A-Level Edexcel Statistics: Case Study Practical Exercises | A-Level Edexcel 统计：案例分析实战演练

📚 A-Level Edexcel Statistics: Case Study Practical Exercises | A-Level Edexcel 统计：案例分析实战演练

Case study questions are a key feature of the Edexcel A-Level Statistics examination, requiring you to apply statistical concepts to real-world scenarios. This article provides a comprehensive set of practical exercises, guiding you through the thought process, methodology, and presentation needed to excel. By working through these examples, you will build confidence in handling data, selecting appropriate models, performing calculations, and interpreting results within context.

案例分析题是Edexcel A-Level统计考试的核心特征，要求你将统计概念应用于真实场景。本文提供了一套全面的实战练习，引导你掌握解题思路、方法和表述，从而取得优异成绩。通过演练这些示例，你将建立起处理数据、选择合适模型、进行计算并结合背景解读结果的信心。

1. Understanding Case Study Format | 理解案例分析题型格式

Edexcel Statistics papers often present a scenario accompanied by a dataset, followed by a series of interconnected questions. You might be asked to summarise data, calculate probabilities, choose a distribution, perform a hypothesis test, or evaluate whether statistical conditions are met. The key is to read the stem carefully, identify the variables and their types, and note any assumptions that need to be checked.

Edexcel统计试卷通常会给出一个情景和数据集，随后提出一系列相互关联的问题。你可能会被要求概括数据、计算概率、选择分布、进行假设检验或评估统计条件是否满足。关键在于仔细阅读题干，识别变量及其类型，并注意任何需要检验的假设。

Always link your statistical conclusions back to the original context. For example, if a hypothesis test leads to rejecting the null hypothesis, explain what that means for the scenario, such as ‘there is sufficient evidence to suggest the new drug improves recovery time’. Never just write ‘reject H₀’ without interpretation.

始终要将统计结论与原始背景联系起来。例如，如果假设检验导致拒绝原假设，要解释这对情景意味着什么，比如“有充分证据表明新药改善了恢复时间”。切勿只写“拒绝H₀”而不加以解释。

2. Probability and Venn Diagram Case Study | 概率与文氏图案例

A survey of 200 students about their subject choices found that 120 study Mathematics (M), 90 study Physics (P), and 50 study both. A student is selected at random. Calculate the probability that the student studies neither subject, and determine whether events M and P are independent.

一项针对200名学生选课情况的调查发现，120人学习数学(M)，90人学习物理(P)，50人同时学习两门。随机选取一名学生。计算该学生既不学数学也不学物理的概率，并判断事件M和P是否独立。

The Venn diagram approach gives n(M only) = 120 – 50 = 70, n(P only) = 90 – 50 = 40, n(both) = 50, so n(neither) = 200 – (70+40+50) = 40. Thus P(neither) = 40/200 = 0.2. For independence, check P(M ∩ P) ?= P(M)×P(P): P(M)=0.6, P(P)=0.45, product = 0.27, while P(M ∩ P)=0.25. Since 0.25 ≠ 0.27, the events are not independent.

文氏图方法得出：仅数学人数=120-50=70，仅物理=90-50=40，两者都选=50，因此都不选的人数=200-(70+40+50)=40。所以P(都不选)=40/200=0.2。对于独立性，检验P(M ∩ P)是否等于P(M)×P(P)：P(M)=0.6，P(P)=0.45，乘积为0.27，而P(M ∩ P)=0.25。由于0.25≠0.27，事件不独立。

In an exam, you must also interpret: ‘The probability a randomly chosen student studies neither is 0.2, and because P(M ∩ P) does not equal P(M)P(P), choosing Mathematics and choosing Physics are dependent events—knowing a student studies one subject changes the likelihood of studying the other.’

考试中你还需解读：“随机选到的学生不学任何一门的概率是0.2，并且由于P(M ∩ P)不等于P(M)P(P)，选择数学和选择物理是相关事件——知道学生学一门会改变学另一门的可能性。”

3. Discrete Random Variables and Expectation | 离散随机变量与期望

A fair six-sided die is rolled twice. Let X be the absolute difference between the two scores. Construct the probability distribution of X and find E(X) and Var(X).

