothesis H₀: μ = 25.0. Alternative hypothesis H₁: μ < 25.0. Significance level α = 0.05. The test statistic will follow a t-distribution with degrees of freedom df = n − 1 = 29.
应采用单侧检验，因为消费者组织怀疑均值低于 25.0。原假设 H₀: μ = 25.0，备择假设 H₁: μ < 25.0。显著性水平 α = 0.05。检验统计量服从自由度为 df = n − 1 = 29 的 t 分布。
5. Calculating Test Statistic and Critical Region | 计算检验统计量与拒绝域
The t statistic is computed as:
t = (x̄ − μ₀) / (s / √n) = (24.3 − 25.0) / (1.8 / √30) ≈ −0.7 / 0.3286 ≈ −2.130
检验统计量 t 的计算式为：
t = (x̄ − μ₀) / (s / √n) = (24.3 − 25.0) / (1.8 / √30) ≈ −0.7 / 0.3286 ≈ −2.130
From t-tables, the critical value for a one-tailed test with df = 29 at α = 0.05 is approximately −1.699 (the negative value because we look at the left tail). Since our computed t = −2.130 is less than −1.699, it falls in the rejection region.
查 t 分布表，df = 29、α = 0.05 的单侧临界值约为 −1.699（取负值是因为看左侧尾部）。我们算得的 t 值为 −2.130，小于 −1.699，落入拒绝域。
6. Drawing Conclusions and Contextualising | 得出结论并联系实际背景
We reject H₀ at the 5% significance level. There is sufficient evidence to support the consumer group’s suspicion that the true mean battery lifetime is less than 25.0 hours. The result is statistically significant. In the context of the investigation, the manufacturer’s claim appears to be overstated, and the consumer group may use this evidence to request corrective action. Always remember to phrase the conclusion in terms of the original problem, not just ‘reject H₀’.
在 5% 显著性水平下，我们拒绝原假设。有充分证据支持消费者组织的怀疑，即电池真实平均寿命低于 25.0 小时。结果具有统计显著性。结合调查背景，制造商的声称可能被夸大，消费者组织可据此要求整改。务必记住，结论要用原始问题的语言来表达，而不只是说“拒绝 H₀”。
7. Case Study 2: Testing Improvement in Recovery Rate | 案例2：检验康复率的提升
A new physiotherapy programme claims to increase the recovery rate from a knee injury beyond the historical rate of 70%. In a trial with 85 patients, 68 recover within the expected timeframe. Test at the 5% significance level whether the new programme has genuinely improved the recovery rate.
一项新的理疗方案声称能将膝关节损伤的康复率从历史水平 70% 提高至更高。在 85 名患者的试验中，有 68 人按时康复。在 5% 的显著性水平下检验新方案是否确实提升了康复率。
8. Binomial Exact Test and Normal Approximation | 二项精确检验与正态近似
Here we have a binomial situation: n = 85, X = number of recoveries, p₀ = 0.70. We test H₀: p = 0.70 vs H₁: p > 0.70. Because n is large, we can use the normal approximation to the binomial. The test statistic is:
z = (p̂ − p₀) / √[p₀(1 − p₀) / n]
这里是一个二项情形：n = 85，X 为康复人数，p₀ = 0.70。需检验 H₀: p = 0.70 vs H₁: p > 0.70。由于 n 很大，可以使用二项分布的正态近似。检验统计量为：
z = (p̂ − p₀) / √[p₀(1 − p₀) / n]
Sample proportion p̂ = 68/85 ≈ 0.80. Standard error = √[0.70 × 0.30 / 85] ≈ √0.0024706 ≈ 0.0497. Then z = (0.80 − 0.70) / 0.0497 ≈ 2.012. The critical z-value for a one-tailed test at 5% is 1.645. Since 2.012 > 1.645, we reject H₀. Evidence suggests the recovery rate has improved. (A continuity correction could be applied but is not required in many AQA mark schemes; always follow the question’s instruction.)
样本比例 p̂ = 68/85 ≈ 0.80。标准误 = √[0.70 × 0.30 / 85] ≈ √0.0024706 ≈ 0.0497。于是 z = (0.80 − 0.70) / 0.0497 ≈ 2.012。单侧检验在 5% 显著性水平下的 z 临界值为 1.645。由于 2.012 > 1.645，我们拒绝 H₀。证据表明康复率有所提升。（可考虑连续性校正，但多数 AQA 评分方案不强制要求，严格按题目要求即可。）
9. Case Study 3: Investigating Correlation between Study Hours and Exam Scores | 案例3：探究学习时间与考试成绩的相关性
A teacher collects data from 10 randomly chosen students: weekly study hours (x) and their recent examination mark (y). The results are summarised as follows: Σx = 85, Σy = 620, Σx² = 825, Σy² = 40600, Σxy = 5675. Assess whether there is a linear relationship between study hours and exam performance by calculating the product moment correlation coefficient (PMCC) and testing its significance at the 1% level.
