Case Study Practice for Year 13 CIE Statistics | Year 13 CIE 统计：案例分析实战演练

📚 Case Study Practice for Year 13 CIE Statistics | Year 13 CIE 统计：案例分析实战演练

Welcome to this comprehensive case study walkthrough designed for Year 13 CIE Statistics students. In this article, you will step into the role of a data analyst working for a supermarket chain. The management wants to examine customer waiting times at checkout, the pattern of customer arrivals, and the relationship between customer satisfaction and the time of day. Using a real-world dataset, we will apply a range of statistical techniques from the CIE A Level Statistics 2 syllabus: hypothesis tests on means, a chi-squared test for a population variance, a Poisson goodness-of-fit test, and a chi-squared test of independence. Each section blends theory with practical computation, followed by clear interpretations, so you can see how these methods are used together to drive evidence-based decision-making.

欢迎来到这篇为 Year 13 CIE 统计学生设计的综合案例分析实战演练。在本文中，你将扮演一家连锁超市的数据分析师。管理层希望检查顾客在收银台的等待时间、顾客到达的模式，以及顾客满意度与时段之间的关系。我们将使用一个真实世界的数据集，应用 CIE A Level 统计 2 大纲中的一系列统计技术：关于均值的假设检验、总体方差的卡方检验、泊松拟合优度检验，以及独立性卡方检验。每一节都将理论与实际计算相结合，并辅以清晰的解释，让你看到这些方法如何共同推动基于证据的决策。

1. Scenario and Data Description | 情景与数据描述

The supermarket operates seven days a week and has recently received complaints about long queues during busy periods. To investigate, the manager collected three sets of data over the course of one month. The first dataset contains the waiting time (in minutes) for a random sample of 40 customers on weekdays and another random sample of 35 customers on weekends. The second dataset records the number of customer arrivals in 100 consecutive 15-minute intervals during opening hours. The third dataset is a contingency table of customer satisfaction levels (satisfied, neutral, dissatisfied) cross-classified by time slot (morning, afternoon, evening). All data are assumed to be randomly sampled and independent. The target waiting time set by the head office is 4.0 minutes, and the store claims that the standard deviation of waiting times is 1.0 minute.

该超市每周七天营业，最近收到了关于高峰期排长队的投诉。为了调查，经理在一个月内收集了三组数据。第一组数据集包含在工作日随机抽取的 40 名顾客和在周末随机抽取的 35 名顾客的等待时间（以分钟计）。第二组数据集记录了在 100 个连续的 15 分钟时段内，营业时间中顾客到达的人数。第三组数据集是按时间段（上午、下午、晚上）交叉分类的顾客满意度（满意、一般、不满意）列联表。所有数据都假设为随机抽样且相互独立。总公司设定的目标等待时间为 4.0 分钟，并且该商店声称等待时间的标准差为 1.0 分钟。

Variable	Weekday sample (n=40)	Weekend sample (n=35)
Mean waiting time	4.2 min	4.8 min
Sample standard deviation	1.1 min	1.3 min

ion 6) and the satisfaction data are presented in a 3×3 contingency table (see Section 7).

每 15 分钟到达人数的频数分布总结在表格中（见第 6 节），满意度数据呈现在 3×3 列联表中（见第 7 节）。

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2. Descriptive Statistics and Visualisation | 描述统计与可视化

Before undertaking any formal hypothesis test, we examine the data graphically and numerically. For the weekday waiting times, the sample mean is 4.2 minutes with a standard deviation of 1.1 minutes. A box plot reveals a slight right skew, but with no extreme outliers. The weekend sample has a higher mean of 4.8 minutes and slightly greater spread (standard deviation 1.3 minutes). Overlaid histograms show that both distributions are roughly bell-shaped, which supports the assumption of normality required for t-tests. Summary statistics such as the median and interquartile range are also calculated to check robustness.

在进行任何正式的假设检验之前，我们先以图形和数值的方式检查数据。对于工作日的等待时间，样本均值是 4.2 分钟，标准差为 1.1 分钟。箱线图显示出轻微的右偏，但没有极端离群值。周末样本的均值更高，为 4.8 分钟，且离散程度稍大（标准差 1.3 分钟）。叠加的直方图显示两者的分布都大致呈钟形，这支持了 t 检验所需的正态性假设。我们还计算了中位数和四分位距等汇总统计量，以检查稳健性。

These initial explorations suggest that the average waiting time may exceed the 4.0-minute target on weekdays, and that weekends appear even busier. Formal tests will confirm whether these differences are statistically significant.

