📚 Year 12 WJEC Statistics: Case Study Walkthrough | WJEC 12年级统计：案例分析实战演练
This article presents a detailed case study that ties together all major topics in the Year 12 WJEC Statistics syllabus. By working through a realistic industrial problem – quality control on a bolt production line – you will practise data collection, sampling, presentation, measures of central tendency and dispersion, probability, the binomial and normal distributions, hypothesis testing, and bivariate analysis. Each section builds on the previous one, just as you would in a real statistical investigation. Read the English explanation first, then the Chinese equivalent, to strengthen both your understanding and your exam confidence.
本文通过一个完整的工业案例——螺栓生产线的质量控制——将 WJEC 12年级统计学课程中的核心知识点串联起来。你将依次练习数据收集、抽样、数据呈现、集中趋势与离散程度的度量、概率、二项分布、正态分布、假设检验以及双变量分析。每个部分都层层递进，模拟真实的统计调查流程。请先阅读英文讲解，再对照中文，以加深理解并增强考试信心。
1. Case Overview | 案例概述
A quality engineer suspects that the automatic machine producing M10 bolts has drifted from its target length of 50.0 mm. She also needs to check whether the bolts’ weight is correlated with their length, since heavier bolts might indicate excess material. A random sample of 30 bolts is taken from one hour’s production. Each bolt’s length (mm) and weight (g) are recorded. Historical records show that bolt lengths follow a normal distribution with known standard deviation σ = 1.5 mm. The factory also reports that the long-term defective rate, defined as a bolt length greater than 50.5 mm, is 5%. This case study will address all WJEC AS Statistics skills.
质量工程师怀疑生产 M10 螺栓的自动机床已偏离 50.0 mm 的目标长度。她还需要检验螺栓的重量是否与长度相关，因为偏重的螺栓可能意味着用料过多。她从一小时的产量中随机抽取了 30 个螺栓，记录每个螺栓的长度 (mm) 和重量 (g)。历史记录显示螺栓长度服从正态分布，且已知总体标准差 σ = 1.5 mm。工厂还报告，长期次品率（定义长度超过 50.5 mm 的螺栓）为 5%。本案例将覆盖 WJEC AS 统计学的全部技能要求。
2. Sampling Method | 抽样方法
The engineer uses a simple random sample: every bolt produced during that hour has an equal chance of being selected. A random number generator picks 30 time stamps, and the corresponding bolts are taken from the conveyor belt. This method minimises bias and makes the sample representative of the production population. However, the sample size of 30 is relatively small, so the engineer must check whether the Central Limit Theorem conditions hold when the population standard deviation is known.
工程师采用简单随机抽样：该小时内生产的每一个螺栓被选中的机会均等。用随机数生成器挑选 30 个时间点，然后从传送带上取出对应的螺栓。这种方法能最大程度减少偏差，使样本能代表生产总体。然而，样本量 30 相对较小，因此在已知总体标准差的条件下，工程师仍需检查中心极限定理的条件是否满足。
3. Data Presentation: Stem-and-Leaf and Box Plots | 数据呈现：茎叶图与箱线图
The recorded lengths (to 0.1 mm) of the 30 bolts are organised into an ordered stem-and-leaf plot. The stem represents the integer part and the leaf shows the tenths. A back-to-back stem plot could also compare this batch with a previous one, but here a single plot is enough to reveal shape.
记录下的 30 个螺栓长度（精确到 0.1 mm）被整理成有序茎叶图。茎为整数部分，叶为十分位。如果需要，可以绘制背靠背茎叶图来对比不同批次，但此处单幅茎叶图足以展示分布形态。
| Stem (tens) | Leaf (tenths) |
|---|---|
| 47 | 8 |
| 48 | 5 9 |
| 49 | 1 3 4 6 7 9 |
| 50 | 0 0 1 1 2 2 2 3 3 4 4 5 5 6 7 8 9 |
| 51 | 0 1 3 5 |
| 52 | 3 |
The distribution appears roughly symmetric and unimodal, centred near 50 mm. A box plot based on the five-number summary (min = 47.8, Q₁ = 49.5, median = 50.15, Q₃ = 50.7, max = 52.3) confirms no extreme outliers, making the normal model appropriate.