一个公平的六面骰子掷两次。设X为两次点数差的绝对值。构建X的概率分布，并求E(X)和Var(X)。

Possible differences range from 0 to 5. Count the 36 equally likely outcomes: X=0 when scores are equal (6 outcomes), X=1 when difference is 1 (e.g. (1,2),(2,1) etc., 10 outcomes), similarly X=2 (8 outcomes), X=3 (6 outcomes), X=4 (4 outcomes), X=5 (2 outcomes). Thus P(X=0)=6/36=1/6, P(X=1)=10/36=5/18, P(X=2)=8/36=2/9, P(X=3)=6/36=1/6, P(X=4)=4/36=1/9, P(X=5)=2/36=1/18.

可能的差值从0到5。数出36个等可能结果：X=0时点数相同(6种结果)，X=1时差1(如(1,2),(2,1)等，10种)，类似地X=2(8种)，X=3(6种)，X=4(4种)，X=5(2种)。因此P(X=0)=6/36=1/6，P(X=1)=10/36=5/18，P(X=2)=8/36=2/9，P(X=3)=6/36=1/6，P(X=4)=4/36=1/9，P(X=5)=2/36=1/18。

E(X) = Σ x·P(X=x) = 0×(1/6) + 1×(5/18) + 2×(2/9) + 3×(1/6) + 4×(1/9) + 5×(1/18) = 35/18 ≈ 1.944

E(X²) = 0²×(1/6)+1²×(5/18)+4×(2/9)+9×(1/6)+16×(1/9)+25×(1/18) = 105/18 = 35/6 ≈ 5.833

Var(X) = E(X²) – [E(X)]² = 35/6 – (35/18)² = 665/324 ≈ 2.052

The expected absolute difference is about 1.94, which makes sense because small differences are more likely. The variance quantifies the spread. Be sure to show clearly your working for E(X²).

绝对差的期望值约为1.94，这合理因为小差值更可能出现。方差量化了离散程度。注意清晰地展示E(X²)的计算过程。

4. Binomial Distribution Application | 二项分布应用

A manufacturer claims that only 8% of its light bulbs are defective. A retailer tests a random sample of 25 bulbs. If the claim is true, find the probability that the sample contains at most 2 defective bulbs. Also, determine the critical region for a two-tailed test at the 10% significance level to examine whether the defect rate has changed.

某制造商声称其灯泡仅有8%为次品。零售商随机检测了25只灯泡。如果声称属实，求样本中次品数不超过2只的概率。同时，确定在10%显著性水平下检验次品率是否改变的双尾检验的临界域。

Let X ~ B(25, 0.08). P(X ≤ 2) = P(X=0)+P(X=1)+P(X=2) using formula or tables. P(X=0)=0.92²⁵ ≈ 0.124, P(X=1)=25×0.08×0.92²⁴ ≈ 0.270, P(X=2)=300×0.08²×0.92²³ ≈ 0.282. Sum = 0.676. So there is about a 67.6% chance of observing ≤2 defectives if the claim is correct.

设X ~ B(25, 0.08)。P(X ≤ 2) = P(X=0)+P(X=1)+P(X=2)，使用公式或查表。P(X=0)=0.92²⁵ ≈ 0.124，P(X=1)=25×0.08×0.92²⁴ ≈ 0.270，P(X=2)=300×0.08²×0.92²³ ≈ 0.282。合计0.676。因此如果声称正确，观察到≤2个次品的概率约为67.6%。

For a two-tailed test at α=0.10, find the lower and upper tails each with probability ≤0.05. Using cumulative binomial tables with p=0.08, P(X ≤ 0)=0.124 >0.05, so lower critical region is empty (no too-small value). For the upper tail, P(X ≥ 5) = 1 – P(X ≤ 4) = 1 – 0.984 = 0.016 < 0.05, and P(X ≥ 4)=1–P(X ≤ 3)≈0.053>0.05. Thus critical region is X ≥ 5. So we would reject the claim if we find 5 or more defectives.

双尾检验中α=0.10，需找到上下尾概率各≤0.05。利用p=0.08的二项累积表，P(X ≤ 0)=0.124>0.05，因此下临界域为空（没有过小的值）。对上尾，P(X ≥ 5)=1–P(X ≤ 4)=1–0.984=0.016<0.05，而P(X ≥ 4)=1–P(X ≤ 3)≈0.053>0.05。因此临界域为X ≥ 5。即若发现5只或更多次品，我们拒绝原声称。

Interpretation: If the sample contains 5 or more defectives, there is significant evidence at the 10% level that the defect rate has increased. Since the lower tail was not triggered, you would only be detecting an increase, not a decrease, at this significance level with this sample size.

