📚 Interdisciplinary Applied Problems for AQA Statistics | AQA 统计跨学科综合题型训练
Interdisciplinary questions in AQA Statistics require you to apply statistical methods — hypothesis tests, correlation, probability distributions — in real-world contexts spanning biology, psychology, business, and beyond. This article provides structured practice across 12 applied scenarios, each carefully matched to the AQA S1 and S2 specification, to help you recognise which technique to use and how to justify your choice under exam conditions.
AQA 统计中的跨学科问题要求你在生物学、心理学、商业等多个真实情境中灵活运用假设检验、相关性、概率分布等统计方法。本文围绕 AQA S1 和 S2 考试大纲设计了 12 个应用场景，帮助你识别不同背景下应选择的统计工具，并学会在考试要求下清晰地表述推理过程。
1. Biology: Paired t‑test for Drug Efficacy | 生物学：使用配对 t 检验比较药物疗效
A pharmaceutical company measures blood pressure reduction (mmHg) before and after a new drug on 12 patients. Differences are calculated as after − before. The mean difference d̄ = 5.8 mmHg, with standard deviation sd = 4.2 mmHg. Test at the 1% significance level whether the drug significantly lowers blood pressure.
一家制药公司在 12 名患者身上测量了服用新药前后的血压降幅 (mmHg)，差值定义为服药后 − 服药前。差值均值 d̄ = 5.8 mmHg，标准差 sd = 4.2 mmHg。在 1% 显著性水平下检验药物是否显著降低血压。
We choose a paired t‑test because the same patients are measured twice. Let μd be the population mean difference. H₀: μd = 0; H₁: μd > 0 (one‑tailed, as we test for a reduction). The test statistic is t = (d̄ − 0) / (sd / √n) = 5.8 / (4.2/√12) ≈ 4.784. Degrees of freedom = 11. The critical value from t‑tables at 1% (one‑tail) is approximately 2.718. Since 4.784 > 2.718, we reject H₀. There is sufficient evidence to conclude the drug lowers blood pressure.
应使用配对 t 检验，因为同一组患者被测量两次。设 μd 为总体平均差值。H₀: μd = 0；H₁: μd > 0（单尾，检验是否降低）。检验统计量 t = (d̄ − 0) / (sd / √n) = 5.8 / (4.2/√12) ≈ 4.784。自由度 = 11。查 t 分布表，1% 单尾临界值约为 2.718。由于 4.784 > 2.718，拒绝 H₀。有足够证据表明该药物能降低血压。
2. Psychology: Correlation between Study Time and Test Scores | 心理学：学习时间与考试分数的相关性
A psychologist collects data from 25 students on revision hours (x) and test marks (y). The summary statistics are: Σx = 520, Σy = 1640, Σx² = 14 200, Σy² = 113 800, Σxy = 36 200. Is the product‑moment correlation coefficient significantly different from zero at the 5% level?
一位心理学家收集了 25 名学生的复习时间 (x) 与考试成绩 (y) 数据。汇总统计量为：Σx = 520，Σy = 1640，Σx² = 14 200，Σy² = 113 800，Σxy = 36 200。检验积矩相关系数在 5% 水平下是否显著不等于零。
First, compute r using the formula. Sxx = Σx² − (Σx)²/n = 14 200 − 520²/25 = 14 200 − 10 816 = 3 384. Syy = 113 800 − 1640²/25 = 113 800 − 107 584 = 6 216. Sxy = 36 200 − (520×1640)/25 = 36 200 − 34 112 = 2 088. Hence r = 2 088 / √(3 384 × 6 216) ≈ 2 088 / √21 034 944 ≈ 0.455. For a two‑tailed test of H₀: ρ = 0 with n = 25, degrees of freedom = 23. The critical value from the PMCC table at 5% is 0.396. Because 0.455 > 0.396, we reject H₀. There is evidence of a linear relationship between revision hours and marks.