分布大致对称、单峰，中心在 50 mm 附近。基于五数概括（最小值 = 47.8，Q₁ = 49.5，中位数 = 50.15，Q₃ = 50.7，最大值 = 52.3）绘制的箱线图也显示没有极端离群值，因此适合使用正态模型。
4. Central Tendency: Mean, Median, Mode | 集中趋势：均值、中位数、众数
Using the raw data, the sample mean length is calculated as:
利用原始数据，样本均值计算如下：
x̄ = Σx / n = 1503.6 / 30 = 50.12 mm
The median is the average of the 15th and 16th ordered values: (50.1 + 50.2) / 2 = 50.15 mm. The mode is the most frequent value, which is 50.2 mm (occurring three times). Because mean ≈ median ≈ mode, the data can be taken as approximately symmetric.
中位数是第 15 和第 16 个有序值的平均：(50.1 + 50.2) / 2 = 50.15 mm。众数是出现频率最高的值 50.2 mm（出现三次）。由于均值、中位数、众数接近相等，可认为数据近似对称。
5. Dispersion: Range, IQR, Standard Deviation | 离散度：极差、四分位距、标准差
Dispersion is measured by several statistics. The range is max − min = 52.3 − 47.8 = 4.5 mm. The interquartile range (IQR) is Q₃ − Q₁ = 50.7 − 49.5 = 1.2 mm. The sample standard deviation, using the unbiased estimator, is:
离散程度由多个统计量衡量。极差 = 最大值 − 最小值 = 52.3 − 47.8 = 4.5 mm。四分位距 (IQR) = Q₃ − Q₁ = 50.7 − 49.5 = 1.2 mm。经过无偏估计的样本标准差为：
s = √[ Σ(x − x̄)² / (n − 1) ] = √[ 63.52 / 29 ] ≈ 1.48 mm
The sample standard deviation is close to the known population σ of 1.5 mm, which adds confidence that the process variability has not changed. A small IQR relative to the range also indicates consistency.
样本标准差与已知的总体标准差 1.5 mm 接近，这增强了我们对工艺变异未发生变化的信心。相对于极差而言，IQR 较小也表明数据较为一致。
6. Normal Distribution and Probability | 正态分布与概率
Since bolt lengths are modelled as X ~ N(50, 1.5²), the engineer can estimate probabilities. For example, what proportion of bolts is expected to have a length less than 48.5 mm?
由于螺栓长度可以建模为 X ~ N(50, 1.5²)，工程师可以估计概率。例如，预期长度低于 48.5 mm 的螺栓占多大比例？
z = (48.5 − 50) / 1.5 = −1.00
From the standard normal table, P(Z < −1.00) ≈ 0.1587. So about 15.9% of bolts would be shorter than 48.5 mm if the process were on target. To find the symmetrical cut-off for the central 95% of lengths, we use z = ±1.96:
查阅标准正态分布表，P(Z < −1.00) ≈ 0.1587。因此如果过程处于目标状态，约有 15.9% 的螺栓长度低于 48.5 mm。要找出中间 95% 螺栓长度的对称界限，使用 z = ±1.96：
x = μ ± zσ = 50 ± 1.96 × 1.5 ⇒ (47.06, 52.94)
These calculations are essential for setting control limits and for the hypothesis test in the next section.
这些计算对于设定控制界限以及下一节的假设检验至关重要。
7. Hypothesis Testing: z-test for a Population Mean | 假设检验：总体均值的 z 检验
The main question: has the true mean μ changed from 50 mm? A two-tailed z-test at the 5% significance level is performed. The hypotheses are:
核心问题：真实均值 μ 是否偏离了 50 mm？在 5% 显著性水平下进行双尾 z 检验。假设为：
H₀: μ = 50 vs H₁: μ ≠ 50
With known σ = 1.5 and n = 30, the test statistic is:
已知 σ = 1.5，n = 30，检验统计量为：
z = (x̄ − μ₀) / (σ / √n) = (50.12 − 50) / (1.5 / √30) ≈ 0.438
The critical values are ±1.96. Since 0.438 lies inside the acceptance region, we do not reject H₀. The p-value for a two‑tailed test is 2 × P(Z > 0.438) ≈ 0.661, which is far above 0.05. There is insufficient evidence to say the machine is off-target.
