📚 Mastering A-Level Cambridge Statistics through Case Study Practice | 通过案例分析实战演练精通 A-Level Cambridge 统计
In A-Level Cambridge Statistics, the true test of understanding lies not in memorising formulas, but in applying statistical reasoning to real-world scenarios. This article walks you through a series of carefully crafted case studies that mirror actual exam-style investigations. Each case tackles a distinct topic—from binomial hypothesis tests to linear regression—with step-by-step guidance, bilingual explanations, and examiner insights. Work through these examples to sharpen your analytical skills and gain the confidence needed to excel in the examination.
在 A-Level Cambridge 统计考试中，真正的理解考验不在于死记公式，而在于将统计推理应用于现实场景。本文通过一系列精心设计的案例分析，带你演练与真实考题风格相似的研究问题。每个案例针对一个不同的主题——从二项假设检验到线性回归——提供分步指导、中英双语解释和考官提示。通过这些实例练习，将有效提升你的分析能力，并建立起冲刺高分的信心。
1. Setting the Stage: The Role of Case Studies in A-Level Statistics | 开场：案例研究在 A-Level 统计中的角色
Examiners frequently embed statistical techniques within a narrative—such as quality control in a factory, medical trials, or environmental monitoring. A case study requires you to identify the correct model, formulate hypotheses, check assumptions, perform calculations, and interpret results in context. Developing a systematic approach is essential for achieving top marks.
考官经常将统计方法嵌入到具体情境中——例如工厂质量控制、医学试验或环境监测。案例分析要求你识别正确的模型、建立假设、检验假定条件、进行计算并在上下文中解释结果。培养一套系统化的解题流程是获取高分的关键。
Your workflow should follow this pattern: (1) Read the scenario and highlight keywords indicating the distribution or test type; (2) Define the random variable and its parameters; (3) State null and alternative hypotheses precisely; (4) Select and calculate the test statistic; (5) Determine the p-value or critical region; (6) Draw a conclusion in everyday language linked to the original problem.
你的解题流程应遵循以下模式：(1) 阅读情境，标出暗示分布或检验类型的关键词；(2) 定义随机变量及其参数；(3) 准确陈述零假设和备择假设；(4) 选择并计算检验统计量；(5) 确定 p 值或临界域；(6) 使用日常语言结合原始问题得出结论。
2. Case Study 1: Testing a Coin’s Fairness – Binomial Distribution & Normal Approximation | 案例一：检验硬币公平性 – 二项分布与正态近似
A student suspects a coin is biased. She tosses it 100 times and observes 35 heads. Using a 5% significance level, test whether the coin is fair. Let X be the number of heads; under the null hypothesis X ~ B(100, 0.5). We test H₀: p = 0.5 against H₁: p ≠ 0.5. Because np = 50 and n(1-p) = 50 are both greater than 5, a normal approximation is suitable: X ~ N(50, 25) approximately, so the test statistic is Z = (35 – 50)/√25 = -3.0. For a two-tailed test at α = 0.05, the critical values are ±1.96. Since -3.0 < -1.96, we reject H₀ and conclude there is sufficient evidence that the coin is biased.
一名学生怀疑一枚硬币不均匀。她抛掷 100 次，观察到 35 次正面。使用 5% 显著性水平，检验这枚硬币是否公平。设 X 为正面次数；在零假设下 X ~ B(100, 0.5)。我们检验 H₀: p = 0.5，H₁: p ≠ 0.5。因为 np = 50 和 n(1-p) = 50 均大于 5，适合使用正态近似：X ~ N(50, 25)（近似），故检验统计量为 Z = (35 – 50)/√25 = -3.0。对于 α = 0.05 的双尾检验，临界值为 ±1.96。由于 -3.0 < -1.96，我们拒绝 H₀，并得出结论：有充分证据表明该硬币不均匀。
If we instead used the exact binomial test, we would calculate the probability of obtaining 35 or fewer heads, and double it for the two-tailed p-value. Using cumulative binomial tables, P(X ≤ 35 | p = 0.5) ≈ 0.0033, so p-value ≈ 0.0066. Because 0.0066 < 0.05, the conclusion remains the same. The normal approximation gives a very similar result, demonstrating its validity for large n.
