📚 Edexcel Year 12 Statistics: Case Study Walkthrough – Light Bulb Quality Control | Edexcel 12年级统计：灯泡质量控制案例分析实战演练
This article presents a full, step-by-step statistical case study based on the Edexcel Year 12 specification. We take the role of a quality control analyst investigating claims made by a light bulb manufacturer. Using real-world-style data, we will move through data collection, graphical representation, calculations of summary statistics, probability modelling with the normal and binomial distributions, and formal hypothesis tests. Every step is designed to reinforce your understanding of key topics and to show how they connect in practice.
本文按照 Edexcel 12 年级统计考纲，完整演示一个统计案例分析。我们将扮演质量控制分析师，调查一家灯泡制造商的声明。利用仿真的真实数据，我们将依次完成数据收集、图形表示、汇总统计量计算、正态分布与二项分布的概率建模，以及正式的假设检验。每一步都旨在巩固你对关键主题的理解，并展示它们在实际中是如何关联的。
1. Case Background and Objectives | 案例背景与分析目标
A company, BrightLux, produces LED bulbs and claims that the mean lifetime of their bulbs is 800 hours with a known standard deviation of 40 hours. They also state that the proportion of bulbs failing before 750 hours (classified as 'defective') is no more than 8%. As a quality control analyst, you have been asked to investigate these claims using a random sample of 50 bulbs. Your objectives are to summarise the sample data, to check both claims using appropriate hypothesis tests at the 5% significance level, and to report your findings with clear statistical reasoning.
一家名为 BrightLux 的公司生产 LED 灯泡，并声称其灯泡的平均寿命为 800 小时，已知标准差为 40 小时。他们还声称在 750 小时之前失效的灯泡（即“次品”）比例不超过 8%。作为质量控制分析师，你被要求抽取 50 只灯泡的随机样本，对这些声明进行调查。你的目标是汇总样本数据，使用合适的假设检验（显著性水平 5%）检验两个声明，并以清晰的统计推理报告结果。
2. Collecting the Sample Data | 收集样本数据
A simple random sample of 50 bulbs was taken from a large production batch and each bulb was tested until failure. The lifetimes (in hours) were recorded. A random sample ensures that every bulb in the batch had an equal chance of being selected, reducing bias. The data were recorded to the nearest hour and are summarised below for convenience, though in practice we would analyse the full raw data.
从一个大生产批次中抽取了 50 只灯泡的简单随机样本，每只灯泡测试至失效。寿命（以小时为单位）被记录下来。随机抽样保证了批次中每只灯泡被选中的机会均等，减少了偏差。为方便起见，数据被汇总如下，但在实际中我们会分析完整的原始数据。
Sample size n = 50. The key observed values: Sample mean, x̄ = 785 hours; sample standard deviation, s = 38.5 hours. The number of bulbs with lifetime less than 750 hours was 5.
样本量 n = 50。关键观测值：样本均值 x̄ = 785 小时；样本标准差 s = 38.5 小时。寿命小于 750 小时的灯泡数量为 5 只。
3. Exploratory Data Analysis: Histogram and Box Plot | 探索性数据分析：直方图与箱线图
Before performing formal tests, we should visualise the data. A histogram of the 50 lifetimes shows a roughly symmetric, bell-shaped distribution centred near 790 hours, with a slight left skew caused by a few lower values. There are no extreme outliers, and the spread is consistent with the claimed standard deviation. This indicates that a normal model is plausible for the population.
在正式检验之前，我们应当先可视化数据。50 个寿命数据的直方图显示出大致对称的钟形分布，中心在 790 小时附近，由于少数低值而略显左偏。没有极端异常值，且散布与声称的标准差相符。这表明正态模型对总体是合理的。
IQR) of 50 hours. The whiskers extend to 718 hours and 852 hours. The median is slightly below 800, and the box is fairly centred. No data point lies beyond 1.5 × IQR from the quartiles, confirming the absence of outliers.
