📚 Case Study in Statistical Inference: Real-World Data Analysis | 统计推断案例分析：真实世界数据分析实战
Statistical inference transforms raw data into actionable insights, enabling organisations to make evidence-based decisions under uncertainty. In this AQA Year 13 masterclass, we walk through a real-world scenario involving LED bulb manufacturing, applying a full suite of inferential techniques – from confidence intervals and hypothesis tests to regression, chi‑squared analysis and control charts.
统计推断将原始数据转化为可操作的见解，使组织能够在不确定条件下做出基于证据的决策。在这堂 AQA 13 年级大师课中，我们将走进一个 LED 灯泡制造的真实场景，全面运用置信区间、假设检验、回归分析、卡方检验以及控制图等整套推断方法。
1. Background and Dataset | 案例背景与数据集
A leading LED manufacturer claims that its new bulb has an average lifetime of at least 1000 hours. To verify this claim, the quality‑engineering team selects a simple random sample of 30 bulbs and records their lifetimes (in hours) under controlled testing conditions.
一家领先的 LED 制造商声称其新灯泡的平均寿命不低于 1000 小时。为了验证这一声明，质量工程团队在受控测试条件下抽取了一个 30 只灯泡的简单随机样本，并记录了它们的寿命（小时）。
The measurement process is carefully standardised, and the data are treated as continuous. Summary statistics are computed as the foundation for all later inference. The sample is large enough to invoke the Central Limit Theorem, ensuring the sampling distribution of the mean is approximately normal.
测量过程经过仔细标准化，数据被视为连续型。计算汇总统计量作为后续所有推断的基础。样本量足够大，可以调用中心极限定理，从而保证均值的抽样分布近似正态。
2. Descriptive Statistics and Data Exploration | 描述性统计与数据探索
From the 30 observations we obtain: sample mean x̄ = 985.2 h, sample standard deviation s = 48.7 h, median = 988 h, lower quartile = 950 h, upper quartile = 1018 h. The range is 138 h and the interquartile range is 68 h. A boxplot would show a roughly symmetric distribution with no extreme outliers.
根据 30 个观测值，我们得到：样本均值 x̄ = 985.2 小时，样本标准差 s = 48.7 小时，中位数 = 988 小时，下四分位数 = 950 小时，上四分位数 = 1018 小时。极差为 138 小时，四分位距为 68 小时。箱线图将显示大致对称的分布，没有极端离群值。
The standard deviation suggests that the lifetimes vary by about 49 hours around the mean. Because the sample mean is below 1000, we already suspect that the manufacturer’s claim may not be fulfilled, but we need formal statistical evidence.
标准差表明，灯泡寿命在均值上下波动约 49 小时。由于样本均值低于 1000 小时，我们已怀疑制造商的声明可能不成立，但还需要正式的统计证据。
| Statistic | Value (hours) |
|---|---|
| Sample Mean (x̄) | 985.2 |
| Sample Std Dev (s) | 48.7 |
| Sample Size (n) | 30 |
| Minimum | 906 |
| Maximum | 1044 |
The summary table provides a quick snapshot of the data’s central tendency and spread, guiding the choice of subsequent inferential methods.
汇总表快速呈现出数据的集中趋势和离散程度，指导着后续推断方法的选择。
3. Confidence Interval for the Population Mean | 总体均值的置信区间
Because the population standard deviation is unknown, we construct a 95% confidence interval for the true mean lifetime μ using the t‑distribution. The standard error is SE = s/√n = 48.7/√30 ≈ 8.891 h. With df = 29, the two‑tailed critical value is t* = 2.045. The margin of error is ME = 2.045 × 8.891 ≈ 18.18 h.
由于总体标准差未知，我们使用 t 分布构建真实平均寿命 μ 的 95% 置信区间。标准误为 SE = s/√n = 48.7/√30 ≈ 8.891 小时。自由度为 29 时，双尾临界值为 t* = 2.045。误差幅度 ME = 2.045 × 8.891 ≈ 18.18 小时。
95% CI for μ: x̄ ± t* × SE = 985.2 ± 18.18 → (967.02, 1003.38)
The interval (967.0, 1003.4) suggests that the plausible values for the true mean include 1000 hours, so the claim cannot be discounted solely by interval estimation. However, the interval leans slightly to the left of 1000, motivating a formal hypothesis test.
该区间 (967.0, 1003.4) 表明真实均值的合理取值包含 1000 小时，因此仅凭区间估计无法否定声称。然而区间略微偏左，这促使我们进行正式的假设检验。
4. One‑Sample t‑Test: Testing the Manufacturer’s Claim | 单样本 t 检验：检验制造商声明
We set up a one‑tailed test to see if the true mean is below 1000 hours: H₀: μ = 1000 versus H₁: μ < 1000. The significance level is α = 0.05.