先用公式计算 r。Sxx = 14 200 − 520²/25 = 3 384，Syy = 113 800 − 1640²/25 = 6 216，Sxy = 36 200 − (520×1640)/25 = 2 088。故 r = 2 088 / √(3 384 × 6 216) ≈ 0.455。双尾检验 H₀: ρ = 0，样本量 n = 25，自由度 23。查积矩相关系数表，5% 临界值为 0.396。因 0.455 > 0.396，拒绝 H₀。有证据表明复习时间与考试分数之间存在线性关系。
3. Business: Time Series Analysis and Moving Averages for Sales | 商业：时间序列分析与移动平均预测销售额
Quarterly sales data (£’000) for a clothing store over two years are: Q1 52, Q2 68, Q3 84, Q4 58, Q1 56, Q2 74, Q3 90, Q4 64. Calculate four‑point centred moving averages to estimate the trend and then derive rough seasonal variations.
一家服装店两年内季度销售额（千英镑）为：Q1 52, Q2 68, Q3 84, Q4 58, Q1 56, Q2 74, Q3 90, Q4 64。计算四点中心化移动平均以估计趋势，并求出粗略的季节性变化。
The first four‑point moving total for Q1–Q4 is 52+68+84+58 = 262, average 65.5. Position this between Q2 and Q3. Next four‑point (Q2–Q1 next year) 68+84+58+56 = 266, average 66.5, positioned between Q3 and Q4. The centred moving average for Q3 is (65.5+66.5)/2 = 66.0. Continue: next total 84+58+56+74 = 272, avg 68.0; then 58+56+74+90 = 278, avg 69.5; centred for Q4: (66.5+68.0)/2=67.25? Wait, careful adjustment. Let’s compute systematically. Moving averages (two‑point centred) for each quarter from Q3 year 1 onwards: Q3 y1: (65.5+66.5)/2=66.0; Q4 y1: (66.5+68.0)/2=67.25; Q1 y2: (68.0+69.5)/2=68.75; Q2 y2: (69.5+71.0)/2=70.25 (next total 56+74+90+64=284, avg 71.0, so 69.5 and 71.0 centred). Then seasonal variation = actual − trend. For Q3, 84 − 66.0 = +18; Q4, 58 − 67.25 = −9.25; Q1, 56 − 68.75 = −12.75; Q2, 74 − 70.25 = +3.75. Average by quarter to refine seasonal effects.
第一个四点移动总和 Q1–Q4 为 262，平均 65.5，定位在 Q2 与 Q3 之间。下一个四点（Q2–下年 Q1）为 266，平均 66.5，定位在 Q3 与 Q4 之间。Q3 的中心化移动平均为 (65.5+66.5)/2=66.0。继续：Q3 y1 66.0；Q4 y1 (66.5+68.0)/2=67.25；Q1 y2 (68.0+69.5)/2=68.75；Q2 y2 (69.5+71.0)/2=70.25。季节性变化 = 实际值 − 趋势值。Q3: 84−66.0=+18；Q4: 58−67.25=−9.25；Q1: 56−68.75=−12.75；Q2: 74−70.25=+3.75。按季度求平均值即可得到季节性因素。
4. Geography: Poisson Distribution for Volcanic Eruptions | 地理：泊松分布与火山喷发次数
A region has experienced an average of 2.4 volcanic eruptions per decade over the last century. Assuming a Poisson model, find the probability that exactly 1 eruption occurs in the next decade, and the probability of more than 3 eruptions.
某地区在过去一百年中平均每十年经历 2.4 次火山喷发。假定服从泊松分布，求下一个十年恰好发生 1 次喷发的概率，以及喷发次数超过 3 次的概率。
Let X ~ Po(2.4). P(X = x) = (e⁻²·⁴ × 2.4ˣ) / x! . P(X = 1) = e⁻²·⁴ × 2.4¹ / 1! ≈ 0.0907 × 2.4 ≈ 0.2177. For P(X > 3), use complement: 1 − P(X ≤ 3). Compute P(0) = e⁻²·⁴ ≈ 0.0907; P(1) = 0.2177; P(2) = e⁻²·⁴ × 2.4² / 2 ≈ 0.0907 × 2.88 ≈ 0.2612; P(3) = e⁻²·⁴ × 2.4³ / 6 ≈ 0.0907 × 13.824/6 ≈ 0.2089. Sum = 0.0907+0.2177+0.2612+0.2089 = 0.7785. So P(X>3) = 1 − 0.7785 = 0.2215. A Poisson model is appropriate for rare, independent events in a fixed interval.