如果我们改为使用精确二项检验，将计算得到 35 次或更少正面的概率，并乘以 2 得到双尾 p 值。查累积二项分布表，P(X ≤ 35 | p = 0.5) ≈ 0.0033，因此 p 值 ≈ 0.0066。由于 0.0066 < 0.05，结论保持不变。正态近似给出了非常相似的结果，说明在大样本下该近似的有效性。
3. Case Study 2: Modelling Emergency Calls – Poisson Distribution | 案例二：紧急呼叫建模 – 泊松分布
An emergency hotline receives an average of 4.2 calls per hour. The manager wants to know if a day with 10 calls in a single hour is unusually high, assuming calls occur randomly and independently. The appropriate model is a Poisson distribution with λ = 4.2. Let Y be the number of calls per hour; we test H₀: λ = 4.2 against H₁: λ > 4.2 at the 1% significance level. The p-value is P(Y ≥ 10 | λ = 4.2) = 1 – P(Y ≤ 9). Using Poisson tables or software, P(Y ≤ 9) ≈ 0.9867, so p-value ≈ 0.0133. Since 0.0133 > 0.01, we do not reject H₀. There is insufficient evidence at the 1% level to claim the hour is abnormal.
某紧急热线平均每小时接到 4.2 个电话。经理想知道在一小时内接到 10 个电话是否属于异常偏高，假设呼叫随机且独立发生。适用的模型是参数 λ = 4.2 的泊松分布。设 Y 为每小时呼叫次数；我们在 1% 显著性水平下检验 H₀: λ = 4.2，H₁: λ > 4.2。p 值为 P(Y ≥ 10 | λ = 4.2) = 1 – P(Y ≤ 9)。查泊松表或使用软件，P(Y ≤ 9) ≈ 0.9867，因此 p 值 ≈ 0.0133。由于 0.0133 > 0.01，我们不拒绝 H₀。在 1% 水平下没有足够证据声称该小时异常。
If a normal approximation were used (λ large enough, λ > 10 commonly required, but some syllabuses accept λ > 5), we would set Y ~ N(4.2, 4.2) and compute Z = (10 – 4.2)/√4.2 ≈ 2.83, giving a p-value of about 0.0023, which would lead to rejection. This discrepancy highlights the importance of checking validity conditions for approximations. With λ = 4.2, the normal approximation is questionable; the exact Poisson calculation is safer and more accurate for small means.
如果使用正态近似（通常要求 λ > 10，但部分大纲接受 λ > 5），我们会设 Y ~ N(4.2, 4.2) 并计算 Z = (10 – 4.2)/√4.2 ≈ 2.83，得到 p 值约 0.0023，这会导致拒绝 H₀。这种差异凸显了检验近似有效条件的重要性。对于 λ = 4.2，正态近似值得怀疑；对于较小的均值，精确泊松计算更为安全且准确。
4. Case Study 3: Confidence Intervals for Mean Lifetime – Normal Distribution | 案例三：平均寿命的置信区间 – 正态分布
A manufacturer tests a random sample of 25 light bulbs and finds a mean lifetime of 1570 hours with a standard deviation of 120 hours. Assume lifetime is normally distributed. Construct a 95% confidence interval for the population mean lifetime μ. Because the population variance is unknown, we use the t-distribution with 24 degrees of freedom. The 95% confidence interval is given by x̄ ± t₂₄,₀.₅₉₂ × (s/√n) where t₂₄,₀.₅₉₂ ≈ 2.064. The margin of error is 2.064 × (120/5) = 2.064 × 24 = 49.536. Thus the interval is 1570 ± 49.5, i.e. (1520.5, 1619.5) hours. We are 95% confident that the true mean lifetime lies within this range.
某制造商测试了 25 只灯泡的随机样本，得到平均寿命 1570 小时，标准差 120 小时。假设寿命服从正态分布。构建总体平均寿命 μ 的 95% 置信区间。由于总体方差未知，我们使用自由度为 24 的 t 分布。95% 置信区间由 x̄ ± t₂₄,₀.₅₉₂ × (s/√n) 给出，其中 t₂₄,₀.₅₉₂ ≈ 2.064。误差边际为 2.064 × (120/5) = 2.064 × 24 = 49.536。因此区间为 1570 ± 49.5，即 (1520.5, 1619.5) 小时。我们有 95% 的把握认为真实的平均寿命落在此范围内。
If the sample size were larger, the t-value would approach the z-value of 1.96. Always remember to state the confidence level explicitly and interpret the interval in context—do not say “there is a 95% probability that μ is in the interval” but rather “we are 95% confident” or “95% of such constructed intervals will contain μ.”
