📚 AS WJEC Statistics: Case Study Practical Drill | AS WJEC 统计：案例分析实战演练
In AS WJEC Statistics, the ability to apply theoretical knowledge to real-world scenarios is essential for achieving high marks. This case study follows a structured investigation into customer waiting times and spending at a coffee shop chain, demonstrating how to move from raw data collection through to formal hypothesis testing. By working through each stage methodically, you will see how descriptive statistics, probability distributions, correlation, regression, and inferential methods combine to form a complete statistical analysis. This walkthrough mirrors the style of questions found in WJEC examination papers, where candidates are expected to interpret data, select appropriate techniques, and justify their conclusions in context.
在 AS WJEC 统计中，将理论知识应用于真实场景的能力对于获得高分至关重要。本案例研究围绕一家咖啡连锁店的顾客等待时间和消费金额展开结构化调查，展示了如何从原始数据收集推进到正式的假设检验。通过有条不紊地完成每个阶段，你将看到描述性统计、概率分布、相关、回归和推断方法如何结合形成一个完整的统计分析。本演练反映了 WJEC 考试试卷中常见的题型风格，考生需要解读数据、选择合适的技巧并在具体情境中论证其结论。
1. Case Background and Data Collection | 案例背景与数据收集
A regional coffee shop chain, Brew & Ground, has received customer feedback suggesting that waiting times during peak morning hours are longer than the company's target of four minutes. Management wishes to investigate whether this perception is supported by evidence and to explore whether longer waiting times are associated with higher spending, as customers might purchase additional items while queuing. A simple random sample of 50 customers was selected over a two-week period between 07:30 and 09:30. For each customer, the waiting time in minutes (from joining the queue to receiving the order) and the total spending in pounds sterling were recorded. The sampling frame consisted of all transaction records during that time window, and a random number generator was used to select 50 unique transactions. This method minimises selection bias and ensures that every customer in the target period had an equal chance of inclusion, satisfying the requirements for simple random sampling as specified in the WJEC specification.
一家名为 Brew & Ground 的区域咖啡连锁店收到顾客反馈，表明早高峰时段的等待时间超过了公司设定的四分钟目标。管理层希望调查这一感知是否有证据支持，并探讨较长的等待时间是否与更高的消费金额相关，因为顾客在排队时可能会购买额外商品。在两周内，于 07:30 至 09:30 之间，采用简单随机抽样方法选取了 50 名顾客。对于每位顾客，记录了等待时间（从加入队列到收到订单，以分钟计）和总消费金额（以英镑计）。抽样框由该时段内所有交易记录组成，使用随机数生成器选取了 50 笔独立交易。该方法最大限度地减少了选择偏差，并确保目标时段内的每位顾客都有均等的被选中的机会，满足了 WJEC 规范中对简单随机抽样的要求。
2. Organising Raw Data and Frequency Tables | 整理原始数据与频数表
Before any analysis can begin, raw data must be organised into a manageable form. Grouped frequency tables allow us to summarise large datasets and identify patterns. For the 50 recorded waiting times, the following grouped frequency distribution was constructed using class intervals of equal width. The table below shows the frequency (f), cumulative frequency (CF), and relative frequency for each class. This structured presentation is the foundation for calculating estimates of the mean and standard deviation from grouped data, a skill explicitly tested in AS WJEC Statistics.
