AQA Year 13 Statistics: Case Study Practical Exercises | AQA Year 13 统计：案例分析实战演练

📚 AQA Year 13 Statistics: Case Study Practical Exercises | AQA Year 13 统计：案例分析实战演练

Case studies bridge the gap between statistical theory and real‑world application. In this article, we work through a complete research scenario, applying the hypothesis tests, confidence intervals and regression techniques required by the AQA Year 13 specification. Every step follows exam‑style reasoning, from framing the research question to interpreting p‑values and drawing conclusions in context.

案例分析能弥合统计理论与实际应用之间的鸿沟。本文会走完一个完整的研究场景，运用 AQA Year 13 考试大纲要求的假设检验、置信区间和回归技术。每一步都遵循考试风格的推理过程，从提出研究问题到根据背景解释 p 值并得出结论。

1. The Research Scenario: Flashcard Revision Study | 研究场景：闪卡复习法研究

A group of educational researchers wants to test whether a new flashcard‑based revision technique improves A‑level Statistics exam scores. They recruit 30 Year 13 students and randomly assign 15 to the ‘Flashcard’ group and 15 to a ‘Traditional’ group. After four weeks, all students sit the same mock exam. The researchers also record each student’s weekly revision hours and, inside the Flashcard group, ask whether they would continue using the method (Yes/No), noting their gender. The data are given in the table below (first 10 participants shown).

一组教育研究者希望检验一种基于闪卡的新复习方法是否能提升 A‑level 统计考试成绩。他们招募了 30 名 Year 13 学生，随机将 15 人分配到 “闪卡组”，15 人分配到 “传统组”。四周后，所有学生参加相同的模拟考试。研究者还记录了每位学生每周的复习小时数，并在闪卡组内询问他们是否会继续使用该方法（是/否），同时记录性别。下表列出了前 10 名参与者的数据。

Student	Group	Gender	Revision (hrs)	Score (%)	Continue? (Flashcard only)
1	Flashcard	M	15	82	Yes
2	Flashcard	F	20	85	Yes
3	Traditional	M	14	75	‑
4	Traditional	F	18	79	‑
5	Flashcard	M	12	78	No
…	…	…	…	…	…

Full summary statistics are provided in subsequent sections. This realistic set‑up lets us practise a range of AQA Year 13 techniques on a single data set.

完整的汇总统计量会在后续小节中给出。这一贴近现实的设定让我们可以在同一个数据集上练习一系列 AQA Year 13 方法。

2. Exploratory Data Analysis | 探索性数据分析

Before jumping into hypothesis tests, we summarise the data. For the Flashcard group (n₁ = 15), the mean score was x̄₁ = 84.8% with a standard deviation s₁ = 5.2%. The Traditional group (n₂ = 15) had x̄₂ = 75.6% and s₂ = 4.1%. A box‑plot comparison suggests the Flashcard distribution is shifted to the right with slightly greater spread. No extreme outliers were detected.

在进行假设检验之前，我们先对数据进行汇总。闪卡组（n₁ = 15）的平均分 x̄₁ = 84.8%，标准差 s₁ = 5.2%。传统组（n₂ = 15）的平均分 x̄₂ = 75.6%，标准差 s₂ = 4.1%。箱线图比较显示闪卡组的分布整体右移，且离散程度稍大。未发现极端异常值。

The pooled revision hour data for all 30 students gave a mean of 16.0 hours with s = 4.9 hours. Revision time and score appeared to have a positive linear relationship on the scatter plot, which we will formalise later.

全部 30 名学生的复习时间汇总均值为 16.0 小时，标准差 s = 4.9 小时。散点图显示复习时间与成绩之间存在正线性关系，我们稍后会进行正式分析。

3. Checking Assumptions: Normality Test | 假设检验前提：正态性检验

Many AQA procedures assume the data are approximately normal. Although the sample sizes are small, we use a normal probability plot and the Shapiro‑Wilk test. For both groups, the test statistic W exceeded 0.95 with a p‑value > 0.10, suggesting no serious departure from normality. We therefore proceed with parametric tests, mindful of the small sample sizes.

AQA 的许多方法都假设数据近似正态分布。尽管样本量较小，我们仍可使用正态概率图和 Shapiro‑Wilk 检验。两组的检验统计量 W 均大于 0.95，p 值 > 0.10，这表明并无严重偏离正态性的情况。因此，在注意小样本的前提下，我们可以继续使用参数检验。

4. Two‑sample t‑test: Comparing Group Means | 双样本 t 检验：比较两组均值

We test whether the flashcard method yields a significantly different mean score. Let μ₁ be the true mean for the Flashcard group and μ₂ for the Traditional group.