我们设定单尾检验，以判断真实均值是否低于 1000 小时：H₀: μ = 1000，H₁: μ < 1000。显著性水平 α = 0.05。
t = (x̄ – μ₀) / (s/√n) = (985.2 – 1000) / 8.891 ≈ –1.665
The test statistic is t ≈ –1.665 with 29 degrees of freedom. The critical value from t‑tables is –1.699 (one‑tailed). Since –1.665 > –1.699, we fail to reject H₀ at the 5% level. The p‑value is approximately 0.053, just above the threshold.
检验统计量 t ≈ –1.665，自由度 29。查 t 分布表得单尾临界值为 –1.699。由于 –1.665 > –1.699，在 5% 水平上未能拒绝 H₀。p 值约为 0.053，刚好高于阈值。
In context, there is insufficient evidence to conclude that the average lifetime is less than 1000 hours. The borderline result suggests that a larger sample size might provide a clearer picture, but the manufacturer’s claim stands for now.
在实际情形中，没有足够证据表明平均寿命低于 1000 小时。这一临界结果说明，增大样本量可能让结论更清晰，但目前制造商声明得以维持。
5. Two‑Sample t‑Test: Comparing Production Lines A and B | 双样本 t 检验：比较 A 和 B 生产线
The factory operates two production lines. A rival hypothesis suggests line B may produce bulbs with a shorter average life. We draw independent random samples: from line A, n₁ = 25, x̄₁ = 1002 h, s₁ = 42 h; from line B, n₂ = 30, x̄₂ = 972 h, s₂ = 55 h. We assume equal population variances after an F‑test (p > 0.10).
工厂运行两条生产线。有人提出，生产线 B 生产的灯泡平均寿命可能更短。我们抽取独立随机样本：A 线 n₁ = 25，x̄₁ = 1002 h，s₁ = 42 h；B 线 n₂ = 30，x̄₂ = 972 h，s₂ = 55 h。F 检验后（p > 0.10），假定总体方差相等。
sₚ² = [(24×42² + 29×55²)] / (25+30–2) ≈ 2454.0 → sₚ ≈ 49.54
t = (1002 – 972) / [49.54 × √(1/25 + 1/30)] ≈ 2.236, df = 53
We test H₀: μ₁ – μ₂ = 0 against H₁: μ₁ – μ₂ ≠ 0 at α = 0.05. The two‑tailed critical value is around 2.006. Since 2.236 > 2.006, we reject H₀. There is significant evidence of a difference in true mean lifetimes between the two lines. The point estimate suggests line A produces bulbs lasting about 30 hours longer on average.
我们检验 H₀: μ₁ – μ₂ = 0 对 H₁: μ₁ – μ₂ ≠ 0，α = 0.05。双尾临界值约为 2.006。因为 2.236 > 2.006，拒绝 H₀。有显著证据表明两条生产线的真实平均寿命存在差异。点估计显示 A 线灯泡平均寿命长约 30 小时。
6. Linear Regression: Modelling Lifetime against Voltage | 线性回归：对寿命与电压建模
Engineers suspect that the applied voltage influences bulb lifetime. Eight prototype bulbs are tested at different voltages (V): 110, 120, 130, 140, 150, 160, 170, 180, yielding lifetimes (h): 1020, 990, 965, 940, 910, 885, 860, 830. A scatterplot suggests a strong negative linear relationship.
工程师怀疑施加的电压会影响灯泡寿命。在 110, 120, 130, 140, 150, 160, 170, 180 V 下测试了 8 只样品的寿命：1020, 990, 965, 940, 910, 885, 860, 830 小时。散点图显示强烈的负线性关系。
Using the least‑squares method we find the fitted model: predicted lifetime = β₀̂ + β₁̂ × Voltage. Calculations give β₁̂ ≈ –2.150 and β₀̂ ≈ 1230. So for every extra volt, the lifetime drops by about 2.15 hours. The R² value is 0.95, indicating the model explains 95% of the variability in lifetime.
用最小二乘法求得拟合模型：预测寿命 = β₀̂ + β₁̂ × 电压。计算得到 β₁̂ ≈ –2.150，β₀̂ ≈ 1230。因此电压每增加 1 伏特，寿命约减少 2.15 小时。R² = 0.95，表明模型解释了寿命 95% 的变异。
The residuals show no obvious pattern, confirming that a linear model is appropriate. This simple regression gives engineers a powerful predictive tool for setting voltage limits.
残差图无明显模式，确认线性模型适用。这一简单回归为工程师设定电压限制提供了有力的预测工具。
7. Inferences for the Regression Slope | 回归斜率的推断
To test whether the slope is genuinely non‑zero, we conduct a two‑tailed t‑test on β₁: H₀: β₁ = 0, H₁: β₁ ≠ 0. The standard error of β₁̂ is SE(β₁̂) ≈ 0.180. The test statistic is t = –2.150 / 0.180 ≈ –11.94 with df = 6. The p‑value is smaller than 0.001, so we strongly reject H₀.
Published by TutorHao | Year 13 统计 Revision Series | aleveler.com
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