📚 SQA Higher Statistics: Case Study Practical Exercises | SQA 高级统计：案例分析实战演练
Welcome to this revision guide focused on case study practical exercises for the SQA Higher Statistics course. Throughout Year 12, you will encounter multi‑step real‑world problems that require you to select appropriate statistical methods, perform calculations correctly, interpret results in context, and communicate your findings clearly. This article walks you through a series of worked case studies mirroring the style of SQA assessments, covering descriptive statistics, probability distributions, confidence intervals, hypothesis tests, correlation, regression, and more. Each section presents a scenario, step‑by‑step working, and a discussion of common pitfalls, all designed to build your confidence in tackling open‑ended statistical investigations.
欢迎阅读这篇为 SQA 高级统计课程准备的案例分析实战演练复习指南。在 Year 12 的学习中，你会遇到多步骤的真实世界问题，要求你选择合适的统计方法、准确进行计算、在上下文中解释结果并清晰地表达你的发现。本文通过一系列与 SQA 考试风格相似的案例，带你走过描述统计、概率分布、置信区间、假设检验、相关与回归等内容。每一节都呈现一个情境、逐步推演，并讨论常见陷阱，旨在帮助你自信地应对开放性的统计调查。
1. Introduction to Case Studies in Statistics | 统计案例分析简介
Case studies in SQA Higher Statistics go beyond simple number crunching. They ask you to act as a data analyst: you identify the problem, plan the analysis, choose graphs and summary measures, test hypotheses, and write a short conclusion. Marks are awarded not only for correct computation but also for interpretation and justification. Typical scenarios include quality control in factories, survey analysis for market research, educational experiments, and environmental monitoring.
SQA 高级统计的案例分析不仅仅是数字运算。它们要求你扮演数据分析师的角色：识别问题、规划分析、选择图表和汇总指标、检验假设，并写出简短的结论。评分时不仅看重计算正确，也看重解释与论证。典型场景包括工厂的质量控制、市场研究的调查分析、教育实验和环境监测。
To succeed, you should practise reading questions carefully, noting which statistical technique is suggested by keywords such as “mean”, “spread”, “association”, “significant difference”, or “predict”. This guide will help you connect theory to application by working through nine detailed case studies, followed by tips on exam technique.
要取得成功，你应当练习仔细阅读题目，注意哪些统计方法被 “均值”、”离散度”、”关联”、”显著差异” 或 “预测” 等关键词暗示。本指南将通过九个详细案例带你将理论与实践联系起来，随后再提供考试技巧。
2. Case Study 1: Descriptive Statistics for a Retail Dataset | 案例一：零售数据集的描述性统计
A small online shop recorded the number of orders received each day for 12 consecutive days: 12, 15, 14, 10, 13, 18, 16, 14, 17, 12, 11, 16. The manager wants to summarise the central tendency and spread of daily orders to understand typical performance and variability.
一家小型网店连续 12 天记录每天收到的订单数：12, 15, 14, 10, 13, 18, 16, 14, 17, 12, 11, 16。经理想概括每日订单的集中趋势和离散程度，以了解典型业绩和波动性。
First, arrange the data in ascending order: 10, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 18. The sample mean is calculated as x̄ = Σx/n = (10+11+12+12+13+14+14+15+16+16+17+18)/12 = 168/12 = 14 orders. The median is the average of the 6th and 7th values: (14+14)/2 = 14 orders. The mode is 12, 14 and 16 (each appears twice), so the dataset is multimodal. The range is 18 − 10 = 8. Quartiles: Q1 is the median of the lower half (10,11,12,12,13,14) → (12+12)/2 = 12; Q3 from the upper half (14,15,16,16,17,18) → (16+16)/2 = 16. Hence the interquartile range (IQR) is 16 − 12 = 4.
