📚 A-Level OCR Statistics: Case Study Practical Workout | A-Level OCR 统计：案例分析实战演练
Welcome to this practical workout designed for OCR A-Level Statistics. In the following case studies, we tackle real-world scenarios that require hypothesis tests, probability distributions, confidence intervals, correlation, regression, and chi‑squared analysis. Each case mirrors the style of exam questions, helping you sharpen your ability to select appropriate statistical methods and interpret results accurately.
欢迎来到为 OCR A-Level 统计学专门设计的实战演练。在下面的案例研究中，我们将面对真实场景，需要运用假设检验、概率分布、置信区间、相关与回归以及卡方分析。每个案例都模拟考试题目风格，助你提高选择合适统计方法并准确解释结果的能力。
1. Testing a Coin’s Fairness Using a Binomial Test | 使用二项检验测试硬币的公平性
Scenario: A casino manager suspects a coin is biased towards heads. She flips it 100 times and records 62 heads. Use a 5% significance level to test whether the coin is fair.
场景：一位赌场经理怀疑一枚硬币偏向正面。她抛掷该硬币100次，记录下62次正面。使用5%的显著性水平检验该硬币是否公平。
Step 1 – Hypotheses: H₀: p = 0.5 (the coin is fair); H₁: p > 0.5 (the coin is biased towards heads). A one‑tailed test is appropriate because the suspicion specifically points to an excess of heads.
步骤1 – 假设：原假设 H₀: p = 0.5（硬币公平）；备择假设 H₁: p > 0.5（硬币偏向正面）。由于怀疑的方向特指正面过多，采用单尾检验是合适的。
Step 2 – Test statistic: Under H₀, X ~ B(100, 0.5). The observed test statistic is x = 62. We calculate the p‑value = P(X ≥ 62). While exact binomial tables could be used, we often apply a normal approximation with continuity correction because n is large.
步骤2 – 检验统计量：在 H₀ 下，X ~ B(100, 0.5)。观测到的检验统计量为 x = 62。我们计算 p 值 = P(X ≥ 62)。尽管可以使用精确二项分布表，但由于 n 很大，我们常用带连续性校正的正态近似。
Step 3 – Normal approximation: Mean μ = np = 100×0.5 = 50; standard deviation σ = √(np(1‑p)) = √(25) = 5. With continuity correction, we use 61.5. Then z = (61.5 ‑ 50) / 5 = 2.30. From tables, P(Z ≥ 2.30) = 1 − 0.9893 = 0.0107.
步骤3 – 正态近似：均值 μ = np = 100×0.5 = 50；标准差 σ = √(np(1‑p)) = √25 = 5。使用连续性校正，取 61.5。于是 z = (61.5 − 50) / 5 = 2.30。查表得 P(Z ≥ 2.30) = 1 − 0.9893 = 0.0107。
Step 4 – Conclusion: The p‑value 0.0107 is less than 0.05, so we reject H₀. There is sufficient evidence at the 5% level to conclude that the coin is biased towards heads. If we had used a two‑tailed test, the p‑value would double, but the conclusion would remain the same at this significance level.
步骤4 – 结论：p 值 0.0107 小于 0.05，因此我们拒绝 H₀。在 5% 显著性水平下，有充分证据表明该硬币偏向正面。若使用双尾检验，p 值会加倍，但在该显著性水平下结论不变。
2. Modelling Traffic Accidents with the Poisson Distribution | 用泊松分布对交通事故建模
Scenario: A small roundabout experiences an average of 0.3 accidents per hour. We are interested in a 10‑hour monitoring period. Authorities want to know the probability of observing at least 5 accidents and whether this would signal an increase in risk.
场景：一个小型环岛平均每小时发生0.3起事故。我们关注一段10小时的监测期。管理部门想知道观察到至少5起事故的概率，以及这是否表明风险上升。
Step 1 – Poisson model: Let X be the number of accidents in 10 hours. Since events occur randomly and independently at a constant average rate, X ~ Po(λ) with λ = 0.3 × 10 = 3.
步骤1 – 泊松模型：令 X 为10小时内的事故数量。由于事件以恒定平均速率随机且独立地发生，X ~ Po(λ)，其中 λ = 0.3 × 10 = 3。
Step 2 – Required probability: P(X ≥ 5) = 1 − P(X ≤ 4). Using cumulative Poisson tables for λ = 3, P(X ≤ 4) = 0.8153, so P(X ≥ 5) = 1 − 0.8153 = 0.1847.
步骤2 – 需求概率： P(X ≥ 5) = 1 − P(X ≤ 4)。查 λ = 3 的泊松累积分布表，P(X ≤ 4) = 0.8153，因此 P(X ≥ 5) = 1 − 0.8153 = 0.1847。
Step 3 – Hypothesis test for a rate change: Suppose we actually observe 5 accidents. We test H₀: λ = 3 against H₁: λ > 3 at the 5% level. The p‑value = P(X ≥ 5) = 0.1847 > 0.05, so we do not reject H₀. A single observation of 5 accidents is not unusual under the current rate.
步骤3 – 速率变化的假设检验：假设我们实际观测到5起事故。在5%显著性水平下检验 H₀: λ = 3 对 H₁: λ > 3。p 值 = 0.1847 > 0.05，因此我们不能拒绝 H₀。在当前事故率下，单次观测到5起事故并不罕见。
Step 4 – Practical insight: If the observed value had been 7, P(X ≥ 7) = 1 – 0.9665 = 0.0335 < 0.05, leading to rejection of H₀. This would warrant a safety review. The Poisson distribution helps distinguish random fluctuation from a genuine increase in risk.