在开始任何分析之前，必须将原始数据整理成易于管理的形式。分组频数表使我们能够总结大型数据集并识别模式。对于 50 个记录的等待时间，使用等宽组距构建了以下分组频数分布。下表显示了每个组别的频数 (f)、累积频数 (CF) 和相对频率。这种结构化的呈现方式是从分组数据中估算均值和标准差的基础，这是 AS WJEC 统计中明确考查的一项技能。
| Waiting Time (min) | Frequency f | Cumulative Frequency | Relative Frequency |
|---|---|---|---|
| 1 ≤ t < 2 | 3 | 3 | 0.06 |
| 2 ≤ t < 3 | 6 | 9 | 0.12 |
| 3 ≤ t < 4 | 10 | 19 | 0.20 |
| 4 ≤ t < 5 | 13 | 32 | 0.26 |
| 5 ≤ t < 6 | 9 | 41 | 0.18 |
| 6 ≤ t < 7 | 5 | 46 | 0.10 |
| 7 ≤ t < 8 | 4 | 50 | 0.08 |
The modal class is 4 ≤ t < 5 minutes, containing 13 customers. The cumulative frequency column reveals that 32 out of 50 customers (64%) waited less than five minutes, while the remaining 18 customers (36%) waited five minutes or longer. Later, the cumulative frequency will be used to estimate quartiles and construct a cumulative frequency curve. When constructing frequency tables, the WJEC mark scheme rewards careful attention to class boundaries and the correct calculation of class midpoints, which are needed for grouped estimates.
众数组为 4 ≤ t < 5 分钟，包含 13 名顾客。累积频数列显示，50 名顾客中有 32 名（64%）等待时间少于五分钟，而其余 18 名顾客（36%）等待了五分钟或更长时间。稍后，累积频数将用于估算四分位数并绘制累积频数曲线。在构建频数表时，WJEC 的评分方案会奖励对组界和组中值的仔细处理，这些对于分组估算是必需的。
3. Graphical Representation: Histograms and Box Plots | 图表表示：直方图与箱线图
am is appropriate because the variable is continuous. Since the class intervals have equal width, the frequency can be plotted directly on the vertical axis. The histogram shows a distribution that is approximately symmetric with a slight positive skew, as the tail extends further towards higher waiting times. Alongside the histogram, a box plot (box-and-whisker diagram) provides a concise five-number summary of the data. From the raw data, the five-number summary was determined as: minimum = 1.6 min, lower quartile Q₁ = 3.7 min, median Q₂ = 4.5 min, upper quartile Q₃ = 5.4 min, maximum = 7.8 min. The interquartile range is IQR = Q₃ - Q₁ = 5.4 - 3.7 = 1.7 minutes.
可视化呈现是传达分布形状、中心和离散程度的有力工具。对于等待时间数据，直方图是合适的，因为变量是连续的。由于组距宽度相等，频数可以直接绘制在纵轴上。直方图显示分布大致对称，略带正偏态，尾部向较高的等待时间延伸。与直方图配合，箱线图提供了数据的简洁五数概括。根据原始数据，五数概括确定为：最小值 = 1.6 分钟，下四分位数 Q₁ = 3.7 分钟，中位数 Q₂ = 4.5 分钟，上四分位数 Q₃ = 5.4 分钟，最大值 = 7.8 分钟。四分位距为 IQR = Q₃ - Q₁ = 5.4 - 3.7 = 1.7 分钟。
QR = 3.7 - 2.55 = 1.15 minutes, and the upper boundary is Q₃ + 1.5 × IQR = 5.4 + 2.55 = 7.95 minutes. Since all 50 observations fall within these boundaries, there are no outliers in the waiting time data. This indicates that the sample is reasonably homogeneous, which strengthens the reliability of subsequent inferential procedures. The WJEC examination frequently asks candidates to compare distributions using box plots drawn on the same scale.
箱线图的一个优点是它能清晰地识别潜在的异常值。使用 1.5 × IQR 规则，异常值的下界为 Q₁ - 1.5 × IQR = 3.7 - 2.55 = 1.15 分钟，上界为 Q₃ + 1.5 × IQR = 5.4 + 2.55 = 7.95 分钟。由于所有 50 个观测值都落在这些界限之内，等待时间数据中没有异常值。这表明样本相当均匀，增强了后续推断程序的可靠性。WJEC 考试经常要求考生使用绘制在相同比例尺上的箱线图来比较分布。
4. Measures of Central Tendency | 集中趋势的测量
l tendency each offer a different perspective on the typical value of a dataset. Using the raw waiting time data, the sample mean was calculated as the sum of all observations divided by the sample size. The total waiting time across all 50 customers summed to 226 minutes, giving a sample mean of x̄ epresenting the middle value when data are ordered, was found to be 4.5 minutes. The mode, or most frequently occurring value, falls within the modal class of 4 to 5 minutes. The closeness of the mean and median confirms the approximate symmetry observed in the histogram. For the spending data, the sample mean was £8.46 and the median was £8.20, again showing consistency between the two measures.
