Case Study Practice for Year 12 CIE Statistics | Year 12 CIE 统计：案例分析实战演练

📚 Case Study Practice for Year 12 CIE Statistics | Year 12 CIE 统计：案例分析实战演练

This article presents a series of applied case studies designed to strengthen your problem-solving skills for the CIE AS-Level Statistics syllabus. Each scenario is crafted to reflect the style of examination questions, integrating multiple topics such as probability, distributions, estimation, and hypothesis testing. By working through these examples and the detailed bilingual explanations, you will learn to deconstruct real-world problems, select appropriate statistical techniques, and interpret your findings accurately. Use these practice cases to consolidate your understanding and build confidence for the final assessment.

本文提供一系列应用型案例分析，旨在帮助你提升 CIE AS 统计课程的问题解决能力。每个场景都模仿考试题目的风格，融合了概率、分布、估计和假设检验等多个主题。通过这些实例和详细的中英双语解析，你将学会如何拆解真实问题、选择合适的统计方法并准确解释分析结果。请利用这些实战练习巩固理解，为最终考试建立信心。

1. Defective Components and Tree Diagrams | 缺陷元件与树形图

A factory produces components from three machines: Machine X supplies 50% of output, Machine Y 30%, and Machine Z 20%. The defect rates are 2% for X, 3% for Y, and 5% for Z. A component is selected at random from a day’s production. We want to determine the probability that it is defective and, if defective, the probability it came from each machine. This case involves conditional probability and Bayes’ theorem, fundamental tools for Year 12 CIE Statistics.

某工厂使用三台机器生产元件：机器 X 占总产量的 50%，机器 Y 占 30%，机器 Z 占 20%。三台机器的缺陷率分别为 2%、3% 和 5%。现从一天的产量中随机抽取一个元件，我们需计算该元件有缺陷的概率，以及已知它是有缺陷的条件下，它来自每台机器的概率。本案例涉及条件概率和贝叶斯定理，是 Year 12 CIE 统计的基础工具。

Let D denote the event that a component is defective. Then P(D|X) = 0.02, P(D|Y) = 0.03, P(D|Z) = 0.05. The total probability of a defective component is:

设 D 表示元件有缺陷的事件。则 P(D|X) = 0.02，P(D|Y) = 0.03，P(D|Z) = 0.05。有缺陷元件的总概率为：

P(D) = P(X)·P(D|X) + P(Y)·P(D|Y) + P(Z)·P(D|Z) = 0.5×0.02 + 0.3×0.03 + 0.2×0.05 = 0.010 + 0.009 + 0.010 = 0.029

Applying Bayes’ theorem, the probability that a defective component came from Machine X is:

应用贝叶斯定理，给定元件有缺陷，它来自机器 X 的概率为：

P(X|D) = [P(X)·P(D|X)] / P(D) = 0.01 / 0.029 ≈ 0.3448 (34.5%)

Similarly, P(Y|D) = 0.009 / 0.029 ≈ 0.3103, and P(Z|D) = 0.01 / 0.029 ≈ 0.3448. Notice how Machine Z’s contribution to defects is proportionally higher than its production share. This case illustrates how prior probabilities are updated with evidence.

类似可得 P(Y|D) ≈ 0.3103，P(Z|D) ≈ 0.3448。注意机器 Z 对缺陷的贡献比例高于其产量比例。本案例展示了先验概率如何随着证据更新。

For exam success, always draw a labelled tree diagram showing branches with probabilities. This visual aid reduces errors in multi-stage probability problems.

为了考试顺利，请一定画出带标注的树形图，在分支上标明概率。这种可视化辅助能减少多阶段概率问题中的错误。

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2. Lottery Winnings and Expected Value | 彩票奖金与期望值

A charity lottery offers 1,000 tickets at £5 each. The prizes are: one first prize of £500, two second prizes of £100, and five third prizes of £20. A participant buys one ticket. Determine the probability distribution of the net gain, its expected value, and the standard deviation. This case tests your ability to work with discrete random variables, a core topic in CIE Statistics.

