📚 Pre-U CCEA Statistics: Case Study Practical Work | Pre-U CCEA 统计：案例分析实战演练
In Pre-U Statistics, the ability to apply statistical methods to real-world scenarios is essential for success. This article walks you through a complete case study, from formulating a research question to interpreting results and evaluating limitations. The example used throughout is an investigation into whether a new targeted revision programme improves performance on CCEA Statistics past-paper scores. By the end, you will have a clear template for tackling your own practical investigations with confidence.
在 Pre-U 统计学习中，将统计方法应用到真实场景是取得好成绩的关键。本文通过一个完整的案例研究，带您从提出研究问题开始，一直到结果解读与局限性评估。贯穿全文的例子是：探讨一种新的针对性复习方案能否提高学生在 CCEA 统计历年真题中的成绩。读完本文，您将掌握一个清晰的模板，能够自信地应对自己的实践探究。
1. Introducing the Case Study | 案例引入
A Pre-U teacher suspects that a structured revision programme focusing on past-paper techniques can raise the mean score of her students. She decides to conduct a study with her two Year 13 Statistics classes of similar prior attainment. Class A receives the standard revision support, while Class B follows the new programme over six weeks. Both groups then sit the same past paper under exam conditions, and their scores out of 100 are recorded. This realistic setup allows us to demonstrate the full statistical cycle.
一位 Pre-U 教师猜想，一套以真题技巧为核心的结构化复习方案可以提升学生的平均分。她决定在自己所教的两个先修统计班级中进行研究，这两个班级之前成绩水平相近。A 班接受常规复习辅导，B 班则进行为期六周的新方案。之后两个班级在考试条件下完成同一份历年真题，满分为 100 分并记录成绩。这个贴近实际的设定能够完整展示统计探究的全过程。
2. Research Question and Hypotheses | 研究问题与假设
The research question is: ‘Does the new revision programme lead to a higher mean score on the past paper compared with standard revision?’ We define population parameter μ₁ as the mean score for all students under standard revision and μ₂ as the mean under the new programme. The null hypothesis H₀: μ₂ – μ₁ = 0 states there is no difference. The alternative is one‑tailed: H₁: μ₂ – μ₁ > 0, because the teacher believes the new programme is beneficial. Using a two‑sample context keeps the analysis rigorous and in line with CCEA assessment expectations.
研究问题是：“与常规复习相比，新的复习方案是否能使真题平均分更高？”我们定义总体参数 μ₁ 为所有接受常规复习学生的平均分，μ₂ 为接受新方案学生的平均分。原假设 H₀: μ₂ – μ₁ = 0 表示没有差异。备择假设为单侧：H₁: μ₂ – μ₁ > 0，因为教师相信新方案有益。采用双样本分析框架既严谨，又符合 CCEA 评估的要求。
3. Study Design and Sampling | 研究设计与抽样
The teacher used two intact classes as a convenience sample. To strengthen the design, she checked that the two groups had very similar means and standard deviations in the previous term’s mock exam. Ideally, a randomised controlled trial would randomly allocate students to groups, but here the practical constraints are acknowledged. Each group contains 16 students, giving us a total of 32 independent observations. We will treat the scores as approximately normally distributed based on past evidence, allowing parametric tests to be used.
教师使用了两个现成班级作为方便样本。为了增强设计可靠性，她检验了两个组在上学期模考中的均值和标准差非常接近。最理想的是随机对照试验，将学生随机分配到各组，但此处承认实际操作上的限制。每组有 16 名学生，共 32 个独立观测值。基于以往经验，我们将成绩视为近似正态分布，从而可以使用参数检验。
4. Exploratory Data Analysis: Descriptive Statistics | 探索性数据分析：描述性统计
Before formal inference, we calculate summary statistics. For Class A (standard): n₁ = 16, sample mean x̄₁ = 62.5, sample standard deviation s₁ = 10.2. For Class B (new programme): n₂ = 16, x̄₂ = 70.8, s₂ = 8.9. The difference in sample means is +8.3 marks. Boxplots show Class B has a higher median and smaller spread, with no obvious outliers. These visual and numerical summaries suggest a potentially meaningful improvement, but we need to test whether this difference is statistically significant.
在进行正式推断前，我们先计算汇总统计量。A 班（常规复习）：n₁ = 16，样本均值 x̄₁ = 62.5，样本标准差 s₁ = 10.2。B 班（新方案）：n₂ = 16，x̄₂ = 70.8，s₂ = 8.9。样本均值差为 +8.3 分。箱线图显示 B 班中位数更高且离散程度更小，没有明显的异常值。这些图形与数值汇总暗示可能存在有意义的提高，但我们需要检验这个差异是否具有统计显著性。
5. Probability Distributions and Assumptions | 概率分布与假设
We assume the exam scores in each population are independent and normally distributed with unknown but possibly equal variances. To verify normality, normal probability plots can be drawn; with 16 points per group, slight deviations are acceptable due to the robustness of the t-test. The assumption of equal variances can be checked using an F‑test. Here, s₁² = 104.0 and s₂² = 79.2, giving F = 104.0/79.2 ≈ 1.31. With degrees of freedom (15,15), the critical value at 5% is about 2.40, so we do not reject equal variances and thus use a pooled two‑sample t‑test.