如果样本量更大，t 值将趋近于 z 值 1.96。务必明确陈述置信水平，并在上下文中解释区间——不要说“μ 落在区间内的概率为 95%”，而应该说“我们有 95% 的置信度”或“如此构造的区间有 95% 会包含 μ”。
5. Case Study 4: Comparing Two Fertilizers – Two-Sample t-Test | 案例四：比较两种肥料 – 双样本 t 检验
A botanist compares two fertilizers, A and B, by applying each to 10 randomly assigned plots and measuring yields (kg). The summary statistics are: Fertilizer A, nₐ = 10, x̄ₐ = 30.2 kg, sₐ = 4.1 kg; Fertilizer B, nᵦ = 10, x̄ᵦ = 35.4 kg, sᵦ = 3.8 kg. Assuming equal population variances, test whether Fertilizer B yields more on average at the 5% significance level. Let μₐ and μᵦ be the population means. H₀: μₐ = μᵦ, H₁: μₐ < μᵦ (one-tailed test). The pooled variance estimate sₚ² = [(9×4.1² + 9×3.8²)/18] = (151.29 + 129.96)/18 = 281.25/18 = 15.625, so sₚ = 3.953. The test statistic is t = (30.2 - 35.4) / (3.953×√(1/10+1/10)) = -5.2 / (3.953×0.4472) = -5.2 / 1.768 = -2.94. The critical t-value for 18 d.f. (one-tailed, 5%) is about -1.734. Since -2.94 < -1.734, we reject H₀ and conclude that Fertilizer B yields a significantly higher mean yield.
一位植物学家比较了两种肥料 A 和 B，每种施用于 10 块随机分配的小区并测量产量（kg）。汇总统计量为：肥料 A，nₐ = 10，x̄ₐ = 30.2 kg，sₐ = 4.1 kg；肥料 B，nᵦ = 10，x̄ᵦ = 35.4 kg，sᵦ = 3.8 kg。假设总体方差相等，在 5% 显著性水平下检验肥料 B 的平均产量是否更高。设 μₐ 和 μᵦ 为总体均值。H₀: μₐ = μᵦ，H₁: μₐ < μᵦ（单尾检验）。合并方差估计 sₚ² = [(9×4.1² + 9×3.8²)/18] = (151.29 + 129.96)/18 = 281.25/18 = 15.625，故 sₚ = 3.953。检验统计量 t = (30.2 - 35.4) / (3.953×√(1/10+1/10)) = -5.2 / (3.953×0.4472) = -5.2 / 1.768 = -2.94。18 自由度的临界 t 值（单尾，5%）约为 -1.734。由于 -2.94 < -1.734，我们拒绝 H₀，并得出结论：肥料 B 显著地具有更高的平均产量。
Examiners often ask you to state assumptions: (1) yields are normally distributed within each group; (2) the variances are equal; (3) observations are independent. If the equal variance assumption is in doubt, you could perform an F-test or use Welch’s t-test without pooling. Explicitly mentioning these checks adds depth to your answer and shows statistical maturity.
考官经常要求陈述假设条件：(1) 各组内产量服从正态分布；(2) 方差相等；(3) 观测值相互独立。若等方差假设存疑，可执行 F 检验或使用不合并方差的 Welch t 检验。明确提及这些检验能增加答案深度，体现统计思维的严谨性。
6. Case Study 5: Association between Smoking and Lung Capacity – Chi-Squared Test for Independence | 案例五：吸烟与肺活量的关联 – 卡方独立性检验
A health survey records smoking status (Never, Former, Current) and lung capacity (Reduced, Normal) for 300 adults. The observed frequencies are shown in the table below. Test at the 1% level whether smoking status and lung capacity are independent.
一项健康调查记录了 300 名成年人的吸烟状况（从不、曾吸、现在吸）和肺活量（减退、正常）。观测频数见下表。检验在 1% 水平下吸烟状况与肺活量是否独立。
| Never / 从不 | Former / 曾吸 | Current / 现在吸 | Total / 合计 | |
|---|---|---|---|---|
| Reduced / 减退 | 20 | 35 | 55 | 110 |
| Normal / 正常 | 80 | 60 | 50 | 190 |
| Total / 合计 | 100 | 95 | 105 | 300 |
Expected frequencies under independence are calculated as (row total × column total) / grand total. For example, the expected number of never-smokers with reduced lung capacity is (110×100)/300 ≈ 36.67. The χ² test statistic is Σ (O – E)² / E. Computing for all cells gives χ² ≈ 30.58. Degrees of freedom = (2-1)×(3-1) = 2. The critical value at 1% significance with 2 d.f. is 9.210. Since 30.58 > 9.210, we reject H₀ and conclude there is a significant association between smoking status and lung capacity.
