Case Study in Action: Analysing Light Bulb Lifetimes | 案例分析实战演练：灯泡寿命分析

📚 Case Study in Action: Analysing Light Bulb Lifetimes | 案例分析实战演练：灯泡寿命分析

A light bulb manufacturer claims that its production process yields bulbs with a mean lifetime of 1500 hours and a standard deviation of 120 hours, following a normal distribution. The quality control team suspects that a recent adjustment to the manufacturing process may have increased the mean lifetime. To investigate, they collect a random sample of 50 bulbs from the new process and record their lifetimes. Additionally, historical data from the old process (n=40) is available. This case study walks through the complete statistical analysis, from descriptive exploration to hypothesis testing, confidence intervals, power analysis, chi-squared goodness-of-fit test, and practical recommendations – all within the Year 13 CIE Statistics syllabus.

一家灯泡制造商声称其生产工艺的灯泡平均寿命为1500小时，标准差为120小时，且服从正态分布。质量控制团队怀疑近期工艺调整可能提高了平均寿命。为验证这一点，他们从新工艺中随机抽取了50只灯泡并记录了寿命。同时，旧工艺的历史数据（样本量40）也可获得。本案例分析将完整演练从描述性探索到假设检验、置信区间、功效分析、卡方拟合优度检验以及管理建议的全流程，全部紧扣 Year 13 CIE 统计课程大纲。

1. Case Background and Objectives | 案例背景与分析目标

The quality control department wants to determine whether the new manufacturing process has significantly increased the average lifetime of bulbs beyond the claimed 1500 hours. They also need to assess whether the variability has changed and whether the lifetimes still follow a normal distribution. Furthermore, management requires a comparison between the old and new processes based on independent samples to decide whether to fully adopt the new method.

质量部门希望确定新工艺是否显著将灯泡的平均寿命提升至1500小时以上。他们还需要评估波动性是否发生变化，以及寿命是否仍然服从正态分布。此外，管理层要求基于独立样本对旧工艺和新工艺进行比较，以决定是否全面采用新方法。

2. Data Presentation: New Process Sample | 数据展示：新工艺样本

The lifetimes (in hours) of the 50 bulbs from the new process are summarised below. The first ten observations are: 1582, 1490, 1623, 1534, 1475, 1558, 1519, 1641, 1487, 1565. The full sample yields a mean of 1540 hours and a standard deviation of 115 hours.

新工艺的50只灯泡的寿命（小时）汇总如下。前10个观测值为：1582, 1490, 1623, 1534, 1475, 1558, 1519, 1641, 1487, 1565。全样本的平均值为1540小时，标准差为115小时。

The old process sample (n=40) has a mean of 1502 hours and a standard deviation of 118 hours. Population standard deviation for both processes is assumed to be 120 hours based on long-term records.

旧工艺的样本（n=40）平均值为1502小时，标准差为118小时。根据长期记录，两种工艺的总体标准差均假设为120小时。

3. Descriptive Statistics and Initial Observations | 描述性统计与初步观察

The sample mean of 1540 hours is 40 hours higher than the claimed mean of 1500 hours. The sample standard deviation of 115 hours is slightly below the assumed 120 hours, suggesting that the dispersion may not have increased. A box plot (not shown) indicates a roughly symmetric distribution with no extreme outliers, supporting the possibility of normality.

样本均值1540小时比声称的均值1500小时高出40小时。样本标准差115小时略低于假设的120小时，表明离散程度可能并未增加。箱线图（未展示）显示分布大致对称且无明显离群值，支持了正态性的可能性。

A frequency distribution for the new process lifetimes, using the intervals below, will later be used for the chi-squared test.

新工艺寿命的频数分布（按以下区间划分）将用于后续的卡方检验。

Interval (hours)	Observed Frequency
< 1400	2
1400 – 1500	11
1500 – 1600	19
1600 – 1700	13
> 1700	5

These observed counts will be compared against expected frequencies from a normal distribution with mean 1500 and standard deviation 120.