我们假设每个总体的考试成绩相互独立且服从正态分布，方差未知但可能相等。为了验证正态性，可以绘制正态概率图；每组 16 个数据点，即使有轻微偏离，由于 t 检验的稳健性也是可以接受的。方差齐性可以通过 F 检验来检查。此处 s₁² = 104.0，s₂² = 79.2，F = 104.0/79.2 ≈ 1.31。自由度为 (15, 15)，5% 水平下的临界值约为 2.40，因此我们不拒绝方差齐性的假设，从而使用合并双样本 t 检验。
6. Confidence Interval for the Difference in Means | 均值差异的置信区间
We construct a 95% confidence interval for μ₂ – μ₁ using the pooled estimate of variance. The pooled standard deviation sₚ = √[((n₁−1)s₁² + (n₂−1)s₂²)/(n₁+n₂−2)] = √[(15×104.0 + 15×79.2)/30] = √[(1560 + 1188)/30] = √[91.6] ≈ 9.57. The standard error of the difference is sₚ × √(1/n₁ + 1/n₂) = 9.57 × √(1/16+1/16) = 9.57 × √(0.125) ≈ 3.38. With 30 degrees of freedom, t₀.₀₂₅ = 2.042. Thus the interval is (70.8 − 62.5) ± 2.042 × 3.38 = 8.3 ± 6.90, giving (1.4, 15.2). Since the interval does not contain zero, it suggests a statistically significant improvement at the 5% level.
我们使用合并方差估计量构建 μ₂ – μ₁ 的 95% 置信区间。合并标准差 sₚ = √[((n₁−1)s₁² + (n₂−1)s₂²)/(n₁+n₂−2)] = √[(15×104.0 + 15×79.2)/30] = √[(1560 + 1188)/30] = √[91.6] ≈ 9.57。差异的标准误为 sₚ × √(1/n₁ + 1/n₂) = 9.57 × √(1/16+1/16) = 9.57 × √(0.125) ≈ 3.38。自由度为 30 时，t₀.₀₂₅ = 2.042。因此区间为 (70.8 − 62.5) ± 2.042 × 3.38 = 8.3 ± 6.90，即 (1.4, 15.2)。该区间不包含零，表明在 5% 水平上新方案的提高具有统计显著性。
7. Two-Sample t‑Test for the Hypothesis | 双样本 t 检验验证假设
We perform a one‑tailed independent‑samples t‑test. The test statistic is t = (x̄₂ − x̄₁) / (sₚ × √(1/n₁+1/n₂)) = 8.3 / 3.38 ≈ 2.455. With 30 degrees of freedom, the critical value for a one‑tailed test at α = 0.05 is 1.697. Since 2.455 > 1.697, we reject H₀. The p‑value is approximately 0.010, meaning the probability of observing a difference of at least 8.3 marks by chance, if the new programme had no true effect, is very low. We conclude there is sufficient evidence that the new revision programme raises mean scores.
我们进行单侧独立样本 t 检验。检验统计量为 t = (x̄₂ − x̄₁) / (sₚ × √(1/n₁+1/n₂)) = 8.3 / 3.38 ≈ 2.455。自由度为 30 时，α = 0.05 下单侧检验的临界值为 1.697。由于 2.455 > 1.697，我们拒绝 H₀。p 值约为 0.010，意味着如果新方案真的无效，观察到至少 8.3 分差异的概率非常低。我们得出结论：有充分证据表明新的复习方案能提高平均分。
8. Checking the Assumptions with a Chi-Squared Test for Normality | 用卡方检验验证正态性
As an additional check, we can perform a chi‑squared goodness‑of‑fit test on the combined or individual samples to confirm normality. For Class B scores grouped into intervals (e.g., ≤60, 61–70, 71–80, >80), the expected frequencies based on a normal distribution with mean 70.8 and s = 8.9 are computed. The observed frequencies are 3, 5, 6, 2. The expected are about 3.8, 4.8, 4.3, 3.1. The χ² statistic = Σ (O−E)²/E ≈ 0.6. With 1 degree of freedom (using pooled groups), the critical value at 5% is 3.84, so the data are consistent with normality. This supports our earlier parametric test choice.
