📚 Case Study Practice: Real-World Statistics for CCEA Year 13 | CCEA Year 13 统计案例分析实战演练
Welcome to an integrated case study that connects the key statistical methods required for the CCEA Year 13 Statistics course. By working through realistic scenarios from a café chain called BeanCount, you will see how discrete distributions, normal approximations, confidence intervals, hypothesis tests, chi-squared analysis, regression, and time series are combined to make data-driven business decisions. Practising these applications will sharpen your ability to identify appropriate models, check assumptions, and interpret results in the exam.
欢迎来到这个综合案例分析，它联系了 CCEA Year 13 统计学课程所要求的关键方法。通过一个名为 BeanCount 的咖啡连锁店的真实场景，你将看到如何结合离散分布、正态近似、置信区间、假设检验、卡方分析、回归和时间序列来做出数据驱动的商业决策。练习这些应用将加强你在考试中识别适当模型、检查假设和解释结果的能力。
1. Case Background and Data Collection | 案例背景与数据收集
BeanCount Café operates three branches in Belfast, Derry, and Lisburn. Management wishes to evaluate the uptake of a new drink ‘Mocha Magic’, compare waiting times, and forecast sales. A survey of 400 randomly selected customers recorded their satisfaction (Satisfied or Not Satisfied), location, whether they purchased Mocha Magic, and the time they waited for service (in minutes). In addition, weekly sales figures and advertising spend were retrieved for the last 24 weeks, and the number of customers arriving per hour was recorded daily during the lunch peak.
BeanCount 咖啡在贝尔法斯特、德里和利斯本经营着三家分店。管理层希望评估新饮品“Mocha Magic”的接受度、比较等待时间并预测销售额。随机抽取了 400 名顾客，记录他们的满意度（满意或不满意）、所在分店、是否购买了 Mocha Magic 以及等待服务的时间（分钟）。此外，还调取了最近 24 周的周销售额和广告支出，并每日记录了午餐高峰时段每小时到达的顾客人数。
2. Descriptive Statistics and Data Exploration | 描述性统计与数据探索
Initial exploration of the waiting times at the Belfast branch (sample size n=50) gave a mean of 3.4 minutes and a standard deviation of 1.2 minutes. The proportion of customers who bought Mocha Magic across all branches was 0.28. A box plot of waiting times revealed a slightly right-skewed distribution, but the sample size is large enough for the Central Limit Theorem to be invoked for the mean.
对贝尔法斯特分店等待时间（样本量 n=50）的初步探索得到均值为 3.4 分钟，标准差为 1.2 分钟。所有分店购买 Mocha Magic 的顾客比例为 0.28。等待时间的箱线图显示分布略呈右偏，但样本量足够大，可以对均值应用中心极限定理。
3. Binomial Distribution: Mocha Magic Uptake | 二项分布：Mocha Magic 的接受度
Let X be the number of customers who purchase Mocha Magic in a random sample of 6 customers, given the overall purchase probability p=0.28. The distribution of X can be modelled as Binomial: X ~ B(6, 0.28). The probability that at least two customers buy the new drink is P(X ≥ 2) = 1 − [P(X=0) + P(X=1)]. We calculate P(X=r) using the formula:
设 X 为随机抽取的 6 名顾客中购买 Mocha Magic 的人数，已知总体购买概率 p=0.28。X 的分布可用二项分布建模：X ~ B(6, 0.28)。至少两名顾客购买新饮品的概率为 P(X ≥ 2) = 1 − [P(X=0) + P(X=1)]。我们使用下列公式计算 P(X=r)：
P(X=r) = nCr pr (1−p)n−r
Using a calculator or tables, P(X=0) = 0.726 ≈ 0.1393, P(X=1) = 6 × 0.28 × 0.725 ≈ 0.3026. Hence P(X ≥ 2) = 1 − 0.4419 = 0.5581. There is a 55.8% chance that at least two out of six randomly chosen customers will buy the new drink.
