📚 Case Study Practical Exercises | 案例分析实战演练
Statistical skills come to life when applied to real-world problems. This case study takes you through a complete data investigation at Tony’s Café, showing how each technique from the Edexcel GCSE Statistics syllabus can be used to make better business decisions. Work through the exercises, and you will not only strengthen your understanding of key methods but also see why statistics is a powerful tool for uncovering insights in everyday situations.
统计技能在解决真实世界问题时才会真正鲜活起来。本案例将带你完整经历Tony咖啡馆的数据调查过程，展示如何运用Edexcel GCSE统计大纲中的每一种技术来做出更好的商业决策。跟随练习一步步做下去，你不仅能加深对核心方法的理解，还会明白为什么统计是揭示日常生活中深层洞见的强大工具。
1. Introducing the Case Study | 案例背景介绍
Tony’s Café has been operating for six months. Tony wants to understand customer satisfaction, sales patterns, and waiting times in order to improve service. He decides to collect data over two weeks. The variables of interest include the number of customers per day, drink sales by type (coffee, tea, hot chocolate, smoothie), satisfaction scores on a scale of 1 to 5, and waiting time in minutes from order to delivery. This gives us a rich dataset to explore the full range of statistical techniques.
托尼咖啡馆已经营业六个月了。托尼希望了解顾客满意度、销售模式以及等待时间，以便改善服务。他决定收集两周的数据。感兴趣的变量包括每日顾客数量、按类型划分的饮品销量（咖啡、茶、热巧克力、冰沙）、1至5分的满意度评分，以及从下单到送达的等待时间（分钟）。这为我们提供了一个丰富的数据集，可以探索统计方法的完整范围。
The dataset for the first week (Monday to Friday) looks like this:
第一周（周一至周五）的数据如下：
Customers per day: 45, 52, 48, 60, 55
Drink sales: Coffee 120, Tea 85, Hot Chocolate 50, Smoothie 45
Satisfaction scores (sample of 20 customers): 4, 5, 3, 4, 5, 2, 4, 3, 5, 4, 4, 3, 5, 4, 4, 5, 3, 4, 5, 1
Waiting times (minutes, same sample): 5.2, 3.8, 7.1, 6.5, 4.0, 8.3, 6.7, 5.5, 4.2, 7.8, 6.1, 5.0, 7.4, 4.9, 6.3, 8.0, 5.8, 6.9, 7.5, 4.5
每周顾客数：45, 52, 48, 60, 55
饮品销量：咖啡 120, 茶 85, 热巧克力 50, 冰沙 45
满意度评分（20位顾客样本）：4,5,3,4,5,2,4,3,5,4,4,3,5,4,4,5,3,4,5,1
等待时间（分钟，同一样本）：5.2,3.8,7.1,6.5,4.0,8.3,6.7,5.5,4.2,7.8,6.1,5.0,7.4,4.9,6.3,8.0,5.8,6.9,7.5,4.5
2. Data Collection Methods | 数据收集方法
Tony needs to decide how to gather his data. He could observe and record every transaction (a census), but this would be time-consuming. Instead, he uses a sample for satisfaction and waiting times. He interviews every fifth customer leaving the café during lunch hours (systematic sampling). For drink sales, he simply uses the till receipts for the entire week – this is effectively a census of all purchases. Understanding the difference between a sample and a population is critical: the population is all customers in that week, while the sample is the 20 selected customers.
托尼需要决定如何收集数据。他可以观察并记录每一笔交易（普查），但这会很耗时。因此，他对满意度和等待时间采用抽样调查。他在午餐时段每五位离开咖啡馆的顾客中采访一位（系统抽样）。对于饮品销售，他直接使用整周的收银小票——这实际上是对所有购买行为的普查。理解样本与总体的区别至关重要：总体是那一周的所有顾客，而样本是选出的20位顾客。
He also designs a short questionnaire for the satisfaction score. To avoid bias, the question is neutral: ‘On a scale of 1 (very dissatisfied) to 5 (very satisfied), how would you rate your experience today?’ The response scale is clearly defined. By collecting data systematically, Tony ensures his findings will be reliable and representative.