独立假设下的期望频数由（行合计 × 列合计）/ 总合计得出。例如，从不吸烟且肺活量减退的期望人数为 (110×100)/300 ≈ 36.67。χ² 检验统计量为 Σ (O – E)² / E。计算所有单元格得到 χ² ≈ 30.58。自由度 = (2-1)×(3-1) = 2。在 1% 显著性水平下，自由度为 2 的临界值为 9.210。由于 30.58 > 9.210，我们拒绝 H₀，并得出结论：吸烟状况与肺活量之间存在显著关联。
The chi-squared test is only valid if no more than 20% of expected frequencies are below 5 and none are below 1. Here all expected values exceed 5, so the test is appropriate. In your answer, always calculate or mention this condition to justify the test choice.
卡方检验仅当不超过 20% 的期望频数低于 5 且没有任何一个低于 1 时才有效。这里所有期望值均大于 5，因而检验适用。在作答时，务必计算或提及这一条件以证明检验选择的合理性。
7. Case Study 6: Predicting Sales from Advertising Spend – Linear Regression & Correlation | 案例六：由广告支出预测销售额 – 线性回归与相关
A company records monthly advertising expenditure (x, in £1000s) and sales revenue (y, in £1000s) over 8 months. Summary statistics: n = 8, Σx = 192, Σy = 480, Σx² = 5088, Σy² = 31200, Σxy = 12560. (i) Calculate the product moment correlation coefficient r. (ii) Find the equation of the regression line y on x. (iii) Estimate sales when £30,000 is spent on advertising. (iv) Test at 5% level whether the population correlation coefficient ρ is zero.
一家公司记录了 8 个月的月度广告支出（x，千英镑）和销售收入（y，千英镑）。汇总统计量：n = 8, Σx = 192, Σy = 480, Σx² = 5088, Σy² = 31200, Σxy = 12560。(i) 计算积矩相关系数 r。(ii) 求 y 对 x 的回归直线方程。(iii) 估计当广告支出为 £30,000 时的销售额。(iv) 在 5% 水平下检验总体相关系数 ρ 是否为零。
Sxx = Σx² – (Σx)²/n = 5088 – 192²/8 = 5088 – 4608 = 480
Syy = Σy² – (Σy)²/n = 31200 – 480²/8 = 31200 – 28800 = 2400
Sxy = Σxy – (Σx)(Σy)/n = 12560 – (192×480)/8 = 12560 – 11520 = 1040
Thus r = Sxy / √(Sxx × Syy) = 1040 / √(480 × 2400) = 1040 / √1152000 = 1040 / 1073.31 ≈ 0.969. This indicates a very strong positive linear relationship. The regression equation is y – ȳ = b(x – x̄) where b = Sxy / Sxx = 1040 / 480 ≈ 2.1667, x̄ = 192/8 = 24, ȳ = 480/8 = 60. Hence y = 60 + 2.1667(x – 24) = 2.1667x + 8.0 (to 1 d.p.). For x = 30, predicted sales = 2.1667×30 + 8 = £73.0 thousand.
因此 r = Sxy / √(Sxx × Syy) = 1040 / √(480 × 2400) = 1040 / √1152000 = 1040 / 1073.31 ≈ 0.969。这表明极强的正线性关系。回归方程为 y – ȳ = b(x – x̄)，其中 b = Sxy / Sxx = 1040 / 480 ≈ 2.1667，x̄ = 192/8 = 24，ȳ = 480/8 = 60。因此 y = 60 + 2.1667(x – 24) = 2.1667x + 8.0（保留一位小数）。当 x = 30 时，预测销售额 = 2.1667×30 + 8 = £73.0 千英镑。
To test H₀: ρ = 0 vs H₁: ρ ≠ 0, we use the t-test: t = r√(n-2) / √(1 – r²) = 0.969×√6 / √(1 – 0.939) = 0.969×2.449 / √0.061 = 2.373 / 0.247 ≈ 9.61. With 6 d.f., the critical t-value at 5% (two-tailed) is 2.447. Clearly |9.61| > 2.447, so we reject H₀ and conclude there is significant linear correlation. Always be careful to interpret the test in the context of x and y, not as proof of causation.