利用计算器或查表，P(X=0) = 0.726 ≈ 0.1393，P(X=1) = 6 × 0.28 × 0.725 ≈ 0.3026。因此 P(X ≥ 2) = 1 − 0.4419 = 0.5581。在随机抽取的六位顾客中，至少两人购买新饮品的概率为 55.8%。
4. Poisson Distribution: Customer Arrivals | 泊松分布：顾客到达人数
During the lunch peak, the number of customers arriving per hour at the Lisburn branch is modelled by a Poisson distribution with mean λ = 5.2. The probability of exactly 3 arrivals in a randomly selected hour is given by:
在午餐高峰时段，利斯本分店每小时到达的顾客人数用均值为 λ = 5.2 的泊松分布建模。在随机选择的一小时内恰好到达 3 名顾客的概率由下式给出：
P(X=r) = (λr e−λ) / r!
Substituting r=3 and λ=5.2: P(X=3) = (5.23 e−5.2) / 6 = (140.608 × 0.005516) / 6 ≈ 0.1293. The Poisson model assumes arrivals occur independently at a constant average rate – a condition that appears reasonable during a stable lunch period.
代入 r=3 和 λ=5.2：P(X=3) = (5.23 e−5.2) / 6 = (140.608 × 0.005516) / 6 ≈ 0.1293。泊松模型假设到达事件独立发生且平均速率恒定——在稳定的午餐时段这一条件似乎合理。
5. Normal Approximation to the Binomial | 二项分布的正态近似
The proportion of satisfied customers in the entire survey is to be examined. Out of 400 respondents, 292 expressed satisfaction, giving a sample proportion p̂ = 0.73. To test whether the true proportion p exceeds 0.70, we can use the normal approximation, since np̂ = 400×0.73 = 292 > 5 and n(1−p̂) = 108 > 5. Under the null hypothesis H₀: p = 0.70, the standardised test statistic is:
需要检查整个调查中满意顾客的比例。在 400 位受访者中，292 位表示满意，样本比例 p̂ = 0.73。为检验真实比例 p 是否超过 0.70，由于 np̂ = 400×0.73 = 292 > 5 且 n(1−p̂) = 108 > 5，我们可以使用正态近似。在原假设 H₀: p = 0.70 下，标准化检验统计量为：
Z = (p̂ − p₀) / √[p₀(1−p₀)/n]
Z = (0.73 − 0.70) / √(0.70×0.30/400) = 0.03 / √(0.000525) = 0.03 / 0.02291 ≈ 1.309. At the 5% significance level, the one-tailed critical value is 1.645. Since 1.309 < 1.645, we do not reject H₀. There is insufficient evidence to claim the true satisfaction proportion is greater than 0.70.
Z = (0.73 − 0.70) / √(0.70×0.30/400) = 0.03 / √(0.000525) = 0.03 / 0.02291 ≈ 1.309。在 5% 显著性水平下，单尾临界值为 1.645。由于 1.309 < 1.645，我们不拒绝 H₀。没有足够证据声称真实满意度比例大于 0.70。
6. Confidence Interval for the Mean Waiting Time | 平均等待时间的置信区间
For the Belfast branch waiting times with n=50, sample mean x̄ = 3.4 min and sample standard deviation s = 1.2 min, a 95% confidence interval for the population mean μ is built using the t-distribution with 49 degrees of freedom because σ is unknown. The t-value for 49 df is 2.010. The interval is:
对于贝尔法斯特分店的等待时间，n=50，样本均值 x̄ = 3.4 分钟，样本标准差 s = 1.2 分钟，由于 σ 未知，我们使用自由度为 49 的 t 分布构建总体均值 μ 的 95% 置信区间。49 自由度的 t 值为 2.010。区间为：
x̄ ± t* × (s/√n)
Margin of error = 2.010 × (1.2/√50) = 2.010 × 0.1697 ≈ 0.341. The 95% CI is (3.4 − 0.341, 3.4 + 0.341) = (3.059, 3.741) minutes. We are 95% confident that the true mean waiting time in Belfast lies between 3.06 and 3.74 minutes.
