Case Study Analysis in Practice | 案例分析实战演练

📚 Case Study Analysis in Practice | 案例分析实战演练

In Year 13 OCR Statistics, case study analysis is a powerful way to connect theoretical knowledge with real-world problem-solving. This article presents a series of applied scenarios that cover the core A2 topics—hypothesis testing, regression, probability distributions, confidence intervals, and more. Each case is worked through step by step, with explanations in both English and Chinese, so you can build confidence and fluency in tackling exam-style questions.

在 Year 13 OCR 统计中，案例分析是将理论知识联系实际解题的有效方式。本文呈现一系列应用场景，涵盖 A2 阶段的核心主题——假设检验、回归、概率分布、置信区间等。每个案例逐步推演，并配有中英双语解释，帮助你建立应对考试题型的信心与熟练度。

1. Two-Sample t-Test for Comparing Means | 双样本 t 检验比较均值

A teacher wants to test whether the mean mathematics score differs between male and female students. A random sample of 30 male students has mean 72 and standard deviation 8; a sample of 35 female students has mean 68 and standard deviation 7. Assume equal population variances and use a 5% significance level.

一位教师想检验男女生的数学平均成绩是否有差异。随机抽取 30 名男生，平均分 72，标准差 8；抽取 35 名女生，平均分 68，标准差 7。假设两总体方差相等，显著性水平为 5%。

Step 1: State the hypotheses. H₀: μ₁ = μ₂, H₁: μ₁ ≠ μ₂ (two-tailed test).

步骤 1：设定假设。H₀: μ₁ = μ₂，H₁: μ₁ ≠ μ₂（双尾检验）。

Step 2: Calculate the pooled variance and test statistic.

步骤 2：计算合并方差和检验统计量。

s²ₚ = [(n₁ − 1)s₁² + (n₂ − 1)s₂²] / (n₁ + n₂ − 2)

Substituting: s²ₚ = [29×64 + 34×49] / 63 = (1856 + 1666) / 63 ≈ 55.905. The standard error is SE = √(s²ₚ × (1/30 + 1/35)) ≈ √(55.905 × 0.06190) ≈ 1.860. Thus t = (72 − 68) / 1.860 ≈ 2.150 with df = 63.

代入计算：s²ₚ = [29×64 + 34×49] / 63 = (1856 + 1666) / 63 ≈ 55.905。标准误 SE = √(s²ₚ × (1/30 + 1/35)) ≈ √(55.905 × 0.06190) ≈ 1.860。于是 t = (72 − 68) / 1.860 ≈ 2.150，自由度 63。

The critical two-tailed t-value at α = 0.05 with df ≈ 60 is about 2.000. Since 2.150 > 2.000, we reject H₀. There is sufficient evidence of a difference in mean scores between genders.

在 α = 0.05 下，df ≈ 60 的临界双尾 t 值约为 2.000。由于 2.150 > 2.000，我们拒绝 H₀。有充分证据表明男女平均成绩存在差异。

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

A school investigates whether eating breakfast is associated with passing an exam. 200 students are surveyed: among students who eat breakfast, 45 passed and 55 failed; among those who skip breakfast, 20 passed and 80 failed. Test at the 5% level whether the two variables are independent.

某学校调查吃早餐与考试及格是否有关联。共调查 200 名学生：吃早餐者中，45 人及格，55 人不及格；不吃早餐者中，20 人及格，80 人不及格。在 5% 水平下检验两变量是否独立。

H₀: Breakfast and exam result are independent. H₁: They are not independent.

H₀：早餐与考试成绩独立。H₁：不独立。

Expected frequencies: (100×65)/200 = 32.5 for breakfast-pass, 67.5 for breakfast-fail, etc. Calculate χ² = Σ (O − E)² / E.

期望频数：早餐-及格 (100×65)/200 = 32.5，早餐-不及格 67.5，依此类推。计算 χ² = Σ (O − E)² / E。

χ² = (45−32.5)²/32.5 + (55−67.5)²/67.5 + (20−32.5)²/32.5 + (80−67.5)²/67.5 ≈ 4.808 + 2.315 + 4.808 + 2.315 = 14.246

Degrees of freedom = (2−1)(2−1) = 1. The critical value χ²₁, ₀.₀₅ = 3.841. Since 14.246 > 3.841, reject H₀. There is a significant association between breakfast and exam performance.

自由度 = (2−1)(2−1) = 1。临界值 χ²₁, ₀.₀₅ = 3.841。由于 14.246 > 3.841，拒绝 H₀。早餐与考试成绩存在显著关联。

3. Linear Regression and Correlation | 线性回归与相关

Ten students recorded study hours (x) and exam scores (y). Data: (5,55), (8,65), (10,72), (4,50), (6,60), (12,80), (7,62), (9,70), (3,48), (11,78). Find the regression line and test if the correlation is significant.

10 名学生记录学习时长 x（小时）与考试成绩 y。数据：(5,55), (8,65), (10,72), (4,50), (6,60), (12,80), (7,62), (9,70), (3,48), (11,78)。求回归直线并检验相关性是否显著。

Compute sums: ∑x = 75, ∑y = 640, ∑x² = 645, ∑y² = 42066, ∑xy = 5091. Then Sxx = 82.5, Syy = 1106, Sxy = 291.

计算和：∑x = 75, ∑y = 640, ∑x² = 645, ∑y² = 42066, ∑xy = 5091。进而 Sxx = 82.5, Syy = 1106, Sxy = 291。

The slope b = Sxy / Sxx = 291 / 82.5 ≈ 3.527; intercept a = ȳ − b x̄ = 64 − 3.527×7.5 ≈ 37.55. Regression equation: ŷ = 37.55 + 3.53x.

斜率 b = 291 / 82.5 ≈ 3.527；截距 a = 64 − 3.527×7.5 ≈ 37.55。回归方程：ŷ = 37.55 + 3.53x。

The product-moment correlation r = S

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

更多咨询请联系16621398022（同微信）