一位老师收集了 10 名随机选择的学生数据：每周学习时间（x）和近期考试成绩（y）。汇总统计量如下：Σx = 85，Σy = 620，Σx² = 825，Σy² = 40600，Σxy = 5675。通过计算积矩相关系数（PMCC）并在 1% 水平下检验其显著性，评估学习时间与考试成绩之间是否存在线性关系。
10. Calculating PMCC and Testing for Significance | 计算PMCC与显著性检验
Using the formula r = [n Σxy − (Σx)(Σy)] / √{[n Σx² − (Σx)²][n Σy² − (Σy)²]}, we substitute n = 10:
r = [10×5675 − 85×620] / √{[10×825 − 85²][10×40600 − 620²]}
计算过程使用公式 r = [n Σxy − (Σx)(Σy)] / √{[n Σx² − (Σx)²][n Σy² − (Σy)²]}，带入 n = 10：
Numerator = 56750 − 52700 = 4050. First denominator part: 10×825 − 7225 = 8250 − 7225 = 1025. Second denominator part: 10×40600 − 384400 = 406000 − 384400 = 21600. Denominator = √(1025 × 21600) = √(22140000) ≈ 4705.32. Hence r ≈ 4050 / 4705.32 ≈ 0.8607.
分子 = 56750 − 52700 = 4050。分母第一部分：10×825 − 7225 = 1025；分母第二部分：10×40600 − 384400 = 21600。分母 = √(1025 × 21600) = √22140000 ≈ 4705.32。因此 r ≈ 4050 / 4705.32 ≈ 0.8607。
To test H₀: ρ = 0 vs H₁: ρ > 0 (or two-tailed depending on context; here we suspect a positive correlation), we compare |r| with the critical value from the PMCC table for n = 10 at 1% significance (two-tailed commonly, but exam may specify one-tailed). For n = 10, the two-tailed 1% critical value is 0.7646. Since 0.8607 > 0.7646, we reject H₀ and conclude there is significant positive correlation between study hours and exam scores.
检验 H₀: ρ = 0 vs H₁: ρ > 0（或根据上下文采用双尾检验；我们预计正相关）。将 |r| 与 PMCC 表中 n = 10、显著性水平 1% 的临界值比较（通常双侧）。n = 10 的双侧 1% 临界值为 0.7646。因 0.8607 > 0.7646，拒绝 H₀，认为学习时间与考试成绩之间存在显著正相关。
11. Interpretation and Limitations | 解释与局限性
A significant PMCC indicates a strong linear association in the sample; it does not prove causation. The scatter plot should be checked for outliers and non-linear patterns. With only 10 observations, the result is sensitive to any anomaly. Also, the sample may not be representative of all students. In an exam answer, you should explicitly mention ‘the evidence suggests an association, but we cannot conclude that increasing study hours causes higher marks without controlled experiments.’
显著的 PMCC 表明样本中存在强烈的线性关联，但并不能证明因果关系。应当通过散点图检查是否存在异常值或非线性模式。仅 10 个观测值使得结果对任何异常都较为敏感。此外，样本可能无法代表所有学生。在考试作答时，应明确指出“证据表明存在关联，但若无控制实验，我们不能断定延长学习时间会导致成绩提高”。
12. Summary of Practical Approaches | 实战方法总结
Across these three case studies, a consistent methodology emerges: (1) Understand the context and state the statistical objective clearly; (2) Identify the parameter(s) and variable type (continuous, binomial); (3) Check assumptions (normality, independence, sample size); (4) Formulate H₀ and H₁ with correct notation; (5) Select and compute the appropriate test statistic; (6) Find the critical value or p-value using tables; (7) Compare and decide, always in context; (8) Discuss limitations or assumptions further if asked. Practise writing conclusions that link back to the wording of the scenario – this is where many students lose marks. In AQA exams, clear presentation, proper labelling of hypotheses, and explicit reference to significance level are essential for full credit.
从这三个案例中可以总结出一套通用的实战方法：(1) 理解背景并清晰陈述统计目标；(2) 确定参数及变量类型（连续、二项）；(3) 检验假设条件（正态性、独立性、样本量）；(4) 用正确符号建立 H₀ 和 H₁；(5) 选择并计算恰当的检验统计量；(6) 查表找出临界值或 p 值；(7) 比较并作出决策，始终联系情境；(8) 如有需要，讨论局限性或假设条件。多加练习写结论，使之与场景描述紧密结合 —— 这是许多同学失分的地方。在 AQA 考试中，清晰的书写、假设的正确标注以及对显著性水平的明确提及，是取得满分的关键。
Published by TutorHao | Statistics Revision Series | aleveler.com
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