作为额外检验，我们可以对合并样本或各组进行卡方拟合优度检验以确认正态性。将 B 班成绩分组为若干区间（如 ≤60, 61–70, 71–80, >80），基于均值 70.8 和 s = 8.9 的正态分布计算期望频数。实际频数为 3, 5, 6, 2。期望频数约为 3.8, 4.8, 4.3, 3.1。χ² 统计量 = Σ (O−E)²/E ≈ 0.6。自由度为 1（合并组后），5% 临界值为 3.84，因此数据与正态分布一致，支持我们之前选用参数检验。
9. Correlation and Simple Linear Regression | 相关分析与简单线性回归
Suppose we also recorded the number of hours each student actively participated in the revision sessions (0 to 30 hours). We want to model the relationship between hours (x) and exam score (y) for the 32 students. The scatterplot shows a positive linear trend. The Pearson correlation coefficient r = 0.647. The regression equation is y = 54.2 + 0.96x. The slope’s 95% confidence interval is (0.45, 1.47), and the coefficient of determination r² = 0.419 indicates that about 41.9% of the variation in scores is explained by session hours. The slope is significantly different from zero (t = 3.89, p < 0.001), confirming that greater participation is associated with higher achievement.
假设我们还记录了每个学生积极参与复习课的时数（0 到 30 小时）。我们想为全部 32 名学生建立学习时数 (x) 与考试成绩 (y) 之间的关系模型。散点图呈现正线性趋势。Pearson 相关系数 r = 0.647。回归方程为 y = 54.2 + 0.96x。斜率的 95% 置信区间为 (0.45, 1.47)，决定系数 r² = 0.419 表明约 41.9% 的成绩变异可由学习时数解释。斜率显著不等于零（t = 3.89, p < 0.001），证实了积极参与与高分之间存在关联。
10. Extension: Non‑parametric Alternative | 拓展：非参数替代方法
If the normality assumption were seriously violated, a Mann–Whitney U‑test could be used. Ranking all 32 scores, the sum of ranks for Class B is 604. The expected rank sum under H₀ is 33×32/4 = 264. The test statistic U = 16×16 + 16×17/2 − 604 = 164, with a standardised z ≈ 2.52, yielding a two‑tailed p‑value ≈ 0.012. This aligns with the t‑test result and illustrates how to choose methods when assumptions are suspect. In the CCEA exam, you may be asked to discuss the advantages and disadvantages of parametric versus non‑parametric procedures.
如果正态性假设严重不满足，则可以采用 Mann–Whitney U 检验。对所有 32 个成绩排序后，B 班的秩和为 604。H₀ 下的期望秩和为 33×32/4 = 264。检验统计量 U = 16×16 + 16×17/2 − 604 = 164，标准化的 z ≈ 2.52，双侧 p 值约为 0.012。这与 t 检验的结果一致，也说明了在假设存疑时如何选择方法。在 CCEA 考试中，可能会要求讨论参数与非参数方法的优缺点。
11. Interpretation and Drawing Conclusions | 结果解读与得出结论
Statistical significance does not automatically imply educational importance. The observed mean difference of 8.3 marks and a confidence interval of (1.4, 15.2) suggest a practically meaningful gain. The effect size, measured by Cohen’s d = (x̄₂ − x̄₁)/sₚ ≈ 8.3/9.57 = 0.87, is considered large, indicating the programme has a substantial impact. Therefore, the teacher can confidently adopt the new programme. The regression analysis further reveals that the benefit increases with more active participation, offering a dose‑response insight that strengthens the argument for the programme’s effectiveness.
统计显著并不自动意味着教育上重要。观察到的 8.3 分平均差异以及 (1.4, 15.2) 的置信区间表明在实际中有意义的提升。以 Cohen’s d = (x̄₂ − x̄₁)/sₚ ≈ 8.3/9.57 = 0.87 衡量的效应量属于大效应，说明方案具有实质性的影响。因此，教师可以自信地采用新方案。回归分析还显示，积极参与时间越长收益越大，这种剂量–反应关系进一步证实了方案的有效性。
12. Critical Evaluation and Limitations | 批判性评价与局限性
No study is without limitations. The use of convenience sampling restricts generalisability to the wider Pre-U population. The quasi‑experimental design means we cannot rule out confounding variables such as teacher enthusiasm or peer effects. Sample size is modest, which influences the power to detect smaller effects. Reliability of the assessment depends on the particular past paper used. In a formal report, these points should be discussed honestly, alongside suggestions for future research, such as randomisation, larger samples, and multiple outcome measures. Acknowledging limitations demonstrates statistical maturity and is highly rewarded in CCEA examinations.
任何研究都有局限性。方便样本的使用限制了结果向更广大的 Pre-U 群体的推广。准实验设计意味着我们不能排除干扰变量的影响，如教师热情或同伴效应。样本量不大，影响了检测较小效应的把握度。评估的信度取决于所使用的特定历年真题。在正式报告中，应诚实讨论这些要点，并提出未来研究建议，如随机分组、扩大样本量和多样化的结局指标。承认局限性体现了统计思维的成熟度，在 CCEA 考试中会得到高度认可。
Published by TutorHao | Pre-U CCEA Statistics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply