📚 Year 13 WJEC Statistics: Case Study Practical Drills | Year 13 WJEC 统计：案例分析实战演练
In WJEC A‑Level Statistics, case study questions demand the ability to translate a real‑world context into a statistical investigation, select the correct model, perform calculations accurately, and communicate findings clearly. This article walks you through three complete case studies that mirror the style and depth required at Year 13, covering two‑sample t‑tests, chi‑squared tests for independence, and linear regression with hypothesis testing. Each step is explained in both English and Chinese, so you can master the terminology and logic in both languages.
在 WJEC A‑Level 统计中，案例分析题要求你将真实情境转化为统计调查，正确选择模型，精准计算并清晰表达结论。本文通过三个完整案例，逐一演练 Year 13 所需的双样本 t 检验、卡方独立性检验以及线性回归与假设检验。每一步都用中英双语讲解，帮助你掌握两语术语和推理逻辑。
1. What Makes a Strong WJEC Case Study Response? | 高分 WJEC 案例分析的关键
A successful WJEC case study answer always starts by identifying the type of data (continuous or categorical), the number of groups, and the hypothesis to be tested. Before you touch a formula, write down the null and alternative hypotheses, the significance level, and any assumptions you are making. Examiners look for methodical working, correct use of statistical tables, and a concluding statement that directly answers the original question.
一份高分的 WJEC 案例分析答案，首先要识别数据类型（连续或分类）、组别数量以及待检验的假设。在动笔计算之前，先写下原假设与备择假设、显著性水平以及所依赖的假设条件。考官看重的是有条理的工作步骤、正确使用统计表格，以及能直接回答原始问题的结论陈述。
Throughout these examples, you will see how to report a test statistic alongside its critical value, how to phrase a conclusion in context, and how to check that your interpretation does not overclaim causality when only association is shown.
在接下来的例子中，你将看到如何同时报告检验统计量与临界值、如何在情境中表述结论，以及如何避免在只显示关联性时错误地推断因果关系。
2. Case Study 1: Comparing Two Teaching Methods | 案例一：比较两种教学方法
Context: A school randomly assigns 30 students to a traditional lecture‑based class (Group A) and 30 students to a new technology‑enhanced class (Group B). After one term, all students sit the same examination. Group A earns a mean score of 75 with a standard deviation of 8; Group B earns a mean of 80 with a standard deviation of 7. The school wants to know if the new method produces a statistically significant difference in mean scores at the 5% level.
情境：某学校将 30 名学生随机分配到传统讲授班（A 组），另外 30 名分配到新科技强化班（B 组）。一个学期后，所有学生参加同一场考试。A 组平均分 75，标准差 8；B 组平均分 80，标准差 7。学校想知道在 5% 显著性水平下，新方法是否带来了统计上显著的平均分差异。
The response variable is a continuous score, the samples are independent, and the population variances are unknown. We therefore apply a two‑sample t‑test. We first assume equal variances and check this assumption with an F‑test; if the equal‑variance assumption is tenable, we will pool the variances.
响应变量是连续分数，样本独立，总体方差未知。因此我们使用双样本 t 检验。首先假定方差相等并用 F 检验核查；若等方差假设成立，则合并方差。
3. Setting Up the Hypotheses and Checking Assumptions | 设定假设与核查条件
Let μₐ be the population mean for Group A and μ_b for Group B. Since the school asks about a difference without specifying a direction, a two‑tailed test is appropriate.
记 μₐ 为 A 组总体均值，μ_b 为 B 组总体均值。学校仅询问是否有差异，未预设方向，因此采用双尾检验。
H₀: μₐ = μ_b
H₁: μₐ ≠ μ_b
Significance level α = 0.05. The sample sizes are equal (nₐ = n_b = 30). We check the assumption of equal variances using an F‑test: H₀: σₐ² = σ_b². The test statistic is F = sₐ² / s_b² (larger variance in numerator). Here sₐ² = 64, s_b² = 49, so F = 64 / 49 ≈ 1.306. With degrees of freedom (29, 29), the two‑tailed critical value at 5% is about 2.10 (from tables). Since 1.306 < 2.10, we do not reject equal variances, so pooled procedures are justified.
