📚 AS OCR Statistics: Case Study Practical Exercises | AS OCR 统计：案例分析实战演练
Welcome to a hands-on case study that ties together the core topics of AS OCR Statistics. Using a real-world scenario—quality control of crisp packets—we will walk through data description, probability distributions, and hypothesis testing step by step. This practical approach will help you see how statistical concepts connect and prepare you for exam-style questions.
欢迎来到这个动手案例研究，它将 AS OCR 统计学的核心主题串联起来。通过一个真实场景——薯片包装的质量控制——我们将逐步演练数据描述、概率分布和假设检验。这种实践方法将帮助你理解统计概念之间的关联，并为应对考试风格的问题做好准备。
1. Case Study Background | 案例背景
A snack company claims that each packet of its popular crisps has a net weight of 200 g. From long-term production records, the filling process is known to be normally distributed with a standard deviation of 5 g. The quality control team randomly selects 16 packets to investigate whether the mean weight is below the stated value.
一家零食公司声称其受欢迎的薯片每包净重为200克。根据长期生产记录，灌装过程服从正态分布，标准差为5克。质量控制团队随机抽取了16包薯片，以调查平均重量是否低于标称值。
2. Data Collection & Sampling | 数据收集与抽样
The 16 packets are weighed on a calibrated scale, giving the following results in grams: 196, 202, 199, 198, 197, 201, 203, 198, 200, 195, 205, 197, 203, 199, 201, 200. This is a simple random sample, ensuring every packet on the line had an equal chance of selection and avoiding bias.
这16包薯片在校准过的秤上称重，得到以下结果（克）：196, 202, 199, 198, 197, 201, 203, 198, 200, 195, 205, 197, 203, 199, 201, 200。这是一个简单随机样本，确保了生产线上每一包都有均等的机会被选中，从而避免偏差。
3. Visualising Data: Stem-and-Leaf Diagram & Box Plot | 数据可视化：茎叶图和箱线图
A stem-and-leaf diagram helps us see the shape of the data. Using stems of 19 and 20 (representing 190–199 and 200–209) gives an ordered display:
茎叶图帮助我们观察数据的形状。以19和20为茎（代表190–199和200–209），可以得到有序展示：
| Stem (19) | Leaf | Stem (20) | Leaf |
|---|---|---|---|
| 19 | 5 6 7 7 8 8 9 9 | 20 | 0 0 1 1 2 3 3 5 |
A box plot can be constructed from the five-number summary: minimum = 195, Q₁ = 197.5, median = 199.5, Q₃ = 201.5, maximum = 205. The plot is roughly symmetric, suggesting no extreme skewness.
可以从五数概括构建箱线图：最小值=195，Q₁=197.5，中位数=199.5，Q₃=201.5，最大值=205。图形大致对称，表明没有极端偏斜。
4. Measures of Location and Spread | 位置的度量和分散度
The sample mean x̄ is computed as Σx/n = 3194/16 = 199.625 g. The sample median is the average of the 8th and 9th ordered values: (199+200)/2 = 199.5 g. For spread, although the population σ is known, we can still report the sample standard deviation s. Using the formula s = √[Σ(x – x̄)²/(n-1)], we get s ≈ 2.99 g.
样本均值 x̄ 计算为 Σx/n = 3194/16 = 199.625 g。样本中位数是第8和第9个有序值的平均数：(199+200)/2 = 199.5 g。就分散度而言，尽管总体 σ 已知，我们仍可报告样本标准差 s。利用公式 s = √[Σ(x – x̄)²/(n-1)]，得到 s ≈ 2.99 g。
5. Sampling Distribution of the Sample Mean | 样本均值的抽样分布
Because the population is normally distributed, the sample mean X̄ also follows a normal distribution with mean μ = 200 g and standard error σ/√n = 5/√16 = 1.25 g. This distribution is the basis for our probability calculations and hypothesis test.
