📚 Year 13 Edexcel Statistics: Case Study Masterclass | 案例分析实战演练
This article walks you through a complete statistical investigation using real-world-style data, covering correlation, regression, hypothesis testing for ρ, Spearman’s rank, two-sample t-test, and chi-squared test of independence – all key topics for Edexcel Year 13 Statistics. By working through the case study, you will see how each technique fits into a coherent analysis and gain confidence for your exam.
本文将带领你使用真实风格的数据完成一个完整的统计调查，涵盖相关、回归、ρ 的假设检验、斯皮尔曼等级相关、双样本 t 检验和卡方独立性检验——这些都是 Edexcel Year 13 统计的核心主题。通过这个案例研究，你将理解每种方法如何融入连贯的分析，并为考试建立信心。
1. Introducing the Case Study | 案例介绍
Twenty Year 13 students participated in a study. Data were collected for weekly revision hours (x), final exam score out of 100 (y), revision method used (Method A or Method B), and gender. The first ten students followed Method A, while the remaining ten followed Method B. We want to explore the relationship between hours and score, test whether the correlation is significant, compare the two methods, and check for association between method and gender.
20 名 Year 13 学生参与了一项研究。收集的数据包括每周复习小时数 (x)、满分 100 的期末考试成绩 (y)、所用的复习方法（方法 A 或方法 B）以及性别。前 10 名学生采用方法 A，后 10 名采用方法 B。我们希望探索学习时间与成绩之间的关系，检验相关性是否显著，比较两种方法的效果，并检验方法与性别之间是否存在关联。
| Student | Hours (x) | Score (y) | Method | Gender |
|---|---|---|---|---|
| 1 | 2 | 50 | A | M |
| 2 | 4 | 65 | A | M |
| 3 | 6 | 72 | A | F |
| 4 | 3 | 54 | A | M |
| 5 | 5 | 68 | A | F |
| 6 | 7 | 78 | A | M |
| 7 | 8 | 82 | A | M |
| 8 | 3 | 56 | A | F |
| 9 | 6 | 74 | A | M |
| 10 | 5 | 70 | A | F |
| 11 | 4 | 64 | B | M |
| 12 | 7 | 80 | B | F |
| 13 | 9 | 85 | B | M |
| 14 | 2 | 52 | B | F |
| 15 | 8 | 84 | B | M |
| 16 | 6 | 72 | B | F |
| 17 | 5 | 66 | B | F |
| 18 | 3 | 53 | B | F |
| 19 | 7 | 76 | B | M |
| 20 | 4 | 63 | B | F |
2. Data Description and Summary Statistics | 数据描述与汇总统计量
First we compute the basic summary measures. For all 20 students, the total hours Σx = 104, so the mean x̄ = 5.2 h. The total score Σy = 1364, giving ȳ = 68.2 marks. These will be used to calculate sums of squares and cross‑products.
首先我们计算基本的汇总测度。对于全部 20 名学生，总学习时间 Σx = 104，因此均值 x̄ = 5.2 小时。总分 Σy = 1364，得出 ȳ = 68.2 分。这些将用于计算平方和与交叉乘积。
The sums of squares are found from raw data: Σx² = 622, Σy² = 95344, and Σxy = 7520. From these we obtain corrected sums:
从原始数据求出平方和：Σx² = 622，Σy² = 95344，Σxy = 7520。由此我们得到校正和：
Sxx = Σx² − (Σx)²/n = 622 − 104²/20 = 81.2
Syy = Σy² − (Σy)²/n = 95344 − 1364²/20 = 2319.2
Sxy = Σxy − (Σx)(Σy)/n = 7520 − 104×1364/20 = 427.2
3. Visualising the Relationship: Scatter Diagram | 可视化关系：散点图
A scatter plot of score against hours shows a strong, positive, linear pattern. As revision hours increase, the exam score rises consistently, suggesting that a linear model is appropriate. The points lie close to an imaginary straight line, which motivates the calculation of the product moment correlation coefficient.
