📚 CCEA Year 12 Statistics: Case Study Practice | CCEA 12年级统计：案例分析实战演练
Welcome to this focused revision guide on applying statistical methods through a coherent case study. We will work step‑by‑step from initial problem definition to drawing conclusions, mirroring the structure of CCEA Year 12 examination questions. Each technique you revise is connected to a real research scenario, helping you see why we use standard deviation, hypothesis testing with the binomial distribution, correlation and regression, and effective data visualisation.
欢迎阅读这篇专门针对CCEA 12年级统计的案例分析实战指南。我们将从问题定义开始，一步步地走到最终结论，这完全贴合CCEA考试题的出题结构。你复习的每一项技术都会与一个真实研究场景连接，让你理解为什么我们需要使用标准差、基于二项分布的假设检验、相关与回归，以及有效的数据可视化。
1. Case Study Introduction and Context | 案例背景与问题设定
A psychologist claims that a single 20‑minute session of aerobic exercise can measurably improve short‑term memory recall in young adults. To test this, 20 participants aged 17‑18 are recruited and randomly split into two groups: an ‘exercise’ group that jogs on a treadmill for 20 minutes, and a ‘control’ group that sits quietly for the same period. Immediately afterwards, each participant is shown a list of 15 common words and, after a 2‑minute distraction task, is asked to recall as many words as possible. The number of correctly remembered words is recorded.
一位心理学家声称，一次20分钟的有氧运动可以显著提高年轻人的短期记忆回忆能力。为了验证这一说法，招募了20名17至18岁的参与者并随机分为两组：“运动组”在跑步机上慢跑20分钟，“对照组”则静坐相同时间。随后，每名参与者会看到一份包含15个常见词的清单，经过2分钟的分心任务后，要求他们尽可能多地回忆单词。记录正确回忆的单词数。
Our job as statisticians is to design a suitable analysis plan, explore the data, test the psychologist’s belief formalised as a binomial hypothesis, and investigate the relationship between exercise duration and memory score using a second dataset. This mirrors the multi‑stage nature of CCEA exam case studies.
作为统计人员，我们的任务是设计合适的分析方案、探索数据、将心理学家的观点形式化为二项假设进行检验，并利用第二份数据集研究运动时长与记忆力得分之间的关系。这正好反映了CCEA考试中案例分析的多阶段特性。
2. Data Collection and Preliminary Cleaning | 数据收集与初步整理
The raw data from the 20 participants are shown below. Each value represents the number of words recalled out of a possible 15. The study used random allocation to reduce bias, and the sample size, though small, is manageable for the statistical tools taught in Year 12.
20名参与者的原始数据如下所示。每个数值代表从15个单词中正确回忆出的个数。研究采用随机分配以减少偏差，样本量虽然不大，但足以运用12年级所学的统计工具。
| Group | Sample size, n | Words recalled |
|---|---|---|
| Exercise | 10 | 7, 9, 8, 10, 6, 11, 7, 8, 9, 10 |
| Control | 10 | 5, 6, 4, 7, 5, 8, 6, 5, 7, 4 |
Before calculating any statistics we check for obvious recording errors or impossible values. All scores lie between 0 and 15, so no cleaning is needed. In an exam, this step shows awareness of data quality.
在进行任何统计计算之前，我们要检查是否存在明显的记录错误或不可能出现的数值。所有得分都在0到15范围内，因此无需清理数据。在考试中，这一步骤体现对数据质量的关注。
3. Descriptive Statistics: Central Tendency and Dispersion | 描述性统计：集中趋势与离散程度
For the exercise group, we first arrange the scores in ascending order: 6, 7, 7, 8, 8, 9, 9, 10, 10, 11. The mean is found by summing the ten values and dividing by 10.
对于运动组，首先将得分按升序排列：6, 7, 7, 8, 8, 9, 9, 10, 10, 11。将十个数值求和后除以10可得到均值。
x̄_ex = (7+9+8+10+6+11+7+8+9+10) ÷ 10 = 85 ÷ 10 = 8.5
The median is the average of the 5th and 6th ordered values: (8 + 9) ÷ 2 = 8.5. The sample standard deviation measures spread around this central value. We use the divisor n−1 for a sample.
中位数是第5个和第6个有序数值的平均值：(8+9)÷2=8.5。样本标准差衡量围绕该中心值的离散程度，我们使用除数n−1。
s_ex = √[ Σ(x − x̄)² ÷ (n−1) ] = √[ ( (6−8.5)²+(7−8.5)²×2+(8−8.5)²×2+(9−8.5)²×2+(10−8.5)²×2+(11−8.5)² ) ÷ 9 ]
Performing the squaring yields deviations: 6.25, 2.25 (×2), 0.25 (×2), 0.25 (×2), 2.25 (×2), 6.25. Sum = 6.25 + 4.5 + 0.5 + 0.5 + 4.5 + 6.25 = 22.5. Hence s_ex = √(22.5 ÷ 9) = √2.5 ≈ 1.58.
计算平方偏差后得到总和22.5，因此 s_ex = √(22.5÷9) = √2.5 ≈ 1.58。
For the control group: sorted data 4, 4, 5, 5, 5, 6, 6, 7, 7, 8. Mean = 57 ÷ 10 = 5.7. Median = (5+6)÷2 = 5.5. Standard deviation s_ctrl ≈ 1.34. These summary statistics immediately suggest higher average recall in the exercise condition.