他还为满意度评分设计了一份简短的问卷。为避免偏差，问题的措辞保持中立：“按1（非常不满意）至5（非常满意）打分，您如何评价今天的体验？”回答的评分刻度界定清晰。通过系统化地收集数据，托尼确保研究结果可靠且具有代表性。
3. Organising and Classifying Data | 数据整理与分类
Before analysing, we must organise raw data. First, the satisfaction scores (ordinal data) can be summarised in a frequency table. Count how many times each score appears among the 20 respondents.
在分析之前，我们必须先整理原始数据。首先，满意度评分（有序数据）可以用频数表进行汇总。数一数每个分数在20位受访者中出现了多少次。
| Score 评分 | Frequency 频数 |
|---|---|
| 1 | 1 |
| 2 | 1 |
| 3 | 4 |
| 4 | 8 |
| 5 | 6 |
We can also organise the waiting times into grouped intervals because the data is continuous. Grouping into intervals like 3.0 ≤ t < 5.0, 5.0 ≤ t < 7.0, etc., will help us draw histograms later.
我们也可以将等待时间整理成分组区间，因为这是一个连续数据。按区间分组，如 3.0 ≤ t < 5.0、5.0 ≤ t < 7.0 等，将有助于我们稍后绘制直方图。
4. Graphical Representation: Bar Charts and Pie Charts | 图表表示：条形图和饼图
To make comparisons easy, Tony creates a bar chart for drink sales and a pie chart for satisfaction scores. For drink sales (discrete categorical data), a bar chart is perfect. The heights show Coffee 120, Tea 85, Hot Chocolate 50, Smoothie 45. The bar chart quickly reveals that coffee is the most popular drink, accounting for 120 out of 300 total drinks (40%).
为了便于比较，托尼制作了饮品销量的条形图和满意度评分的饼图。对于饮品销量（离散分类数据），条形图非常适合。柱子的高度分别为咖啡120、茶85、热巧克力50、冰沙45。条形图迅速揭示出咖啡是最受欢迎的饮品，在总计300杯饮品中占了120杯（40%）。
A pie chart for satisfaction scores uses the frequencies to calculate angles: total frequency 20, so 360° ÷ 20 = 18° per observation. Score 4 (frequency 8) gets 8 × 18° = 144°. Score 5: 6 × 18° = 108°; Score 3: 4 × 18° = 72°; Scores 1 and 2 each get 18°. The pie chart shows at a glance that more than two-thirds of customers gave scores 4 or 5, indicating high satisfaction.
满意度评分的饼图用频数计算角度：总频数20，因此每个观测值对应 360° ÷ 20 = 18°。分数4（频数8）对应 8 × 18° = 144°；分数5：6 × 18° = 108°；分数3：4 × 18° = 72°；分数1和2各对应18°。饼图一目了然地显示出超过三分之二的顾客给出4或5分，表明满意度较高。
5. Central Tendency: Mean, Median, Mode | 中心趋势：均值、中位数、众数
Tony calculates the average satisfaction score to get one number that summarises customer feeling. The mean is sum of all scores divided by 20. Sum = 4+5+3+4+5+2+4+3+5+4+4+3+5+4+4+5+3+4+5+1 = 79. Mean x̄ = 79 ÷ 20 = 3.95. This confirms customers are generally satisfied, slightly below ‘satisfied’ level 4.
托尼计算了满意度评分的平均数，以便用一个数字概括顾客的感受。均值是所有分数之和除以20。总和 = 4+5+3+4+5+2+4+3+5+4+4+3+5+4+4+5+3+4+5+1 = 79。均值 x̄ = 79 ÷ 20 = 3.95。这证实顾客总体满意，略低于“满意”的4分水平。
The median (middle value) is found by ordering scores. Ordered: 1,2,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,5. With even n=20, median is between the 10th and 11th values: both are 4, so median = 4. The mode is 4, occurring 8 times. Mean, median and mode are all around 4, suggesting the distribution is roughly symmetric and centred on a positive experience.
