📚 AS CCEA Statistics: Case Study Practical Workout | AS CCEA 统计：案例分析实战演练
In this article we work through a complete case study that mirrors the type of investigation you might tackle in the CCEA AS Statistics examination. We will follow a real-world scenario — evaluating the effectiveness of a new teaching method — while applying descriptive statistics, graphical methods, confidence intervals, and hypothesis tests. This practical workout will help you consolidate key statistical concepts and develop the analytical thinking required for AS‑level success.
本文将带你完整走过一个案例分析，模拟 CCEA AS 统计学考试中可能遇到的调查研究。我们将围绕一个真实场景——评估一种新教学方法的有效性——依次运用描述统计、图形方法、置信区间和假设检验。这场实战演练旨在帮助你巩固关键的统计学概念，培养 AS 阶段所需的分析思维能力。
1. Case Background | 案例背景
A mathematics teacher wants to determine whether a new interactive teaching approach genuinely raises students’ test scores. Thirty Year 12 pupils took the same standardised test before and after a four‑week intervention. The teacher recorded each student’s score out of 50 marks. The main question is: does the average improvement differ significantly from zero? This single‑sample problem encourages us to work with paired differences and to perform a complete inferential analysis.
一位数学教师希望判断一种新的互动式教学方法是否能真实地提高学生的测验成绩。三十名 12 年级学生在四周干预期前后参加了同一份标准化测验（满分 50 分）。教师记录了每位学生的得分。核心问题是：平均提升幅度是否显著异于零？这个单样本问题促使我们使用配对差值并进行完整的推断分析。
The raw data consist of 30 pairs of scores. For each student we compute the improvement d = post‑test − pre‑test. The resulting 30 difference values become the focus of our statistical investigation. Throughout the case study we maintain a 5 % significance level and rely on the CCEA AS Statistics formulae booklet.
原始数据由 30 对分数组成。我们为每名学生计算提升值 d = 后测 − 前测，所得的 30 个差值即成为本次统计研究的核心对象。在整个案例分析中，我们保持 5 % 的显著性水平，并参考 CCEA AS 统计学公式手册。
2. Data Collection & Design | 数据收集与设计
The study followed a repeated‑measures design: the same individuals are measured twice, so each student acts as their own control. This paired setup removes between‑subject variability and often yields a more powerful test of the intervention effect. The teacher ensured that both tests were administered in comparable conditions and that no other major learning initiatives occurred during the four weeks.
本研究采用重复测量设计：同一群个体被测量两次，因此每位学生均作为自身对照。这种配对设计消除了受试者间的变异性，通常能给出更有效的干预效应检验。教师确保两次测验在可比条件下进行，且四周内未开展其他重大学习活动。
Before the analysis we checked the dataset for outliers and data‑entry errors. Any score exceeding 50 or below 0 would have been flagged, but the data were clean. The 30 difference scores ranged from 0 to 9, with most values clustering between 2 and 7. This range already suggests a positive shift, but a formal test is required.
分析之前我们检查了数据以排除异常值及录入错误。任何超过 50 或低于 0 的分数都会被标记，不过数据是干净的。30 个差值分数介于 0 至 9 之间，大多数数值集中在 2 到 7。这一范围已暗示存在正向变化，但仍需进行正式的统计检验。
3. Summarising the Difference Scores | 差值分数汇总
We begin with descriptive statistics for the improvement values d. Let n = 30. The sum of the differences Σd = 138, hence the sample mean improvement is:
我们首先计算提升值 d 的描述统计量。样本量 n = 30，差值总和 Σd = 138，因此样本平均提升为：
d̄ = Σd / n = 138 ÷ 30 = 4.6 marks
The standard deviation of the differences, calculated using the formula s = √[Σ(d − d̄)²/(n−1)], came to 1.98 (to 3 s.f.). The median improvement was 4.5, the first quartile 3.0 and the third quartile 6.0. A five‑number summary is displayed in the table below.