设 X ~ Po(2.4)。P(X = 1) = e⁻²·⁴ × 2.4 / 1! ≈ 0.2177。对于 P(X > 3)，使用补集：1 − P(X ≤ 3)。计算 P(0)≈0.0907，P(1)≈0.2177，P(2)≈0.2612，P(3)≈0.2089，总和为 0.7785。故 P(X>3)=0.2215。泊松模型适用于固定区间内稀有的、独立的事件。
5. Economics: Chi‑Squared Test of Independence for Employment Sectors | 经济学：卡方独立性检验与就业领域
A survey classifies 300 workers by sector (Manufacturing, Services, Agriculture) and job satisfaction (Satisfied, Neutral, Dissatisfied). Observed frequencies are given in the table. Test whether satisfaction is independent of sector at the 5% level.
| Manuf. | Services | Agri. | Total | |
|---|---|---|---|---|
| Satisfied | 42 | 55 | 23 | 120 |
| Neutral | 28 | 30 | 22 | 80 |
| Dissatisfied | 30 | 35 | 35 | 100 |
| Total | 100 | 120 | 80 | 300 |
Expected frequencies are (row total × column total) / grand total. For satisfied in manufacturing: 120×100/300 = 40. Similarly, compute all expected values, then the chi‑squared statistic χ² = Σ (O−E)²/E. The degrees of freedom = (r−1)(c−1) = 2×2 = 4. With given O and E, χ² ≈ 6.42. The 5% critical value for χ²(4) is 9.488. Since 6.42 < 9.488, we do not reject H₀. There is insufficient evidence to conclude satisfaction depends on sector.
期望频数 = (行合计 × 列合计) / 总合计。制造业满意的期望值 = 120×100/300 = 40。类似计算其他期望值，得 χ² ≈ 6.42。自由度 = (3−1)×(3−1) = 4。在 5% 水平下 χ²(4) 的临界值为 9.488。因 6.42 < 9.488，不拒绝 H₀。没有足够证据表明满意度与行业有关。
6. Sports Science: Normal Distribution and Performance Percentiles | 体育科学：正态分布与运动表现百分位数
The times for 100 m sprint in a youth athletics squad are normally distributed with mean 11.2 seconds and standard deviation 0.5 s. The top 10% of the squad receive advanced training. What is the maximum time required to qualify?
某青年田径训练队的 100 米短跑时间服从正态分布，均值为 11.2 秒，标准差为 0.5 秒。前 10% 的队员将接受高级训练，需要的最慢达标时间是多少？
Let X ~ N(11.2, 0.5²). We need the 10th percentile, i.e. the time t such that P(X < t) = 0.10. Standardise: P(Z < (t − 11.2)/0.5) = 0.10. From tables, the z‑value corresponding to Φ(z)=0.10 is approximately −1.2816. So (t − 11.2)/0.5 = −1.2816 → t = 11.2 − 1.2816×0.5 = 10.559 seconds. Thus athletes must run faster than 10.56 s to be in the top 10%.
设 X ~ N(11.2, 0.5²)。求第 10 百分位数，即 P(X < t) = 0.10 的 t。标准化得 P(Z < (t−11.2)/0.5) = 0.10。查表得 z = −1.2816。于是 t = 11.2 − 1.2816×0.5 = 10.559 秒。因此运动员需跑进 10.56 秒以内方可进入前 10%。
7. Medicine: Bayes’ Theorem and Diagnostic Testing | 医学：贝叶斯定理与诊断检测
A disease affects 0.5% of a population. A test gives a positive result for 98% of those with the disease and a positive result for 3% of those without. If a randomly chosen person tests positive, what is the probability they actually have the disease?