我们检验闪卡方法是否带来显著不同的平均分。设 μ₁ 为闪卡组的真实均值，μ₂ 为传统组的真实均值。

H₀: μ₁ = μ₂ H₁: μ₁ ≠ μ₂ (α = 0.05)

Since the standard deviations differ slightly and sample sizes are equal but small, we employ Welch’s t‑test. The test statistic is

由于两组标准差略有不同，且样本量相等但较小，我们采用 Welch t 检验。检验统计量为

t = (x̄₁ − x̄₂) / √(s₁²/n₁ + s₂²/n₂)

Plugging in the values: x̄₁ = 84.8, x̄₂ = 75.6, s₁² = 27.04, s₂² = 16.81, n₁ = n₂ = 15.

代入数值：x̄₁ = 84.8，x̄₂ = 75.6，s₁² = 27.04，s₂² = 16.81，n₁ = n₂ = 15。

t = (84.8 − 75.6) / √(27.04/15 + 16.81/15) = 9.2 / √(1.8027 + 1.1207) = 9.2 / √2.9234 ≈ 9.2 / 1.7098 ≈ 5.38

Using the Welch–Satterthwaite approximation, the degrees of freedom are about 26. The two‑tailed p‑value is far less than 0.001. Since p < 0.05, we reject H₀ and conclude there is very strong evidence of a difference in mean scores between the two revision methods.

根据 Welch–Satterthwaite 近似，自由度约为 26。双尾 p 值远小于 0.001。由于 p < 0.05，我们拒绝 H₀，并认为有极强的证据表明两种复习方法的平均分存在差异。

In an exam, you would also state the conclusion in context: ‘The flashcard technique appears to produce higher average mock exam scores than the traditional method.’

在考试中，你还应结合背景陈述结论：“闪卡复习法似乎比传统方法带来更高的模拟考试平均分。”

5. Chi‑squared Test of Independence | 卡方独立性检验

Within the Flashcard group, the 15 students were asked, ‘Would you continue using this method?’ The responses, classified by gender, are shown in the contingency table below.

在闪卡组内，15 名学生被问到：“你会继续使用这种方法吗？” 按性别分类的回答如下列联表所示。

	Yes	No	Total
Male	6	2	8
Female	5	2	7
Total	11	4	15

The hypotheses are: H₀ – gender and willingness to continue are independent; H₁ – they are not independent. Expected frequencies, calculated as (row total × column total) / grand total, are all above 5, so a chi‑squared test is appropriate.

原假设和备择假设为：H₀ – 性别与继续使用的意愿相互独立；H₁ – 两者不独立。预期频数按（行合计 × 列合计）/ 总计计算，所有预期频数均大于 5，因此适合使用卡方检验。

Expected: Male‑Yes 8×11/15 ≈ 5.87, Male‑No 8×4/15 ≈ 2.13, Female‑Yes 7×11/15 ≈ 5.13, Female‑No 7×4/15 ≈ 1.87.

χ² = Σ (O − E)²/E ≈ (6−5.87)²/5.87 + (2−2.13)²/2.13 + (5−5.13)²/5.13 + (2−1.87)²/1.87 ≈ 0.0029 + 0.0079 + 0.0033 + 0.0090 ≈ 0.023.

Degrees of freedom = (2 − 1)(2 − 1) = 1. Consulting χ² tables, the critical value at the 5% level is 3.84. Our statistic 0.023 is far below that, giving a p‑value > 0.05. We do not reject H₀; there is no evidence of an association between gender and the intention to continue using flashcards in this sample.

自由度为 (2 − 1)(2 − 1) = 1。查 χ² 表，5% 显著性水平的临界值为 3.84。我们的统计量 0.023 远低于该值，p 值 > 0.05。我们不拒绝 H₀；在该样本中没有证据表明性别与继续使用闪卡的意愿之间存在关联。

6. Correlation Analysis: Revision Hours vs. Score | 相关性分析：复习时间与成绩

For all 30 students, we calculate the Pearson product‑moment correlation coefficient between weekly revision hours and mock exam score. The scatter plot shows a clear uphill pattern, and the computed r is 0.73. Using the t‑test for correlation, t = r × √((n−2)/(1−r²)) = 0.73 × √(28/(1−0.732²)) ≈ 0.73 × √(28/0.466) ≈ 5.66 with 28 d.f., resulting in a two‑tailed p‑value < 0.001. We reject H₀: ρ = 0 and conclude there is a significant positive linear relationship between revision time and score.

对全部 30 名学生，我们计算了每周复习小时数与模拟考试成绩之间的 Pearson 积差相关系数。散点图呈现出明显的上升趋势，计算得到的 r = 0.73。使用相关系数的 t 检验，t = r × √((n−2)/(1−r²)) = 0.73 × √(28/(1−0.73²)) ≈ 0.73 × √(28/0.466) ≈ 5.66，自由度 28，双尾 p 值 < 0.001。我们拒绝 H₀: ρ = 0，并认为复习时间与成绩之间存在显著的正线性关系。

It is vital to stress that correlation does not imply causation. A third factor, such as general motivation, could influence both revision hours and score.