中位数通过将评分排序后得到。排序后：1,2,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,5。由于n=20为偶数，中位数在第10和11个值之间：两者均为4，故中位数为4。众数是出现8次的4。均值、中位数和众数都在4附近，表明分布大致对称且集中在正面的体验上。
6. Measures of Dispersion: Range, Quartiles and Standard Deviation | 离散程度：范围、四分位数和标准差
Averages alone can hide variation. Tony looks at the range of satisfaction scores: 5 – 1 = 4. The interquartile range (IQR) is more robust. Lower quartile Q1 is the median of the first 10 values: between 5th and 6th ordered values (3 and 3) → Q1 = 3. Upper quartile Q3 is median of the last 10: between 15th and 16th values (5 and 5) → Q3 = 5. IQR = 5 – 3 = 2. This tells us the middle 50% of scores lie within 2 points of each other, showing consistency.
单凭平均数可能掩盖变异。托尼查看了满意度评分的范围：5 – 1 = 4。四分位数间距（IQR）更为稳健。下四分位数Q1是前10个数据的中位数：在第5和第6个排序值（3和3）之间 → Q1 = 3。上四分位数Q3是后10个的中位数：在第15和第16个值（5和5）之间 → Q3 = 5。IQR = 5 – 3 = 2。这告诉我们中间50%的评分集中在2分范围之内，显示了评分的一致性。
For waiting times, standard deviation shows the average distance from the mean. First, find mean waiting time: sum = 5.2+3.8+7.1+6.5+4.0+8.3+6.7+5.5+4.2+7.8+6.1+5.0+7.4+4.9+6.3+8.0+5.8+6.9+7.5+4.5 = 121.6, so x̄ = 6.08 min. Then compute the squared deviations, sum them, divide by n – 1 (19), and take square root. The result is approximately s = 1.45 min. This means typical waiting times vary by about 1.45 minutes around the average, a useful measure for improving service consistency.
对于等待时间，标准差展示了偏离均值的平均距离。先求等待时间均值：总和 = 5.2+3.8+7.1+6.5+4.0+8.3+6.7+5.5+4.2+7.8+6.1+5.0+7.4+4.9+6.3+8.0+5.8+6.9+7.5+4.5 = 121.6，故 x̄ = 6.08分钟。然后计算离差平方，求和，除以 n – 1（19），再开平方根。结果约为 s = 1.45分钟。这意味着典型的等待时间在平均值附近波动约1.45分钟，对于改善服务的一致性是一个有用的度量。
7. Box Plots and Outliers | 箱线图与异常值
A box plot for waiting times visualises the five-number summary. Minimum 3.8, Q1 = 5.0 (the median of the lower half), Median (Q2) = 6.2, Q3 = 7.4, Maximum 8.3. We can check for outliers: IQR = 7.4 – 5.0 = 2.4. Fences: lower fence Q1 – 1.5×IQR = 5.0 – 3.6 = 1.4, upper fence Q3 + 1.5×IQR = 7.4 + 3.6 = 11.0. No values fall outside these fences, so there are no outliers in waiting times. The box plot confirms the data is slightly skewed to the left (median closer to Q3).
等待时间的箱线图将五数汇总可视化。最小值3.8，Q1 = 5.0（下半部分的中位数），中位数（Q2）= 6.2，Q3 = 7.4，最大值8.3。我们可以检查异常值：IQR = 7.4 – 5.0 = 2.4。阈值：下阈值 Q1 – 1.5×IQR = 5.0 – 3.6 = 1.4，上阈值 Q3 + 1.5×IQR = 7.4 + 3.6 = 11.0。没有数值落在此阈值之外，因此等待时间中没有异常值。箱线图确认了数据略微左偏（中位数更靠近Q3）。
For satisfaction scores, the box plot would have Q1=3, Median=4, Q3=5, min=1, max=5. The lower whisker extends to 1, but with IQR=2, the lower fence is 3 – 3 = 0, so score 1 is not an outlier. However, the presence of a score 1 might still warrant investigation: why was that customer so dissatisfied?
