📚 Year 13 WJEC Statistics: Mastering Case Study Analysis | A Level WJEC 统计：案例分析实战演练
In Year 13 WJEC Statistics, the ability to apply a wide range of inferential techniques to a single real-world scenario is a crucial skill. This practical drill walks you through a comprehensive case study, demonstrating how hypotheses are formulated, assumptions checked, and appropriate tests selected and interpreted. Mastering this integrated approach not only prepares you for examination questions but also deepens your understanding of statistical reasoning in context.
在 WJEC 的 Year 13 统计课程中，将各种推断技术灵活运用于同一个真实情境是一项关键能力。本篇实战演练带你深入一个完整的案例研究，展示如何构建假设、检验前提条件、选择合适的检验方法并加以解读。掌握这种综合分析方法，不仅有助于应对考试题目，也能加深你对统计推理在具体情境中应用的理解。
1. Case Introduction and Data Overview | 案例介绍与数据概览
A sixth-form college introduced an online learning platform as a supplement to normal mathematics lessons. To evaluate its effect, 20 Year 13 students were randomly selected. They sat a pre-test before using the platform and a similar post-test after eight weeks. Additional data were recorded for each student: gender, average weekly hours spent on the platform, and a satisfaction survey completed at the end. The satisfaction survey asked students to choose from 'Very Satisfied', 'Satisfied', 'Neutral' or 'Dissatisfied'. Usage level was categorised as Low (less than 3 hours per week), Medium (3 to 6 hours per week) and High (more than 6 hours per week).
某高中六年级学院引入了一个在线学习平台，作为数学课堂教学的补充。为评估其效果，随机选取了 20 名 Year 13 学生。他们在使用平台前参加了一次前测，八周后参加了一次类似的后测。此外，还记录了每名学生的性别、每周在平台上的平均学习时长，以及期末完成的满意度调查。满意度调查要求学生在 'Very Satisfied'、'Satisfied'、'Neutral'、'Dissatisfied' 四个选项中选择。使用水平分为三档：Low（每周少于 3 小时）、Medium（每周 3 至 6 小时）、High（每周超过 6 小时）。
Both tests were marked out of 100. The pre-test mean was 72.3 marks. The post-test mean was 76.8 marks with a standard deviation of 7.9. Pairing each student's two scores gives a mean difference of 4.5 marks with a standard deviation of 5.2.
两次测验满分均为 100 分。前测均值为 72.3 分，后测均值为 76.8 分，标准差为 7.9。将每名学生的两次成绩配对后，得到平均差为 4.5 分，差值的标准差为 5.2。
d = post-test score − pre-test score d̄ = Σd ÷ n = 4.5 sd = 5.2
| Variable 变量 | Type 类型 | Role in analysis 分析角色 |
| Pre-test score 前测分 | Continuous 连续 | Paired observation 配对观测 |
| Post-test score 后测分 | Continuous 连续 | Paired observation 配对观测 |
| Difference d 差值 | Continuous 连续 | Response for paired t-test 配对 t 检验的响应变量 |
| Gender 性别 | Nominal binary 二分类名义 | Grouping variable 分组变量 |
| Weekly hours 每周时长 | Continuous 连续 | Explanatory variable 解释变量 |
| Usage level 使用水平 | Ordinal 有序分类 | Grouping for chi-squared 卡方分组 |
| Satisfaction 满意度 | Ordinal 有序分类 | Categorical response 分类响应 |
2. Formulating Hypotheses for the Paired Design | 为配对设计构建假设
oach is a paired t-test on the differences. The parameter of interest is the population mean difference, μ_d, not the difference of the two sample means.
第一个分析问题是：平台是否提升了成绩。由于每名学生提供两个相互关联的成绩，最自然的做法是对差值做配对 t 检验。我们关心的参数是总体平均差 μ_d，而不是两个样本均值之差。
Because the college wants to know specifically whether scores rose, a one-tailed alternative is defensible. If the question simply asks whether there is any change, use a two-tailed alternative. State the hypotheses before looking at the data.