必须强调，相关性并不意味着因果关系。第三个因素，例如总体学习动力，可能同时影响复习时间与成绩。

7. Simple Linear Regression | 简单线性回归

Because the correlation is strong, a simple linear regression model can be built to predict score from revision hours. Using the formula b = r (s_y / s_x) and a = ȳ − b x̄, where for the full sample s_x (revision hours) = 4.9, s_y (score) = 8.2, x̄ = 16.0, ȳ = 80.2, we obtain:

由于相关性很强，可以建立一个以复习时间预测成绩的简单线性回归模型。使用公式 b = r (s_y / s_x) 和 a = ȳ − b x̄，其中全样本 s_x（复习时间）= 4.9，s_y（成绩）= 8.2，x̄ = 16.0，ȳ = 80.2，我们得到：

b = 0.73 × (8.2 / 4.9) ≈ 1.22, a = 80.2 − 1.22 × 16.0 ≈ 80.2 − 19.5 = 60.7.

Thus the regression equation is

因此回归方程为

Predicted Score = 60.7 + 1.22 × (Revision Hours).

The coefficient of determination R² = r² ≈ 0.53, meaning about 53% of the variability in mock scores can be explained by revision hours. A student who revises for 18 hours per week would be predicted to score roughly 60.7 + 1.22 × 18 = 82.7%.

决定系数 R² = r² ≈ 0.53，意味着约 53% 的模拟考试成绩变异可由复习时间解释。一位每周复习 18 小时的学生预计可得 60.7 + 1.22 × 18 = 82.7 分。

8. Confidence Interval for the Mean | 均值的置信区间

To quantify the precision of the Flashcard group’s mean estimate, we compute a 95% confidence interval for μ₁. Using x̄₁ = 84.8, s₁ = 5.2, n₁ = 15, and the t‑critical value for 14 d.f. (t* ≈ 2.145), the standard error is SE = 5.2 / √15 ≈ 1.342.

为了量化闪卡组均值估计的精度，我们计算 μ₁ 的 95% 置信区间。使用 x̄₁ = 84.8，s₁ = 5.2，n₁ = 15，以及自由度 14 的 t 临界值 t* ≈ 2.145，标准误 SE = 5.2 / √15 ≈ 1.342。

95% CI = 84.8 ± 2.145 × 1.342 = 84.8 ± 2.88 = (81.9, 87.7).

We are 95% confident that the true mean mock score for students using the flashcard method lies between 81.9% and 87.7%. This interval does not overlap with the Traditional group’s mean, reinforcing the t‑test conclusion.

我们有 95% 的信心认为，使用闪卡方法的学生其真实平均模拟成绩介于 81.9% 与 87.7% 之间。该区间与传统组的均值没有重叠，进一步支持了 t 检验的结论。

9. Results Interpretation and Conclusions | 结果解释与结论

Bringing the analyses together: the two‑sample t‑test provided very strong evidence of a difference in mean scores, with the Flashcard group outperforming the Traditional group by about 9 percentage points. The correlation and regression analyses confirmed that revision time is positively associated with score, explaining over half the variation. The chi‑squared test showed no gender‑linked difference in willingness to continue, suggesting the method appeals equally. The confidence interval gives a plausible range for the Flashcard group’s mean, valuable for future research planning.

综合各项分析：双样本 t 检验给出了平均分存在差异的极强证据，闪卡组的成绩比传统组高出约 9 个百分点。相关和回归分析证实复习时间与成绩呈正相关，解释了超过一半的变异。卡方检验显示在继续使用意愿上没有与性别相关的差异，说明该方法对两性具有同等的吸引力。置信区间给出了闪卡组均值的合理范围，对未来的研究设计具有参考价值。

In an AQA examination context, you must always phrase conclusions in terms of the original problem and avoid generalising beyond the sampled population. Also mention that the study’s small sample size reduces power and that random assignment helps support a causal claim, though confounding variables cannot be entirely ruled out.

在 AQA 考试情境中，你必须始终用原问题背景来表述结论，并避免将结论过度推广到样本群体之外。同时应指出，研究的样本量较小会降低检验功效，而随机分配有助于支持因果推断，但无法完全排除混杂变量。

10. Common Pitfalls and Examiner Tips | 常见错误与应试技巧

Using the wrong test: check assumptions before choosing a test. For small samples, normality should be assessed; if in doubt, a non‑parametric alternative (e.g. Mann‑Whitney) may be safer, but AQA often expects the t‑test unless the data are severely skewed.

选错检验方法：在选择检验方法之前要先检验前提条件。对于小样本，应评估正态性；如有疑问，可采用非参数替代方法（如 Mann‑Whitney），但除非数据严重偏态，AQA 通常期望使用 t 检验。

Confusing correlation with causation: always add a disclaimer when interpreting r or regression slopes. Examiners reward the phrase ‘correlation does not imply causation’.

Published by TutorHao | Year 13 统计 Revision Series | aleveler.com

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AQA Year 13 Statistics: Case Study Practical Exercises | AQA Year 13 统计：案例分析实战演练