误差边界 = 2.010 × (1.2/√50) = 2.010 × 0.1697 ≈ 0.341。95% 置信区间为 (3.4 − 0.341, 3.4 + 0.341) = (3.059, 3.741) 分钟。我们有 95% 的信心认为贝尔法斯特真实的平均等待时间介于 3.06 到 3.74 分钟之间。
7. One-Sample t-Test for Service Time | 单样本 t 检验：服务时间
Management claims that the mean waiting time at the Derry branch is less than 3.0 minutes. A sample of 30 visits yielded x̄ = 2.85 min and s = 0.90 min. We perform a one-tailed t-test at the 1% level. H₀: μ = 3.0, H₁: μ < 3.0. Degrees of freedom = 29, critical t-value ≈ −2.462. Test statistic:
管理层声称德里分店的平均等待时间低于 3.0 分钟。一个包含 30 次访问的样本得到 x̄ = 2.85 分钟，s = 0.90 分钟。我们在 1% 水平下进行单尾 t 检验。H₀: μ = 3.0，H₁: μ < 3.0。自由度 = 29，临界 t 值 ≈ −2.462。检验统计量：
t = (x̄ − μ₀) / (s/√n) = (2.85 − 3.0) / (0.90/√30)
t = −0.15 / 0.1643 ≈ −0.913. Because −0.913 > −2.462, we do not reject H₀. The sample does not provide sufficient evidence at the 1% level to support the claim of a mean below 3 minutes.
t = −0.15 / 0.1643 ≈ −0.913。因为 −0.913 > −2.462，我们不拒绝 H₀。在 1% 水平下，样本没有提供足够的证据支持平均等待时间低于 3 分钟的说法。
8. Two-Sample t-Test Comparing Branches | 双样本 t 检验：分店比较
We compare the average waiting times of Belfast (n₁=50, x̄₁=3.40, s₁=1.20) and Lisburn (n₂=45, x̄₂=3.10, s₂=1.05). Assuming equal population variances, the pooled variance estimate is used. The two-tailed test at 5% significance examines if there is any difference. The pooled standard deviation sp = √{[(49×1.2²)+(44×1.05²)]/(50+45−2)} = √(119.01/93) ≈ √1.2797 ≈ 1.131. The test statistic:
我们比较贝尔法斯特（n₁=50, x̄₁=3.40, s₁=1.20）和利斯本（n₂=45, x̄₂=3.10, s₂=1.05）的平均等待时间。假设总体方差相等，使用合并方差估计。在 5% 显著性水平下的双尾检验考察是否存在任何差异。合并标准差 sp = √{[(49×1.2²)+(44×1.05²)]/(50+45−2)} = √(119.01/93) ≈ √1.2797 ≈ 1.131。检验统计量：
t = (x̄₁ − x̄₂) / [sp √(1/n₁ + 1/n₂)]
t = (3.40 − 3.10) / [1.131 × √(1/50 + 1/45)] = 0.30 / [1.131 × √(0.04222)] = 0.30 / (1.131 × 0.2055) = 0.30 / 0.2324 ≈ 1.291. With 93 df, the critical t-value is 1.986. Since 1.291 < 1.986, we do not reject H₀. There is no significant difference in mean waiting times between the two branches.
t = (3.40 − 3.10) / [1.131 × √(1/50 + 1/45)] = 0.30 / [1.131 × √(0.04222)] = 0.30 / (1.131 × 0.2055) = 0.30 / 0.2324 ≈ 1.291。在 93 自由度下，临界 t 值为 1.986。由于 1.291 < 1.986，我们不拒绝 H₀。两家分店的平均等待时间没有显著差异。
9. Chi-Squared Test for Independence | 卡方独立性检验
The survey data for satisfaction and location are summarised below. We wish to test whether satisfaction is independent of branch location.
关于满意度和分店位置的调查数据总结如下。我们希望检验满意度是否独立于分店位置。
| Location / 位置 | Satisfied / 满意 | Not Satisfied / 不满意 | Total / 总计 |
|---|---|---|---|
| Belfast | 105 | 45 | 150 |
| Derry | 92 | 28 | 120 |
| Lisburn | 95 | 35 | 130 |
| Total | 292 | 108 | 400 |
Under the null hypothesis of independence, expected frequencies are calculated as (row total × column total)/grand total. For example, E for Belfast & Satisfied = (150×292)/400 = 109.5. The chi-squared statistic is:
在独立性的原假设下，期望频数按（行合计 × 列合计）/ 总计 计算。例如，贝尔法斯特且满意的期望值 = (150×292)/400 = 109.5。卡方统计量为：
χ² = Σ (O−E)² / E
Computing all six cells yields χ² ≈ 2.47. Degrees of freedom = (3−1)×(2−1) = 2. The 5% critical value from χ²₂ is 5.991. Since 2.47 < 5.991, we do not reject H₀. Satisfaction appears to be independent of location.
