📚 Case Study Practice for GCSE OCR Statistics | GCSE OCR 统计：案例分析实战演练
GCSE OCR Statistics Paper 2 often features a substantial case study that requires you to analyse a real-world dataset using the full range of statistical techniques from the specification. This article offers a step-by-step walkthrough of a typical case study, emphasising how to structure your response, perform calculations accurately, and write meaningful interpretations. Our example investigates whether revision time predicts exam marks among 15 students.
GCSE OCR 统计学试卷二通常包含一个大型案例分析，要求你使用考纲中的全套统计技术分析真实世界的数据集。本文将通过一个典型案例的逐步演练，强调如何组织回答、准确计算并写出有意义的解释。我们的例子将探究15名学生的复习时间是否能预测考试成绩。
1. Understanding the Case Study Task | 理解案例分析任务
Read the context carefully. Identify the explanatory variable (revision hours) and the response variable (exam marks %). The task will include instructions to explore the relationship, model it, and possibly make predictions or test hypotheses. Always note the type of data: hours are continuous numerical, marks are discrete numerical but treated as continuous.
仔细阅读背景信息。识别解释变量（复习小时数）和响应变量（考试成绩%）。题目通常会包括探索关系、建立模型、可能还要做出预测或检验假设。始终要注意数据类型：小时数是连续型数值，成绩虽然本质上是离散型但可按连续型处理。
2. Organising the Data and Identifying Variables | 整理数据与识别变量
Before any analysis, present the data in a clear table. For our 15 students, the paired data are:
在进行任何分析前，用清晰的表格呈现数据。我们15名学生的配对数据如下：
| Student | Hours (x) | Marks % (y) |
|---|---|---|
| 1 | 1.0 | 30 |
| 2 | 2.0 | 35 |
| 3 | 2.5 | 42 |
| 4 | 3.0 | 45 |
| 5 | 4.0 | 50 |
| 6 | 4.5 | 48 |
| 7 | 5.0 | 55 |
| 8 | 5.5 | 58 |
| 9 | 6.0 | 60 |
| 10 | 6.5 | 65 |
| 11 | 7.0 | 68 |
| 12 | 8.0 | 72 |
| 13 | 9.0 | 75 |
| 14 | 10.0 | 80 |
| 15 | 12.0 | 85 |
Always check for missing values or outliers. Here the data appear complete and with a plausible range.
务必检查缺失值或异常值。这里的数据看起来完整，且取值范围合理。
3. Constructing a Scatter Diagram | 绘制散点图
Plot each pair (x, y) on graph paper or using software. Label axes ‘Revision Hours’ and ‘Exam Mark (%)’. The scatter diagram shows a clear upward linear trend, meaning that students who revise more tend to score higher marks.
在方格纸或软件中绘制每一对 (x, y) 数据点。坐标轴分别标注“复习小时数”和“考试成绩(%)”。散点图呈现出明显的向上线性趋势，说明复习时间较长的学生往往得分更高。
Take care to use appropriate scales and add a title. The plot also helps you spot any data points that deviate significantly from the overall pattern.
注意使用合适的刻度并添加标题。该图还能帮助你发现严重偏离整体趋势的数据点。
4. Calculating the Product-Moment Correlation Coefficient | 计算积差相关系数
For a linear relationship, use Pearson’s r. The formula is: r = Σ(x – x̄)(y – ȳ) / √[ Σ(x – x̄)² × Σ(y – ȳ)² ]. First compute the means: x̄ = Σx / n = 86 / 15 = 5.733, and ȳ = Σy / n = 868 / 15 = 57.867.
对于线性关系，使用皮尔逊相关系数 r。公式为：r = Σ(x – x̄)(y – ȳ) / √[ Σ(x – x̄)² × Σ(y – ȳ)² ]。先计算均值：x̄ = Σx / n = 86 / 15 = 5.733，ȳ = Σy / n = 868 / 15 = 57.867。
Then calculate the three key sums:
然后计算三个关键的平方和与乘积和：
-
Σ(x – x̄)² = 144.933
Σ(x – x̄)² = 144.933
-
Σ(y – ȳ)² = 3857.733
Σ(y – ȳ)² = 3857.733
-
Σ(x – x̄)(y – ȳ) = 745.733
Σ(x – x̄)(y – ȳ) = 745.733
Substituting into the formula gives r = 745.733 / √(144.933 × 3857.733) = 745.733 / 748.1 ≈ 0.997 (to 3 significant figures). This is extremely close to +1, indicating a very strong positive linear association.
代入公式得 r = 745.733 / √(144.933 × 3857.733) = 745.733 / 748.1 ≈ 0.997（保留三位有效数字）。该值极其接近 +1，表明存在极强的正向线性关系。
5. Interpreting Correlation and Its Limitations | 解释相关性及其局限性
A correlation coefficient of 0.997 tells you there is a strong tendency for marks to increase as revision hours increase. However, correlation does not imply causation. It could be that more motivated students choose to revise longer and also perform better for other reasons.
相关系数为 0.997 说明，随着复习时间增加，考试成绩有很强的上升趋势。但相关不代表因果。也可能是学习动力更强的学生选择更长时间复习，并且由于其他原因表现更好。
Also discuss the effect of sample size: with only 15 data points, the result is less reliable for predicting beyond this group. Always mention possible confounding variables and the need for further investigation.
同时要讨论样本量的影响：仅有15个数据点，该结果用于推广到更大群体时可靠性较低。总要提及可能的混淆变量以及进一步调查的必要性。
6. Fitting a Least Squares Regression Line | 拟合最小二乘回归线
To model the relationship, find the line of best fit y = a + bx. The gradient b = Σ(x – x̄)(y – ȳ) / Σ(x – x̄)² = 745.733 / 144.933 ≈ 5.143. The intercept a = ȳ – b x̄ = 57.867 – 5.143×5.733 ≈ 28.3. Thus the regression equation is: y = 28.3 + 5.14x (values rounded).
为建立关系模型，求最佳拟合线 y = a + bx。斜率 b = Σ(x – x̄)(y – ȳ) / Σ(x – x̄)² = 745.733 / 144.933 ≈ 5.143。截距 a = ȳ – b x̄ = 57.867 – 5.143×5.733 ≈ 28.3。因此回归方程为：y = 28.3 + 5.14x（已四舍五入）。
The gradient of 5.14 means that for every extra hour of revision, the predicted exam mark increases by about 5.14 percentage points. The intercept would be the predicted mark for zero revision hours, but caution is needed when extrapolating.
斜率为 5.14 意味着每增加一小时复习，预期考试成绩提高约 5.14 个百分点。截距表示复习时间为零时的预测成绩，但外推时需谨慎。
7. Making Predictions and Calculating Residuals | 进行预测与计算残差
Use the regression line to make predictions within the data range. For a student who revises 8 hours, predicted mark y = 28.3 + 5.14×8 = 69.42%. The actual mark for such a student in our data is 72%, so the residual is 72 – 69.42 = 2.58%.
利用回归线在数据范围内进行预测。对于复习8小时的学生，预测成绩 y = 28.3 + 5.14×8 = 69.42%。数据中该学生的实际成绩是72%，因此残差
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