同时也绘制了箱线图。中位数为 791 小时，下四分位数为 761 小时，上四分位数为 811 小时，四分位距 (IQR) 为 50 小时。须线延伸至 718 小时和 852 小时。中位数略低于 800，箱体大致居中。没有数据点超出四分位数的 1.5 × IQR 范围，确认无异常值。
4. Measures of Location and Spread | 位置与离散度量
We calculate the main summary statistics for the sample:
我们计算样本的主要汇总统计量：
| Statistic | Value (hours) |
| Mean (x̄) | 785 |
| Median | 791 |
| Sample standard deviation (s) | 38.5 |
| Lower quartile (Q₁) | 761 |
| Upper quartile (Q₃) | 811 |
| IQR (Q₃ − Q₁) | 50 |
y by a few shorter lifetimes. The sample standard deviation (38.5) is close to the assumed population σ = 40, providing some support for using the known σ in our tests.
样本均值 785 小时比声称的 800 小时低 15 小时。中位数更接近 800，这表明均值可能被少数较短的寿命略微拉低。样本标准差 (38.5) 接近假设的总体 σ = 40，这为我们在检验中使用已知 σ 提供了一定的支持。
5. Normal Probability: Lifespan Below a Threshold | 正态分布概率：低于阈值的寿命
Using the manufacturer's claimed distribution, we can calculate the theoretical probability that a randomly chosen bulb fails before 750 hours. Let X ~ N(800, 40²) be the lifetime. We standardise:
利用制造商声称的分布，我们可以计算随机抽取的一只灯泡在 750 小时前失效的理论概率。令 X ~ N(800, 40²) 为寿命。我们进行标准化：
z = (750 − 800) / 40 = −1.25
750) = Φ(−1.25) = 1 − Φ(1.25) ≈ 1 − 0.8944 = 0.1056. So, if the claim is true, about 10.56% of bulbs would be classified as defective. This is higher than the stated maximum of 8%, so the defect rate claim already looks questionable.
P(X < 750) = Φ(−1.25) = 1 − Φ(1.25) ≈ 1 − 0.8944 = 0.1056。因此，若声明为真，约有 10.56% 的灯泡会被归类为次品。这高于所声称的最高 8%，因此次品率声明看起来已经值得怀疑。
6. Hypothesis Test for the Mean | 均值的假设检验
We first test the claim about the mean lifetime. Because the population standard deviation is given as 40, we use a one-sample z-test, even though we also have the sample standard deviation.
我们首先检验关于平均寿命的声明。因为总体标准差已知为 40，我们使用单样本 z 检验，尽管我们也有样本标准差。
800; H₁: μ < 800 (one-tailed test, as we are checking if the mean is lower than claimed).
第 1 步：陈述假设。H₀: μ = 800；H₁: μ < 800（单尾检验，因为我们要检验均值是否低于声称值）。
ep 2: Significance level α = 0.05. Critical value from normal tables: z_critical = −1.645.
第 2 步：显著性水平 α = 0.05。从正态分布表查得临界值：z_critical = −1.645。
Step 3: Test statistic. x̄ = 785, n = 50, σ = 40. Standard error = σ / √n = 40 / √50 ≈ 5.657. Then z = (785 − 800) / 5.657 ≈ −2.65.
第 3 步：检验统计量。x̄ = 785, n = 50, σ = 40。标准误 = σ / √n = 40 / √50 ≈ 5.657。于是 z = (785 − 800) / 5.657 ≈ −2.65。
Step 4: Decision. Since −2.65 < −1.645, the test statistic falls in the rejection region. We reject H₀. There is sufficient evidence at the 5% level to conclude that the true mean lifetime is less than 800 hours.
第 4 步：决策。因为 −2.65 < −1.645，检验统计量落在拒绝域中。我们拒绝 H₀。在 5% 水平上有足够证据表明真实平均寿命小于 800 小时。
is P(Z < −2.65) ≈ 0.0040, which is well below 0.05, confirming strong evidence against H₀.
此检验的 p 值为 P(Z < −2.65) ≈ 0.0040，远低于 0.05，证实了反对 H₀ 的强证据。
7. Binomial Distribution and Defect Rate Hypothesis Test | 二项分布与次品率假设检验
manufacturer claims the proportion of defective bulbs (lifetime < 750 hours) is at most 8% est H₀: p = 0.08 against H₁: p > 0.08. Let the number of defectives in a sample of 50 be D ~ B(50, 0.08) under H₀.