计算所有六个单元格得到 χ² ≈ 2.47。自由度 = (3−1)×(2−1) = 2。χ²₂ 在 5% 的临界值为 5.991。由于 2.47 < 5.991，我们不拒绝 H₀。满意度似乎与分店位置无关。
10. Linear Regression: Advertising and Sales | 线性回归：广告与销售额
Using weekly data from the past 24 weeks for the Belfast branch, with x = advertising spend (£100s) and y = weekly sales (£1000s), the least squares regression line was fitted: y = 8.2 + 1.45x. The coefficient 1.45 indicates that for every extra £100 spent on advertising, weekly sales increase by approximately £1450. The coefficient of determination R² = 0.78 suggests that 78% of the variation in sales can be explained by advertising spend.
利用贝尔法斯特分店过去 24 周的周数据，以 x = 广告支出（百英镑）和 y = 周销售额（千英镑），拟合了最小二乘回归线：y = 8.2 + 1.45x。系数 1.45 表示广告支出每增加 100 英镑，周销售额大约增加 1450 英镑。决定系数 R² = 0.78 表示销售额 78% 的变异可由广告支出来解释。
When the spend is £2000 (x=20), the predicted sales are ŷ = 8.2 + 1.45×20 = 37.2, i.e., £37,200. The assumptions of linearity, constant variance, and independence of residuals were checked via a residual plot and appeared reasonable.
当支出为 2000 英镑（x=20）时，预测销售额为 ŷ = 8.2 + 1.45×20 = 37.2，即 37,200 英镑。通过残差图检查了线性、等方差性和残差独立性等假设，这些假设似乎合理。
11. Time Series Analysis and Forecasting | 时间序列分析与预测
Monthly sales (£1000s) for the entire chain over the last 24 months showed a clear upward trend and strong seasonality with a peak in December. A 4-point centred moving average was calculated to smooth the series and isolate the trend. The seasonal variation for December, after averaging detrended values, was +12.5 (£12,500). If the trend value for next December is forecast to be 45.0, the sales prediction becomes 45.0 + 12.5 = 57.5 (£57,500). Forecasting beyond the immediate future must consider the uncertainty bands.
整个连锁店过去 24 个月的月销售额（千英镑）显示出明显的上升趋势和强烈的季节性，12 月达到峰值。计算了 4 点中心移动平均以平滑序列并分离出趋势。在将去趋势值平均后，12 月的季节性差异为 +12.5（12,500 英镑）。如果预测明年 12 月的趋势值为 45.0，则销售额预测变为 45.0 + 12.5 = 57.5（57,500 英镑）。对较远的未来进行预测必须考虑不确定性区间。
The moving average approach assumes that the past pattern of trend and seasonality will continue, an assumption that should be critically evaluated in a business setting. Identifying outliers or level shifts is also essential before finalising a forecast.
移动平均法假设过去的趋势和季节性模式将持续下去，这一假设在商业环境中应进行批判性评估。在最终确定预测之前，识别异常值或水平变化也至关重要。
12. Exam Preparation Tips and Conclusion | 考试准备技巧与总结
Real-world case studies demand that you choose the correct model, state hypotheses clearly, check conditions (e.g., np>5 for normal approximation, random sampling for chi-squared, equal variances for pooled t-test), and interpret results in context, not just provide numerical answers. Practice using language such as ‘there is insufficient evidence at the x% level to reject H₀’ rather than simply ‘accept H₀’.
真实世界的案例分析要求你选择正确的模型，清晰地陈述假设，检查条件（例如正态近似的 np>5，卡方检验的随机抽样，合并 t 检验的等方差），并在情境中解释结果，而不只是给出数字答案。练习使用诸如“在 x% 水平下没有足够证据拒绝 H₀”之类的表述，而不是简单地说“接受 H₀”。
This collection of interconnected problems shows how statistics is a unified toolkit. From binomial and Poisson probability to hypothesis tests, regression, and time series, each technique builds on a common foundation: understanding variation and making inference under uncertainty. Work through each stage of the case, ensure you can replicate the calculations with your formula sheet, and always link your conclusion to the original problem. Good luck in your CCEA Statistics exam.
这一系列相互关联的问题展示了统计学是一个统一的工具包。从二项和泊松概率到假设检验、回归和时间序列，每种技术都建立在共同的基础上：理解变异性并在不确定性下进行推断
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