Pre-U OCR Statistics: Case Study Practical Exercises | Pre-U OCR 统计：案例分析实战演练

📚 Pre-U OCR Statistics: Case Study Practical Exercises | Pre-U OCR 统计：案例分析实战演练

In Pre-U OCR Statistics, the ability to apply statistical techniques to real-world problems is tested through extended case studies. This article works through a realistic scenario – comparing two revision methods – to demonstrate how the full statistical toolkit is used in practice. You will see how data are described, models are fitted, hypotheses are tested, and conclusions are drawn in context.

在 Pre-U OCR 统计考试中，通过扩展案例分析来考查将统计技术应用于实际问题的能力。本文通过一个真实场景——比较两种复习方法——来展示如何综合运用统计工具。你将看到如何描述数据、拟合模型、检验假设并在情境中得出结论。

1. Setting the Scene | 场景设定

A school wants to investigate whether a new interactive online revision platform (Method B) leads to higher test scores than the traditional paper-based revision (Method A). Sixty Year 12 students are randomly allocated to two groups of 30. After a revision period, both groups sit the same 100-mark test. The data and the context drive every statistical decision that follows.

某学校想研究新的互动式在线复习平台（方法B）是否比传统纸质复习（方法A）带来更高的测验成绩。60 名 12 年级学生被随机分成两组，每组 30 人。复习期结束后，两组参加相同的 100 分测验。数据与情境驱动着随后所有的统计决策。

The raw scores for each group are recorded. For Method A, the sample mean x̄A = 62.4 and sample standard deviation sA = 9.7; for Method B, x̄B = 68.1 and sB = 8.3. The school also categorises each student’s result as ‘below 60′, ’60–79′ or ’80 and above’, and notes the number of hours each student spent revising.

每组的原始分数被记录下来。对于方法 A，样本均值 x̄A = 62.4，样本标准差 sA = 9.7；对于方法 B，x̄B = 68.1，sB = 8.3。学校还将每位学生的成绩归为“60 以下”“60–79”或“80 及以上”三个等级，并记录了每位学生的复习时长。

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

We start by summarising and visualising the data. Boxplots for each group highlight the median, quartiles and potential outliers. The median for A is 63, and for B is 69. The interquartile ranges (IQRs) are 13 and 11 respectively, suggesting slightly more variability in the traditional method.

我们从数据汇总和可视化开始。各组箱线图展示了中位数、四分位数和可能的异常值。方法 A 的中位数为 63，方法 B 的中位数为 69。四分位距（IQR）分别为 13 和 11，表明传统方法组的变异性略大。

Descriptive statistics such as the mean, standard deviation, and skewness are calculated. Both distributions are approximately symmetric, with skewness values of −0.2 and 0.1, supporting the assumption of normality later on.

计算均值、标准差、偏度等描述统计量。两个分布近似对称，偏度值分别为 −0.2 和 0.1，这为后续正态性假设提供了支持。

3. Probability Models for Scores | 成绩的概率模型

We model the test scores in each group using a normal distribution. For Method A, we assume scores follow N(μA, σ²A) with μA estimated by 62.4 and σA by 9.7. For B, N(68.1, 8.3²).

我们用正态分布对每组测验成绩建模。对于方法 A，假设成绩服从 N(μA, σ²A)，其中 μA 用 62.4 估计，σA 用 9.7 估计；方法 B 服从 N(68.1, 8.3²)。

These models allow us to compute probabilities, such as the chance a randomly chosen student from Group A scores above 70: P(X > 70) = 1 − Φ((70−62.4)/9.7) ≈ 1 − Φ(0.784) ≈ 0.216.

这些模型使我们能够计算概率，例如，从方法 A 组随机挑选一名学生成绩超过 70 的概率：P(X > 70) = 1 − Φ((70−62.4)/9.7) ≈ 1 − Φ(0.784) ≈ 0.216。

4. Sampling Distributions and the CLT | 抽样分布与中心极限定理

Even if individual scores were not exactly normal, the Central Limit Theorem ensures that the sample mean x̄ is approximately normally distributed for large enough samples. With n=30, the sampling distribution of x̄ under each method has mean μ and standard error σ/√n, estimated by s/√30.