对于满意度评分，箱线图的Q1=3，中位数=4，Q3=5，最小值=1，最大值=5。下触须延伸到1，但由于IQR=2，下阈值为0，所以1分并非异常值。不过，出现了1分仍值得探究：为什么那位顾客如此不满意？
8. Cumulative Frequency and Percentiles | 累积频率与百分位数
Tony wants to know what percentage of customers waited more than 7 minutes. Using the grouped waiting time data, we can build a cumulative frequency table and draw a cumulative frequency graph. Suppose we group waiting times as: 3.0- (4), 5.0- (8), 7.0- (6), 8.0-8.9 (2). The cumulative frequencies become 4, 12, 18, 20. The 90th percentile is the time below which 90% of customers fall. Estimate from the graph: 90% of 20 = 18th value, which lies in the 7.0–7.9 group, around 7.8 minutes. This means only 10% of customers wait longer than 7.8 minutes, which is acceptable for Tony’s target.
托尼想知道等待超过7分钟的顾客占多大比例。使用分组后的等待时间数据，我们可以构建累积频率表并绘制累积频率图。假设我们将等待时间分组为：3.0– (4), 5.0– (8), 7.0– (6), 8.0–8.9 (2)。累积频率依次为 4, 12, 18, 20。第90百分位数是指90%的顾客所需等待时间。根据图表估算：20的90% = 第18个值，落在7.0–7.9组，大约为7.8分钟。这意味着只有10%的顾客等待超过7.8分钟，对于托尼的目标来说是可以接受的。
Cumulative frequency also lets us estimate the median and quartiles precisely. The median is at cumulative frequency 10, which falls in the 5.0–6.9 group. Using proportional interpolation, the median is approximately 6.2 minutes, matching our earlier calculation.
累积频率还能让我们精确估计中位数和四分位数。中位数位于累积频率10，落在5.0–6.9组。通过比例插值，中位数约为6.2分钟，与我们之前的计算相符。
9. Scatter Diagrams and Correlation | 散点图与相关性
Tony suspects that on days with more customers, average waiting time increases. He records daily customer numbers and average waiting times over 10 days: (45, 5.8), (52, 6.2), (48, 5.9), (60, 7.0), (55, 6.5), (50, 6.0), (58, 6.8), (62, 7.3), (40, 5.5), (53, 6.4). Plotting these on a scatter diagram, we see an upward trend. The points cluster around an imaginary straight line with positive slope, indicating positive correlation. Tony calculates the product moment correlation coefficient (PMCC) r. Using the formula, r ≈ 0.92, which is very close to +1, confirming a strong positive linear relationship.
托尼猜想顾客人数较多的日子里，平均等待时间也会增加。他记录了10天的每日顾客数和平均等待时间：(45,5.8), (52,6.2), (48,5.9), (60,7.0), (55,6.5), (50,6.0), (58,6.8), (62,7.3), (40,5.5), (53,6.4)。将这些点绘制在散点图上，我们看到上升趋势。这些点聚集在一条具有正斜率的假想直线附近，表明正相关。托尼计算了积矩相关系数（PMCC）r。通过公式计算，r ≈ 0.92，非常接近+1，证实了强正线性相关。
Although correlation does not imply causation, it strongly suggests that reducing customer wait times might require managing peak-time customer flow. Tony can use a line of best fit to predict the average waiting time if customer numbers reach, say, 65. Extrapolation must be treated with caution.