这些观测频数将与均值为1500、标准差为120的正态分布下的期望频数进行比较。

4. One-Sample Hypothesis Test for the Mean (z-test) | 单样本均值的假设检验（z 检验）

We test whether the new process mean lifetime has increased. Let μ be the true mean lifetime under the new process. The hypotheses are:

我们检验新工艺的平均寿命是否有所提高。设 μ 为新工艺的真实平均寿命。假设如下：

H₀: μ = 1500
H₁: μ > 1500

Significance level α = 0.05. The test statistic is calculated using the known population standard deviation σ = 120 hours.

显著性水平 α = 0.05。检验统计量使用已知总体标准差 σ = 120 小时进行计算。

z = (x̄ – μ₀) / (σ / √n) = (1540 – 1500) / (120 / √50) ≈ 40 / 16.97 ≈ 2.357

The critical value for a one-tailed z-test at α = 0.05 is z₀.₀₅ = 1.645. Since 2.357 > 1.645, we reject H₀. There is sufficient evidence at the 5% level to conclude that the new process has increased the mean lifetime.

在 α = 0.05 下的单尾 z 检验临界值为 z₀.₀₅ = 1.645。由于 2.357 > 1.645，我们拒绝原假设。在 5% 显著性水平下，有充分证据表明新工艺提高了平均寿命。

5. p-value Interpretation | p 值的解读

The p-value is the probability of observing a sample mean as extreme as 1540 hours, given that H₀ is true. Using standard normal tables, P(Z > 2.357) = 1 – Φ(2.357) ≈ 0.0092. This p-value of 0.0092 is less than α = 0.05, confirming the rejection of H₀. The result is statistically significant even at the 1% level.

p 值是在 H₀ 为真的条件下，观察到样本均值高达1540小时（或更极端）的概率。查标准正态表，P(Z > 2.357) = 1 – Φ(2.357) ≈ 0.0092。该 p 值 0.0092 小于 α = 0.05，确认拒绝原假设。即使在 1% 显著性水平下，结果依然显著。

A very small p-value indicates strong evidence against the null hypothesis, implying that the observed increase is unlikely to be due to random sampling variation.

极小的 p 值表明有很强的证据反对原假设，意味着观测到的寿命提高不太可能由随机抽样变异导致。

6. Confidence Interval for the Population Mean | 总体均值的置信区间

We construct a 95% confidence interval for the true mean lifetime under the new process. With σ known, the formula is:

我们构建新工艺下真实平均寿命的 95% 置信区间。由于 σ 已知，公式为：

x̄ ± zₐ/₂ × (σ / √n) = 1540 ± 1.96 × (120 / √50) ≈ 1540 ± 33.27

Thus the 95% confidence interval is (1506.73, 1573.27) hours. Since the interval lies entirely above 1500, it corroborates the hypothesis test conclusion that the mean has increased. We are 95% confident that the true mean lifetime of bulbs from the new process is between 1506.7 and 1573.3 hours.

因此 95% 置信区间为 (1506.73, 1573.27) 小时。由于整个区间都位于1500以上，这进一步支持了假设检验的结论，即均值有所提高。我们有 95% 的把握认为新工艺灯泡的真实平均寿命在 1506.7 至 1573.3 小时之间。

7. Comparing Two Production Processes: Two-Sample z-Test | 两个工艺过程的比较：双样本 z 检验

Management also wants to directly compare the old process (population mean μ₁) and the new process (population mean μ₂). Independent random samples give:

管理层还希望直接比较旧工艺（总体均值 μ₁）和新工艺（总体均值 μ₂）。独立随机样本信息如下：

Old process: n₁ = 40, x̄₁ = 1502, σ₁ = 120 (assumed known)
New process: n₂ = 50, x̄₂ = 1540, σ₂ = 120 (assumed known)

We test H₀: μ₁ = μ₂ against H₁: μ₂ > μ₁. The combined standard error is:

检验假设 H₀: μ₁ = μ₂ 对比 H₁: μ₂ > μ₁。合并标准误为：

SE = √(σ₁²/n₁ + σ₂²/n₂) = √(120²/40 + 120²/50) = √(360 + 288) = √648 ≈ 25.46

The test statistic is z = (x̄₂ – x̄₁) / SE = (1540 – 1502) / 25.46 ≈ 1.493. The critical value at α = 0.05 (one-tailed) is 1.645. Since 1.493 < 1.645, we fail to reject H₀ at the 5% level. The p-value is approximately 0.0677. Although the new process gave a higher sample mean, the evidence is not quite strong enough to conclude superiority when compared directly with the old process at α = 0.05.

