Case Study: Analysing Student Test Scores | 案例分析：分析学生考试成绩

📚 Case Study: Analysing Student Test Scores | 案例分析：分析学生考试成绩

In this IGCSE Statistics case study, we put theory into practice by analysing a real dataset: the test scores of 20 students in a mathematics exam. We will follow the OCR specification step by step, applying techniques from data collection and frequency distributions to measures of central tendency, dispersion, cumulative frequency, box plots, probability, bivariate analysis and sampling. Each section demonstrates how statistical reasoning helps uncover patterns and supports data-driven conclusions.

在这篇 IGCSE 统计案例分析中，我们将理论付诸实践，分析一个真实的数据集：20 名学生的数学测验成绩。我们将按照 OCR 考试大纲逐步操作，应用从数据收集、频数分布到集中趋势、离散程度、累积频率、箱线图、概率、双变量分析和抽样等各种技巧。每一节都展示了统计推理如何揭示规律并支持数据驱动的结论。

1. Introducing the Dataset | 数据集介绍

Twenty Year 11 students sat a 100-mark mathematics test. Their raw scores are: 42, 48, 54, 60, 65, 68, 70, 72, 75, 78, 80, 82, 85, 88, 90, 92, 94, 96, 98, 100. In addition, each student reported the number of hours they had spent revising for the test — this second variable will be examined in the bivariate section. Our goal is to summarise the performance, evaluate the spread of scores, make probability statements and investigate the relationship between study hours and achievement.

20 名十一年级学生参加了一次满分 100 分的数学测验。他们的原始分数为：42, 48, 54, 60, 65, 68, 70, 72, 75, 78, 80, 82, 85, 88, 90, 92, 94, 96, 98, 100。此外，每位学生还报告了他们为测验复习的小时数——这个第二个变量将在双变量部分中分析。我们的目标是概括成绩表现、评估分数的离散程度、作出概率表述并探究学习时间与成绩之间的关系。

2. Organising the Data: Frequency Distribution | 整理数据：频数分布

To see the overall shape, we group the scores into class intervals of width 10. The frequency table below shows how many students fall into each band.

为了看清整体形状，我们将分数划分为宽度为 10 的组距。下面的频数表展示了每个区间有多少名学生。

Notice that the modal class is 90–99 with 5 students. The distribution appears slightly skewed to the left, as more scores are clustered at the higher end.

请注意，众数所在的组是 90–99，有 5 名学生。数据的分布似乎略微左偏，因为更多的分数集中在高端。

3. Measures of Central Tendency | 集中趋势的度量

The mean is calculated by summing all scores and dividing by the number of students. For our dataset, Σx = 1537 and n = 20, so the mean x̄ = 1537 ÷ 20 = 76.85.

均值通过将所有分数相加再除以学生人数来计算。在我们的数据集中，Σx = 1537，n = 20，因此均值 x̄ = 1537 ÷ 20 = 76.85。

To find the median, we locate the two middle scores, the 10th and 11th in the ordered list. These are 78 and 80, giving a median of (78 + 80) ÷ 2 = 79.

为找中位数，我们找出有序列表中位于中间的两个分数，即第 10 和第 11 个。它们分别是 78 和 80，因此中位数为 (78 + 80) ÷ 2 = 79。

There is no single mode because every score appears only once. In grouped data the modal class is 90–99, but for raw data we conclude the distribution is unimodal by class rather than by individual value.

由于每个分数只出现一次，因此没有单一的众数。在分组数据中，众数所在的组是 90–99，但对于原始数据，我们认定该分布是按组距而非个别值呈现单峰的。

x̄ = ∑x / n

中位数位置 = (n+1)/2

4. Measures of Dispersion | 离散程度的度量

The range, the simplest measure of spread, is 100 – 42 = 58. While easy to compute, the range is sensitive to extreme values.

范围是最简单的离散度量，为 100 – 42 = 58。虽然易于计算，但它对极端值敏感。

We find the lower quartile (Q₁) at position 0.25 × (20+1) = 5.25, so Q₁ lies between the 5th and 6th values: 65 + 0.25 × (68 – 65) = 65.75. The upper quartile (Q₃) is at position 15.75, between 90 and 92: 90 + 0.75 × (92 – 90) = 91.5. The interquartile range (IQR) is 91.5 – 65.75 = 25.75.

我们求得下四分位数 (Q₁) 的位置为 0.25×(20+1)=5.25，因此 Q₁ 介于第 5 和第 6 个值之间：65 + 0.25×(68 – 65)=65.75。上四分位数 (Q₃) 的位置为 15.75，介于第 15 和第 16 个值之间：90+0.75×(92–90)=91.5。四分位数间距 (IQR) 为 91.5 – 65.75 = 25.75。

The sample standard deviation provides a more complete picture of spread. Using the formula s = √[∑(x – x̄)²/(n–1)], we compute ∑(x – x̄)² = 5420.55, thus s = √(5420.55 ÷ 19) = √285.2921 ≈ 16.9 (to 3 s.f.).

样本标准差能更全面地反映分散情况。使用公式 s = √[∑(x – x̄)²/(n–1)]，我们计算出 ∑(x – x̄)² = 5420.55，因此 s = √(542

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