某种疾病在人群中的患病率为 0.5%。患者中有 98% 检测呈阳性，非患者中有 3% 检测呈阳性。若随机一人检测呈阳性，此人确实患病的概率是多少？
Let D be “has disease” and + be “positive test”. P(D)=0.005, P(+|D)=0.98, P(+|D′)=0.03. By Bayes, P(D|+) = [P(+|D)P(D)] / [P(+|D)P(D) + P(+|D′)P(D′)] = (0.98×0.005) / (0.98×0.005 + 0.03×0.995) = 0.0049 / (0.0049 + 0.02985) ≈ 0.141. Remarkably, even with a positive test, there is only about 14.1% chance of having the disease, highlighting the effect of low prevalence.
设 D 为患病事件，+ 为检测阳性。P(D)=0.005，P(+|D)=0.98，P(+|D′)=0.03。由贝叶斯定理，P(D|+) = (0.98×0.005) / (0.98×0.005 + 0.03×0.995) ≈ 0.141。即使检测呈阳性，实际患病的概率仅约 14.1%，可见低患病率对结果的影响。
8. Engineering: Binomial Distribution for Component Reliability | 工程：二项分布与组件可靠性
A circuit has 8 identical components, each with a 0.95 probability of functioning properly. The circuit works if at least 6 components function. Find the probability the circuit fails.
一个电路包含 8 个相同组件，每个组件正常工作的概率为 0.95。至少 6 个组件工作时电路才正常。求电路失灵的概率。
Let X ~ B(8, 0.95) count functioning components. Circuit fails if X ≤ 5. P(fail) = P(X ≤ 5) = 1 − P(X ≥ 6). Using tables or computation: P(6) = C(8,6)×0.95⁶×0.05² ≈ 0.0515; P(7) = C(8,7)×0.95⁷×0.05 ≈ 0.2793; P(8) = 0.95⁸ ≈ 0.6634. Sum of P(6)+P(7)+P(8) ≈ 0.9942, so P(fail) ≈ 1 − 0.9942 = 0.0058. The probability of failure is extremely small, about 0.58%.
设 X ~ B(8, 0.95) 为工作正常的组件数。电路失灵当 X ≤ 5。P(失灵) = 1 − P(X≥6)。计算 P(6)≈0.0515，P(7)≈0.2793，P(8)≈0.6634，总和 ≈0.9942，故失灵概率 ≈0.0058，仅为 0.58%。
9. Environmental Science: Mann–Whitney U Test for Soil pH | 环境科学：Mann–Whitney U 检验比较土壤 pH 值
Two independent soil samples from different forests yield pH readings: Forest A: 5.2, 4.8, 5.5, 4.9; Forest B: 6.1, 5.7, 6.4, 5.9, 6.0. Use a Mann–Whitney U test at the 5% level to determine whether the pH differs between forests.
两个独立土壤样本的 pH 值：森林 A – 5.2, 4.8, 5.5, 4.9；森林 B – 6.1, 5.7, 6.4, 5.9, 6.0。采用 Mann–Whitney U 检验在 5% 水平下判断两森林的土壤 pH 是否不同。
Combine and rank all 9 observations: 4.8(1), 4.9(2), 5.2(3), 5.5(4), 5.7(5), 5.9(6), 6.0(7), 6.1(8), 6.4(9). Sum of ranks for Forest A (nA=4): RA = 1+2+3+4 = 10. UA = nAnB + nA(nA+1)/2 − RA = 4×5 + 4×5/2 − 10 = 20 + 10 − 10 = 20. UB = nAnB − UA = 20 − 20 = 0. The test statistic U = min(20, 0) = 0. For a two‑tailed test at 5% with nA=4, nB=5, the critical value from tables is 1. Since 0 ≤ 1, we reject H₀. There is a significant difference in pH.
合并排序：4.8(1), 4.9(2), 5.2(3), 5.5(4), 5.7(5), 5.9(6), 6.0(7), 6.1(8), 6.4(9)。森林 A 的秩和 RA=10。UA = 4×5 + 4×5/2 − 10 = 20。U = min(UA, UB) = 0。查表，nA=4, nB=5 双尾 5% 临界值为 1。0 ≤ 1，拒绝 H₀。两森林 pH 存在显著差异。
10. Sociology: Sampling and Confidence Interval for a Proportion | 社会学：抽样方法与比例的置信区间
A sociologist surveys a random sample of 200 households and finds that 64 subscribe to a streaming service. Construct a 95% confidence interval for the true proportion of households in
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