📚 Case Study in Year 13 OCR Statistics: A Practical Drill | Year 13 OCR 统计：案例分析实战演练
This article immerses you in a realistic statistical investigation that ties together the core techniques of the Year 13 OCR Statistics specification. Through a continuous case study, you will see how hypothesis testing, confidence intervals, Chi‑squared tests, regression, and distributions are applied to genuine data questions, just as they appear in the exam.
本文将通过一个真实的统计调查案例，综合运用 Year 13 OCR 统计课程的核心方法。贯穿整个案例分析，你会看到假设检验、置信区间、卡方检验、回归分析以及各类概率分布如何被用来解决实际数据问题，就像考试中出现的情景一样。
1. Overview of the Case Study | 案例概述
BrewBetter, a chain of coffee shops, has introduced a new espresso machine in one of its branches. The management wants to know whether the new machine reduces the average service time for a latte, whether customer satisfaction has improved, and whether the number of customers arriving per 15‑minute interval follows a Poisson distribution. Data were collected over two weeks for the new machine and a comparable period for the old machine.
连锁咖啡店 BrewBetter 在其中一家分店引入了新的意式咖啡机。管理层想了解新机器是否减少了拿铁的平均服务时间、顾客满意度是否提高，以及每 15 分钟到达的顾客数量是否服从泊松分布。数据在新机器运行的连续两周内收集，并与旧机器在相似时段的数据进行对比。
2. Data Collection and Summary | 数据收集与汇总
A sample of 30 latte orders was timed for the old machine (mean 135.2 seconds, standard deviation 18.5 seconds) and 30 for the new machine (mean 124.6 seconds, standard deviation 16.3 seconds). Customer satisfaction was recorded as ‘satisfied’ or ‘not satisfied’ for 200 customers under the new machine, with 148 satisfied. Arrival counts were noted in 80 fifteen‑minute slots; the observed frequencies are given in a table.
对旧机器记录 30 份拿铁订单的服务时间（均值 135.2 秒，标准差 18.5 秒），对新机器同样记录 30 份（均值 124.6 秒，标准差 16.3 秒）。在新机器下，200 名顾客中 148 人表示满意。到达人数在 80 个 15 分钟时段内被记录，观察频数列于表格中。
3. Binomial Distribution Model | 二项分布模型
Before conducting a test on the satisfaction proportion, the company assumes that each customer independently has a constant probability p of being satisfied. With 200 customers, the number satisfied X follows a B(200, p) distribution. The observed value 148 gives a point estimate p̂ = 148/200 = 0.74.
在对满意比例进行检验之前，公司假设每位顾客独立地以恒定概率 p 感到满意。在 200 名顾客中，满意人数 X 服从 B(200, p)。观察值 148 给出点估计 p̂ = 148/200 = 0.74。
4. Hypothesis Test for a Proportion | 比率的假设检验
The management claims that the new machine increases the satisfaction rate above the historical level of 65%. We set up H₀: p = 0.65 against H₁: p > 0.65 and use a binomial test at the 5% significance level. Under H₀, X ~ B(200, 0.65). Using a normal approximation (np = 130, np(1–p) = 45.5), we compute the test statistic z = (148 – 130)/√45.5 ≈ 2.67. The critical value for a one‑tailed test is 1.645, so we reject H₀. There is sufficient evidence that the satisfaction rate has increased.
管理层声称新机器将满意率提高到了历史水平 65% 以上。我们建立原假设 H₀: p = 0.65 对备择假设 H₁: p > 0.65，并在 5% 显著性水平下进行二项检验。在 H₀ 下，X ~ B(200, 0.65)。借助正态近似（np = 130，np(1–p) = 45.5），计算检验统计量 z = (148 – 130)/√45.5 ≈ 2.67。单侧临界值为 1.645，因此拒绝 H₀。有足够证据表明满意率已经提高。
5. Poisson Distribution Application | 泊松分布应用
The number of customers arriving every 15 minutes was recorded across 80 intervals. The total number of arrivals was 600, giving a mean rate λ̂ = 600/80 = 7.5. The management suspects that arrivals follow a Poisson distribution with this mean. This will be tested using a goodness‑of‑fit test.
每 15 分钟到达的顾客数在 80 个时段内被记录。到达总数为 600 人，因此平均到达率 λ̂ = 600/80 = 7.5。管理层猜测到达人数服从均值为 7.5 的泊松分布。这一假设将通过拟合优度检验加以验证。
6. Goodness‑of‑Fit Chi‑Squared Test | 拟合优度卡方检验
We group arrival counts into categories: 0–4, 5–6, 7–8, 9–10, 11+. Expected frequencies are calculated from a Poisson(7.5) distribution. The observed and expected frequencies are shown in the table below. The test statistic χ² = Σ(O–E)²/E is computed. With 4 degrees of freedom (after estimating λ) and a 5% critical value of 9.488, the obtained χ² is 5.23, so we do not reject H₀. The Poisson model fits adequately.
将到达人数分成 0–4、5–6、7–8、9–10、11+ 五个类别，由 Poisson(7.5) 计算期望频数。观察与期望频数见下表。计算检验统计量 χ² = Σ(O–E)²/E。由于估计了 λ，自由度为 4，在 5% 水平下临界值为 9.488；实际 χ² 为 5.23，因此不拒绝原假设。泊松模型拟合良好。
| Category | O | E |
|---|---|---|
| 0–4 | 12 | 10.5 |
| 5–6 | 18 | 19.2 |
| 7–8 | 24 | 22.8 |
| 9–10 | 16 | 17.4 |
| 11+ | 10 | 10.1 |
7. Normal Approximation to Binomial | 二项分布的正态近似
In the proportion test above, the normal approximation to the binomial was employed because n is large and p is not extreme. We checked that np = 130 > 5 and n(1–p) = 70 > 5. A continuity correction can refine the result, but even without it, the conclusion remains valid. For a two‑tailed test, the critical region would be |z| > 1.96.