某个慈善彩票发行 1000 张票，每张售价 5 英镑。奖项设置为：一等奖 1 个，奖金 500 英镑；二等奖 2 个，奖金 100 英镑；三等奖 5 个，奖金 20 英镑。某人购买一张彩票。求净收益的概率分布、期望值和标准差。本案例考查离散随机变量的处理能力，是 CIE 统计的核心主题。

Define the net gain X (winnings minus ticket cost). If a person wins first prize, net gain = 500 – 5 = £495; second prize: 100 – 5 = £95; third prize: 20 – 5 = £15; no win: -£5. The probabilities: P(X=495) = 1/1000, P(X=95) = 2/1000, P(X=15) = 5/1000, P(X=-5) = 992/1000.

定义净收益 X（奖金减去票价）。若中一等奖，净收益为 495 英镑；二等奖：95 英镑；三等奖：15 英镑；未中奖：-5 英镑。对应概率：P(X=495)=1/1000，P(X=95)=2/1000，P(X=15)=5/1000，P(X=-5)=992/1000。

The expected value E(X) is:

期望值 E(X) 为：

E(X) = 495×(1/1000) + 95×(2/1000) + 15×(5/1000) + (-5)×(992/1000) = (495 + 190 + 75 – 4960) / 1000 = -4200 / 1000 = -£4.20

On average, a ticket buyer loses £4.20. This negative expectation is typical for fund-raising lotteries. The variance Var(X) is calculated as E(X²) – [E(X)]², where E(X²) = 495²×(1/1000) + 95²×(2/1000) + 15²×(5/1000) + (-5)²×(992/1000).

平均而言，购票者损失 4.20 英镑。这种负期望在筹款彩票中很典型。方差 Var(X) 通过 E(X²) – [E(X)]² 计算，其中 E(X²) 将各值平方后按概率加权。

Work through the arithmetic: E(X²) = (245025 + 18050 + 1125 + 24800) / 1000 = 289000 / 1000 = 289. Then Var(X) = 289 – (-4.2)² = 289 – 17.64 = 271.36, so the standard deviation σ = √271.36 ≈ £16.47. In your answer, always state the units and interpret the results, e.g., the high standard deviation indicates considerable variability in net gain.

计算过程：E(X²) = (245025 + 18050 + 1125 + 24800)/1000 = 289; Var(X) = 289 – 17.64 = 271.36；标准差 σ ≈ 16.47 英镑。答题时务必注明单位并解释结果，例如高标准差表明净收益波动很大。

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3. Binomial Distribution in Exam Success | 二项分布在考试通过率中的应用

A multiple-choice test consists of 12 questions, each with 4 options and only one correct answer. A student guesses every answer randomly. Let Y be the number of correct answers. Then Y ~ B(12, 0.25). Find the probability the student gets exactly 5 correct; at least 3 correct; and find the expected number and variance of correct answers. This case reinforces binomial probability calculations and the use of cumulative distribution tables.

某场选择题考试有 12 题，每题 4 个选项且仅有一个正确。某学生对全部题目随机猜测。设 Y 为答对题数，则 Y ~ B(12, 0.25)。求该学生恰好答对 5 题的概率；至少答对 3 题的概率；并求答对题数的期望和方差。本案例强化二项概率计算和累积分布表的使用。

The probability mass function is:

概率质量函数为：

P(Y = y) = C(12, y) × 0.25ʸ × 0.75¹²⁻ʸ

For y=5: C(12,5) = 792, so P(Y=5) = 792 × (0.25)⁵ × (0.75)⁷. Compute step by step: 0.25⁵ = 1/1024 ≈ 0.0009766, 0.75⁷ ≈ 0.1334839, product ≈ 0.0009766 × 0.1334839 ≈ 0.0001303, multiplied by 792 gives approximately 0.1032. So there is about a 10.3% chance of exactly 5 correct by pure guessing.

