📚 A-Level WJEC Statistics: Case Study Practical Drills | A-Level WJEC 统计：案例分析实战演练
This article walks you through a complete statistical investigation, demonstrating how the core techniques from the WJEC A-Level Statistics syllabus are applied to a real-world dataset. By working step by step from raw data to final conclusions, you will strengthen your ability to select appropriate methods, carry out calculations, and interpret findings in context – exactly the skills examined in case-study questions.
本文将带您完成一项完整的统计调查，展示如何将 WJEC A-Level 统计学大纲的核心技术应用于真实数据集。通过从原始数据到最终结论的逐步操作，您将强化选择合适方法、进行计算以及在具体情境中解读结果的能力——这正是案例分析题考查的关键技能。
1. Introduction to the Case Study | 案例研究介绍
A team of student researchers wanted to investigate the weekly screen time (hours) of 16–18-year-olds in a large sixth-form college. They were particularly interested in whether the mean screen time exceeded 28 hours, a figure quoted by a national health campaign. They also wished to compare screen habits between males and females, and to explore any relationship between screen time and sleep duration. A stratified random sample of 50 students was taken, with proportional allocation by gender. Each participant logged their total screen time over a seven-day period and self-reported average nightly sleep hours.
一组学生研究员希望调查一所大型高中16至18岁青少年每周屏幕使用时间（小时）。他们尤其关心平均屏幕时间是否超过28小时（一项全国健康宣传活动引用的数据）。同时，他们还希望比较男女生之间的屏幕习惯，并探索屏幕时间与睡眠时长之间的任何关系。调查采用按性别比例分配的分层随机抽样，共抽取50名学生。每位参与者记录了自己七天的总屏幕时间，并自我报告了平均每晚睡眠小时数。
2. Data Summary and Descriptive Statistics | 数据汇总与描述统计
The recorded data were collated in a spreadsheet. The full sample of 50 students gave a mean screen time of 30.2 hours with a standard deviation of 6.5 hours. Summary statistics by gender are shown in the table below.
收集到的数据汇总在电子表格中。全部50名学生的样本平均屏幕时间为30.2小时，标准差为6.5小时。按性别划分的汇总统计如下表所示。
| Group | Sample size (n) | Mean (hours) | Standard deviation (hours) |
| All students | 50 | 30.2 | 6.5 |
| Male | 24 | 31.7 | 6.1 |
| Female | 26 | 28.9 | 6.8 |
The summary suggests that the overall average is above 28 hours, with males reporting a higher mean than females. However, the variability within each group is similar. Before drawing inferences, we must check whether the sample data meet the assumptions required for parametric tests.
汇总结果表明总体平均时间高于28小时，男生的平均报告时间高于女生。不过，各组内部的变异程度相似。在做出推断之前，我们必须检查样本数据是否满足参数检验所需的假设条件。
3. Graphical Exploration of the Data | 数据的图形化探索
Before formal testing, side-by-side box plots were drawn for male and female screen times, along with a histogram of the combined sample. The box plots showed slightly higher medians for males, with the female distribution exhibiting one mild outlier at the upper end. The histogram of all 50 screen times appeared roughly bell-shaped, with a slight positive skew, but not severe enough to reject normality outright. A normal probability plot (Q–Q plot) was also constructed; points fell reasonably close to the diagonal line, supporting the assumption of approximate normality for the population of screen times.
在正式检验之前，我们绘制了男女生屏幕时间的并列箱线图，以及合并样本的直方图。箱线图显示男生的中位数略高，女生分布在上端存在一个温和异常值。全部50个屏幕时间的直方图大致呈钟形，略有正偏，但未严重到完全拒绝正态性。我们还构建了正态概率图（Q–Q 图）；各点相当接近对角线，支持屏幕时间总体近似正态分布的假设。
- Box plots help compare centres, spreads and identify outliers.
- Histograms and Q–Q plots are essential for assessing distribution shape.
- 箱线图有助于比较中心、离散度并识别异常值。
- 直方图和 Q–Q 图对评估分布形态至关重要。
4. Modelling Screen Time with the Normal Distribution | 用正态分布对屏幕时间建模
Since the histogram and Q–Q plot suggested approximate normality, we may model the sample mean using the normal distribution. For the whole sample of size n = 50, the sample mean x̄ = 30.2 hours and sample standard deviation s = 6.5 hours. According to the Central Limit Theorem, the sampling distribution of the mean is approximately normal with mean μ and standard error σ/√n, even if the population is not perfectly normal, because n is large. Here we estimate the standard error by s/√n = 6.5/√50 ≈ 0.919 hours.