首先将数据按升序排列：10, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 18。样本均值计算为 x̄ = Σx/n = (10+11+12+12+13+14+14+15+16+16+17+18)/12 = 168/12 = 14 单。中位数是第 6 和第 7 个值的平均：(14+14)/2 = 14 单。众数为 12、14 和 16（各出现两次），因此数据是多峰的。极差为 18 − 10 = 8。四分位数：Q1 是下半部分 (10,11,12,12,13,14) 的中位数 → (12+12)/2 = 12；Q3 来自上半部分 (14,15,16,16,17,18) → (16+16)/2 = 16。因此四分位距 (IQR) 为 16 − 12 = 4。
To measure dispersion more fully, compute the sample standard deviation. Using the formula s = √[Σ(x−x̄)²/(n−1)], first find the squared deviations: (10−14)²=16, (11−14)²=9, (12−14)²=4 twice, (13−14)²=1, (14−14)²=0 twice, (15−14)²=1, (16−14)²=4 twice, (17−14)²=9, (18−14)²=16. Sum = 16+9+4+4+1+0+0+1+4+4+9+16 = 68. Then s = √(68/11) ≈ √6.1818 ≈ 2.49 orders. A boxplot would show a symmetric distribution with no outliers, consistent with the mean equalling the median.
为了更全面地度量离散度，计算样本标准差。使用公式 s = √[Σ(x−x̄)²/(n−1)]，首先求离差平方：(10−14)²=16，(11−14)²=9，(12−14)²=4 两次，(13−14)²=1，(14−14)²=0 两次，(15−14)²=1，(16−14)²=4 两次，(17−14)²=9，(18−14)²=16。总和 = 68。于是 s = √(68/11) ≈ √6.1818 ≈ 2.49 单。箱线图将显示出对称分布且无异常值，与均值等于中位数一致。
The manager can report that the typical daily order number is 14 with a moderate spread (IQR=4, s≈2.5). In an exam, always state your findings in the context of the problem: ‘On average, the shop receives 14 orders per day, with half the days falling between 12 and 16 orders.’
经理可以报告称，每日订单数的典型值为 14，具有中等离散度 (IQR=4, s≈2.5)。在考试中，务必在问题情境下陈述你的发现：”该店平均每天收到 14 单订单，有一半天数的订单数在 12 至 16 单之间。”
3. Case Study 2: Probability and Expected Value in Games | 案例二：游戏中的概率与期望值
A charity stall offers a dice game: you pay £2 to roll two fair six‑sided dice. If the sum is 8, you win £8; if both dice show the same number (a double), you win £3; otherwise you win nothing. The organiser wants to know if the game is likely to make a profit over many trials.
一个慈善摊位推出骰子游戏：支付 2 英镑掷两个公平的六面骰子。如果点数和为 8，赢得 8 英镑；如果两个骰子出现相同点数（对子），赢得 3 英镑；否则没有奖金。组织者想知道多次试验后该游戏是否可能盈利。
First, list the sample space (36 equally likely outcomes). Sum of 8 occurs for (2,6), (3,5), (4,4), (5,3), (6,2) — 5 outcomes. Doubles (excluding (4,4) already counted in sum 8? Check rules: do we win both if sum=8 and double? The problem says ‘if the sum is 8, you win £8; if both dice show the same number, you win £3’. Usually such games pay the higher prize only. Let’s assume the rules specify that the highest prize applies; thus (4,4) would pay £8, not £3. So align with typical SQA wording: ‘A player wins the highest applicable prize.’ Then doubles: (1,1), (2,2), (3,3), (5,5), (6,6) — 5 outcomes, plus (4,4) already covered. Therefore, winning £8: 5 outcomes, winning £3: 5 outcomes, winning £0: 26 outcomes. Probabilities: P(£8) = 5/36, P(£3) = 5/36, P(£0) = 26/36.
首先列出样本空间（36 个等概率结果）。和为 8 的组合有 (2,6)、(3,5)、(4,4)、(5,3)、(6,2) — 5 种。对子（不含已在和为 8 中计算的 (4,4)？检查规则：如果 “和为 8 则赢 8 英镑；如果两个骰子相同则赢 3 英镑”。通常这类游戏只支付较高奖金。假设规则规定适用最高奖金；因此 (4,4) 支付 8 英镑而非 3 英镑。符合典型 SQA 措辞：”玩家赢得适用的最高奖金。” 于是对子有：(1,1)、(2,2)、(3,3)、(5,5)、(6,6) — 5 种，加上已涵盖的 (4,4)。因此赢得 8 英镑：5 种结果；赢得 3 英镑：5 种结果；赢得 0 英镑：26 种结果。概率：P(8英镑) = 5/36，P(3英镑) = 5/36，P(0英镑) = 26/36。
Now define net gain X = prize − cost = prize − 2. The possible net gains are £6 (from £8 prize), £1 (from £3 prize), and −£2 (no win). The expected net gain per game is E(X) = 6×(5/36) + 1×(5/36) + (−2)×(26/36) = (30 + 5 − 52)/36 = −17/36 ≈ −£0.47. This negative expected value means the charity can expect to make about 47p profit per game on average, which is good for fundraising.