检验统计量 z = (x̄₂ – x̄₁) / SE = (1540 – 1502) / 25.46 ≈ 1.493。α = 0.05 下的单尾临界值为 1.645。因为 1.493 < 1.645，在 5% 显著性水平下无法拒绝原假设。p 值约为 0.0677。尽管新工艺的样本均值更高，但与旧工艺直接比较时，在 α = 0.05 下证据尚不足以断言新工艺更优。

8. Type I and Type II Errors and Power Analysis | 第Ⅰ类与第Ⅱ类错误及功效分析

In the one-sample test, rejecting H₀ when μ is truly 1500 would be a Type I error. With α = 0.05, this risk is controlled. A Type II error occurs if we fail to reject H₀ when μ is actually greater than 1500, say 1530 hours. We compute the power of the test to detect such an increase.

在单样本检验中，若 μ 真为1500却拒绝原假设，会犯第Ⅰ类错误，通过设定 α = 0.05 控制了该风险。若 μ 实际上大于1500（例如1530小时）却未能拒绝原假设，则发生第Ⅱ类错误。我们计算检验对此增量的功效。

For μ = 1530, σₓ̄ = 120/√50 ≈ 16.97. The rejection region is x̄ > 1500 + 1.645 × 16.97 ≈ 1527.94.
Power = P(X̄ > 1527.94 | μ = 1530) = P(Z > (1527.94 – 1530)/16.97) = P(Z > -0.121) ≈ 0.548.

This means there is only a 54.8% chance of detecting a true increase of 30 hours with n = 50. To increase power, we could increase the sample size or accept a higher α. The power analysis helps in planning future studies and highlights the limitations of the current sample.

这意味着使用样本量50时，仅约有 54.8% 的概率能检测出真实的30小时提升。为提高功效，可以增大样本量或接受更高的 α。功效分析有助于规划未来研究，也凸显了当前样本的局限性。

9. Chi-Squared Goodness-of-Fit Test for Normality | 正态性的卡方拟合优度检验

We verify whether the new process lifetime data follow a normal distribution with mean 1500 and standard deviation 120 (the original specification). Using the observed frequencies from Section 3, we calculate expected probabilities and expected frequencies.

我们验证新工艺寿命数据是否服从均值为1500、标准差为120的正态分布（即原始规格）。利用第3节中的观测频数，我们计算期望概率和期望频数。

For each interval, we find the cumulative probabilities. For N(1500, 120²):

针对每个区间求累积概率。对于 N(1500, 120²)：

P(X < 1400): standardising (1400-1500)/120 = -0.8333, probability ≈ 0.2023. Expected frequency = 50 × 0.2023 = 10.12.
1400-1500: P(X<1500) = 0.5, so P(1400
1500-1600: P(X<1600): (1600-1500)/120 = 0.8333, probability ≈ 0.7977, so interval prob = 0.7977 - 0.5 = 0.2977, expected = 14.89.
1600-1700: P(X<1700): (1700-1500)/120 = 1.6667, prob ≈ 0.9522, interval prob = 0.9522 - 0.7977 = 0.1545, expected = 7.72.
>1700: 1 – 0.9522 = 0.0478, expected = 2.39.

Combining the first and last two intervals to ensure expected frequencies >5 gives three categories: <1500, 1500-1600, ≥1600. Observed: 13, 19, 18. Expected: 25.01, 14.89, 10.10 (recalculated to maintain total 50). Chi-squared statistic:

将首尾区间合并以确保期望频数大于5，得到三个类别：<1500, 1500-1600, ≥1600。观测频数：13, 19, 18。期望频数：25.01, 14.89, 10.10（重新计算以使合计为50）。卡方统计量：

χ² = Σ (O – E)²/E = (13-25.01)²/25.01 + (19-14.89)²/14.89 + (18-10.10)²/10.10 ≈ 5.77 + 1.13 + 6.16 = 13.06

Degrees of freedom = 3 – 1 = 2 (no parameters estimated, as we tested a fully specified distribution). From tables, the critical value at α = 0.05 is 5.991. Since 13.06 > 5.991, we reject the hypothesis that the data come from N(1500, 120²). This suggests the new process may have altered the distribution’s shape, possibly shifting the mean and increasing the right tail – consistent with an improvement.