差值的标准差通过公式 s = √[Σ(d − d̄)²/(n−1)] 算得为 1.98（保留三位有效数字）。提升中位数为 4.5，第一四分位数 3.0，第三四分位数 6.0。下表展示了五数概括。
| Minimum | 0 |
| Q₁ | 3.0 |
| Median | 4.5 |
| Q₃ | 6.0 |
| Maximum | 9 |
The gap between Q₃ and the maximum is larger than the gap between the minimum and Q₁, hinting at a mild positive skew. However, with 30 observations the distribution remains sufficiently symmetric for our parametric tests.
Q₃ 与最大值之间的跨度大于最小值与 Q₁ 之间的跨度，提示分布有轻微正偏态。不过，30 个观测值的分布仍然足够对称，可以安全使用参数检验。
4. Visualising the Improvement | 可视化提升情况
A histogram of the 30 difference scores reveals a single peak around 4–5 marks, with frequencies falling away on both sides. Overlaying a normal curve shows a reasonable fit, supporting the use of a t‑test. We also constructed a box plot that highlights the median at 4.5 and the interquartile range from 3 to 6, with no outliers beyond the whiskers.
30 个差值分数的直方图显示出位于 4–5 分附近的单峰，两侧频数逐渐降低。叠加上正态曲线后两者较为吻合，这支持了 t 检验的使用。我们还绘制了箱线图，突出中位数 4.5 以及从 3 到 6 的四分位距，须须之外并无异常值。
Visual checks are essential in AS Statistics because they help you assess assumptions before jumping into calculations. The symmetric shape and lack of extreme values reassure us that the sample mean is a reliable measure and that the sampling distribution of the mean will be approximately normal, thanks to the Central Limit Theorem.
在 AS 统计学中，图形检查不可或缺，因为它能帮你在计算前先评估假设。对称的形状和无极端值让我们确信：样本均值是可靠的度量，且根据中心极限定理，均值的抽样分布近似正态。
5. Checking Normality Assumptions | 检查正态性假设
For small to moderate sample sizes, we verify that the population of differences is roughly normal. With n = 30, the t‑test is fairly robust to mild departures. We inspected a normal Q‑Q plot: the points hugged the diagonal, confirming linearity. The skewness coefficient was +0.38, well within the range that permits a normal‑based test.
对于小到中等样本量，我们需要验证差值总体大致服从正态。由于 n = 30，t 检验对轻微偏离相当稳健。我们检查了正态 Q‑Q 图：各点紧贴对角线，证实了线性。偏度系数为 +0.38，完全在允许使用正态检验的范围内。
Had the distribution been clearly non‑normal, we might have considered a transformation or a non‑parametric alternative such as the Wilcoxon signed‑rank test. However, given the evidence, we proceed with the classic one‑sample t‑test, which is the AS‑level standard.
如果分布明显非正态，我们或许要考虑数据变换，或改用非参数替代方法，例如 Wilcoxon 符号秩检验。但根据目前证据，我们继续使用经典的、符合 AS 标准的单样本 t 检验。
6. Confidence Interval for Mean Improvement | 平均提高的置信区间
A 95 % confidence interval for the true mean improvement μd is built from the sample statistics. With 29 degrees of freedom, the critical t‑value (two‑tailed) from the table is t0.025,29 = 2.045. The standard error of the mean is s / √n = 1.98 / √30 ≈ 0.3615.
真实平均提升 μd 的 95 % 置信区间由样本统计量构建。自由度为 29 时，查表得双侧临界 t 值 t0.025,29 = 2.045。均值的标准误为 s / √n = 1.98 / √30 ≈ 0.3615。
95 % CI = d̄ ± t0.025,29 × (s/√n) = 4.6 ± 2.045 × 0.3615 ≈ (3.86, 5.34)
We are 95 % confident that the long‑run average improvement lies between 3.86 and 5.34 marks. Because the entire interval is above zero, we have strong evidence that the teaching method produces a genuine positive effect.
我们有 95 % 的信心认为，长期平均提升落在 3.86 至 5.34 分之间。由于整个区间都高于零，我们有强有力的证据表明该教学方法产生了真实的正向效果。
7. One‑Sample t‑Test | 单样本 t 检验
We now formalise the test of significance. The null and alternative hypotheses are:
现在我们正式进行显著性检验。原假设与备择假设为：
H₀: μd = 0; H₁: μd ≠ 0
The test statistic is calculated using the standard formula:
检验统计量使用标准公式计算：
t = (d̄ − 0) / (s/√n) = 4.6 / 0.3615 ≈ 12.72
With df = 29, the critical values at the 5 % level are ±2.045. Our observed t = 12.72 falls far into the rejection region. Alternatively, the p‑value is extremely small (p < 0.001). Therefore we reject H₀ and conclude that the mean improvement is statistically significant.