现在定义净收益 X = 奖金 − 成本 = 奖金 − 2。可能的净收益为：6 英镑（来自 8 英镑奖金）、1 英镑（来自 3 英镑奖金）和 −2 英镑（无奖金）。每局期望净收益为 E(X) = 6×(5/36) + 1×(5/36) + (−2)×(26/36) = (30+5−52)/36 = −17/36 ≈ −£0.47。该负期望值意味着慈善机构平均每局可盈利约 47 便士，有利于筹款。
When answering SQA questions, always state: ‘In the long run, the game is expected to lose 47p per play for the participant,’ and link back to the context. If asked about fairness, you could note that a fair game would have E(X)=0, which could be achieved by adjusting the cost or prizes.
回答 SQA 问题时，务必说明：”从长期看，参与者每玩一次预期损失 47 便士”，并联系上下文。如果被问及公平性，你可以指出公平游戏的 E(X)=0，这可通过调整参与费或奖金实现。
4. Case Study 3: Normal Distribution – Quality Control in Manufacturing | 案例三：正态分布 – 制造业质量控制
A drinks company fills bottles labelled as containing 330 ml. The actual volume follows a normal distribution with mean μ = 332 ml and standard deviation σ = 4 ml. A bottle is underfilled if it contains less than 325 ml. Calculate the probability that a randomly selected bottle is underfilled, and find the proportion of bottles expected to contain between 328 ml and 336 ml.
一家饮料公司灌装标称 330 毫升的瓶子。实际容量服从均值为 μ = 332 毫升、标准差为 σ = 4 毫升的正态分布。如果容量低于 325 毫升则属灌装不足。计算随机抽取一瓶灌装不足的概率，并求预期容量在 328 毫升到 336 毫升之间的瓶子比例。
For underfilling, let X ~ N(332, 4²). We need P(X < 325). Standardise: Z = (X − μ)/σ = (325 − 332)/4 = −7/4 = −1.75. Using standard normal tables, P(Z < −1.75) = 1 − Φ(1.75) ≈ 1 − 0.9599 = 0.0401. So about 4.0% of bottles are underfilled.
对于灌装不足，设 X ~ N(332, 4²)。需要求 P(X < 325)。标准化：Z = (X − μ)/σ = (325 − 332)/4 = −7/4 = −1.75。查标准正态表，P(Z < −1.75) = 1 − Φ(1.75) ≈ 1 − 0.9599 = 0.0401。因此约 4.0% 的瓶子灌装不足。
For the interval, P(328 < X < 336) requires two Z‑values: Z₁ = (328−332)/4 = −1.00, Z₂ = (336−332)/4 = 1.00. P(−1.00 < Z < 1.00) = Φ(1.00) − Φ(−1.00) = Φ(1.00) − (1 − Φ(1.00)) = 2Φ(1.00) − 1 ≈ 2×0.8413 − 1 = 0.6826. Thus, about 68.3% of bottles have volumes within 4 ml of the mean, consistent with the 68‑95‑99.7 rule.
对于区间，P(328 < X < 336) 需要两个 Z 值：Z₁ = (328−332)/4 = −1.00，Z₂ = (336−332)/4 = 1.00。P(−1.00 < Z < 1.00) = Φ(1.00) − Φ(−1.00) = 2Φ(1.00) − 1 ≈ 2×0.8413 − 1 = 0.6826。因此约 68.3% 的瓶子容量在均值 ±4 毫升内，符合 68‑95‑99.7 规则。
In a quality‑control context, the company may want to reduce the standard deviation or increase the mean to minimise the proportion of underfilled bottles. Always comment on practical implications: ‘The current process results in roughly 1 in 25 bottles being below 325 ml, which may violate regulations.’