即使个体成绩不完全服从正态分布，中心极限定理保证对于足够大的样本，样本均值 x̄ 近似正态分布。在 n=30 时，每一种方法下 x̄ 的抽样分布均值为 μ，标准误为 σ/√n，用 s/√30 来估计。

For Method A, the estimated standard error is 9.7/√30 ≈ 1.77; for B, 8.3/√30 ≈ 1.52. This knowledge underpins all subsequent confidence intervals and hypothesis tests.

对于方法 A，估计标准误为 9.7/√30 ≈ 1.77；方法 B 为 8.3/√30 ≈ 1.52。这一知识是所有后续置信区间和假设检验的基础。

5. Confidence Intervals for the Mean | 均值的置信区间

To express uncertainty about the true population mean, we construct 95% confidence intervals. Using the t-distribution with 29 degrees of freedom (t₀.₀₂₅,₂₉ ≈ 2.045), the interval for Method A is 62.4 ± 2.045 × 1.77, giving (58.8, 66.0). For Method B, it is 68.1 ± 2.045 × 1.52, giving (65.0, 71.2).

为表达对真实总体均值的不确定性，我们构建 95% 置信区间。使用自由度为 29 的 t 分布（t₀.₀₂₅,₂₉ ≈ 2.045），方法 A 的区间为 62.4 ± 2.045 × 1.77，即 (58.8, 66.0)；方法 B 为 68.1 ± 2.045 × 1.52，即 (65.0, 71.2)。

Notice the intervals do not overlap, suggesting a significant difference. However, a formal two-sample test is required to confirm this.

注意到两个区间并不重叠，这暗示存在显著差异。然而，仍需通过正式的双样本检验来确认。

6. One-Sample t-Test: Is the Mean Different from a Target? | 单样本 t 检验：均值是否不同于目标值？

The head teacher suspects that the traditional method yields a population mean of 60. We test H₀: μA = 60 against H₁: μA ≠ 60 at the 5% significance level. The test statistic is t = (62.4 − 60) / 1.77 ≈ 1.356.

校长怀疑传统方法的总体均值为 60。我们在 5% 显著性水平下检验 H₀: μA = 60 对 H₁: μA ≠ 60。检验统计量为 t = (62.4 − 60) / 1.77 ≈ 1.356。

The critical value from t-distribution with 29 df is ±2.045. Since 1.356 is not in the critical region, we do not reject H₀. There is insufficient evidence to say the mean differs from 60.

来自自由度为 29 的 t 分布临界值为 ±2.045。由于 1.356 不在拒绝域内，我们不拒绝 H₀。没有足够证据表明均值不同于 60。

7. Two-Sample t-Test: Comparing Means | 双样本 t 检验：比较均值

The core question is whether Method B genuinely leads to higher scores. We assume equal population variances (confirmed by a variance test) and conduct a pooled two-sample t-test. H₀: μA = μB vs H₁: μA < μB.

核心问题是方法 B 是否真的带来更高成绩。我们假设总体方差相等（经方差检验确认），并进行合并双样本 t 检验。H₀: μA = μB 对 H₁: μA < μB。

The pooled standard deviation is sp = √((29×9.7² + 29×8.3²) / 58) ≈ 9.03. The standard error of the difference is 9.03×√(1/30+1/30) ≈ 2.33. The test statistic t = (68.1−62.4)/2.33 ≈ 2.45.

合并标准差为 sp = √((29×9.7² + 29×8.3²) / 58) ≈ 9.03。差值的标准误为 9.03×√(1/30+1/30) ≈ 2.33。检验统计量 t = (68.1−62.4)/2.33 ≈ 2.45。

With 58 df, the 5% one-tailed critical value is about 1.671. Since 2.45 > 1.671, we reject H₀. There is significant evidence that Method B improves performance.

自由度为 58 时，5% 单侧临界值约为 1.671。由于 2.45 > 1.671，我们拒绝 H₀。有显著证据表明方法 B 提高了成绩。

8. Chi-Squared Test for Independence | 卡方独立性检验

The school also categorises results into grade bands and wonders if the grade distribution is independent of the revision method. Observed frequencies are:

学校还将成绩分为等级段，想知道成绩分布是否与复习方法独立。观测频数为：

	Below 60	60–79	80+	Total
Method A	12	14	4	30
Method B	5	16	9	30
Total	17	30	13	60

Expected frequencies under independence are calculated as (row total × column total)/grand total. For example, Method A & ‘Below 60’: (30×17)/60 = 8.5.