解读：如果样本中包含5只或更多次品，则在10%显著性水平下有显著证据表明次品率上升。由于下尾未被触发，在此显著性水平和样本量下仅能检测到上升，而非下降。

5. Normal Distribution and Approximations | 正态分布与近似

The weight of cereal in a box is normally distributed with mean 500 g and standard deviation 8 g. A quality control inspector selects a random sample of 40 boxes. Find the probability that the sample mean weight is less than 497 g. Also, use the normal approximation to the binomial to solve a problem: if each box has a 5% chance of being underfilled (<490 g), find the approximate probability that in a batch of 200 boxes, fewer than 15 are underfilled.

某谷物盒装重量服从均值为500克、标准差为8克的正态分布。质检员随机选取40盒。求样本均重小于497克的概率。同时，利用正态近似二项分布解决如下问题：若每盒有5%的概率装填不足（<490克），求在200盒的批次中，装填不足的盒数少于15的近似概率。

For sample mean, X̄ ~ N(500, 8²/40) i.e. N(500, 1.28). Standardise: Z = (497 – 500)/√1.28 = –3/1.131 ≈ –2.65. P(Z < –2.65) = Φ(–2.65) ≈ 0.0040. So the probability is about 0.4%.

对于样本均值，X̄ ~ N(500, 8²/40) 即 N(500, 1.28)。标准化：Z = (497 – 500)/√1.28 = –3/1.131 ≈ –2.65。P(Z < –2.65) = Φ(–2.65) ≈ 0.0040。因此概率约为0.4%。

For the approximation: Let Y ~ B(200, 0.05). μ = np = 10, σ² = np(1–p) = 9.5. Using continuity correction, P(Y < 15) ≈ P(Y ≤ 14.5) after standardising. Z = (14.5 – 10)/√9.5 ≈ 4.5/3.082 ≈ 1.46. P(Z < 1.46) ≈ 0.9279. So approx 92.8% chance.

对于近似：设Y ~ B(200, 0.05)。μ = np = 10，σ² = np(1–p) = 9.5。使用连续性校正，P(Y < 15) ≈ P(Y ≤ 14.5) 标准化后。Z = (14.5 – 10)/√9.5 ≈ 4.5/3.082 ≈ 1.46。P(Z < 1.46) ≈ 0.9279。所以约92.8%的概率。

Always remember to check the conditions for normal approximation: np=10 > 5 and n(1–p)=190 > 5, so approximation is reasonable. Also, apply continuity correction unless told otherwise.

始终记住检查正态近似的条件：np=10>5且n(1–p)=190>5，因此近似合理。另外，除非另有说明，否则使用连续性校正。

6. Hypothesis Testing for a Mean | 均值假设检验

A gym claims that its new training programme increases the average weight a person can bench press by more than 10 kg. A random sample of 18 participants had a mean increase of 12.2 kg with a sample standard deviation of 3.5 kg. Test at the 5% significance level.

某健身房声称其新训练计划能使人平均卧推重量增加超过10公斤。随机抽取18名参与者，平均增重为12.2公斤，样本标准差为3.5公斤。在5%显著性水平下进行检验。

One-sample t-test because population s.d. is unknown. H₀: μ = 10, H₁: μ > 10 (upper-tailed). Test statistic: t = (x̄ – μ) / (s/√n) = (12.2 – 10) / (3.5/√18) = 2.2 / 0.825 ≈ 2.667. Degrees of freedom = 17. Critical value at 5% (one-tailed) from t-table is 1.740. Since 2.667 > 1.740, we reject H₀.

使用单样本t检验，因总体标准差未知。H₀: μ = 10，H₁: μ > 10（上尾检验）。检验统计量：t = (x̄ – μ) / (s/√n) = (12.2 – 10) / (3.5/√18) = 2.2 / 0.825 ≈ 2.667。自由度为17。查t表得5%单尾临界值为1.740。由于2.667 > 1.740，拒绝H₀。

Conclusion: There is sufficient evidence at the 5% significance level to support the claim that the mean increase in bench press weight exceeds 10 kg. It is important to state the context: ‘The sample provides significant evidence that the new programme adds more than 10 kg on average.’