对于 y=5：C(12,5)=792，P(Y=5) = 792 × 0.25⁵ × 0.75⁷。逐步计算：0.25⁵ = 1/1024 ≈ 0.0009766，0.75⁷ ≈ 0.1334839，乘积约 0.0001303，乘以 792 得约 0.1032。纯猜测恰好答对 5 题的概率约 10.3%。

For P(Y ≥ 3), it is easier to use the complement: P(Y ≥ 3) = 1 – P(Y ≤ 2). From binomial tables or manual calculation: P(Y=0) = 0.75¹² ≈ 0.0317, P(Y=1) = 12 × 0.25 × 0.75¹¹ ≈ 12 × 0.25 × 0.0422 ≈ 0.1267, P(Y=2) = 66 × 0.0625 × 0.75¹⁰ ≈ 66 × 0.0625 × 0.0563 ≈ 0.232. Sum ≈ 0.3904. Thus P(Y ≥ 3) ≈ 1 – 0.3904 = 0.6096. The expected value E(Y) = np = 12 × 0.25 = 3, and variance Var(Y) = np(1-p) = 12 × 0.25 × 0.75 = 2.25.

求 P(Y ≥ 3) 时宜用补集：P(Y ≥ 3) = 1 – P(Y ≤ 2)。查二项分布表或手动计算：P(Y=0)≈0.0317，P(Y=1)≈0.1267，P(Y=2)≈0.232，总和约 0.3904。因此 P(Y ≥ 3) ≈ 0.6096。期望值 E(Y)=np=3，方差 Var(Y)=np(1-p)=2.25。

Always verify conditions for binomial: fixed number of trials, independent trials, constant probability of success, and two possible outcomes. For guessing exams, these hold reasonably well, making the binomial a good model.

务必验证二项分布条件：试验次数固定、各次独立、成功概率恒定、结果只有两种。对于随机猜测的考试，这些条件基本满足，二项分布是一个合适的模型。

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4. Normal Distribution and Light Bulb Lifetime | 正态分布与灯泡寿命

A manufacturer claims that its light bulbs have a mean lifetime of 1,200 hours with a standard deviation of 150 hours. Assume lifetimes are normally distributed. Calculate the proportion of bulbs expected to last more than 1,400 hours; between 1,000 and 1,350 hours; and find the lifetime that 90% of bulbs will exceed. This case practices normal distribution applications and inverse normal function.

某制造商声称其灯泡的平均寿命为 1200 小时，标准差 150 小时。假设寿命服从正态分布。计算预期寿命超过 1400 小时的灯泡比例；寿命在 1000 至 1350 小时之间的比例；并求出 90% 灯泡都能超过的寿命值。本案例练习正态分布应用和逆正态功能。

Let L ~ N(1200, 150²). Standardise to Z = (L – 1200)/150. For L > 1400, Z > (1400-1200)/150 = 1.333. Using normal tables, P(Z > 1.333) = 1 – Φ(1.333) ≈ 1 – 0.9088 = 0.0912. So about 9.12% of bulbs last more than 1,400 hours.

设 L ~ N(1200, 150²)。标准化 Z = (L – 1200)/150。对于 L > 1400，Z > 1.333，查表得 P(Z > 1.333) ≈ 1 – 0.9088 = 0.0912，约 9.12% 的灯泡寿命超过 1400 小时。

For 1000 < L < 1350, we have Z₁ = (1000-1200)/150 = -1.333, Z₂ = (1350-1200)/150 = 1.000. Then P(-1.333 < Z < 1.000) = Φ(1.000) - Φ(-1.333) = Φ(1.000) - [1 - Φ(1.333)] ≈ 0.8413 - (1 - 0.9088) = 0.8413 - 0.0912 = 0.7501. So 75.01% of bulbs fall within this range.

