📚 Case Study Practice for Year 13 Edexcel Statistics | 案例分析实战演练
In the Year 13 Edexcel Statistics specification, the ability to apply a range of statistical techniques to a real-world scenario is essential for both exam success and further study. This article walks you through a complete case study, from initial data exploration to final conclusions, demonstrating how each method is selected, justified, and interpreted. The case revolves around the effectiveness of a new training programme on employee productivity in a retail company.
在 Year 13 Edexcel 统计课程中，能够将一系列统计技术应用于真实场景对考试和未来学习都至关重要。本文将通过一个完整的案例研究，从初始数据探索到最终结论，展示如何选择、证明和解释每种方法。案例围绕一家零售公司新培训计划对员工生产力的有效性展开。
1. The Scenario and Data Description | 场景与数据描述
The company ‘ShopSmart’ has introduced a three-week digital skills training programme for 50 randomly selected store associates. Another 50 associates serve as a control group with no training. Productivity is measured as the average daily sales value (in £) over the four weeks following the training. Additional variables collected include years of experience, previous training score (out of 100), and whether the associate works full-time or part-time.
公司“ShopSmart”为随机选择的 50 名店员推行了一项为期三周的数字技能培训计划。另外 50 名店员作为对照组，未接受培训。生产力用培训后四周内平均日销售额（单位：英镑）衡量。收集的其他变量还包括工作经验年限、之前的培训评分（满分 100）以及店员是全职还是兼职。
We also survey the associates about whether they feel more confident after the training, resulting in a binary yes/no response. This mix of continuous and categorical data allows us to practise a wide variety of statistical tests.
我们还调查了店员在培训后是否感到更自信，得到了二元的“是/否”回答。这种连续数据和分类数据的混合让我们可以练习多种统计检验。
2. Exploratory Data Analysis | 探索性数据分析
Before formal testing, we produce summary statistics for the trained and control groups. For the trained group, the mean daily sales is £342 with a standard deviation of £38; for the control group, the mean is £315 with a standard deviation of £45. Box plots suggest a roughly symmetric distribution in both groups, with one potential outlier in the control group corresponding to an associate with 20 years of experience.
在正式检验之前，我们为培训组和对照组生成汇总统计量。培训组的平均日销售额为 £342，标准差为 £38；对照组的均值为 £315，标准差为 £45。箱线图显示两组数据大致对称，对照组中有一个可能的异常值，对应一位拥有 20 年经验的店员。
Using a histogram and Q-Q plot, we assess normality. The Shapiro-Wilk test gives a p-value of 0.08 for the trained group and 0.15 for the control group, so the assumption of normality is reasonable at the 5% level. We note that sample sizes of 50 are large enough to rely on the Central Limit Theorem for the difference in means.
使用直方图和 Q-Q 图，我们评估正态性。Shapiro-Wilk 检验给出培训组的 p 值为 0.08，对照组为 0.15，因此在 5% 水平下正态性假设是合理的。我们注意到样本量为 50，足以依赖中心极限定理来处理均值之差。
3. Formulating Hypotheses for the Primary Outcome | 针对主要结果建立假设
The primary research question is: ‘Does the new training programme increase average daily sales?’ We set up a two-sample t-test for independent means. H₀: μₜᵣₐᵢₙₑd = μcₒₙₜᵣₒₗ, and H₁: μₜᵣₐᵢₙₑd > μcₒₙₜᵣₒₗ (one-tailed). The significance level is α = 0.05.
主要研究问题是：“新培训计划是否提高了平均日销售额？”我们建立独立均值双样本 t 检验。H₀：μₜᵣₐᵢₙₑd = μcₒₙₜᵣₒₗ，H₁：μₜᵣₐᵢₙₑd > μcₒₙₜᵣₒₗ（单尾）。显著性水平 α = 0.05。
We must also check the assumption of equal variances. The sample standard deviations are 38 and 45, and Levene’s test yields a p-value of 0.23, so we can assume homogeneity and use the pooled variance t-test.