自由度 df = 29，5 % 水平下的临界值为 ±2.045。我们观测到的 t = 12.72 远远落入拒绝域。另一种方式，p 值极小（p < 0.001）。因此我们拒绝 H₀，得出平均提升在统计上显著异于零的结论。
The calculations above assume a two‑tailed test; in reality, because we expected improvement, a one‑tailed test could also be justified. The conclusion would be equally strong.
以上计算基于双侧检验；实际上，由于我们预期分数会提升，使用单侧检验也是合理的。结论会同样有力。
8. Interpreting the Results | 结果解读
Statistical significance tells us that the observed improvement is unlikely to be due to chance. However, practical significance is equally important: a gain of about 4.6 marks on a 50‑mark test represents a meaningful 9.2 % increase. The confidence interval narrows the estimate to between 7.7 % and 10.7 % of the total marks.
统计显著性告诉我们，观测到的提升不太可能源于偶然。但实际显著性同样重要：在一份满分为 50 分的测验中，提升约 4.6 分相当于有实际意义的 9.2 % 增幅。置信区间将这一估计收窄至总分的 7.7 % 到 10.7 % 之间。
We should remind ourselves that association does not imply causation without a carefully controlled design. Nevertheless, the paired design, the clear temporal sequence, and the absence of other known interventions strengthen the causal interpretation.
我们需要提醒自己，如果没有精心控制的设计，相关性并不意味着因果性。然而，本研究中的配对设计、明确的时间顺序以及没有其他已知干预措施，这些都增强了因果解释的可信度。
9. Extending the Analysis: Two‑Sample Comparison | 扩展分析：双样本比较
Suppose the teacher also wants to compare the new method against a traditional lecture approach. The 30 students could be split into two independent groups: 15 who received the interactive method (Group A) and 15 who were taught conventionally (Group B). Group A’s mean improvement was 5.2 with standard deviation 1.5; Group B’s mean improvement was 4.0 with standard deviation 1.9.
假设教师还想将新方法与传统的讲授法进行比较。30 名学生可分为两个独立组：15 人接受互动式教学（A 组），15 人接受常规教学（B 组）。A 组的平均提升为 5.2，标准差 1.5；B 组的平均提升为 4.0，标准差 1.9。
A two‑sample t‑test (assuming equal variances) could be applied to test H₀: μA − μB = 0. The pooled variance and the appropriate t‑statistic would be calculated. This extension illustrates how the same investigative framework adapts to comparative studies, a skill often examined in AS papers.
可以应用双样本 t 检验（假设方差相等）来检验 H₀: μA − μB = 0。需要计算合并方差及相应的 t 统计量。这一扩展展示了相同的分析框架如何适配比较研究，这也是 AS 考试中常见的能力要求。
10. Conclusion & Reflection | 结论与反思
This case study has walked you through the full statistical pipeline: posing a question, collecting paired data, summarising with graphs and numbers, checking assumptions, constructing a confidence interval, and performing a hypothesis test. The evidence strongly supports the effectiveness of the interactive teaching method.
本案例带你走完了统计分析的完整流程：提出问题、收集配对数据、用图形和数字进行汇总、检查假设、构建置信区间并实施假设检验。证据有力地支持了互动教学方法的有效性。
As you prepare for your CCEA AS Statistics exam, practise extracting the key steps from any scenario you encounter. Look for the variable of interest, decide whether data are paired or independent, choose an appropriate inferential method, and always interpret your findings in context. Building this disciplined analytical routine will serve you well on the exam and beyond.
在备战 CCEA AS 统计学考试时，请练习从任何情境中提取关键步骤。寻找感兴趣变量，判断数据是配对还是独立，选择合适的推断方法，并始终结合情境解读你的发现。养成这种严谨的分析习惯，不仅有助于考试，更会让你长期受益。
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