对于 1000 < L < 1350，得 Z₁ = -1.333，Z₂ = 1.000。P(-1.333 < Z < 1.000) = Φ(1.000) - [1 - Φ(1.333)] ≈ 0.8413 - 0.0912 = 0.7501，即 75.01% 的灯泡在此区间。

To find the 90th percentile (the lifetime exceeded by 90%), we need the 10th percentile because the area to the left is 0.10. The Z-value for lower tail area 0.10 is approximately -1.2816. Thus L = 1200 + (-1.2816)(150) ≈ 1200 – 192.2 = 1007.8 hours. So 90% of bulbs will last longer than about 1,008 hours. Always sketch a normal curve and shade the region of interest to avoid confusion between upper and lower tails.

要求 90% 灯泡都能超过的寿命，即求第 10 百分位数，左侧面积 0.10 对应的 Z 值约 -1.2816。因此 L = 1200 – 1.2816×150 ≈ 1007.8 小时。所以 90% 的灯泡寿命超过约 1008 小时。务必画正态曲线并标出目标区域，以避免上下尾混淆。

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5. Sampling Distribution of the Mean | 样本均值的抽样分布

The waiting time at a bank follows a distribution with mean μ = 4.2 minutes and standard deviation σ = 1.6 minutes. Random samples of size n = 36 customers are selected. Find the probability that the sample mean waiting time exceeds 4.8 minutes. This case explores the Central Limit Theorem (CLT) and how sample means behave, a key concept for confidence intervals and hypothesis testing.

某银行等待时间服从均值 μ=4.2 分钟、标准差 σ=1.6 分钟的分布。随机抽取 n=36 名顾客构成样本。求样本平均等待时间超过 4.8 分钟的概率。本案例探讨中心极限定理以及样本均值的性质，是掌握置信区间和假设检验的关键概念。

By the CLT, the sample mean X̄ is approximately normally distributed with mean μₓ̄ = μ = 4.2 and standard error σₓ̄ = σ/√n = 1.6/√36 = 1.6/6 ≈ 0.2667. We need P(X̄ > 4.8). Standardise: Z = (4.8 – 4.2) / 0.2667 ≈ 0.6 / 0.2667 = 2.25. Then P(Z > 2.25) = 1 – Φ(2.25) ≈ 1 – 0.9878 = 0.0122. So there is only a 1.22% chance of observing a sample mean greater than 4.8 minutes if the population mean is 4.2.

根据中心极限定理，样本均值 X̄ 近似服从正态分布，均值为 μ=4.2，标准误 σₓ̄ = 1.6/√36 ≈ 0.2667。需计算 P(X̄ > 4.8)。标准化：Z = (4.8 – 4.2)/0.2667 = 2.25。P(Z > 2.25) ≈ 1 – 0.9878 = 0.0122。因此，若总体均值确为 4.2，观察到样本均值超过 4.8 的概率仅 1.22%。

This low probability might suggest that the population mean could be higher, a line of reasoning used in hypothesis testing. Remember that the CLT requires a sufficiently large sample size (usually n ≥ 30) and that the population standard deviation is known. In exam problems, always state the distribution of the sample mean explicitly and check the conditions.

这一低概率可能暗示总体均值实际上更高，这正是假设检验的推理思路。切记中心极限定理要求样本量足够大（通常 n≥30），且已知总体标准差。在考试题中，务必明确写出样本均值的分布并检查条件。

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

A health researcher measures the body mass index (BMI) of 25 randomly chosen adults from a large city. The sample mean is 24.6 and the sample standard deviation s = 3.8. Construct a 95% confidence interval for the population mean BMI. Since the population standard deviation is unknown and the sample size is small, this case employs the t-distribution, common in Year 12 CIE Statistics.

