📚 Year 12 Cambridge Statistics: Case Study Practical Training | Year 12 剑桥统计：案例分析实战演练
Statistical case studies are the bridge between abstract theory and real‑world decision‑making. This article works through a complete investigation of light bulb lifetimes, illustrating the key topics from the Cambridge AS Statistics syllabus step by step. You will see how to collect and display data, calculate summary statistics, apply the normal and binomial distributions, perform a hypothesis test, and explore correlation and regression—all within one coherent context.
统计案例分析是连接抽象理论与现实决策的桥梁。本文将完整演示一个灯泡寿命的调查，逐步展示剑桥AS统计大纲中的核心主题。你将看到如何收集和展示数据、计算汇总统计量、应用正态分布和二项分布、进行假设检验，以及探索相关与回归——全部在一个连贯的情境中完成。
1. Data Collection and the Research Question | 数据收集与研究问题
A lighting company manufactures two types of bulbs: the standard Process A and the improved Process B. The research question is whether the new process genuinely increases the average lifetime beyond the current specification of 1500 hours. A random sample of 30 bulbs from Process A is tested in a laboratory, and their lifetimes (in hours) are recorded below.
一家照明公司生产两种灯泡：标准工艺A和改良工艺B。研究问题是新工艺是否真的能将平均寿命提高到现行规格1500小时以上。从工艺A中随机抽取30个灯泡在实验室进行测试，记录其寿命（小时）如下。
| Process A lifetimes (hours) / 工艺A寿命（小时） | |||||
|---|---|---|---|---|---|
| 1448 | 1462 | 1476 | 1485 | 1490 | 1495 |
| 1500 | 1503 | 1506 | 1510 | 1512 | 1518 |
| 1520 | 1522 | 1525 | 1530 | 1532 | 1535 |
| 1538 | 1540 | 1542 | 1545 | 1550 | 1556 |
| 1562 | 1570 | 1580 | 1592 | 1605 | 1620 |
The data form the basis of all subsequent analysis. Our objectives are to describe the distribution of Process A lifetimes, estimate its central values and spread, and then compare it with a sample from Process B to decide if the improvement is statistically significant.
这些数据构成了后续所有分析的基础。我们的目标是描述工艺A寿命的分布，估计其中心值和离散程度，然后将其与工艺B的样本进行比较，以判断改善是否具有统计显著性。
2. Stem‑and‑Leaf Diagram and Histogram | 茎叶图与直方图
A stem‑and‑leaf diagram organises the raw data while preserving every value. Using the last two digits as leaves and the hundreds as stems gives the display below. The key 14|48 represents 1448 hours.
茎叶图在不丢失任何数值的前提下整理原始数据。将最后两位数字作为叶，百位作为茎，得到如下展示。图例14|48表示1448小时。
| Stem / 茎 | Leaves / 叶 |
|---|---|
| 14 | 48, 62, 76, 85, 90, 95 |
| 15 | 00, 03, 06, 10, 12, 18, 20, 22, 25, 30, 32, 35, 38, 40, 42, 45, 50, 56, 62, 70, 80, 92 |
| 16 | 05, 20 |
For a histogram, we group the data into intervals of width 20 hours: 1440–, 1460–, …, 1620–. The frequency density ensures that area is proportional to frequency.
对于直方图，我们将数据按组距20小时分组：1440–、1460–、…、1620–。频率密度确保面积与频率成正比。
| Interval / 区间 | Frequency / 频数 | Frequency density / 频率密度 |
|---|---|---|
| 1440–1459 | 2 | 0.10 |
| 1460–1479 | 2 | 0.10 |
| 1480–1499 | 3 | 0.15 |
| 1500–1519 | 6 | 0.30 |
| 1520–1539 | 6 | 0.30 |
| 1540–1559 | 5 | 0.25 |
| 1560–1579 | 2 | 0.10 |
| 1580–1599 | 2 | 0.10 |
| 1600–1619 | 1 | 0.05 |
| 1620–1639 | 1 | 0.05 |
Drawing the histogram confirms a roughly symmetric, bell‑shaped pattern – a hint that the normal model may be appropriate later.
绘制直方图可确认数据呈大致对称的钟形模式——这暗示后续或许可以采用正态模型。
3. Measures of Central Tendency | 集中趋势的度量
The sample mean is calculated using x̄ = Σx / n. Summing the 30 observations gives Σx = 45869, so the mean lifetime is approximately 1529.0 hours. The median, found from the ordered list, is the average of the 15th and 16th values: (1530 + 1532)/2 = 1531 hours. The mode lies in the class 1520–1539.
样本均值用x̄ = Σx / n计算。将30个观测值相加得到Σx = 45869，因此平均寿命约为1529.0小时。从排序列表中找出的中位数是第15个和第16个值的平均：(1530 + 1532)/2 = 1531小时。众数落在1520–1539组内。
x̄ = 45869 / 30 ≈ 1529.0 hours
Because the mean and median are very close, the distribution is fairly symmetric. This reinforces the impression gained from the histogram.
由于均值和中位数非常接近，分布相当对称。这进一步强化了从直方图中获得的印象。
4. Measures of Dispersion | 离散程度的度量
To measure spread, we compute the range, interquartile range (IQR), and standard deviation. The range is 1620 − 1448 = 172 hours. For the IQR, the first quartile Q₁ is the 8th value = 1506, and the third quartile Q₃ is the 23rd value = 1556, so IQR = 50 hours.
为了度量离散程度，我们计算极差、四分位距（IQR）和标准差。极差为1620 − 1448 = 172小时。对于IQR，第一四分位数Q₁是第8个值 = 1506，第三四分位数Q₃是第23个值 = 1556，因此IQR = 50小时。
The sample variance s² = Σ(x − x̄)² / (n − 1). Subtracting the mean from each value, squaring, summing, and dividing by 29 gives s² ≈ 2816.5, hence the sample standard deviation s ≈ 53.1 hours.
样本方差s² = Σ(x − x̄)² / (n − 1)。将每个值减去均值、平方、求和、再除以29，得到s² ≈ 2816.5，因此样本标准差s ≈ 53.1小时。
s = √[Σ(x − x̄)² / (n − 1)] ≈ 53.1 hours
A relatively small standard deviation compared with the mean indicates that the lifetimes are reasonably consistent.
与均值相比，标准差相对较小，表明寿命相当一致。
5. Cumulative Frequency and Percentiles | 累积频率与百分位数
Using the grouped frequency table, we build a cumulative frequency column and plot the curve. This allows us to estimate any percentile without the raw data.
利用分组频数表，我们建立累积频率列并绘制曲线。这使我们能够在没有原始数据的情况下估计任何百分位数。
| Upper boundary / 上界 | Cumulative frequency / 累积频数 |
|---|---|
| 1459 | 2 |
| 1479 | 4 |
| 1499 | 7 |
| 1519 | 13 |
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