由于总体服从正态分布，样本均值 X̄ 也服从正态分布，其均值为 μ = 200 g，标准误差为 σ/√n = 5/√16 = 1.25 g。这个抽样分布是我们计算概率和进行假设检验的基础。
6. Applying the Normal Distribution | 正态分布的应用
Using the sampling distribution, we can answer questions such as: “What is the probability that a sample of 16 packets has a mean weight less than 199 g?” Standardising: z = (199 – 200) / 1.25 = -0.8. From the standard normal table, P(Z < -0.8) = 0.2119. So there is about a 21.2% chance of observing such a low mean even if the true mean is 200 g.
利用抽样分布，我们可以回答诸如“16包薯片的样本均值低于199克的概率是多少？”的问题。标准化：z = (199 – 200) / 1.25 = -0.8。查标准正态表，P(Z < -0.8) = 0.2119。因此，即使真实均值为200克，观察到如此低的均值的概率约为21.2%。
7. Setting Up the Hypothesis Test | 假设检验的设立
Quality control wants to test if there is evidence that the mean weight is below 200 g. We set up the hypotheses as: H₀: μ = 200 (null – no underfilling) vs H₁: μ < 200 (alternative – underfilling suspected). A significance level of α = 0.05 is chosen for a one-tailed test.
质量控制部门希望检验是否有证据表明平均重量低于200克。我们设定假设：H₀: μ = 200（原假设——没有少装）vs H₁: μ < 200（备择假设——怀疑少装）。选择显著水平 α = 0.05 进行单尾检验。
8. Test Statistic and P-value Calculation | 检验统计量与P值计算
The test statistic is z = (x̄ – μ₀) / (σ/√n) = (199.625 – 200) / 1.25 = -0.3. Using the standard normal distribution, the p-value for a left-tailed test is P(Z < -0.3) = 0.3821. This p-value is much larger than 0.05.
检验统计量为 z = (x̄ – μ₀) / (σ/√n) = (199.625 – 200) / 1.25 = -0.3。对于左尾检验，利用标准正态分布计算的p值为 P(Z < -0.3) = 0.3821。该p值远大于0.05。
9. Conclusion and Interpretation | 结论与解释
Since the p-value (0.3821) > 0.05, we do not reject H₀. There is insufficient evidence to conclude that the mean weight is below 200 g. The small shortfall observed in the sample is likely due to random variation. The result does not warrant adjusting the filling machinery.
由于p值（0.3821）> 0.05，我们不拒绝 H₀。没有足够的证据推断平均重量低于200克。样本中观察到的小幅短缺很可能源于随机波动。这一结果并不需要调整灌装设备。
10. Confidence Interval (Extension) | 置信区间（拓展）
A 95% confidence interval for μ is given by x̄ ± z* × σ/√n, where z* = 1.96 for a two-tailed 95% level. This yields 199.625 ± 1.96 × 1.25, i.e. (197.175, 202.075). The interval contains 200, consistent with the hypothesis test.
μ 的95%置信区间为 x̄ ± z* × σ/√n，其中对于双侧95%水平 z* = 1.96。计算得199.625 ± 1.96 × 1.25，即 (197.175, 202.075)。该区间包含200，与假设检验结果一致。
11. Common Pitfalls in Statistical Analysis | 统计分析的常见陷阱
When conducting tests, students often misstate hypotheses (e.g. putting a sample statistic in the hypotheses), use a two-tailed test when only one direction matters, or misinterpret a non-significant result as “proving H₀ is true”. Another common error is forgetting to check normality assumptions or using the wrong standard error.
在进行检验时，学生经常错误表述假设（如将样本统计量放入假设中），在仅关注一个方向时误用双尾检验，或将不显著的结果误解为“证明H₀为真”。另一个常见错误是忘记检查正态性假设或使用错误的标准误。
12. Exam Tips and Summary | 考试技巧与总结
Always show full working: state hypotheses, write the test statistic formula with substituted values, sketch the distribution with the critical region, and give a conclusion in context. Remember that when σ is known and the population is normal, the z-test is appropriate. Practise linking descriptive statistics with inferential procedures to build confidence.
始终展示完整过程：陈述假设，写出带代入值的检验统计量公式，绘制分布草图并标出拒绝域，并在上下文中给出结论。记住，当 σ 已知且总体正态时，z检验是合适的。通过将描述统计与推断步骤联系起来进行练习，以建立信心。
Published by TutorHao | Statistics Revision Series | aleveler.com
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