由于学院想明确知道成绩是否上升，采用单侧备择假设是有依据的。如果题目只问是否存在差异，则应使用双侧备择假设。务必在查看数据之前把假设写清楚。
H₀: μ_d = 0 H₁: μ_d > 0 (one-tailed 单侧)
or 或 H₁: μ_d ≠ 0 (two-tailed 双侧)
ypotheses in words as well as symbols. A common examiner expectation is a sentence such as 'the mean improvement in the population of Year 13 students is zero'. Always define μ_d in the context of the case study.
假设除了符号形式，也要用文字表述。考官常要求写出类似"Year 13 学生总体的平均提升量为零"的句子。一定要结合本案例情境定义 μ_d。
3. Checking Assumptions Before Testing | 检验前的前提条件核查
tion of differences should be approximately normally distributed. With n = 20 the normality condition matters, because the central limit theorem only rescues us comfortably at larger sample sizes.
配对 t 检验需要三个条件：配对必须真实成立；各差值之间相互独立；差值的总体应近似服从正态分布。由于 n = 20 并不算大，正态性条件就很重要，因为中心极限定理只在样本量较大时才能让人放心。
- Independence 独立性: the 20 students were randomly selected, so their differences can be treated as independent.
- 随机抽取保证了学生之间独立，因此差值可视为相互独立。
- Paired structure 配对结构: each pre-test is tied to the same student's post-test, so the observations are not independent, but the differences are.
- 每个前测与同一名学生的后测绑定，因此观测值本身不独立，但"差值"是独立的。
- Normality 正态性: examine a histogram or box plot of the 20 differences, or use a normal probability plot. Look for strong skewness or extreme outliers.
- 正态性：画出 20 个差值的直方图或箱线图，或使用正态概率图，重点看是否存在严重偏斜或极端离群值。
If normality is clearly violated, report the Wilcoxon signed-rank test instead. It tests the same hypothesis about the median difference and does not require normality, only symmetry of the differences.
若正态性明显不满足，应改用 Wilcoxon 符号秩检验。它检验的是中位数差是否为零，只要求差值分布大致对称，不要求正态。
4. Carrying Out the Paired t-Test | 执行配对 t 检验
The test statistic divides the observed mean difference by its standard error. The standard error of d̄ is sd divided by the square root of n, and the degrees of freedom are n − 1.
检验统计量等于观测到的平均差除以其标准误。d̄ 的标准误为 sd 除以 √n，自由度为 n − 1。
t = (d̄ − 0) ÷ (sd ÷ √n)
t = 4.5 ÷ (5.2 ÷ √20) = 4.5 ÷ 1.1628 = 3.87
df = n − 1 = 19
The critical values at the 5% ther case. The observed significance level is roughly p ≈ 0.001, which is well below 0.05.
5% 显著水平下的临界值，单侧为 1.729，双侧为 2.093。由于 3.87 大于两者，无论单侧还是双侧都应拒绝 H₀。实际显著性水平约为 p ≈ 0.001，远低于 0.05。
Note that the standard error uses sd, the standard deviation of the differences, not the two separate standard deviations reported for pre- and post-tests. Mixing these up is one of the most frequent errors in this topic.
注意标准误使用的是差值的标准差 sd，而不是前测和后测各自的标准差。把这两者混用是本专题最常见的错误之一。
5. Interpreting the p-value and Effect Size | 解读 p 值与效应量
A small p-value tells us that a mean difference of 4.5 marks would be unlikely if the platform truly had no effect. It does not measure how large the effect is. To judge practical importance we compute an effect size, most commonly Cohen's d for paired data.
p 值很小说明：如果平台真的毫无效果，那么出现 4.5 分的平均差是极不可能的。但 p 值并不衡量效应有多大。要判断实际重要性，需要计算效应量，配对数据最常用的是 Cohen's d。
Cohen's d = d̄ ÷ sd = 4.5 ÷ 5.2 = 0.87
A value of 0.87 is conventionally described as a large effect. In context, the platform is associated with an improvement of about 0.87 standard deviations of the paired differences, which for a mathematics intervention is substantial.