我们还必须检查方差齐性假设。样本标准差为 38 和 45，Levene 检验的 p 值为 0.23，因此我们可以假设方差齐性并使用合并方差的 t 检验。
4. Performing the Two-Sample t-test | 执行双样本 t 检验
Calculating the pooled standard deviation: sₚ = √[((n₁-1)s₁² + (n₂-1)s₂²) / (n₁+n₂-2)] = √[((49×1444)+(49×2025))/98] = √[(70756+99225)/98] = √(169981/98) ≈ √1734.5 ≈ 41.65.
计算合并标准差：sₚ = √[((n₁-1)s₁² + (n₂-1)s₂²) / (n₁+n₂-2)] = √[((49×1444)+(49×2025))/98] = √[(70756+99225)/98] = √(169981/98) ≈ √1734.5 ≈ 41.65。
The test statistic t = (x̄₁ – x̄₂) / (sₚ√(1/n₁ + 1/n₂)) = (342 – 315) / (41.65 × √(1/50+1/50)) = 27 / (41.65 × √0.04) = 27 / (41.65 × 0.2) = 27 / 8.33 ≈ 3.24. The degrees of freedom = 98. The critical t-value for a one-tailed test at α=0.05 is about 1.66. Since 3.24 > 1.66, we reject H₀. The p-value is approximately 0.0008, indicating strong evidence that the training increased productivity.
检验统计量 t = (x̄₁ – x̄₂) / (sₚ√(1/n₁ + 1/n₂)) = (342 – 315) / (41.65 × √(1/50+1/50)) = 27 / (41.65 × √0.04) = 27 / (41.65 × 0.2) = 27 / 8.33 ≈ 3.24。自由度为 98。α=0.05 的单尾临界 t 值约为 1.66。由于 3.24 > 1.66，我们拒绝 H₀。p 值约等于 0.0008，说明有强有力的证据表明培训提高了生产力。
5. Confidence Interval for the Difference | 差异的置信区间
To quantify the effect, we compute a 95% confidence interval for the difference in means. Using the t-value for 98 df (approximately 1.984): CI = (27) ± 1.984 × 8.33 = 27 ± 16.53, giving (£10.47, £43.53). This interval does not contain zero, consistent with the hypothesis test.
为了量化效果，我们计算均值差的 95% 置信区间。使用自由度为 98 的 t 值（约 1.984）：CI = (27) ± 1.984 × 8.33 = 27 ± 16.53，得出 (£10.47, £43.53)。该区间不包含零，与假设检验结果一致。
Interpretation: We are 95% confident that the training programme increases average daily sales by between £10.47 and £43.53. This is a practically meaningful improvement for the retail chain.
解释：我们有 95% 的把握认为培训计划将平均日销售额提高了 £10.47 到 £43.53。对于这家零售连锁店来说，这是一个有实际意义的提升。
6. Investigating the Role of Experience | 调查工作经验的作用
Does previous experience moderate the effect? We split the trained group into novices (less than 2 years) and experienced (2+ years). The novice subset (n=22) has a mean daily sales of £335, while experienced (n=28) has a mean of £348. A t-test between these subgroups returns a p-value of 0.18, suggesting the difference is not significant at α=0.05. However, the power may be limited due to reduced sample sizes.
先前的工作经验是否会调节这个效果？我们将培训组分为新手（少于 2 年）和老手（2 年及以上）。新手子集（n=22）的日均销售额为 £335，老手（n=28）为 £348。对这些子组进行 t 检验得到的 p 值为 0.18，表明在 α=0.05 下差异不显著。但由于样本量减小，检验效能可能受限。
We also use a scatter plot of years of experience against daily sales and compute the Pearson correlation coefficient. The overall correlation is r = 0.31 with a p-value of 0.002, indicating a weak but significant positive linear relationship. This suggests that more experienced associates tend to have slightly higher sales, independently of the training.