一位健康研究者测量了某大城市随机抽取的 25 名成年人的体重指数 (BMI)。样本均值 24.6，样本标准差 s=3.8。构建总体平均 BMI 的 95% 置信区间。由于总体标准差未知且样本量较小，本案例使用 t 分布，这在 Year 12 CIE 统计中很常见。

The confidence interval for μ (σ unknown) is: X̄ ± t* × (s/√n). Degrees of freedom df = n – 1 = 24. For 95% confidence, the critical t-value t* from t-table is 2.064. The standard error is s/√n = 3.8/√25 = 3.8/5 = 0.76. The margin of error = 2.064 × 0.76 ≈ 1.57. Therefore, the 95% CI is (24.6 – 1.57, 24.6 + 1.57) = (23.03, 26.17). We are 95% confident that the true population mean BMI lies between 23.03 and 26.17.

σ 未知时均值的置信区间为：X̄ ± t* × (s/√n)。自由度 df = 24，95% 置信水平对应的 t 临界值 t* = 2.064。标准误 s/√n = 3.8/5 = 0.76。边际误差 = 2.064 × 0.76 ≈ 1.57。因此 95% 置信区间为 (23.03, 26.17)。我们有 95% 的把握认为总体平均 BMI 介于 23.03 与 26.17 之间。

Interpretation is crucial: it does not mean that 95% of individual BMIs fall in this interval; it refers to the process producing intervals that capture the true mean 95% of the time. Also, verify that the underlying population is roughly normal or that the sample size is large enough for the CLT to apply. For BMI data, a normal assumption is often plausible with a sample of 25.

正确解释至关重要：这并不意味着 95% 的个体 BMI 落在此区间内，而是指用这样的方法构造的区间有 95% 的把握包含总体真均值。同时应验证总体近似正态，或样本量足够大以使中心极限定理有效。对于 BMI 数据，样本量为 25 时正态假设通常是合理的。

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7. Hypothesis Test for a Proportion | 比例假设检验

A pharmaceutical company claims that a new drug relieves symptoms in 70% of patients. A consumer group suspects the proportion is lower and conducts a trial with 200 patients, finding that 128 experienced relief. Test the claim at the 5% significance level. This case illustrates a one-tailed test for a proportion, a frequent CIE exam scenario.

某制药公司声称一种新药对 70% 的患者能缓解症状。一个消费者权益组织怀疑真实比例更低，对 200 名患者进行试验，发现 128 人症状缓解。在 5% 显著性水平下检验该声称。本案例演绎了比例的单侧检验，是 CIE 考试常见题型。

Step 1: Set hypotheses. H₀: p = 0.70; H₁: p < 0.70 (one-tailed). Step 2: Check conditions: n is large, both np₀ = 140 and n(1-p₀) = 60 exceed 10, so a normal approximation is appropriate. Step 3: Test statistic: sample proportion p̂ = 128/200 = 0.64. Under H₀, the standard error of p̂ is √[p₀(1-p₀)/n] = √(0.70×0.30/200) = √(0.21/200) = √0.00105 ≈ 0.0324. Then Z = (0.64 - 0.70) / 0.0324 ≈ -1.85.

第 1 步：建立假设。H₀: p=0.70；H₁: p<0.70（单侧）。第 2 步：检查条件：n 很大，np₀=140 和 n(1-p₀)=60 均大于 10，因此正态近似适用。第 3 步：检验统计量：样本比例 p̂ = 128/200 = 0.64。在 H₀ 下，p̂ 的标准误 = √(0.70×0.30/200) ≈ 0.0324。Z = (0.64-0.70)/0.0324 ≈ -1.85。

Step 4: The critical value for a lower-tail test at α=0.05 is -1.645 (from normal tables). Since -1.85 < -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 proportion of relief is less than 70%. Always provide the p-value: P(Z < -1.85) ≈ 0.0322, which is less than 0.05, confirming the conclusion.