结论：在5%显著性水平下有充分证据支持平均卧推增重超过10公斤的说法。结合背景陈述很重要：“样本提供了显著证据表明新计划平均增加超过10公斤。”

7. Hypothesis Testing for Proportions | 比例假设检验

A school headteacher believes that the proportion of students achieving grade A or above in mathematics has increased from the national average of 0.28. In a sample of 150 students, 52 achieved A or above. Test at the 1% significance level.

某学校校长认为该校学生数学获得A及以上等级的比例已从全国平均水平0.28上升。150名学生的样本中有52名获得A及以上。在1%显著性水平下检验。

H₀: p = 0.28, H₁: p > 0.28. Sample proportion p̂ = 52/150 ≈ 0.3467. Under H₀, standard error = √[0.28×0.72/150] ≈ 0.0366. Test statistic z = (0.3467 – 0.28) / 0.0366 ≈ 1.823. For one-tailed test at 1%, critical z = 2.326. Since 1.823 < 2.326, we do not reject H₀.

H₀: p = 0.28，H₁: p > 0.28。样本比例 p̂ = 52/150 ≈ 0.3467。在H₀下，标准误 = √[0.28×0.72/150] ≈ 0.0366。检验统计量z = (0.3467 – 0.28) / 0.0366 ≈ 1.823。对于1%单尾检验，临界z = 2.326。由于1.823 < 2.326，不拒绝H₀。

Interpretation: There is insufficient evidence at the 1% level to conclude that the school’s proportion has increased above 0.28. The observed difference may be due to sampling variation. Always check the condition np₀=150×0.28=42 > 10 and n(1–p₀)=108 > 10, so normal approximation is valid.

解读：在1%显著性水平下，没有足够证据表明该校比例已超过0.28。观察到的差异可能由抽样波动引起。记得检查条件：np₀=150×0.28=42>10且n(1–p₀)=108>10，因此正态近似有效。

8. Chi-Squared Tests | 卡方检验

A market researcher surveys 200 people to see if beverage preference (Tea, Coffee, Soft Drink) is associated with age group (Under 30, 30-50, Over 50). Observed frequencies are given in a table. Test at the 5% significance level whether preference and age group are independent.

一位市场研究员调查了200人，以了解饮品偏好（茶、咖啡、软饮）是否与年龄组（30岁以下、30-50岁、50岁以上）有关。列联表给出了观察频数。在5%显著性水平下检验偏好与年龄组是否独立。

	Tea	Coffee	Soft Drink	Total
Under 30	25	35	30	90
30-50	30	25	15	70
Over 50	18	10	12	40
Total	73	70	57	200

Expected frequencies are calculated as (row total × column total)/grand total. For example, expected for Under 30 Tea = (90×73)/200 = 32.85. Compute all expected values, then χ² = Σ (O–E)²/E. For this table, χ² ≈ 8.43 (workings omitted for brevity). Degrees of freedom = (rows–1)(cols–1) = 2×2 = 4. Critical χ²(4) at 5% is 9.488. Since 8.43 < 9.488, we do not reject H₀.

期望频数通过（行合计×列合计）/总和计算。例如，30岁以下茶饮期望频数 = (90×73)/200 = 32.85。计算所有期望值，然后χ² = Σ (O–E)²/E。该表格χ² ≈ 8.43（计算过程从略）。自由度 = (3-1)(3-1) = 4。5%临界χ²(4) = 9.488。由于8.43 < 9.488，不拒绝H₀。

Conclusion: There is insufficient evidence to suggest an association between age group and beverage preference at the 5% level. Ensure you state that all expected frequencies are above 5, validating the test.

结论：在5%水平下，没有足够证据表明年龄组与饮品偏好之间存在关联。务必说明所有期望频数都大于5，从而该检验有效。

9. Correlation and Regression | 相关与回归

A study records the number of hours spent studying (x) and the marks scored in a test (y) for 10 students. The summary statistics are: Σx = 120, Σy = 750, Σx² = 1800, Σy² = 58500, Σxy = 9750. Find the product moment correlation coefficient and the equation of the regression line of y on x. Predict the mark for a student who studies 15 hours.