这些初步的探索表明，在工作日平均等待时间可能超过了 4.0 分钟的目标，而周末似乎更加繁忙。正式的检验将确认这些差异是否具有统计学显著性。

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3. One-Sample t-Test for Mean Waiting Time | 单样本t检验：平均等待时间

The first question is: 'Is the mean weekday waiting time significantly greater than the 4.0-minute target?' Because the population variance is unknown, we use a one-sample t-test. Let μ be the true mean waiting time on weekdays. We set up the hypotheses:

第一个问题是：“工作日的平均等待时间是否显著大于 4.0 分钟的目标？”由于总体方差未知，我们使用单样本 t 检验。设 μ 为工作日等待时间的真实均值。我们建立假设：

H₀ : μ = 4.0
H₁ : μ > 4.0

The test statistic is

检验统计量为

t = (x̄ – μ₀) / (s / √n)

where x̄ = 4.2, μ₀ = 4.0, s = 1.1, n = 40. Substituting gives t = (4.2 – 4.0) / (1.1 / √40) = 0.2 / (1.1/6.3249) ≈ 0.2 / 0.1739 ≈ 1.15. The degrees of freedom are df = n – 1 = 39. Using a one-tailed t-table or software, the critical value at the 5% ere is insufficient evidence to conclude that the mean weekday waiting time exceeds 4.0 minutes.

其中 x̄ = 4.2，μ₀ = 4.0，s = 1.1，n = 40。代入得 t = (4.2 – 4.0) / (1.1 / √40) = 0.2 / (1.1/6.3249) ≈ 0.2 / 0.1739 ≈ 1.15。自由度为 df = n – 1 = 39。查阅单尾 t 值表或使用软件，5% 显著性水平下的临界值约为 1.685。由于 1.15 < 1.685，我们不能拒绝原假设。p 值约为 0.13，大于 0.05。因此，没有足够的证据表明工作日的平均等待时间超过 4.0 分钟。

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4. Two-Sample t-Test: Weekday vs Weekend | 双样本t检验：工作日与周末比较

ation variances are equal. This allows us to use the pooled two-sample t-test. Let μ_wk and μ_we denote the true mean waiting times for weekdays and weekends respectively. The hypotheses are:

接下来，我们比较工作日和周末的平均等待时间。我们假设两个总体近似正态，并且未知的总体方差相等。这使我们能够使用合并双样本 t 检验。设 μ_wk 和 μ_we 分别表示工作日和周末的真实平均等待时间。假设为：

H₀ : μ_wk = μ_we
H₁ : μ_wk < μ_we (one-tailed test)

The pooled variance s²_p is calculated as:

合并方差 s²_p 的计算公式为：

s²_p = [ (n₁ – 1)s₁² + (n₂ – 1)s₂² ] / (n₁ + n₂ – 2)

in s²_p = [39×1.21 + 34×1.69] / (73) = (47.19 + 57.46) / 73 = 104.65 / 73 ≈ 1.433. Hence s_p = √1.433 ≈ 1.197. The test statistic is:

代入 n₁=40，s₁=1.1，n₂=35，s₂=1.3，得到 s²_p = [39×1.21 + 34×1.69] / (73) = (47.19 + 57.46) / 73 = 104.65 / 73 ≈ 1.433。因此 s_p = √1.433 ≈ 1.197。检验统计量为：

t = (x̄₁ – x̄₂) / (s_p × √1/n₁ + 1/n₂)

So t = (4.2 – 4.8) / (1.197 × √1/40 + 1/35) = (-0.6) / (1.197 × √0.025 + 0.02857) = -0.6 / (1.197 × 0.2315) ≈ -0.6 / 0.277 ≈ -2.166. The degrees of freedom = 40+35–2 = 73. The one-tailed critical value at 5% s significant evidence that the mean waiting time on weekends is greater than on weekdays.