独立假设下的期望频数为（行合计 × 列合计）/ 总计。例如，方法 A 与“60 以下”： (30×17)/60 = 8.5。

The χ² statistic is Σ (O−E)²/E. We compute χ² = (12−8.5)²/8.5 + (14−15)²/15 + (4−6.5)²/6.5 + (5−8.5)²/8.5 + (16−15)²/15 + (9−6.5)²/6.5 ≈ 1.44 + 0.07 + 0.96 + 1.44 + 0.07 + 0.96 = 4.94.

χ² 统计量为 Σ (O−E)²/E。计算得 χ² = (12−8.5)²/8.5 + (14−15)²/15 + (4−6.5)²/6.5 + (5−8.5)²/8.5 + (16−15)²/15 + (9−6.5)²/6.5 ≈ 1.44 + 0.07 + 0.96 + 1.44 + 0.07 + 0.96 = 4.94。

Degrees of freedom: (2−1)×(3−1)=2. The 5% critical value is χ²₂,₀.₀₅ = 5.991. Since 4.94 < 5.991, we do not reject the null hypothesis. There is no significant association between revision method and grade band.

自由度为 (2−1)×(3−1)=2。5% 临界值为 χ²₂,₀.₀₅ = 5.991。由于 4.94 < 5.991，我们不拒绝原假设，即复习方法与成绩等级无显著关联。

9. Correlation and Regression | 相关与回归

We explore the relationship between total revision hours (x) and test score (y) for the 60 students. The scatterplot shows a positive linear trend. The product-moment correlation coefficient is r = 0.72. Testing H₀: ρ = 0, the test statistic t = r√(n−2)/√(1−r²) ≈ 0.72×√58/√(1−0.5184) ≈ 5.48/0.694 ≈ 7.90, which is highly significant.

我们探究全部 60 名学生的复习总时数（x）与测验成绩（y）之间的关系。散点图呈正线性趋势。积矩相关系数为 r = 0.72。检验 H₀: ρ = 0，检验统计量 t = r√(n−2)/√(1−r²) ≈ 0.72×√58/√(1−0.5184) ≈ 5.48/0.694 ≈ 7.90，高度显著。

The least-squares regression line is y = 38.2 + 2.65x. For every additional hour of revision, the predicted score increases by about 2.65 marks. The coefficient of determination R² = 0.518 means that revision hours explain about 51.8% of the variability in scores.

最小二乘回归直线为 y = 38.2 + 2.65x。每增加一小时复习，预测成绩约提高 2.65 分。决定系数 R² = 0.518 表示复习时数解释了成绩变异的约 51.8%。

10. Exam-Style Synthesis and Interpretation | 考试风格的综合与解释

In the Pre-U examination, you must interpret your findings in the original context. When reporting the two-sample t-test, state: ‘The difference of 5.7 marks is statistically significant, suggesting that the online platform is genuinely more effective.’ Be precise about the significance level used and mention the assumption of equal variances.

在 Pre-U 考试中，你必须将结论置于原始情境中解释。当报告双样本 t 检验时，应指出：“5.7 分的差异具有统计显著性，表明在线平台确实更有效。”要精确说明所用的显著性水平，并提及方差相等的假设。

Examiners expect critical evaluation of models: e.g., ‘The normal model appears reasonable, but a larger sample would confirm.’ Also discuss potential confounding variables, such as student motivation or prior ability, that may affect the conclusion.

考官期望对模型进行批判性评估，例如：“正态模型看似合理，但更大样本才能确认。”同时讨论潜在的混淆变量，如学生动机或先前能力，它们可能影响结论。

Always link back to the problem statement, use appropriate technical vocabulary, and present calculations clearly. Practice writing coherent statistical narratives – this is what elevates a Pre-U response.

务必回扣问题陈述，使用恰当的专业词汇，并清晰地展示计算过程。练习撰写连贯的统计叙述——这正是提升 Pre-U 答题水平的关键。

Published by TutorHao | Statistics Revision Series | aleveler.com

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Pre-U OCR Statistics: Case Study Practical Exercises | Pre-U OCR 统计：案例分析实战演练