步骤4 – 实际洞察：若观测值为7，P(X ≥ 7) = 1 – 0.9665 = 0.0335 < 0.05，将导致拒绝 H₀，从而需要启动安全审查。泊松分布有助于区分随机波动与真正的风险上升。
3. Confidence Intervals for a Population Mean | 总体均值的置信区间
Scenario: A battery manufacturer claims their AA cells last 50 hours. A consumer group tests a random sample of 36 batteries, finding a sample mean of 48 hours with a sample standard deviation of 5 hours. Construct a 95% confidence interval for the true mean lifetime and comment on the claim.
场景：某电池制造商声称其AA电池可持续50小时。一个消费者组织随机抽取36节电池进行测试，发现样本均值为48小时，样本标准差为5小时。构建真实平均寿命的95%置信区间，并对该声明进行评述。
Step 1 – Model and assumptions: The sample size n = 36 is large enough to invoke the Central Limit Theorem. Although σ is unknown, we estimate it with s = 5 and use the z‑distribution (or t₃₅, which gives a very similar critical value).
步骤1 – 模型与假设：样本量 n = 36 足够大，可以引用中心极限定理。尽管 σ 未知，我们用 s = 5 进行估计，并使用 z 分布（或 t₃₅，其临界值极为相近）。
Step 2 – Formula and calculation: The 95% confidence interval is x̄ ± z* × (s/√n). Here z* = 1.96. Standard error = 5/√36 = 5/6 ≈ 0.8333. Margin of error = 1.96 × 0.8333 ≈ 1.633. So the interval is 48 ± 1.633, giving (46.37, 49.63) hours.
步骤2 – 公式与计算：95%置信区间为 x̄ ± z* × (s/√n)。此处 z* = 1.96。标准误 = 5/√36 = 5/6 ≈ 0.8333。误差边际 = 1.96 × 0.8333 ≈ 1.633。因此区间为 48 ± 1.633，即 (46.37, 49.63) 小时。
Step 3 – Interpretation: The interval does not contain the claimed 50 hours. Therefore we have evidence, at the 95% confidence level, that the true mean lifetime is less than 50 hours. This is consistent with a one‑tailed hypothesis test where H₀: μ = 50 is rejected at the 5% level.
步骤3 – 解释：该区间不包含所声称的50小时。因此，在95%置信水平下，我们有证据表明真实平均寿命低于50小时。这与单尾假设检验 H₀: μ = 50 在5%水平下被拒绝是一致的。
4. Correlation and Regression: Study Hours vs. Exam Scores | 相关与回归：学习时间与考试成绩
Scenario: A teacher records the weekly study hours (x) and final exam scores (y, out of 100) for five students. The data are shown in the table below. We wish to quantify the strength of the linear relationship and predict the score for a student who studies 8 hours per week.
场景：一位老师记录了五名学生每周学习时间（x）与期末考试分数（y，满分100）。数据如下表所示。我们希望量化线性关系的强度，并预测每周学习8小时的学生的分数。
| Student | Study Hours (x) | Score (y) |
|---|---|---|
| A | 3 | 45 |
| B | 4 | 50 |
| C | 6 | 65 |
| D | 7 | 70 |
| E | 10 | 85 |
Step 1 – Summary statistics: Σx = 30, Σy = 315, Σx² = 210, Σy² = 21875, Σxy = 2080. Using the PMCC formula: r = [nΣxy − (Σx)(Σy)] / √[ (nΣx²−(Σx)²) (nΣy²−(Σy)²) ]. Substituting gives r ≈ 0.993, indicating a very strong positive linear correlation.
步骤1 – 汇总统计量： Σx = 30，Σy = 315，Σx² = 210，Σy² = 21875，Σxy = 2080。使用积矩相关系数公式：r = [nΣxy − (Σx)(Σy)] / √[ (nΣx²−(Σx)²)(nΣy²−(Σy)²) ]。代入计算得 r ≈ 0.993，显示极强的正线性相关。
Step 2 – Regression line: Slope b = [nΣxy – (Σx)(Σy)] / [nΣx² – (Σx)²] = (5×2080 – 30×315) / (5×210 – 30²) = (10400 – 9450) / (1050 – 900) = 950/150 ≈ 6.333. Intercept a = ȳ – b x̄ = (315/5) – 6.333×(30/5) = 63 – 6.333×6 = 63 – 38 = 25. So regression equation: y = 25 + 6.33x.
步骤2 – 回归直线：斜率 b = [nΣxy – (Σx)(Σy)] / [nΣx² – (Σx)²] = 950 / 150 ≈ 6.333。截距 a = ȳ – b x̄ = 63 – 38 = 25。因此回归方程为 y = 25 + 6.33x。
Step 3 – Prediction: For x = 8, predicted score = 25 + 6.333×8 ≈ 75.7. Because the data range is 3–10 hours, this prediction is reliable. Always caution against extrapolation beyond the observed range.
步骤3 – 预测：当 x = 8 时，预测分数 = 25 + 6.333×8 ≈ 75.7。由于数据范围为3–10小时，该预测是可靠的。始终提醒避免在观测范围之外进行外推。
5. Chi‑Squared Test for Independence: Education Level and Smoking | 卡方独立性检验：教育水平与吸烟
Scenario: A public health study classifies 300 adults by education level (Low, Medium, High) and smoking status. Observed frequencies are shown. We test whether smoking habits are independent of education level at the 1% significance level.
场景：一项公共卫生研究将300名成年人按教育水平（低、中、高）与吸烟状态分类。观测频数如下。我们在1%显著性水平下检验吸烟习惯是否与教育水平独立。
| Observed | Smoker | Non‑smoker | Total |
|---|---|---|---|
| Low Education | 45 | 55 | 100 |
| Medium | 30 | 70 | 100 |
| High | 15 | 85 | 100 |
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