因此 t = (4.2 – 4.8) / (1.197 × √1/40 + 1/35) = (-0.6) / (1.197 × √0.025 + 0.02857) = -0.6 / (1.197 × 0.2315) ≈ -0.6 / 0.277 ≈ -2.166。自由度为 40+35–2 = 73。5% 单尾临界值（负侧）约为 -1.666。由于 -2.166 < -1.666，我们拒绝原假设。有显著证据表明周末的平均等待时间大于工作日。

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5. Chi-Squared Test for Population Variance | 总体方差的卡方检验

inute. Using the weekday sample (n=40, s=1.1), we can test this claim. The hypotheses are:

商店经理声称等待时间的标准差正好是 1.0 分钟。利用工作日样本 (n=40, s=1.1)，我们可以检验这一声称。假设为：

H₀ : σ² = 1.0² = 1.0
H₁ : σ² > 1.0

The test statistic follows a chi-squared distribution:

检验统计量服从卡方分布：

χ² = (n – 1)s² / σ₀²

ituting: χ² = 39 × 1.1² / 1.0 = 39 × 1.21 = 47.19. The degrees of freedom are 39. At the 5% ere is insufficient evidence to say the standard deviation has increased beyond 1.0 minute.

代入：χ² = 39 × 1.1² / 1.0 = 39 × 1.21 = 47.19。自由度是 39。在 5% 显著性水平下，χ²₃₉ 的临界值约为 54.6。由于 47.19 < 54.6，我们不拒绝原假设。p 值约为 0.17。没有足够的证据表明标准差已增加到超过 1.0 分钟。

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6. Poisson Goodness-of-Fit Test for Arrivals | 到达人数的泊松拟合优度检验

The manager recorded the number of customer arrivals in 100 randomly selected 15-minute intervals. The observed frequencies are shown in the table below. We wish to test whether the arrivals follow a Poisson distribution with a constant mean rate.

经理记录了在 100 个随机选择的 15 分钟时段内到达的顾客人数。观察频数如下表所示。我们希望检验到达人数是否服从具有恒定平均率的泊松分布。

Arrivals per interval	0	1	2	3	4	5 or more
Observed frequency	10	25	30	20	10	5

son model with λ=2.1, we compute expected frequencies for 0,1,2,3,4 and then pool the tail '5 or more' to ensure all expected values are at least 5.

样本均值 λ 的估计值为 (0×10+1×25+2×30+3×20+4×10+5×5)/100 = 210/100 = 2.1。使用 λ=2.1 的泊松模型，我们计算 0,1,2,3,4 的期望频数，然后将尾部“5 或以上”合并，以确保所有期望值至少为 5。

The Poisson probabilities are: P(X=0)=e⁻²·¹×2.1⁰/0! P(X=1)=0.2572, P(X=2)=0.2700, P(X=3)=0.1890, P(X=4)=0.0992. The combined probability for 0–4 is 0.9379, so P(X≥5)=1– 5, 25.72, 27.00, 18.90, 9.92, 6.21 (for ≥5). All are above 5. The chi-squared statistic is

泊松概率为：P(X=0)=e⁻²·¹×2.1⁰/0! ≈ 0.1225, P(X=1)=0.2572, P(X=2)=0.2700, P(X=3)=0.1890, P(X=4)=0.0992。0–4 的合并概率为 0.9379，因此 P(X≥5)=0.0621。乘以 100 得到期望频数：12.25, 25.72, 27.00, 18.90, 9.92, 6.21（对于 ≥5）。所有值都大于 5。卡方统计量为

χ² = Σ (O – E)² / E

Computing: (10–12.25)²/12.25 + (25–25.72)²/25.72 + (30–27)²/27 + (20–18.9)²/18.9 + (10–9.92)²/9.92 + (5– 1 + 0.236 = 1.067. With 6 categories and 1 parameter estimated (λ), degrees of freedom = 6 – 1 – 1 = 4. The 5% 9.488, we accept the null hypothesis. The data are consistent with a Poisson distribution.

计算：(10–12.25)²/12.25 + (25–25.72)²/25.72 + (30–27)²/27 + (20–18.9)²/18.9 + (10–9.92)²/9.92 + (5–6.21)²/6.21 ≈ 0.413 + 0.020 + 0.333 + 0.064 + 0.001 + 0.236 = 1.067。共有 6 个类别，估计了 1 个参数 (λ)，自由度 = 6 – 1 – 1 = 4。χ²₄ 在 5% 的临界值为 9.488。由于 1.067 < 9.488，我们接受原假设。数据与泊松分布一致。

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7. Chi-Squared Test of Independence: Satisfaction vs Time Slot | 独立性卡方检验：满意度与时段

Finally, we examine whether customer satisfaction level is independent of the time of day. A random sample of customers was asked to rate their experience, yielding the following contingency table:

最后，我们检验顾客满意度水平是否与一天中的时段独立。随机抽取的顾客被要求评价他们的体验，得到了下面的列联表：

Published by TutorHao | Year 13 统计 Revision Series | aleveler.com

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