第 4 步：α=0.05 的下尾临界值为 -1.645。由于 -1.85 < -1.645，检验统计量落入拒绝域。拒绝 H₀，在 5% 显著性水平下有充分证据表明真实缓解率低于 70%。答题时还应给出 p 值：P(Z < -1.85) ≈ 0.0322 < 0.05，支持结论。

Clearly state your decision in context. In CIE examinations, marks are awarded for correct hypotheses, test statistic, comparison, and a non-ambiguous conclusion. Avoid common mistakes like using a two-tailed test when the alternative is directional.

务必结合上下文清晰陈述决策。CIE 考试中，评分点包括正确假设、检验统计量、比较和明确的结论。避免常见错误，例如当备择假设有方向时误用双尾检验。

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8. Hypothesis Test for a Mean with Unknown Variance | 未知方差下的均值假设检验

A school introduces a new teaching method, and the exam scores of 20 students are recorded. The mean score is 68, with a standard deviation of 12. The national average is 64. Test at the 1% significance level whether the new method has increased the mean score. This is a one-sample t-test, reinforcing the use of t-distribution for small samples when σ is unknown.

某学校引入新教学法，记录了 20 名学生的考试成绩。平均分 68，标准差 12。全国平均分为 64。在 1% 显著性水平下检验新方法是否提高了平均分。这是一个单样本 t 检验，强化了在 σ 未知时小样本情况下 t 分布的应用。

Hypotheses: H₀: μ = 64; H₁: μ > 64 (upper-tail test). Sample size n=20, degrees of freedom = 19. The t statistic: t = (X̄ – μ₀) / (s/√n) = (68 – 64) / (12/√20) = 4 / (12/4.472) = 4 / 2.683 ≈ 1.491. Critical t-value for df=19, upper tail α=0.01 is 2.539 (from t-table). Since 1.491 < 2.539, we fail to reject H₀. There is not enough evidence at the 1% level to claim an increase.

建立假设：H₀: μ=64；H₁: μ>64（上尾检验）。样本量 n=20，自由度 19。t 统计量：t = (68-64)/(12/√20) = 4 / 2.683 ≈ 1.491。查 t 分布表得 df=19、α=0.01 上尾临界值 2.539。由于 1.491 < 2.539，无法拒绝 H₀。在 1% 显著性水平下无充分证据证明分数提高。

This result does not mean the new method is ineffective—only that the observed difference could plausibly be due to sampling variability given the strict significance level. Report the p-value using technology: P(t₁₉ > 1.491) ≈ 0.076, which further shows the result is not significant at 1% but would be significant at 8%. Such nuance is important for interpretation.

该结果并不意味着新方法无效——只是说在严格的显著性水平下，观察到的差异可能来自抽样波动。使用技术工具报告 p 值：P(t₁₉ > 1.491) ≈ 0.076，进一步表明结果在 1% 下不显著，但在 8% 水平下显著。解释时需注意这种细微差别。

When performing t-tests, always assume the population is normally distributed or approximately normal. The process becomes robust for moderate departures, but you should mention this assumption.

进行 t 检验时，需假设总体服从正态分布或近似正态。该方法对中等程度的偏离是稳健的，但应在答案中提及该假设。

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9. Chi-Squared Goodness-of-Fit Test: Genetics Case | 卡方拟合优度检验：遗传学案例

According to Mendelian genetics, a cross of two hybrid pea plants for flower colour should yield offspring with a 3:1 ratio of purple to white flowers. A biologist observes 152 purple and 48 white offspring from 200 seeds. Test whether the data are consistent with the theoretical ratio at the 5% significance level. This case introduces the chi-squared statistic for categorical data.