0.87 通常被描述为大效应。结合情境，该平台带来的提升约为配对差值标准差的 0.87 倍，对数学干预而言幅度相当可观。
- Statistical significance 统计显著性: p ≈ 0.001, reject H₀.
- 统计显著性：p ≈ 0.001，拒绝 H₀。
- Practical significance 实际显著性: mean gain of 4.5 marks out of 100, roughly half a grade boundary on some specifications.
- 实际显著性：满分 100 分中平均提升 4.5 分，在部分考试局标准下约相当于半个等级分界。
- Caution 谨慎: a pre-test / post-test design without a control group cannot separate the platform effect from normal teaching, maturation or practice effects.
- 注意：没有对照组的"前测—后测"设计，无法把平台效应与常规教学、学生自然成长或练习效应区分开。
6. Confidence Interval for the Mean Difference | 平均差的置信区间
a plausible range of values for μ_d. Use the t distribution with n − 1 degrees of freedom.
置信区间通常比单纯检验更有信息量，因为它给出了 μ_d 的合理取值范围。应使用自由度为 n − 1 的 t 分布。
d̄ ± t₍₀.₉₇₅, ₁₉₎ × (sd ÷ √n)
4.5 ± 2.093 × 1.1628 = 4.5 ± 2.43
95% CI: (2.07, 6.93)
Interpretation: we are 95% confident that the true mean improvement for Year 13 students using the platform for eight weeks lies between about 2.1 and 6.9 marks. Because the interval excludes zero, it agrees with the two-tailed test decision at the 5% level.
解读：我们有 95% 的把握认为，使用该平台八周的 Year 13 学生，其总体平均提升量介于约 2.1 分至 6.9 分之间。由于区间不包含 0，这与 5% 水平下的双侧检验结论一致。
Always write the confidence level, the parameter and the units. A common mark is lost for saying '95% of students improved', which confuses the parameter with individual outcomes.
务必写清置信水平、参数对象和单位。写成"95% 的学生都进步了"会失分，因为这把总体参数与个体结果混淆了。
7. Comparing Two Independent Groups by Gender | 按性别比较两个独立组
s is appropriate. Suppose the males (n₁ = 11) had mean 78.4 with standard deviation 7.1, and the females (n₂ = 9) had mean 74.8 with standard deviation 8.4, both in marks.
第二个问题是：男生与女生的后测平均分是否存在差异。此时两个样本相互独立，应使用两独立样本均值 t 检验。假设男生（n₁ = 11）均值为 78.4，标准差为 7.1；女生（n₂ = 9）均值为 74.8，标准差为 8.4，单位均为分。
0 against H₁: μ₁ − μ₂ ≠ 0. Assuming equal population variances, pool the sample variances.
假设为 H₀: μ₁ − μ₂ = 0，备择假设为 H₁: μ₁ − μ₂ ≠ 0。若假设两总体方差相等，则需合并样本方差。
s²ₚ = [(n₁ − 1)s₁² + (n₂ − 1)s₂²] ÷ (n₁ + n₂ − 2)
s²ₚ = (10 × 50.41 + 8 × 70.56) ÷ 18 = 59.37 sₚ = 7.71
SE = sₚ × √1/n₁ + 1/n₂ = 7.71 × √0.0909 + 0.1111 = 3.46
t = (78.4 − 74.8) ÷ 3.46 = 1.04 df = 18
The critical value at the 5% two-tailed level with 18 degrees of freedom is 2.101. Since 1.04 < 2.101, there is insufficient evidence to conclude that the mean post-test scores of males and females differ. The p-value is roughly 0.31.
自由度 18、5% 双侧水平的临界值为 2.101。由于 1.04 < 2.101，没有足够证据认为男生与女生的后测平均分存在差异。p 值约为 0.31。
Also note the pre-test scores should be checked. If the groups differed at pre-test, a gain-score or analysis of covariance would be a fairer comparison, which is a strong evaluative point to make.
同时应注意核查前测成绩。如果两组在前测时已有差异，用增益分数或协方差分析会更公平——这是很好的评价性加分点。
8. Chi-squared Test for
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