📚 Year 13 SQA Statistics: Case Study Analysis and Practice | Year 13 SQA 统计：案例分析实战演练
This article provides a complete case-study walkthrough for SQA Advanced Higher Statistics (Year 13). We simulate a quality-testing scenario using real-world data, applying descriptive analysis, t-tests, confidence intervals, chi-squared goodness-of-fit, and effect sizes. Every step mirrors the structured approach required in the examination, helping you master statistical inquiry and report writing.
本文为 SQA 高等统计（Year 13）提供了一整套案例分析演练。我们模拟一个真实的质量检测场景，运用描述性分析、t 检验、置信区间、卡方拟合优度以及效应量等工具。每个步骤都呼应考试中要求的结构化探究方法，帮助你掌握统计调查与报告撰写。
1. Case Overview | 案例背景
A lighting manufacturer has developed a new energy-efficient LED bulb and claims that its mean lifetime exceeds 8000 hours. As a Year 13 statistics student, you are asked to evaluate this claim using sample data. You will also compare the new bulb with a standard model and investigate consumer colour preferences.
一家照明制造商新开发了一款节能 LED 灯泡，声称其平均寿命超过 8000 小时。作为 Year 13 统计学生，你需要利用样本数据对该声明进行评估。你还将比较新灯泡与标准型号的差异，并调查消费者对颜色的偏好。
The study is structured into three mini-analyses: (i) inference about a single population mean, (ii) comparison of two independent means, and (iii) a categorical analysis using chi-squared goodness-of-fit. Every conclusion must be supported by appropriate visualisations, calculations and interpretive text, just as in the SQA exam.
本案例设计为三个小分析：(i) 关于单个总体均值的推断，(ii) 两个独立均值的比较，以及 (iii) 使用卡方拟合优度进行类别分析。所有结论都必须以适当的可视化、计算和解释性文字作为支撑，正如同 SQA 考试所要求的那样。
2. Data Collection & Exploration | 数据收集与探索
A quality engineer randomly selects 10 new bulbs and records their lifetimes (hours): 8120, 7950, 8300, 7880, 8050, 8190, 8270, 8010, 7930, 8150. For comparison, 10 standard bulbs are sampled under identical conditions, yielding: 7880, 7790, 7950, 7700, 7850, 7900, 7720, 7810, 7830, 7760.
一名质量工程师随机选取了 10 个新灯泡并记录其寿命（小时）：8120, 7950, 8300, 7880, 8050, 8190, 8270, 8010, 7930, 8150。为了进行比较，在相同条件下抽取了 10 个标准灯泡，数据为：7880, 7790, 7950, 7700, 7850, 7900, 7720, 7810, 7830, 7760。
| Group | Lifetime data (hours) |
|---|---|
| New bulb | 8120, 7950, 8300, 7880, 8050, 8190, 8270, 8010, 7930, 8150 |
| Standard bulb | 7880, 7790, 7950, 7700, 7850, 7900, 7720, 7810, 7830, 7760 |
Before formal inference, we examine the data with box plots and summary statistics. Outliers are not apparent, and the spread appears reasonable for normal-based methods. All bulbs were tested until failure, ensuring interval data.
在正式推断之前，我们先通过箱线图和汇总统计量来检查数据。数据中没有明显的异常值，离散程度在基于正态分布的方法下看起来是合理的。所有灯泡都测试至失效，保证了连续型数据。
Additionally, a separate survey of 50 customers is collected to assess colour preference for the bulb casing: White (23), Warm Yellow (18), Daylight (9). This will be used later for a chi-squared test.
此外，还收集了 50 名顾客对灯泡外壳颜色的偏好调查：白色 (23)，暖黄 (18)，日光色 (9)。这将在后续用于卡方检验。
3. Descriptive Statistics Summary | 描述性统计汇总
For the new bulb sample: mean lifetime x̄ = (8120 + 7950 + … + 8150)/10 = 8085 hours. The sample standard deviation s is calculated from the squared deviations. s² = Σ(x − x̄)²/(n−1) = 188050/9 ≈ 20894.44, giving s ≈ 144.55 hours.
对于新灯泡样本：平均寿命 x̄ = (8120 + 7950 + … + 8150)/10 = 8085 小时。样本标准差 s 由平方偏差计算得出。s² = Σ(x − x̄)²/(n−1) = 188050/9 ≈ 20894.44，因此 s ≈ 144.55 小时。
The standard error of the mean is SE = s/√n = 144.55/√10 ≈ 45.71 hours. For the standard bulb group, analogous calculations give x̄_std = 7819 hours and s_std ≈ 79.51 hours (s² ≈ 6321.11).
均值的标准误为 SE = s/√n = 144.55/√10 ≈ 45.71 小时。对于标准灯泡组，类似的计算得到 x̄_std = 7819 小时，s_std ≈ 79.51 小时（s² ≈ 6321.11）。
| Statistic | New bulb (n=10) | Standard bulb (n=10) |
|---|---|---|
| Mean (x̄) | 8085 hr | 7819 hr |
| Standard deviation (s) | 144.55 hr | 79.51 hr |
| Standard error (SE) | 45.71 hr | 25.14 hr |
These summaries highlight a 266-hour advantage for the new bulb. Variability is greater in the new bulb sample, which will affect the precision of inference.