分数对学习时间的散点图显示出强正线性模式。随着复习时间增加，考试成绩稳步上升，表明线性模型是合适的。数据点紧靠一条假想的直线，这促使我们计算积矩相关系数。
Even without a printed graph, you can verify the trend by sorting the data – the highest scores (85, 84) correspond to the highest hours (9, 8), while the lowest scores (50, 52) correspond to the lowest hours (2, 2).
即使没有打印的图形，你也可以通过排序数据来验证趋势——最高分 (85, 84) 对应最多小时数 (9, 8)，而最低分 (50, 52) 对应最少小时数 (2, 2)。
4. Calculating the Product Moment Correlation Coefficient (PMCC) | 计算积矩相关系数
The Pearson PMCC, denoted by r, measures the strength and direction of a linear relationship. The formula is:
皮尔逊积矩相关系数记为 r，衡量线性关系的强度和方向。其公式为：
r = Sxy / √(Sxx × Syy)
Substituting the values we have already computed:
代入我们已经算出的值：
r = 427.2 / √(81.2 × 2319.2) = 427.2 / √188318.24 ≈ 427.2 / 433.96 ≈ 0.984
A value of r = 0.984 is extremely close to +1, indicating a very strong positive linear correlation. This suggests that hours of revision explain a large proportion of the variation in exam scores.
r = 0.984 非常接近 +1，表明极强的正线性相关。这意味着复习小时数解释了考试成绩变异的很大一部分。
5. Hypothesis Test for the Population Correlation Coefficient ρ | 总体相关系数 ρ 的假设检验
We now formally test whether the linear correlation is significant at the 5% level. Let ρ be the population correlation coefficient.
现在我们正式检验线性相关在 5% 水平上是否显著。设 ρ 为总体相关系数。
Hypotheses: H₀: ρ = 0, H₁: ρ ≠ 0 (two‑tailed test).
假设：H₀: ρ = 0，H₁: ρ ≠ 0（双尾检验）。
The test statistic under H₀ is t = r√(n−2) / √(1−r²), which follows a t‑distribution with n−2 = 18 degrees of freedom.
在 H₀ 下检验统计量为 t = r√(n−2) / √(1−r²)，服从自由度为 n−2 = 18 的 t 分布。
t = 0.984 × √18 / √(1 − 0.984²) = 0.984 × 4.2426 / √0.031744 ≈ 4.174 / 0.1782 ≈ 23.43
The critical value for t(18) at the 5% two‑tailed level is approximately 2.101. Since 23.43 ≫ 2.101, we reject H₀. There is extremely strong evidence of a linear correlation between revision hours and exam score.
在 5% 双尾水平下，t(18) 的临界值约为 2.101。因为 23.43 远大于 2.101，我们拒绝 H₀。有极强的证据表明复习小时数与考试成绩之间存在线性相关。
6. Spearman’s Rank Correlation Coefficient | 斯皮尔曼等级相关系数
As a check against non‑linearity and outliers, we compute Spearman’s rank correlation, rs. First, we rank both hours and score (using average ranks for ties, though there are none here). Then we compute the differences d between the ranks of each pair and find Σd².
为了检查非线性和异常值，我们计算斯皮尔曼等级相关系数 rs。首先，分别对小时数和分数进行排序（如有相同值则用平均秩，这里无相同值）。然后计算每对秩的差值 d 并求出 Σd²。
Using the data we obtain Σd² = 14 (the differences are mostly 0 or ±1 due to the strong monotonic relationship). The formula for rs when there are no tied ranks is:
利用数据我们得到 Σd² = 14（由于强单调关系，差值大多为 0 或 ±1）。无相同秩时 rs 的公式为：
rs = 1 − (6 Σd²) / (n(n²−1))
rs = 1 − (6 × 14) / (20 × (400−1)) = 1 − 84 / (20×399) = 1 − 84/7980 ≈ 1 − 0.01053 ≈ 0.9895
This value is also very close to 1, confirming a strong, positive monotonic association. A hypothesis test
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