根据孟德尔遗传学，两种花色杂交的豌豆植株后代中紫花与白花的理论比例为 3:1。一位生物学家从 200 粒种子中观察到 152 株紫花和 48 株白花。在 5% 显著性水平下检验数据是否与理论比例一致。本案例引入用于分类数据的卡方统计量。

H₀: The population follows a 3:1 ratio (p_purple = 0.75, p_white = 0.25). H₁: The population does not follow this ratio. Expected counts: purple = 200×0.75 = 150, white = 200×0.25 = 50. The chi-squared statistic χ² = Σ[(O – E)² / E] = (152-150)²/150 + (48-50)²/50 = 4/150 + 4/50 = 0.0267 + 0.08 = 0.1067. Degrees of freedom = number of categories – 1 = 2 – 1 = 1. From χ² tables, the critical value at α=0.05, df=1 is 3.841. Since 0.1067 < 3.841, we do not reject H₀. The data are consistent with the Mendelian 3:1 ratio.

H₀：总体比例为 3:1（p_purple=0.75，p_white=0.25）；H₁：不符合此比例。期望频数：紫花 150，白花 50。卡方统计量 χ² = Σ[(O-E)²/E] = (152-150)²/150 + (48-50)²/50 = 0.0267 + 0.08 = 0.1067。自由度 = 类别数 – 1 = 1。查 χ² 表得 α=0.05 下的临界值 3.841。因 0.1067 < 3.841，不拒绝 H₀，数据与 3:1 比例一致。

For validity, ensure expected frequencies are at least 5 in each category. Here they are well above 5, so the test is appropriate. This case shows how even small deviations do not automatically lead to rejection; statistical significance requires sufficiently large discrepancies relative to sample size.

进行卡方检验时，需确保每个类别的期望频数至少为 5。本例中均远大于 5，试验有效。此案例表明，即使有略微偏差也不一定导致拒绝；统计显著性需要相对样本量而言足够大的差异。

Always write the null hypothesis in words and symbols, and interpret the result in context, e.g., “There is insufficient evidence to suggest that the Mendelian model is incorrect for this sample.”

务必用文字和符号书写原假设，并结合情境解释结论，例如“没有充分证据表明该样本不符合孟德尔模型”。

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10. Integrated Case: Delivery Service Performance | 综合案例：快递服务表现

A courier company states that its standard delivery time is 2.5 days on average, with standard deviation 0.8 days. An independent survey of 45 deliveries yields a mean of 2.75 days and a standard deviation of 0.9 days. The investigator wants to know if the delivery time has increased, and also wishes to estimate the current mean delivery time with 99% confidence. This final case combines hypothesis testing and confidence intervals, demonstrating how they complement each other in statistical analysis.

某快递公司声称其标准配送时间的平均值为 2.5 天，标准差 0.8 天。一项独立调查抽取了 45 次配送，得到样本均值 2.75 天，样本标准差 0.9 天。调查人员希望知道配送时间是否增加，同时以 99% 置信水平估计当前平均配送时间。本综合案例结合假设检验与置信区间，展示两者如何在统计分析中相互补充。

First, test H₀: μ = 2.5 vs H₁: μ > 2.5 at α=0.01. Despite σ being stated from the company, using the sample s=0.9 with n=45 is acceptable due to large n (t approximates Z). Standard error = 0.9/√45 ≈ 0.1342. Test statistic: Z ≈ (2.75 – 2.5) / 0.1342 ≈ 1.863. Critical Z for one-tail 0.01 is 2.326. Since 1.863 < 2.326, we fail to reject H₀; the increase is not statistically significant at 1%.

首先进行假设检验，H₀: μ=2.5 vs H₁: μ>2.5，α=0.01。尽管公司给出了总体标准差 σ，但使用样本 s=0.9、n=45 是可接受的（大样本下 t 近似于 Z）。标准误 ≈ 0.1342。检验统计量 Z ≈ 1.863。单侧 1% 临界值 2.326，1.863 < 2.326，无法拒绝 H₀。在 1% 水平下该增加不显著。

Next, construct a 99% confidence interval for μ using t₄₄ (or Z since n is large). Using Z* = 2.576 for 99%: CI = 2.75 ± 2.576 × 0.1342 ≈ 2.75 ± 0.346, giving (2.404, 3.096). Notice that the

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