我们还用散点图展示了工作年限与日销售额的关系，并计算了皮尔逊相关系数。整体相关系数 r = 0.31，p 值为 0.002，表明存在微弱但显著的正线性关系。这意味着经验更丰富的店员往往销量稍高，与培训无关。
7. Linear Regression Model | 线性回归模型
We build a multiple linear regression model to predict daily sales using training status (1=trained, 0=control), years of experience, and previous training score. The fitted equation is:
我们建立一个多元线性回归模型，用培训状态（1=培训，0=对照）、工作年限和之前培训评分来预测日销售额。拟合方程为：
Sales = 290 + 24(Trained) + 5.2(Experience) + 0.35(PrevScore)
All coefficients have t-ratios with p-values below 0.05. The R² is 0.44, meaning the model explains 44% of the variation in daily sales. The adjusted R² is 0.42. The residual standard error is 36.1. The F-test for overall significance yields F=25.1 on 3 and 96 df, p < 0.001.
所有系数的 t 比率对应的 p 值都低于 0.05。R² 为 0.44，意味着模型解释了日销售额 44% 的变异。调整后 R² 为 0.42。残差标准误为 36.1。整体显著性 F 检验给出 F=25.1（3,96 自由度），p < 0.001。
Interpretation: Holding other factors constant, being trained adds about £24 to daily sales. Each extra year of experience adds £5.20, and each extra point on the previous training score adds £0.35. The constant term £290 represents the baseline for an untrained associate with zero experience and zero prev score (extrapolation may not be meaningful).
解释：在其他因素不变的情况下，接受培训会使日销售额增加约 £24。每多一年工作经验增加 £5.20，之前培训评分每多一分增加 £0.35。截距项 £290 表示未接受培训、零经验、零之前评分的店员的基线水平（外推可能没有实际意义）。
8. Regression Diagnostics | 回归诊断
We check residuals for normality using a histogram and a Q-Q plot; the points lie close to the diagonal. A Shapiro-Wilk test on residuals gives p=0.21. The plot of residuals against fitted values shows no obvious pattern, and the Breusch-Pagan test for heteroscedasticity gives p=0.67. Thus the assumptions of constant variance and independence appear satisfied.
我们使用直方图和 Q-Q 图检查残差的正态性；点大致沿对角线分布。对残差的 Shapiro-Wilk 检验给出 p=0.21。残差对拟合值的散点图没有显示出明显模式，Breusch-Pagan 异方差检验的 p 值为 0.67。因此，常数方差和独立性假设似乎得到满足。
We also compute Variance Inflation Factors (VIF) to check multicollinearity. All VIFs are below 2, indicating no serious collinearity issues. The model is deemed adequate for inference.
我们还计算了方差膨胀因子（VIF）以检查多重共线性。所有 VIF 均低于 2，表明没有严重的共线性问题。该模型被认为适合进行推断。
9. Analysing Categorical Data: Confidence After Training | 分类数据分析：培训后的自信度
The survey asks trained associates: ‘Do you feel more confident in your job after the training?’ 38 out of 50 answered ‘Yes’. We test whether the proportion of ‘Yes’ is significantly greater than 0.5 using a one-sample proportion z-test. H₀: p = 0.5, H₁: p > 0.5. The sample proportion p̂ = 38/50 = 0.76. The test statistic z = (0.76 – 0.5) / √(0.5×0.5/50) = 0.26 / √(0.25/50) = 0.26 / √0.005 = 0.26 / 0.0707 ≈ 3.68. The one-tailed p-value is 0.0001, strongly rejecting H₀. A 95% Wilson confidence interval for the true proportion is (0.62, 0.86).
调查询问了受训店员：“你在培训后是否对工作感到更自信？”50 人中有 38 人回答“是”。我们使用单样本比例 z 检验来检测“是”的比例是否显著大于 0.5。H₀：p = 0.5，H₁：p > 0.5。样本比例 p̂ = 38/50 = 0.76。检验统计量 z = (0.76 – 0.5) / √(0.5×0.5/50) = 0.26 / √(0.25/50) = 0.26 / √0.005 = 0.26 / 0.0707 ≈ 3.68。单尾 p 值为 0.0001，强烈拒绝 H₀。真实比例的 95% Wilson 置信区间为 (0.62, 0.86)。
Next, we cross-tabulate confidence status (Yes/No) with employment type (full-time/part-time) using a chi-squared test. The observed frequencies are: Full-time Yes=22, No=8; Part-time Yes=16, No=4. Expected frequencies under independence: Full-time Yes=21.6, No=8.4; Part-time Yes=16.4, No=3.6. The chi-squared statistic (with Yates’ correction) is 0.03, df=1, p=0.86. There is no association between confidence and employment type.