在质量控制情境中，公司可能希望减少标准差或提高均值，以最小化灌装不足瓶子的比例。务必评论实际意义：”当前过程导致大约每 25 瓶中有 1 瓶低于 325 毫升，这可能会违反法规。”
5. Case Study 4: Confidence Interval for a Population Mean | 案例四：总体均值的置信区间
A botanist measures the heights of a random sample of 20 mature oak trees from a large forest. The sample mean is 22.5 m, and the sample standard deviation is 4.8 m. Construct a 95% confidence interval for the population mean height of oak trees in the forest, stating any assumptions.
一位植物学家测量了一片大森林中随机选取的 20 棵成熟橡树的高度。样本均值为 22.5 米，样本标准差为 4.8 米。构建该森林橡树总体平均高度的 95% 置信区间，并说明假设。
Because the population standard deviation is unknown and the sample size is small (n=20), we use the t‑distribution with degrees of freedom df = n − 1 = 19. For a 95% confidence interval, the critical t‑value is t(0.025,19) = 2.093 (from tables). The formula for the confidence interval is x̄ ± t* × (s/√n). The standard error SE = s/√n = 4.8/√20 ≈ 4.8/4.4721 ≈ 1.073. The margin of error is 2.093 × 1.073 ≈ 2.245. Hence the interval is 22.5 ± 2.245, giving (20.255 m, 24.745 m).
由于总体标准差未知且样本量较小 (n=20)，我们使用自由度为 df = n − 1 = 19 的 t 分布。对于 95% 置信区间，临界 t 值为 t(0.025,19) = 2.093（查表）。置信区间公式为 x̄ ± t* × (s/√n)。标准误 SE = s/√n = 4.8/√20 ≈ 1.073。误差边际为 2.093 × 1.073 ≈ 2.245。因此区间为 22.5 ± 2.245，即 (20.255 米, 24.745 米)。
The botanist can be 95% confident that the true mean height of all oak trees in the forest lies between 20.3 m and 24.7 m. Assumptions: the sample is random, the heights are approximately normally distributed (or the sample comes from a normal population — with n=20 this assumption is important; a boxplot of the sample data could be used to check symmetry).
植物学家可以有 95% 的信心认为森林中所有橡树的真实平均高度介于 20.3 米到 24.7 米之间。假设：样本随机，高度近似服从正态分布（或样本来自正态总体 — 在 n=20 时这一假设很重要；可用样本箱线图检验对称性）。
In an SQA exam, always interpret the confidence level correctly: it means that if we repeated the sampling process many times, 95% of the calculated intervals would capture the true mean. It does not mean there is a 95% probability that the mean lies in this specific interval.
在 SQA 考试中，务必正确解释置信水平：它意味着如果多次重复抽样过程，95% 计算出的区间将包含真实均值。这并不意味着真实均值有 95% 的概率位于这个特定区间内。
6. Case Study 5: Hypothesis Testing – Is a New Teaching Method Effective? | 案例五：假设检验 – 新教学方法有效吗？
A school trial a new mathematics teaching strategy with a class of 25 pupils. Previously, the national average score on a standardised test was 65. After the trial, the class mean is 69.2 with a sample standard deviation of 7.5. Test at the 5% significance level whether the new method improves scores. Use a one‑sample t‑test, as the population standard deviation is unknown.
一所学校对 25 名学生试点新的数学教学策略。此前，标准化测试的全国平均分为 65。试点后，班级均值为 69.2，样本标准差为 7.5。在 5% 显著性水平下检验新方法是否提高了成绩。使用单样本 t 检验，因为总体标准差未知。
Step 1: State hypotheses. H₀: μ = 65 (no improvement); H₁: μ > 65 (improvement). This is a one‑tailed test. Step 2: Test statistic t = (x̄ − μ₀)/(s/√n) = (69.2 − 65)/(7.5/√25) = 4.2 / (7.5/5) = 4.2 / 1.5 = 2.80. Step 3: Degrees of freedom = 24. At α = 0.05 one‑tailed, the critical value from t‑table is t(0.05,24) = 1.711. Step
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