自由度为 3 – 1 = 2（未估计参数，因为检验的是完全指定的分布）。查表得 α = 0.05 下的临界值为 5.991。由于 13.06 > 5.991，我们拒绝数据来自 N(1500, 120²) 的原假设。这表明新工艺可能改变了分布形态，可能使均值右移并增厚右尾——这与质量改善一致。

10. Summary and Managerial Recommendations | 总结与管理建议

The one-sample z-test strongly suggests the new process mean exceeds 1500 hours (p ≈ 0.009). However, the two-sample comparison with the old process yields a p-value of 0.068, falling short of significance at the 5% level, possibly due to the smaller sample sizes for the old process and the overlapping standard errors. The confidence interval indicates the new mean is likely between 1507 and 1573 hours, which yields a practical benefit. The chi-squared test indicates that the distribution is no longer exactly N(1500, 120²), which is expected if the mean has shifted. A larger sample for the old process or a paired design could resolve the discrepancy in significance.

单样本 z 检验有力地证明新工艺平均寿命超过1500小时（p ≈ 0.009）。然而，与旧工艺的双样本比较所得 p 值为 0.068，在 5% 水平下不显著，这可能是由于旧工艺样本量较小及标准误叠加所致。置信区间表明新工艺均值很可能在1507至1573小时之间，具有实际效益。卡方检验显示寿命分布不再精确为 N(1500, 120²)，这在均值发生移动时是意料之中的。增大旧工艺样本量或采用配对设计有望解决显著性不一致的问题。

We recommend that management adopts the new process but also commits to a larger-scale validation study with at least 100 bulbs per process to achieve higher statistical power and confirm the economic value of the lifetime increase.

我们建议管理层启用新工艺，同时承诺开展大规模验证研究，每工艺至少抽取100只灯泡，以获得更高的统计功效，确认寿命增加的经济价值。

11. Extension: Testing Proportions of Defective Bulbs | 拓展：灯泡缺陷率的检验

Suppose the company also tracks whether a bulb fails before 1300 hours. Historically, the proportion of such early failures was 0.05. In the new sample of 50 bulbs, 1 bulb failed before 1300 hours. We can test whether the defect proportion has changed using a one-sample proportion test.

假设公司还监测灯泡是否在1300小时前失效。历史早期失效率为 0.05。新抽样的50只灯泡中，有1只在1300小时前失效。我们可以通过单样本比例检验来判断缺陷率是否发生了变化。

H₀: p = 0.05 vs H₁: p < 0.05. Sample proportion p̂ = 1/50 = 0.02.
Test statistic: z = (p̂ – p₀) / √(p₀(1-p₀)/n) = (0.02 – 0.05) / √(0.05 × 0.95 / 50) ≈ -0.03 / 0.0308 ≈ -0.974.

p-value for one-tailed test is about 0.165. At the 5% level, we cannot conclude a significant reduction in defect rate, though the direction is favourable.

单尾检验的 p 值约为 0.165。在 5% 水平下，尚不能得出缺陷率显著降低的结论，尽管方向是有利的。

12. Key Takeaways and CIE Exam Relevance | 核心要点与 CIE 考试关联

This case study integrated many core techniques from the Year 13 CIE Statistics syllabus: formulating hypotheses, performing z-tests, interpreting p-values, constructing confidence intervals, comparing two means, calculating power, and applying the chi-squared goodness-of-fit test. Always remember to state assumptions (known σ, normality, random sampling), report conclusions in context, and evaluate practical significance alongside statistical significance.

本案例分析融合了 Year 13 CIE 统计课程中的多项核心技能：设立假设、执行 z 检验、解读 p 值、构建置信区间、比较两个均值、计算功效以及应用卡方拟合优度检验。务请牢记说明假设（已知 σ、正态性、随机抽样），结合情境报告结论，并兼顾实用意义与统计显著性。

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Case Study in Action: Analysing Light Bulb Lifetimes | 案例分析实战演练：灯泡寿命分析