接下来，我们使用卡方检验对自信状况（是/否）与就业类型（全职/兼职）进行交叉制表。观测频数为：全职-是=22，否=8；兼职-是=16，否=4。独立性假设下的期望频数：全职-是=21.6，否=8.4；兼职-是=16.4，否=3.6。卡方统计量（经耶茨校正）为 0.03，自由度为 1，p=0.86。自信状况与就业类型之间没有关联。
10. Non-parametric Check: Mann-Whitney U Test | 非参数验证：Mann-Whitney U 检验
If we were not confident about normality, we could validate the t-test result with a Mann-Whitney U test on the original daily sales values. The U statistic is calculated, and the p-value is 0.0012 (one-tailed), which again supports a significant difference. The Hodges-Lehmann estimator for the shift between the two groups is £25.80, very close to the parametric estimate.
如果我们对正态性没有把握，可以使用 Mann-Whitney U 检验对原始日销售额值进行验证。计算 U 统计量后，p 值为 0.0012（单尾），再次支持了显著差异。两组之间位移的 Hodges-Lehmann 估计值为 £25.80，非常接近参数估计结果。
This robustness check strengthens our conclusion that the training effect is genuine and not an artifact of distributional assumptions.
这一定性检验强化了我们的结论，即培训效果是真实的，并非分布假设所致。
11. Power Analysis and Sample Size Considerations | 效能分析与样本量考量
Given the observed effect size of about 0.65 (Cohen’s d = 27/41.65), a retrospective power analysis shows that with n=50 per group, we had over 90% power to detect a difference at α=0.05. Had the company only recruited 30 per group, the power would have dropped to 75%, increasing the risk of a Type II error.
鉴于观测到的效应量约为 0.65（Cohen’s d = 27/41.65），回顾性效能分析表明，每组 50 人的样本量在 α=0.05 下检测差异的效能超过 90%。如果公司每组仅招募 30 人，效能将降至 75%，增加第二类错误的风险。
This demonstrates why careful planning of sample size is critical before conducting an experiment. Using pilot data, we can estimate the required sample size for future studies to ensure adequate power.
这说明了为什么在实验前仔细规划样本量至关重要。使用预试验数据，我们可以估计未来研究所需的样本量，以确保足够的效能。
12. Conclusion and Report Writing | 结论与报告撰写
The case study provides clear evidence that the digital skills training programme significantly increases daily sales, by an estimated £10 to £44 (95% CI). Experience also predicts sales, but does not interact significantly with training. Associates report high confidence after training, and this is not influenced by employment type. All statistical assumptions have been checked and, where helpful, non-parametric alternatives confirm the findings. The company should consider rolling out the training to all associates, while collecting further data to monitor long-term impacts.
本案例研究提供了明确证据，表明数字技能培训计划显著增加了日销售额，估计增幅在 £10 到 £44 之间（95% CI）。工作经验也能预测销售额，但与培训之间没有显著的交互作用。店员在培训后报告了高度自信，且不受就业类型影响。所有统计假设都经过了检验，在必要时非参数替代方法也证实了发现。公司应考虑向所有店员推广该培训，同时收集更多数据以监测长期影响。
When writing a statistical report, always state the research question, describe the data, justify the choice of test, present both test statistics and p-values with confidence intervals, and discuss limitations and practical significance. This structured approach aligns with the Edexcel assessment objectives and prepares you for real-world data analysis.
在撰写统计报告时，务必陈述研究问题、描述数据、证明检验方法的选择、同时给出检验统计量和 p 值及置信区间，并讨论局限性和实际意义。这种结构化的方法与 Edexcel 评估目标一致，并为你进行真实世界的数据分析做好准备。
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