由于直方图和 Q–Q 图均显示近似正态性，我们可以用正态分布对样本均值进行建模。对于全部 50 个样本，样本均值 x̄ = 30.2 小时，样本标准差 s = 6.5 小时。根据中心极限定理，即使总体并非完全正态，由于样本量 n 较大，样本均值的抽样分布也近似服从均值为 μ、标准误为 σ/√n 的正态分布。此处我们估计的标准误为 s/√n = 6.5/√50 ≈ 0.919 小时。
SE = s / √n = 6.5 / 7.071 ≈ 0.919
This standard error is vital for constructing confidence intervals and conducting hypothesis tests.
这一标准误对构建置信区间和进行假设检验至关重要。
5. Confidence Interval for the Mean Screen Time | 屏幕时间均值的置信区间
A 95% confidence interval for the population mean screen time μ is given by x̄ ± t* × (s/√n). With n – 1 = 49 degrees of freedom, the appropriate t-critical value (two-tailed) is approximately 2.0096 (using t-distribution tables). The interval is computed as:
总体屏幕时间均值 μ 的 95% 置信区间公式为 x̄ ± t* × (s/√n)。自由度为 n – 1 = 49 时，相应的双侧 t 临界值约为 2.0096（查 t 分布表）。计算得到的区间为：
30.2 ± 2.0096 × 0.919 = 30.2 ± 1.847
This yields an interval from 28.35 hours to 32.05 hours. We are 95% confident that the true mean weekly screen time for all such students lies between 28.4 and 32.0 hours (1 d.p.). Notice that the value 28 hours lies below the lower bound, suggesting the population mean may indeed be greater than 28 hours.
由此得到置信区间为 28.35 小时至 32.05 小时。我们 95% 确信该群体所有学生的真实平均每周屏幕时间介于 28.4 到 32.0 小时之间（保留一位小数）。注意到 28 小时位于下限之下，这表明总体均值可能确实大于 28 小时。
6. One-Sample t-Test for a Population Mean | 单样本 t 检验：总体均值检验
We now formally test whether the mean screen time exceeds 28 hours using a one-sample t-test at the 5% significance level.
现在我们在 5% 显著性水平下，使用单样本 t 检验正式检验平均屏幕时间是否超过 28 小时。
Hypotheses: H₀: μ = 28 versus H₁: μ > 28 (right-tailed test).
假设：H₀: μ = 28 对比 H₁: μ > 28（右尾检验）。
t = (x̄ – μ₀) / (s/√n) = (30.2 – 28) / 0.919 ≈ 2.395
Degrees of freedom = 49. From t-tables, the critical value for a one-tailed test at the 5% level is about 1.677. Because 2.395 > 1.677, we reject H₀. There is sufficient evidence at the 5% level to conclude that the mean weekly screen time for the population of sixth-form students is greater than 28 hours.
自由度为 49。查 t 分布表，5% 水平的单尾临界值约为 1.677。因为 2.395 > 1.677，我们拒绝 H₀。在 5% 显著性水平下，有充分证据表明该高中学生群体的平均每周屏幕时间大于 28 小时。
The p-value associated with t = 2.395 and df = 49 is approximately 0.01 (using software), which is less than 0.05, reinforcing our conclusion.
与 t = 2.395、df = 49 对应的 p 值约为 0.01（使用软件计算），小于 0.05，进一步支持我们的结论。
7. Comparing Two Group Means – Independent Samples t-Test | 两组均值比较 – 独立样本 t 检验
To compare male and female mean screen times, we employ a two-sample t-test assuming unequal variances (Welch’s t-test), as the sample standard deviations are somewhat similar but not identical. Let μ₋ be the male population mean and μ₊ the female population mean.
为了比较男女生平均屏幕时间，我们采用假设方差不等的双样本 t 检验（Welch t 检验），因为样本标准差略有不同。设 μ₋ 为男生总体均值，μ₊ 为女生总体均值。
Hypotheses: H₀: μ₋ = μ₊ versus H₁: μ₋ ≠ μ₊ (two-tailed).
假设：H₀: μ₋ = μ₊ 对比 H₁: μ₋ ≠ μ₊（双尾检验）。
t = (x̄₋ – x̄₊) / √(s₋²/n₋ + s₊²/n₊)
Here x̄₋ = 31.7, s₋ = 6.1, n₋ = 24; x̄₊ = 28.9, s₊ = 6.8, n₊ = 26.
此处 x̄₋ = 31.7, s₋ = 6.1, n₋ = 24；x̄₊ = 28.9, s₊ = 6.8, n₊ = 26。
SE = √(6.1²/24 + 6.8²/26) = √(1.550 + 1.778) = √3.328 ≈ 1.824
t = (31.7 – 28.9) / 1.824 ≈ 1.535
The degrees of freedom (using Welch–Satterthwaite formula) are approximately 47. The two-tailed critical value at 5% for df ≈ 47 is about 2.012. Because 1.535 < 2.012, we do not reject H₀. There is insufficient evidence at the 5% level to claim a difference in mean screen time between males and females. The observed gap may be due to sampling variability.