显著性水平 α = 0.05。样本量相等（nₐ = n_b = 30）。我们用 F 检验核查等方差假设：H₀: σₐ² = σ_b²。检验统计量 F = sₐ² / s_b²（较大方差在分子）。此处 sₐ² = 64，s_b² = 49，F = 64 / 49 ≈ 1.306。自由度为 (29, 29)，5% 双尾临界值约为 2.10（查表）。因为 1.306 < 2.10，我们不拒绝等方差假设，可采用合并方差法。
4. Computing the Pooled t‑Statistic | 计算合并 t 统计量
We calculate the pooled variance s²ₚ:
我们先计算合并方差 s²ₚ：
s²ₚ = ((nₐ − 1)sₐ² + (n_b − 1)s_b²) / (nₐ + n_b − 2)
= (29 × 64 + 29 × 49) / (30 + 30 − 2)
= (1856 + 1421) / 58 = 3277 / 58 ≈ 56.50
Thus the pooled standard deviation sₚ = √56.50 ≈ 7.517. The standard error of the difference in means is:
因此合并标准差 sₚ = √56.50 ≈ 7.517。均值差的标准误为：
SE = sₚ √(1/nₐ + 1/n_b) = 7.517 × √(1/30 + 1/30) = 7.517 × √(0.0667) ≈ 7.517 × 0.2582 ≈ 1.941
The t‑statistic is:
进而计算 t 统计量：
t = (x̄ₐ − x̄_b) / SE = (75 − 80) / 1.941 ≈ −2.575
We take the absolute value for comparison with the critical value.
与临界值比较时我们取绝对值。
5. Critical Value, Decision, and Confidence Interval | 临界值、决策与置信区间
Degrees of freedom df = nₐ + n_b − 2 = 58. From the t‑distribution table, the two‑tailed critical value at α = 0.05 is t(0.025, 58) ≈ 2.001 (using df = 60 as a close approximation). Since |t| = 2.575 > 2.001, we reject the null hypothesis. There is sufficient evidence to conclude that the new teaching method produces a mean score statistically different from the traditional method.
自由度 df = nₐ + n_b − 2 = 58。查 t 分布表，α = 0.05 的双尾临界值 t(0.025, 58) ≈ 2.001（使用 df = 60 的近似值）。由于 |t| = 2.575 > 2.001，我们拒绝原假设。有充分的证据表明，新教学法导致的平均分与传统方法存在统计显著差异。
A 95% confidence interval for the difference in means (μ_b − μₐ) can be constructed as (x̄_b − x̄ₐ) ± t_crit × SE = 5 ± 2.001 × 1.941 ≈ (1.12, 8.88). This interval does not contain zero, consistent with the hypothesis test.
均值差 (μ_b − μₐ) 的 95% 置信区间可构建为 (x̄_b − x̄ₐ) ± t_crit × SE = 5 ± 2.001 × 1.941 ≈ (1.12, 8.88)。该区间不包含零，与假设检验结论一致。
6. Case Study 2: Customer Satisfaction and Gender | 案例二：顾客满意度与性别
Context: A supermarket surveys 200 customers, recording their gender (Male/Female) and satisfaction with the new layout (Satisfied/Not Satisfied). The observed frequencies are summarised in the table below.
情境：一家超市调查了 200 名顾客，记录性别与对新布局的满意度。观测频数汇总如下。
| Satisfied | Not Satisfied | Total | |
|---|---|---|---|
| Male | 60 | 40 | 100 |
| Female | 80 | 20 | 100 |
| Total | 140 | 60 | 200 |
The research question is whether gender and satisfaction are independent. Because both variables are categorical, a chi‑squared test of independence is the correct choice.
研究问题是性别与满意度是否独立。由于两个变量都是分类变量，正确的选择是卡方独立性检验。
7. Computing Expected Frequencies and the Chi‑Squared Statistic | 计算期望频数与卡方统计量
Under the null hypothesis of independence, the expected frequency for each cell is (row total × column total) ÷ grand total. For example, the expected number of Males who are Satisfied is:
在独立原假设下，每个单元格的期望频数为 (行合计 × 列合计) ÷ 总计。例如，“男性且满意”的期望频数为：
E = (100 × 140) / 200 = 70
The complete table of expected frequencies:
完整的期望频数表：
- Male, Satisfied: 70 | 男性，满意：70
- Male, Not Satisfied: 30 | 男性，不满意：30
- Female, Satisfied: 70 | 女性，满意：70
- Female, Not Satisfied: 30 | 女性，不满意：30
The test statistic χ² is calculated as Σ (O − E)² / E:
检验统计量 χ² 计算如下：
χ² = (60−70)²/70 + (40−30)²/30 + (80−70)²/70 + (20−30)²/30
= 100/70 + 100/30 + 100/70 + 100/
Published by TutorHao | Year 13 统计 Revision Series | aleveler.com
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