对照组的排序数据为4,4,5,5,5,6,6,7,7,8，均值5.7，中位数5.5，标准差约1.34。这些汇总统计量立刻表明运动条件下的平均回忆词数更高。
| Statistic | Exercise | Control |
|---|---|---|
| Mean | 8.5 | 5.7 |
| Median | 8.5 | 5.5 |
| Std dev (s) | 1.58 | 1.34 |
4. Data Visualisation: Box Plots and Interpretation | 数据可视化：箱线图与解读
A five‑number summary forms the basis of a box plot. For the exercise group: Minimum = 6, Q1 = 7, Median = 8.5, Q3 = 9.75, Maximum = 11. The interquartile range (IQR) = Q3 − Q1 = 2.75.
五数概括是绘制箱线图的基础。对于运动组：最小值6，下四分位数Q1=7，中位数8.5，上四分位数Q3=9.75，最大值11。四分位距IQR = 9.75 − 7 = 2.75。
We check for outliers using the 1.5×IQR rule: lower fence = 7 − 1.5×2.75 = 2.875, upper fence = 9.75 + 4.125 = 13.875. No data points fall outside these fences, so the box plot would show no outliers. The control group gives Min=4, Q1=5, Median=5.5, Q3=6.75, Max=8, IQR=1.75, again with no outliers.
用1.5×IQR检验离群值：下界=7−4.125=2.875，上界=9.75+4.125=13.875。所有数据点均在界内，因此箱线图不会出现离群值。对照组给出最小值4、Q1=5、中位数5.5、Q3=6.75、最大值8、IQR=1.75，同样无离群值。
When drawn side by side, the exercise box is noticeably shifted upwards with a larger IQR, indicating both a higher typical recall and slightly greater spread compared with the control group. CCEA examiners often ask you to compare medians, IQRs and comment on skew. Here both data sets appear roughly symmetric.
将两幅箱线图并列时，运动组的箱体明显上移且IQR更大，反映出回忆词数典型值更高且离散程度略大。CCEA考官常要求比较中位数、四分位距并评价偏态。此处两组数据均大致对称。
5. Basic Probability Concepts in Context | 案例中的基本概率概念
Let event A be ‘participant is from the exercise group’ and event B be ‘words recalled is at least 9’. From the frequency table, P(A) = 10/20 = 0.5, and P(B) = (number of participants scoring ≥9) ÷ 20. The exercise group has five scores ≥9 (9,9,10,10,11); the control group has none. So P(B) = 5/20 = 0.25.
设事件A为“参与者来自运动组”，事件B为“回忆词数至少为9”。由频数表知P(A)=0.5，P(B)=5/20=0.25。
The conditional probability P(B|A) = 5/10 = 0.5, meaning half of the exercise participants achieved a high score. In contrast, P(B|not A) = 0/10 = 0. Exam questions often embed such ideas within larger case studies—practising conditional probabilities deepens your understanding of independence and association.
条件概率P(B|A)=0.5，相比之下P(B|非A)=0。考题常将这类概念嵌入到大案例中，练习条件概率能加深你对独立性和相关性的理解。
6. Binomial Distribution Application | 二项分布应用
To formalise the psychologist’s claim, we define a ‘success’ as recalling more than 8 words. In the exercise group, 5 out of the 10 participants succeeded. We model the number of successes X by a binomial distribution: X ~ B(10, 0.5) under the null hypothesis that exercise has no effect (the success probability remains at the guessed baseline of 0.5).
为将心理学家的主张形式化，我们将“成功”定义为回忆超过8个词。运动组中有5人成功。我们使用二项分布对成功次数建模：在零假设（运动无效，成功概率维持基线0.5）下，X ~ B(10, 0.5)。
P(X = 5) = C(10,5) × (0.5)⁵ × (0.5)⁵ = 252 × 0.5¹⁰ = 252 × 0.0009766 ≈ 0.2461
To assess whether the observed 5 successes are unusually high under the null, we need the cumulative probability P(X ≥ 5) = 1 − P(X ≤ 4). Using binomial tables or formula: P(X ≤ 4) = [C(10,0)+C(10,1)+C(10,2)+C(10,3)+C(10,4)] × 0.5¹⁰ = (1+10+45+120+210) × 0.0009766 = 386 × 0.0009766 ≈ 0.3770. Hence P(X ≥ 5) = 1 − 0.3770 = 0.6230.
为评估观察到的5次成功在零假设下是否异常偏高，需要累积概率P(X≥5)=1−P(X≤4)≈0.6230。
This high probability indicates that observing 5 or more successes is quite likely even if the true success probability is 0.5. This calculation sets the stage for a formal hypothesis test in the next section.
这一高概率表明，即便真实成功概率为0.5，观察到5次或更多成功也颇有可能。此计算为下一节的正式假设检验做好了铺垫。
7. Hypothesis Testing Using the Binomial Distribution | 利用二项分布的假设检验
We now structure the test strictly as required by CCEA mark schemes. Let p be the probability that an exercise‑group participant recalls more than 8 words.
现在严格按照CCEA评分方案的要求构建检验。令p为运动组参与者回忆超过8个词的概率。
H₀: p = 0.5
H₁: p > 0.5 (one‑tailed test)
Significance level α = 0.05. Test statistic X = number of successes, X ~ B(10, 0.5) under H₀. Observed value x = 5.
显著水平α=0.05。检验统计量X为成功次数，在H₀下X~B(10,0.5)。观测值x=5。
P‑value = P(X ≥ 5 | p=0.5) = 0.6230 (as computed earlier). Since 0.6230 > 0.05, we do not reject H₀. There is insufficient evidence at the 5% level to conclude that the success probability exceeds 0.5.
p值=0.6230>0.05，不拒绝H₀。在5%显著性水平下，没有足够证据表明成功
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