三个主要的集中趋势度量各自从不同的角度反映了数据集的典型值。使用原始等待时间数据，样本均值计算为所有观测值之和除以样本量。50 名顾客的总等待时间合计为 226 分钟，得出样本均值 x̄ = 226 / 50 = 4.52 分钟。中位数代表数据排序后的中间值，为 4.5 分钟。众数，即出现频率最高的值，落在 4 至 5 分钟的众数组内。均值与中位数的接近证实了在直方图中观察到的近似对称性。对于消费数据，样本均值为 £8.46，中位数为 £8.20，再次显示了两个度量之间的一致性。
When deciding which measure to use, context is crucial. The mean incorporates every data point and is mathematically convenient for further statistical calculations, such as the standard deviation and hypothesis tests. However, the median is more robust to extreme values. Since no outliers are present, either measure is appropriate for this dataset. In WJEC answers, it is advisable to state both the mean and the median, and to comment on what their relationship reveals about skewness. A mean noticeably greater than the median suggests positive skew, while a mean noticeably smaller suggests negative skew.
在决定使用哪个度量时，情境至关重要。均值包含了每个数据点，并且在进一步的统计计算（如标准差和假设检验）中具有数学上的便利性。然而，中位数对极端值更具鲁棒性。由于没有异常值存在，两个度量对本数据集都适用。在 WJEC 的答案中，建议同时陈述均值和中位数，并评论它们之间的关系揭示了怎样的偏态。均值明显大于中位数表明正偏态，而均值明显小于中位数则表明负偏态。
5. Measures of Dispersion and Variability | 离散度与变异性的测量
e centre. The sample standard deviation s for the waiting times was calculated using the formula s = √[Σ(x - x̄ d deviation. The sum of squared deviations from the mean was found to be 68.31, giving s = √68.31 / 49 = √1 tion was s = 2.94 pounds sterling, reflecting greater variability in purchasing behaviour.
仅靠集中趋势无法完整描绘数据集的特征。离散度度量量化了观测值在中心周围的分散程度。等待时间的样本标准差 s 使用公式 s = √[Σ(x - x̄)² / (n - 1)] 计算，其中除以 n - 1 提供了对总体标准差的无偏估计。各观测值与均值的离差平方和为 68.31，得到 s = √68.31 / 49 = √1.394 = 1.18 分钟。因此方差 s² 为 1.39 分钟平方。对于消费数据，标准差为 s = 2.94 英镑，反映了购买行为中更大的变异性。
The coefficient of variation (CV), defined as (s / x̄) × 100% etween datasets with different units or means. For waiting times, CV = (1.18 / 4.52) × 100% ≈ 26.1%, while for spending, CV = (2.94 / 8.46) × 100% ≈ 34.8% airly consistent queue lengths but highly variable customer expenditure, which has implications for revenue forecasting and staffing decisions. WJEC questions on dispersion often require candidates to interpret what the standard deviation means in the given context.