为检验 H₀: ρ = 0，H₁: ρ ≠ 0，我们使用 t 检验：t = r√(n-2) / √(1 – r²) = 0.969×√6 / √(1 – 0.939) = 0.969×2.449 / √0.061 = 2.373 / 0.247 ≈ 9.61。自由度为 6 时，5% 水平下的临界 t 值（双尾）为 2.447。显然 |9.61| > 2.447，因此我们拒绝 H₀，并得出结论：存在显著的线性相关。务必注意在 x 与 y 的语境下解释检验结果，不可视为因果证明。
8. Common Pitfalls and Examiner Tips | 常见陷阱与考官提示
(1) Hypothesis writing: Always define the parameter in words before using symbols; e.g. “Let μ be the population mean waiting time.” Never write H₀ = … without specifying the parameter. (2) One-tail vs two-tail: Read the question carefully – words like “increase”, “reduced”, “more than” signal a one-tailed test. (3) Degrees of freedom: For t-tests, use n-1 for one sample, n₁+n₂-2 for two-sample pooled, and Welch’s formula if variances are not pooled. (4) In regression, checking residuals and understanding extrapolation risks can earn evaluation marks. (5) In chi-squared tests, always comment on the validity of the approximation (expected frequencies condition). (6) Round appropriately and keep at least three significant figures during calculation, but present the final statistic to a sensible precision.
(1) 假设书写：始终在用符号前先用文字定义参数，例如“设 μ 为总体平均等待时间”。切勿只写 H₀ = … 而不指定参数。(2) 单尾与双尾：仔细读题——“增加”、“降低”、“高于”等词暗示单尾检验。(3) 自由度：对于 t 检验，单样本使用 n-1，合并双样本使用 n₁+n₂-2，若方差不合并则使用 Welch 公式。(4) 在回归中，检查残差和理解外推风险可赢得评估分数。(5) 在卡方检验中，务必评论近似的有效性（期望频数条件）。(6) 计算过程中适度四舍五入并至少保留三位有效数字，但最终呈现检验统计量时要精度合理。
9. Bringing It All Together: A Mixed Problem | 综合演练：一个混合问题
Consider this exam-style scenario: A researcher investigates whether a new teaching method improves test scores. She gives a pre-test and post-test to 12 students. The score improvements are: 3, 7, -1, 5, 4, 0, 6, 2, 8, 3, 1, 5. (a) Suggest an appropriate test and state its assumptions. (b) Carry out the test at the 5% significance level. (c) Also, the researcher notes that 9 out of 12 students showed improvement. Using a sign test, test the hypothesis that the method has no effect. (d) Comment on the differences between the two test conclusions.
设想一道考试风格的情境：某研究者考察一种新教学法是否能提高考试成绩。她对 12 名学生进行了前测与后测。分数提高值如下：3, 7, -1, 5, 4, 0, 6, 2, 8, 3, 1, 5。(a) 建议一种合适的检验方法并陈述假设条件。(b) 在 5% 显著性水平下执行该检验。(c) 同时，研究者注意到 12 名学生中有 9 名有所提高。使用符号检验，检验该教学法无效的假设。(d) 评论两种检验结论的差异。
(a) Because the data are paired and improvements can be assumed approximately normal, a paired t-test is appropriate. Assumptions: the differences are independent and come from a normal distribution; the sample is random. (b) Calculate d̄ = (3+7-1+5+4+0+6+2+8+3+1+5)/12 = 43/12 ≈ 3.583, s_d = 2.610 (after computing sum of squares). t = 3.583 / (2.610/√12) ≈ 4.75. With 11 d.f., critical t (two-tailed, 5%) = 2.201. Since 4.75 > 2.201, reject H₀; there is significant improvement. (c) For the sign test, H₀: median difference = 0; H₁: median > 0. Number of positive signs = 9 (ignoring the zero). Under H₀, the number of positives follows B(11, 0.5) (excluding the tie). P(X ≥ 9) = 1 – P(X ≤ 8) = 1 – 0.9673 = 0.0327. Since 0.0327 < 0.05, reject H₀. (d) Both tests indicate a significant improvement, but the t-test uses magnitude of differences while the sign test only uses direction, leading to lower power. The sign test does not rely on normality, making it more robust to outliers, but here both agree.
(a) 因为数据是配对的
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