尽管相关性并不意味着因果关系，但它强有力地暗示，若要缩短顾客等待时间，可能需要管理高峰时段的客流。托尼可以使用最佳拟合线预测当顾客数达到65时的平均等待时间。此时外推须谨慎对待。
10. Basic Probability Application | 概率基础应用
From the satisfaction data, Tony estimates probabilities. The probability a randomly selected customer gave a score of 5 is P(5) = 6/20 = 0.3. The probability of a score at least 4 is P(4 or 5) = (8+6)/20 = 14/20 = 0.7. These are experimental probabilities based on observed frequencies.
根据满意度数据，托尼估计概率。随机选择一位顾客，其评分为5的概率为 P(5) = 6/20 = 0.3。评分至少为4的概率为 P(4或5) = (8+6)/20 = 14/20 = 0.7。这些都是基于观测频数的实验概率。
If Tony selects two customers at random without replacement, the probability both gave score 5 is (6/20) × (5/19) = 30/380 ≈ 0.0789. These simple probability calculations help Tony set realistic service benchmarks and understand the likelihood of extreme outcomes.
如果托尼不放回地随机选取两位顾客，两人都打了5分的概率为 (6/20) × (5/19) = 30/380 ≈ 0.0789。这些简单的概率计算帮助托尼设定切合实际的服务基准，并理解极端结果的可能性。
11. Sampling Methods | 抽样方法
Tony’s systematic sampling (every 5th customer) was quick and easy, but could it be biased? If lunchtime customers are systematically different (e.g., more rushed) than morning customers, the sample may not represent the whole day. A stratified sample might be better: divide the day into time slots (morning, lunch, afternoon), and randomly select customers within each in proportion to total traffic. Stratified sampling ensures each part of the day is fairly represented, increasing accuracy for estimating overall satisfaction.
托尼的系统抽样（每5位顾客）快速简便，但可能存在偏差吗？如果午餐时段的顾客与其他时段（如早晨）存在系统性差异（比如更匆忙），那么样本就可能无法代表全天。分层抽样可能更好：将一天划分为若干时段（早晨、午餐、下午），然后按各时段客流量比例随机抽取顾客。分层抽样能确保一天中各时段都被公平地代表，从而提高估计整体满意度的准确性。
When planning a survey, it is also important to consider sample size. A larger sample reduces sampling error. Tony’s sample of 20 is small; collecting 50 or more responses would yield more precise estimates. The choice of sampling method and size directly affects the reliability of conclusions drawn from the data.
在规划调查时，考虑样本量也很重要。较大的样本可以减少抽样误差。托尼的20个样本偏小；收集50份或更多反馈会得出更精确的估计。抽样方法和样本量的选择直接影响从数据中得出结论的可靠性。
12. Conclusion and Reflection | 结论与反思
This case study has applied a range of statistical techniques to the everyday operation of a café. We started with data collection methods, organised raw numbers into tables and graphs, calculated averages and measures of spread, visualised distributions with box plots, modelled relationships with scatter diagrams, and used probability to evaluate likelihoods. Each step built on the previous one to form a complete statistical investigation.
本次案例研究将一系列统计技术应用于咖啡馆的日常运营中。我们从数据收集方法开始，将原始数据整理成表格和图形，计算了平均数和离散度量，用箱线图可视化分布，用散点图建立关系模型，并用概率评估可能性。每一步都建立在之前步骤的基础上，形成了一次完整的统计调查。
The skills practised here are directly transferable to any GCSE Statistics exam question where you are given a dataset and asked to analyse it. Remember to always question the data: Is the sample representative? Are there any outliers? What does the context tell you about the results? By thinking like a statistician, you can turn numbers into meaningful stories and sound decisions.
这里练习的技能可以直接迁移到任何 GCSE 统计考试题目中，当你拿到一个数据集并被要求分析时。请记住要对数据提出质疑：样本有代表性吗？是否存在异常值？背景信息告诉你什么？像统计学家那样思考，你就能将数字转化为有意义的故事和可靠的决策。
Published by TutorHao | Statistics Revision Series | aleveler.com
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