变异系数 (CV) 定义为 (s / x̄) × 100%，它允许比较具有不同单位或均值的数据集之间的相对变异性。对于等待时间，CV = (1.18 / 4.52) × 100% ≈ 26.1%，而对于消费金额，CV = (2.94 / 8.46) × 100% ≈ 34.8%。这表明消费金额比等待时间表现出相对更大的离散程度。在实际意义上，咖啡店可以预期排队长度相当一致，但顾客支出却高度可变，这对收入预测和人员配置决策有影响。关于离散度的 WJEC 题目通常要求考生解释标准差在给定情境中的含义。
6. Fitting a Normal Distribution | 拟合正态分布
thesis testing, we must check whether the sample data plausibly come from a normal distribution. The waiting time data, with a mean of 4.52 minutes and a standard deviation of 1.18 minutes, appear unimodal and roughly symmetric. One informal check involves calculating the proportion of observations falling within one, two, and three standard deviations of the mean. According to the empirical rule for normal distributions, approximately 68% of data should lie within x̄ ± s, 95% within x̄ ± 2s, and 99.7% within x̄ ± 3s. In this sample, 35 out of 50 customers (70%) fell within (4.52 ± 1.18) = (3.34, 5.70) minutes, which is close to the expected 68%. Within two standard deviations, 48 out of 50 (96%) were captured, aligning well with the theoretical 95%.
许多统计推断程序都假设基础总体近似服从正态分布。在进行假设检验之前，我们必须检查样本数据是否可能来自正态分布。等待时间数据，均值为 4.52 分钟，标准差为 1.18 分钟，呈现出单峰且大致对称的特征。一种非正式的检查方法涉及计算落在均值的一倍、两倍和三倍标准差范围内的观测值比例。根据正态分布的经验法则，大约 68% 的数据应落在 x̄ ± s 范围内，95% 落在 x̄ ± 2s 范围内，99.7% 落在 x̄ ± 3s 范围内。在本样本中，50 名顾客中有 35 名（70%）落在 (4.52 ± 1.18) = (3.34, 5.70) 分钟内，接近预期的 68%。在两倍标准差范围内，50 名中有 48 名（96%）被覆盖，与理论值 95% 吻合良好。
A more rigorous approach involves a normal probability plot, which plots ordered data values against theoretical quantiles from a normal distribution. If the points lie approximately along a straight diagonal line, normality can be assumed. For the waiting time data, the probability plot confirmed no systematic curvature, supporting the use of normal-based methods. In WJEC assessments, candidates are not typically required to construct such plots from scratch but should be able to comment on whether a given plot indicates normality and to recognise departures such as heavy tails or skewness.
一种更严格的方法是使用正态概率图，它将有序数据值相对于正态分布的理论分位数进行绘制。如果点大致沿一条对角直线分布，则可以假定正态性。对于等待时间数据，概率图确认了没有系统性的弯曲，支持使用基于正态的方法。在 WJEC 评估中，通常不要求考生从头构建此类图形，但应能够评论给定的图形是否表明正态性，并能识别诸如厚尾或偏态等偏离情况。
7. Correlation Analysis Between Waiting Time and Spending | 等待时间与消费金额的相关分析
To investigate this relationship, a scatter diagram was plotted with waiting time on the horizontal axis and spending on the vertical axis. The scatter plot revealed a moderate positive association: as waiting time increased, spending tended to rise, though with considerable scatter around the general trend. The strength of this linear relationship is quantified by the Pearson product-moment correlation coefficient r. Using the formula r = Sxy / √Sxx × Syy, where Sxy = Σ(x - x̄)(y - ȳ), Sxx = Σ(x - x̄)², and Syy = Σ(y - ȳ)², the correlation coefficient was calculated to be r = 0.62.
Brew & Ground 管理层假设等待时间更长的顾客往往消费更多，可能是由于排队时产生的冲动购买。为探究这一关系，绘制了以等待时间为横轴、消费金额为纵轴的散点图。散点图显示出中等程度的正相关：随着等待时间增加，消费金额趋于上升，尽管在总体趋势附近存在相当大的散布。这一线性关系的强度由皮尔逊积矩相关系数 r 量化。使用公式 r = Sxy / √Sxx × Syy，其中 Sxy = Σ(x - x̄)(y - ȳ)，Sxx = Σ(x - x̄)²，Syy = Σ(y - ȳ)²，计算得出相关系数为 r = 0.62。
nst H₁: ρ ≠ 0, where ρ is the population correlation coefficient. The test statistic t = r√n - 2 / √1 - r². Substituting r = 0.62 and n = 50 gives t = 0.62 × √48 / √1 - 0.3844 = 0.62 × 6.928 / √0 tely 2.011. Since 5.47 far exceeds 2.011, we reject H₀ and conclude that there is a statistically significant linear correlation between waiting time and spending. The WJEC specification requires candidates to be able to perform such tests using both critical values and p-values.