这些汇总数据显示新灯泡有 266 小时的优势。新灯泡样本的变异性更大，这会影响推断的精度。
4. Checking Normality Assumption | 检查正态性假设
With small samples (n=10), t-procedures assume that the population distribution is approximately normal. We can inspect a normal Q-Q plot of the new bulb lifetimes. The points fall roughly along the diagonal, and there is no strong curvature or systematic pattern. A Shapiro-Wilk test (not shown) yields a p-value of 0.38, so we do not reject normality.
对于小样本（n=10），t 程序要求总体大致服从正态分布。我们可以检查新灯泡寿命的正态 Q-Q 图。点大致落在对角线上，没有强烈的弯曲或系统性模式。Shapiro-Wilk 检验（未展示）给出的 p 值为 0.38，因此我们不拒绝正态性假设。
The standard bulb sample also appears symmetric and has no extreme outliers. Consequently, it is reasonable to proceed with parametric t-tests. However, if normality were suspect, a non-parametric alternative such as the Mann-Whitney test should be considered.
标准灯泡样本也表现出对称性，且没有极端离群值。因此，继续使用参数 t 检验是合理的。不过，如果正态性存疑，就应考虑诸如 Mann-Whitney 检验这样的非参数替代方法。
5. Constructing Confidence Intervals | 构建置信区间
A 95% confidence interval for the true mean lifetime of the new bulb uses the t-distribution with 9 degrees of freedom. The critical value t* = 2.262 (two-tailed).
新灯泡真实平均寿命的 95% 置信区间使用自由度为 9 的 t 分布。临界值 t* = 2.262（双尾）。
CI: x̄ ± t* × SE = 8085 ± 2.262 × 45.71 = (7981.6, 8188.4) hours
This interval contains the hypothesised value of 8000 hours. Therefore, at the 5% significance level for a two‑sided test, we would not have enough evidence to say the mean differs from 8000. However, the manufacturer’s claim is one-sided (greater than 8000). This is explored next.
该区间包含了假设值 8000 小时。因此，在 5% 显著性水平下的双侧检验中，我们还没有足够的证据说均值不同于 8000。然而，制造商的声明是单侧的（超过 8000）。这一点接下来会探讨。
For the standard bulb, a 95% CI is 7819 ± 2.262 × 25.14 = (7819 ± 56.86) → (7762.1, 7875.9) hours. The intervals do not overlap, hinting at a real difference between the two bulb types.
标准灯泡的 95% 置信区间为 7819 ± 2.262 × 25.14 = (7819 ± 56.86) → (7762.1, 7875.9) 小时。两个区间不重叠，暗示两类灯泡之间存在真实差异。
6. One-Sample t-Test (Right-Tailed) | 单样本 t 检验（右尾）
We address the company’s claim: H₀: μ ≤ 8000 versus H₁: μ > 8000, where μ is the true mean lifetime of the new bulb. Using the sample data, the test statistic is:
我们处理公司的声明：H₀: μ ≤ 8000 对 H₁: μ > 8000，其中 μ 是新灯泡的真实平均寿命。利用样本数据，检验统计量为：
t = (x̄ − μ₀) / (s / √n) = (8085 − 8000) / 45.71 ≈ 1.86
With df = 9, the one‑tailed critical value at α = 0.05 is t_crit = 1.833. Since 1.86 > 1.833, we reject H₀. The p‑value (one-tailed) is approximately 0.048, computed from the t(9) distribution.
自由度为 9 时，α = 0.05 下的单尾临界值为 t_crit = 1.833。因为 1.86 > 1.833，我们拒绝 H₀。根据 t(9) 分布计算得到的单尾 p 值大约为 0.048。
This provides moderate evidence that the new bulb’s mean lifetime exceeds 8000 hours. The result is statistically significant at the 5% level, although the margin is small. In practice, we would report the p‑value alongside the confidence interval to give a complete picture.
这为新灯泡的平均寿命超过 8000 小时提供了中等程度的证据。结果在 5% 水平上统计显著，尽管幅度很小。在实践中，我们会同时报告 p 值和置信区间，以呈现完整的图景。
7. Interpreting p-values and Decisions | 解读 p 值与决策
The apparent contradiction between the two-sided CI and the one‑sided test is common. The 95% CI corresponds to a two‑tailed test with α = 0.05 split into 0.025 in each tail; a one‑tailed test puts the entire α in the direction of the alternative. Hence the one‑tailed test can be significant even when the two‑sided interval barely includes the null value.
双侧置信区间与单侧检验之间的表面矛盾很常见。95% 的置信区间对应的是双尾检验，α = 0.05 被分配到左右两尾各 0.025；而单尾检验将整个 α 放在备择假设的方向上。因此，即使双侧区间刚刚包含原假设值，单侧检验也可能是显著的。
A p‑value of 0.048 means that, assuming the true mean is exactly 8000, the chance of observing a sample mean of 8085 or higher is 4.8%. Because this is below 5%, we deem the result unlikely under H₀. Always remember to state the conclusion in context: ‘There is sufficient evidence at the 5% level to support the manufacturer’s claim.’
p 值为 0.048 意味着，假设真实均值恰好为 8000，观察到样本均值为 8085 或更高的概率是 4.8%。由于该概率低于 5%，我们认为在 H₀ 下这一结果不太可能发生。始终记得在上下文里陈述结论：“在 5% 水平上有充分证据支持制造商的声明。”
8. Two-Sample Comparison Setup | 双样本比较的准备
To evaluate whether the new bulb genuinely outperforms the standard one, we shift to a two‑independent‑samples design. The research question: ‘Is there a significant difference in mean lifetimes between the two bulb types?’ Hypotheses are H₀:
Published by TutorHao | Year 13 统计 Revision Series | aleveler.com
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