制造商声称次品灯泡（寿命 < 750 小时）的比例最多为 8%，即 p ≤ 0.08。我们检验 H₀: p = 0.08 对 H₁: p > 0.08。设在 H₀ 下 50 个样本中的次品数为 D ~ B(50, 0.08)。
Observed number of defectives d = 5. We compute the p-value = P(D ≥ 5 | p = 0.08). Using the binomial formula or cumulative tables:
观测到的次品数 d = 5。我们计算 p 值 = P(D ≥ 5 | p = 0.08)。利用二项分布公式或累积表：
P(D ≥ 5) = 1 − P(D ≤ 4) = 1 − [P(D=0) + P(D=1) + P(D=2) + P(D=3) + P(D=4)]
his p-value is much larger than 0.05. Therefore we do not reject H₀. There is insufficient evidence to say the defect rate exceeds 8%.
通过技术工具或表格，P(D ≤ 4) ≈ 0.629，因此 p 值 ≈ 1 − 0.629 = 0.371。该 p 值远大于 0.05。因此我们不拒绝 H₀。尚无足够证据表明次品率超过 8%。
We could also perform a normal approximation: under H₀, D ~ e mean = np = 4, variance = np(1−p) = 3.68. With continuity correction, z ≈ (5 − 0.5 − 4)/√3.68 ≈ 0.5/1.918 ≈ 0.26, p-value ≈ 0.397, which leads to the same conclusion.
我们也可以进行正态近似：在 H₀ 下，D 近似服从 N(4, 3.68)，因为均值 = np = 4，方差 = np(1−p) = 3.68。使用连续性校正，z ≈ (5 − 0.5 − 4)/√3.68 ≈ 0.5/1.918 ≈ 0.26，p 值 ≈ 0.397，得出相同结论。
hat the theoretical defect probability from the claimed mean and standard deviation (10.56%) is higher than the tested 8%, but because our sample proportion is 5/50 = 10%, the binomial test does not have enough power to detect a difference with only 50 bulbs.
注意，由声称的均值和标准差得出的理论次品概率 (10.56%) 高于检验的 8%，但由于样本比例仅为 5/50 = 10%，在仅有 50 只灯泡的情况下二项检验的功效不足以检测出差异。
8. Discussion: Interpretation, Limitations and Conclusion | 讨论：结果解释、局限性与结论
The hypothesis test on the mean provided strong evidence that the true average lifetime is below 800 hours. However, the test on the defect rate was inconclusive. This illustrates that different aspects of quality can give different statistical messages. The mean test used all 50 data values and was sensitive to the consistent downward shift in lifetimes. The binomial test reduced the data to a simple pass/fail count and lost information, making it harder to reject H₀.
均值的假设检验提供了有力证据表明真实平均寿命低于 800 小时。然而，次品率的检验却无定论。这说明质量的不同方面会给出不同的统计信息。均值检验使用了全部 50 个数据值，并对寿命整体的下移敏感。二项检验将数据简化为简单的通过/不通过计数，丢失了信息，从而导致更难拒绝 H₀。
ce is exactly known. In practice, we would use a t-test if σ were estimated from the sample, though with n=50 the difference would be minor. We should also check that the bulbs were tested under controlled, identical conditions and that the sample was truly random. The histogram suggested approximate normality, justifying the use of the z-test.
本研究的主要局限包括：二项检验的样本量相对较小 (n=50)，以及假设总体方差确切已知。在实践中，若 σ 由样本估计，我们会使用 t 检验，不过在 n=50 时差异不大。我们还应检查灯泡是否在受控、一致的条件下测试，以及样本是否真正随机。直方图表明近似正态，支持了 z 检验的使用。
In conclusion, while the mean lifetime appears significantly lower than advertised, the evidence regarding the defect rate is not strong enough to draw a firm conclusion. A larger sample would be advisable to investigate the defect rate further. This case study has demonstrated how to integrate descriptive statistics, probability distributions, and hypothesis testing into a coherent statistical investigation, which is exactly the skill expected in Edexcel Year 12 Statistics.
总之，尽管平均寿命看起来显著低于宣传值，但关于次品率的证据尚不足以得出确凿结论。建议使用更大样本进一步调查次品率。本案例演示了如何将描述性统计、概率分布和假设检验整合为连贯的统计调查，这正是 Edexcel 12 年级统计所要求的技能。
Published by TutorHao | Statistics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图