r 值为 0.62 表示中等程度的正相关。为确定统计显著性，我们检验 H₀: ρ = 0 对 H₁: ρ ≠ 0，其中 ρ 是总体相关系数。检验统计量 t = r√n - 2 / √1 - r²。代入 r = 0.62 和 n = 50，得 t = 0.62 × √48 / √1 - 0.3844 = 0.62 × 6.928 / √0.6156 = 4.295 / 0.785 = 5.47。自由度为 48，在 α = 0.05（双尾）下的临界值约为 2.011。由于 5.47 远超 2.011，我们拒绝 H₀，并得出结论：等待时间与消费金额之间存在统计上显著的线性相关。WJEC 规范要求考生能够使用临界值和 p 值进行此类检验。
8. Linear Regression and Predictive Modelling | 线性回归与预测建模
the relationship and make predictions. The least squares regression line takes the form ŷ = a + bx, where b = Sxy / Sxx and a = ȳ - bx̄. Using the summary statistics, Sxy was calculated as 107.8, and Sxx tercept is a = 8.46 - 1.58 × 4.52 = 8.46 - 7.14 = 1.32. Thus, the regression equation is ŷ = 1.32 + 1.58x. This model predicts that for each additional minute of waiting, a customer's spending increases by approximately £1.58 on average. The intercept of £ trapolation to x = 0 lies outside the observed range and should be interpreted cautiously.
在确认了显著的相关性之后，线性回归使我们能够为这种关系建立模型并进行预测。最小二乘回归线的形式为 ŷ = a + bx，其中 b = Sxy / Sxx，a = ȳ - bx̄。利用汇总统计量，计算得 Sxy 为 107.8，Sxx = 68.31。得出斜率 b = 107.8 / 68.31 = 1.58（保留两位小数）。截距 a = 8.46 - 1.58 × 4.52 = 8.46 - 7.14 = 1.32。因此，回归方程为 ŷ = 1.32 + 1.58x。该模型预测，每增加一分钟等待时间，顾客的消费金额平均增加约 £1.58。截距 £1.32 表示当等待时间为零时的预测消费金额，尽管外推到 x = 0 超出了观测范围，应谨慎解释。
To assess the model' ient of determination R² = r² = 0.62² = 0.3844 is used. This means that approximately 38.4% spending can be explained by the linear relationship with waiting time. The remaining 61.6% ze, or promotional offers. For a customer waiting 5.5 minutes, the predicted spending is ŷ = 1.32 + 1.58 × 5.5 = 1.32 + 8.69 = £ for prediction intervals, which are wider than confidence intervals for the mean response.
为评估模型的解释力，使用决定系数 R² = r² = 0.62² = 0.3844。这意味着消费金额中大约 38.4% 的变异可以由与等待时间的线性关系来解释。其余 61.6% 的变异归因于其他因素，如一天中的时段、顾客年龄、团体规模或促销活动。对于等待了 5.5 分钟的顾客，预测消费金额为 ŷ = 1.32 + 1.58 × 5.5 = 1.32 + 8.69 = £10.01。此类预测应伴随对残差的考量以及预测区间的可能性，预测区间比均值响应的置信区间更宽。
9. Hypothesis Testing on the Population Mean | 关于总体均值的假设检验
ne-sample t-test. We formulate the hypotheses as H₀: μ = 4.0 minutes against H₁: μ > 4.0 minutes, establishing a one-tailed test at the 5% significance level. The test statistic is t = (x̄ - μ₀) /
Published by TutorHao | AS 统计 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图