自由度（采用 Welch–Satterthwaite 公式）约为 47。在 5% 显著性水平下，df ≈ 47 的双尾临界值约为 2.012。由于 1.535 < 2.012，我们不拒绝 H₀。在 5% 水平上，没有足够证据表明男女生平均屏幕时间存在差异。观察到的差异可能源于抽样变异性。
8. Correlation and Regression between Screen Time and Sleep | 屏幕时间与睡眠时间的相关与回归
Each student also reported average nightly sleep duration. The researchers wished to quantify the linear relationship between screen time (explanatory variable) and sleep (response). Summary calculations gave the Pearson product–moment correlation coefficient r = –0.47. A scatter plot indicated a moderate negative association: as screen time increased, sleep duration tended to decrease.
每位学生还报告了平均每晚睡眠时长。研究者希望量化屏幕时间（解释变量）与睡眠时间（响应变量）之间的线性关系。汇总计算得到皮尔逊积矩相关系数 r = –0.47。散点图显示中等程度的负相关：屏幕时间增加时，睡眠时长趋于减少。
r = –0.47, r² = 0.2209
A t-test for correlation returned t = r√(n-2) / √(1-r²) = –0.47√48 / √(1-0.2209) ≈ –3.68 with df = 48. The critical two-tailed value at 5% is about 2.01; hence the correlation is significantly different from zero. The least-squares regression line of sleep (y) on screen time (x) was calculated as:
相关性 t 检验：t = r√(n-2) / √(1-r²) = –0.47√48 / √(1-0.2209) ≈ –3.68，df=48。5% 双尾临界值约为 2.01；因此该相关系数显著不等于零。睡眠时间 (y) 对屏幕时间 (x) 的最小二乘回归线计算为：
y = 9.2 – 0.08x
The slope suggests that for each additional hour of weekly screen time, average nightly sleep decreases by about 0.08 hours (roughly 5 minutes). However, the r² value of 0.22 indicates that only 22% of the variation in sleep is explained by screen time; other factors clearly play a role. Residual plots showed no obvious pattern, supporting the linear model.
斜率表明，每周屏幕时间每增加一小时，平均每晚睡眠约减少 0.08 小时（约 5 分钟）。然而，r² 值 0.22 说明睡眠变量中只有 22% 的变异可由屏幕时间解释；其他因素显然也起作用。残差图无明显模式，支持线性模型。
9. Chi-Squared Test for Independence – Screen Time Category and Exercise | 卡方独立性检验 – 屏幕时间类别与锻炼
Students were also asked whether they participated in regular physical exercise (yes/no). Screen time was categorised as ‘Low’ (< 25 hours), 'Moderate' (25–35 hours) or 'High' (> 35 hours). The resulting 3 × 2 contingency table is shown below.
学生还回答了是否定期参加体育锻炼（是/否）。屏幕时间被分为“低”（< 25 小时）、“中”（25–35 小时）和“高”（> 35 小时）三类。得到的 3 × 2 列联表如下所示。
| Screen category | Exercise: Yes | Exercise: No | Row total |
| Low | 12 | 6 | 18 |
| Moderate | 10 | 14 | 24 |
| High | 2 | 6 | 8 |
| Column total | 24 | 26 | 50 |
The chi-squared test statistic is computed as Σ (O – E)² / E. The expected frequencies under independence show a marked difference in the ‘High’ group. The test statistic χ² ≈ 6.52 with (3–1)×(2–1) = 2 degrees of freedom. The critical value at the 5% level is 5.991. Since 6.52 > 5.991, we reject the null hypothesis of independence. There is evidence of an association between screen time category and participation in exercise. The contribution to χ² is largest in the ‘High’ screen time cells, where fewer students exercise than expected.
卡方检验统计量按 Σ (O – E)² / E 计算。独立性假设下的期望频数显示“高”屏幕时间组存在明显差异。检验统计量 χ² ≈ 6.52，自由度为 (3–1)×(2–1) = 2。5% 水平下的临界值为 5.991。由于 6.52 > 5.991，我们拒绝独立性零假设。有证据表明屏幕时间类别与参与锻炼之间存在关联。对 χ² 贡献最大的是“高”屏幕时间单元格，其中锻炼的学生少于期望值。
10. Bringing Findings Together – Report and Reflection | 汇总发现 – 报告与反思
The investigation yielded clear evidence that the mean weekly screen time exceeds 28 hours, which could inform health campaigns. Although males appeared to have a higher mean, the difference was not statistically significant, suggesting the need for larger samples to detect smaller effects. Screen time showed a moderate negative correlation with sleep, but the regression model highlighted that other lifestyle factors are likely influential. The chi-squared test flagged a possible link between high screen usage and reduced exercise frequency, though causation cannot be inferred from this observational data.
该项调查明确显示平均每周屏幕时间超过 28 小时，这可为健康宣传活动提供依据。虽然男生的均值看起来更高，但差异不具有统计显著性，说明需要更大的
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