临界值为 ±1.96。由于 0.438 落在接受域内，我们不拒绝 H₀。双尾检验的 p 值为 2 × P(Z > 0.438) ≈ 0.661，远大于 0.05。没有充分证据表明机床偏离了目标。
8. Binomial Distribution: Defective Items | 二项分布：次品检验
The engineer defines a bolt as defective if its length exceeds 50.5 mm. The long-term defective rate is p = 0.05. In a box of 20 bolts, what is the probability of finding exactly two defectives? The number of defectives X follows a binomial distribution B(20, 0.05).
工程师规定长度超过 50.5 mm 为次品。长期次品率 p = 0.05。在一盒 20 个螺栓中，恰好查出两个次品的概率是多少？次品数 X 服从二项分布 B(20, 0.05)。
P(X = 2) = ²⁰C₂ (0.05)² (0.95)¹⁸ ≈ 190 × 0.0025 × 0.3972 ≈ 0.1887
More practically, the probability of at least one defective in the box is:
更具实际意义的是，盒中至少有一个次品的概率为：
P(X ≥ 1) = 1 − P(X = 0) = 1 − (0.95)²⁰ ≈ 1 − 0.3585 = 0.6415
This high chance justifies the need for regular sampling inspection. Binomial calculations also help determine acceptance sampling plans.
这么高的概率说明定期抽检是必要的。二项分布计算还有助于制定验收抽样方案。
9. Bivariate Data: Correlation and Regression | 双变量数据：相关与回归
For the same 30 bolts, the weight (g) is measured alongside length. The engineer wants to see if weight can be predicted from length. The summary statistics are:
对同批 30 个螺栓，还测量了重量 (g)。工程师想了解是否可以用长度预测重量。汇总统计量如下：
| n = 30 | x̄ = 50.12 | ȳ = 24.8 |
| Sₓₓ = 63.52 | Sᵧᵧ = 54.40 | Sₓᵧ = 46.85 |
The product moment correlation coefficient is:
积矩相关系数为：
r = Sₓᵧ / √(Sₓₓ × Sᵧᵧ) = 46.85 / √(63.52 × 54.40) ≈ 46.85 / 58.77 ≈ 0.797
An r of 0.80 suggests a strong positive linear association. Testing H₀: ρ = 0 against H₁: ρ ≠ 0 at the 5% level for n−2 = 28 d.f. gives a critical value of about 0.361. Since 0.797 > 0.361, we reject H₀ and conclude significant correlation.
相关系数约 0.80 表明存在较强的正线性关联。对 28 个自由度在 5% 水平下检验 H₀: ρ = 0 与 H₁: ρ ≠ 0，临界值约为 0.361。由于 0.797 > 0.361，拒绝 H₀，认为相关性显著。
The least squares regression line of weight on length is:
重量对长度的最小二乘回归线为：
y = a + bx, where b = Sₓᵧ / Sₓₓ = 46.85 / 63.52 ≈ 0.737, a = ȳ − b x̄ = 24.8 − 0.737×50.12 ≈ −12.1
Thus, weight ≈ −12.1 + 0.737 length. For a bolt of length 50.5 mm, predicted weight is about 25.1 g. The residuals are reasonably random, supporting the linear model.
因此，重量 ≈ −12.1 + 0.737 × 长度。对一个长度 50.5 mm 的螺栓，预测重量约为 25.1 g。残差图大致随机，支持线性模型。
10. Conclusions and Limitations | 结论与局限
The z-test found no evidence that the mean bolt length has changed from 50 mm. The binomial analysis indicated a fairly high probability of finding at least one over-length bolt in a box of 20, even when the process is in control –
Published by TutorHao | Year 12 统计 Revision Series | aleveler.com
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