一项研究记录了10名学生花在学习上的时间(x)与考试分数(y)。汇总统计量为：Σx = 120，Σy = 750，Σx² = 1800，Σy² = 58500，Σxy = 9750。求积矩相关系数以及y对x的回归线方程。预测学习15小时的学生的分数。

Sxx = Σx² – (Σx)²/n = 1800 – 120²/10 = 1800 – 1440 = 360

Syy = 58500 – 750²/10 = 58500 – 56250 = 2250

Sxy = 9750 – (120×750)/10 = 9750 – 9000 = 750

r = Sxy / √(Sxx × Syy) = 750 / √(360×2250) = 750 / √810000 = 750/900 = 0.833 (approx)

This strong positive correlation suggests a linear relationship. The regression line: y = a + bx where b = Sxy/Sxx = 750/360 = 2.0833, a = ȳ – b·x̄ = 75 – 2.0833×12 = 75 – 25 = 50. So equation: y = 50 + 2.083x. Predicted mark for 15 hours: 50 + 2.083×15 ≈ 81.25, so about 81 marks.

这个强正相关表明存在线性关系。回归线：y = a + bx，其中b = Sxy/Sxx = 750/360 = 2.0833，a = ȳ – b·x̄ = 75 – 2.0833×12 = 75 – 25 = 50。所以方程为：y = 50 + 2.083x。15小时的预测分数：50 + 2.083×15 ≈ 81.25，约81分。

Remember to interpret the slope: For every extra hour studied, the mark is expected to increase by about 2.1 marks. Use the regression line only for predictions within the range of observed x (interpolation), not extrapolation.

记得解读斜率：每多学习一小时，分数预计提高约2.1分。回归线仅适用于观测范围内x的内插预测，而非外推。

10. Sampling and Data Collection | 抽样与数据收集

A case study description might read: ‘A researcher wants to estimate the mean height of adult males in a large city. They stand outside a basketball stadium and measure the height of every tenth man leaving the venue.’ Identify the sampling method, discuss bias, and suggest an improvement.

一个案例描述可能是：“研究者想估算某大城市成年男性的平均身高。他们站在篮球场外，测量每第十个离开场馆男性的身高。”识别抽样方法，讨论偏差，并提出改进建议。

This is systematic sampling with a potential for severe selection bias. Because the sample is taken at a basketball stadium, the individuals are likely to be taller than the general adult male population. The sample is not representative. A better design would be stratified random sampling across various locations (shopping centres, offices, public transport) or a simple random sample from the city’s electoral register.

这是系统抽样，但存在严重选择偏差的可能性。由于样本选择在篮球场，个体可能比一般成年男性人群更高。样本不具代表性。更好的设计是在不同地点（购物中心、办公室、公共交通）进行分层随机抽样，或从该市选民登记册中进行简单随机抽样。

Also, note that systematic sampling can be valid if the starting point is random and the list is not ordered in a pattern related to the variable of interest. But here, the sampling frame is inherently biased.

此外，注意如果起点随机且名单没有与所关心变量相关的模式，系统抽样可以是有效的。但此处抽样框本身就带有偏差。

In Edexcel exams, you must identify the sampling technique (simple random, stratified, systematic, quota, opportunity) and comment on advantages and disadvantages relative to the context.

在Edexcel考试中，必须识别抽样技术（简单随机、分层、系统、配额、便利），并结合背景评述其优缺点。

11. Tips for Tackling Case Study Questions | 处理案例题技巧

Read the entire question before writing anything. Underline key information: sample size, population parameter claims, distribution assumptions, and the specific questions asked. Check whether conditions for tests or models are satisfied (normality, large counts, independence, etc.) and comment on them explicitly in your answer.

下笔前通读整题。在关键信息下划线：样本量、总体参数声称、分布假设以及具体提问。检查检验或模型的条件是否满足（正态性、大计数、独立性等），并在答案中明确说明。

Show all steps clearly: state hypotheses with correct notation, write formula, substitute values, obtain test statistic, compare with critical value or find p-value, and write a contextual conclusion. For probability questions, define the random variable first, then state the distribution with parameters.

清晰展示所有步骤：用正确符号陈述假设，写出公式，代入数值，得到检验统计量，与临界值比较或求p值，并写出结合背景的结论。对于概率问题，先定义随机变量，然后说明分布及参数。

If a question asks you to ‘comment’ or ‘evaluate’, discuss reliability, possible limitations, and the practical significance—not just statistical significance. Use phrases like ‘assuming the sample is random and independent’ and ‘the result is significant at the 5% level, thus we have evidence to suggest…’.

如果问题要求“评述”或“评估”，需要讨论可靠性、可能局限和实际显著性——而不仅是统计显著性。使用诸如“假设样本随机且独立

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A-Level Edexcel Statistics: Case Study Practical Exercises | A-Level Edexcel 统计：案例分析实战演练