在上述比例检验中，由于 n 较大且 p 不极端，使用了二项分布的正态近似。我们验证了 np = 130 > 5 且 n(1–p) = 70 > 5。连续性校正可使结果更精确，但即使不加校正，结论仍然有效。若是双侧检验，拒绝域为 |z| > 1.96。
8. Confidence Intervals for Means | 均值的置信区间
For the new machine’s service time, a 95% confidence interval for the population mean μ is constructed using the t‑distribution, since the population standard deviation is estimated. With x̄ = 124.6, s = 16.3, n = 30, the 95% CI is x̄ ± t₂₉(0.025) × s/√n. Using t₂₉(0.025) = 2.045, we obtain [118.5, 130.7] seconds. The old machine’s interval is [128.3, 142.1]. The non‑overlap suggests a significant difference.
对于新机器的服务时间，由于总体标准差未知，构造总体均值 μ 的 95% 置信区间采用 t 分布。已知 x̄ = 124.6，s = 16.3，n = 30，95% CI = x̄ ± t₂₉(0.025) × s/√n。查表得 t₂₉(0.025) = 2.045，区间为 [118.5, 130.7] 秒。旧机器的区间为 [128.3, 142.1]。两者不重叠，暗示差异显著。
9. Two‑Sample t‑Test | 双样本 t 检验
To formally compare the mean service times, we perform a two‑sample t‑test assuming unequal variances (Welch’s test). H₀: μ₁ = μ₂ against H₁: μ₁ < μ₂ (new machine faster). The test statistic is t = (124.6 – 135.2) / √(16.3²/30 + 18.5²/30) ≈ –2.41. The approximate degrees of freedom are 56, with a critical value of –1.673 at the 5% level. Since –2.41 < –1.673, we reject H₀ and conclude the new machine significantly reduces service time.
为正式比较平均服务时间，采用假设方差不等的双样本 t 检验（Welch 检验）。H₀: μ₁ = μ₂，H₁: μ₁ < μ₂（新机器更快）。检验统计量 t = (124.6 – 135.2) / √(16.3²/30 + 18.5²/30) ≈ –2.41。近似自由度为 56，5% 水平的单侧临界值为 –1.673。由于 –2.41 < –1.673，拒绝 H₀，得出结论：新机器显著缩短了服务时间。
10. Correlation and Regression Analysis | 相关与回归分析
The company also recorded the temperature of the espresso shot (°C) and the extraction time (seconds) for 20 shots. The product‑moment correlation coefficient was r = –0.72. A hypothesis test for ρ = 0 uses the test statistic t = r√(n–2)/√(1–r²) = –0.72√18/√(1–0.5184) ≈ –4.39. With 18 degrees of freedom, the two‑tailed critical value is 2.101, so we reject H₀; there is evidence of a negative linear relationship. The regression line extraction time = 45 – 0.34 × temperature can be used for prediction.
公司同时记录了 20 杯咖啡的萃取温度（°C）和萃取时间（秒）。积差相关系数 r = –0.72。对 ρ = 0 的假设检验使用 t = r√(n–2)/√(1–r²) = –0.72√18/√(1–0.5184) ≈ –4.39。自由度 18 的双侧临界值为 2.101，故拒绝 H₀；有证据表明存在负线性关系。回归线 萃取时间 = 45 – 0.34 × 温度 可用于预测。
11. Contingency Table and Chi‑Squared Test of Independence | 列联表与独立性卡方检验
The branch manager wants to know whether satisfaction depends on the day part (morning or afternoon). A 2×2 contingency table was formed: 90 morning customers (78 satisfied) and 110 afternoon customers (70 satisfied). The test statistic χ² = Σ(O–E)²/E yields 4.12. With 1 degree of freedom and a 5% critical value of 3.841, we reject H₀ of independence. Satisfaction appears associated with time of day, perhaps higher in the morning.
分店经理想知道满意度是否与时段（上午或下午）有关。建立 2×2 列联表：上午 90 名顾客（78 人满意），下午 110 名顾客（70 人满意）。计算 χ² = Σ(O–E)²/E 得 4.12。自由度为 1，5% 临界值 3.841，因此拒绝独立性的原假设。满意度似乎与时段相关，上午可能更高。
12. Concluding Remarks and OCR Exam Tips | 总结与 OCR 考试技巧
This case study demonstrates how the statistical techniques required in the OCR Year 13 specification are applied in a unified context. In the exam, carefully define hypotheses, check conditions such as np ≥ 5 or E ≥ 5, and present your method step by step. State your conclusion in the context of the problem, and remember that OCR often awards marks for interpreting results correctly, not just for calculations.
本案例展示了 OCR Year 13 统计所需的技术如何在一个统一的背景下应用。在考试中，需要仔细定义假设，检查诸如 np ≥ 5 或 E ≥ 5 等条件，并逐步呈现解题过程。在问题情境中陈述结论，记住 OCR 通常给正确解释结果的动作加分，而不仅仅是计算正确。
Published by TutorHao | Statistics Revision Series | aleveler.com
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