📚 A-Level Eduqas Statistics: Interdisciplinary Integrated Question Practice | A-Level Eduqas 统计：跨学科综合题型训练
In A-Level Statistics, especially for the Eduqas specification, students are encouraged to apply statistical methods across different disciplines. This article presents ten carefully designed interdisciplinary questions, each integrating core statistical techniques with contexts from biology, psychology, economics, physics, environmental science, sociology, medicine, engineering, geography, and sports science. Every section includes a problem statement, a step-by-step solution, and bilingual explanations. Working through these will strengthen your ability to transfer statistical knowledge to real-world scenarios and prepare you thoroughly for exam questions that blend theory with application.
在A-Level统计学（Eduqas大纲）中，学生需要将统计方法应用于不同学科。本文精心设计了十道跨学科综合题，涵盖生物学、心理学、经济学、物理学、环境科学、社会学、医学、工程学、地理学和运动科学。每节包含一道题目、分步解答和双语讲解。通过练习这些题目，你可以提升将统计知识迁移到真实情境的能力，为考试中的综合应用题做好全面准备。
1. Biology: Comparing Drug Effectiveness | 生物学：药物有效性比较
A pharmaceutical company claims that a new drug is more effective than the standard treatment. In a trial, 210 out of 300 patients recovered using the new drug, while 190 out of 300 recovered using the standard treatment. Test at the 5% significance level whether the new drug has a higher recovery rate. State the hypotheses, test statistic, and conclusion clearly.
一家制药公司声称新药比标准治疗更有效。在一项试验中，使用新药的300名患者中有210人康复，而使用标准治疗300名患者中有190人康复。在5%显著性水平下检验新药的康复率是否更高。清晰陈述假设、检验统计量和结论。
Let p₁ be the recovery rate for the new drug and p₂ for the standard treatment. H₀: p₁ = p₂, H₁: p₁ > p₂. The pooled proportion p̂ = (210+190)/(300+300) = 400/600 = 2/3. The test statistic is z = (p̂₁ − p̂₂) / √(p̂(1−p̂)(1/n₁ + 1/n₂)) where p̂₁ = 0.7, p̂₂ = 0.6333. Calculation gives z = (0.7 − 0.6333) / √((2/3 × 1/3)(1/300 + 1/300)) ≈ 1.73. The critical value for a one-tailed test at 5% is 1.645. Since 1.73 > 1.645, we reject H₀. There is sufficient evidence that the new drug has a higher recovery rate.
设新药康复率为p₁，标准治疗为p₂。H₀: p₁ = p₂，H₁: p₁ > p₂。合并比例p̂ = (210+190)/(300+300) = 400/600 = 2/3。检验统计量z = (p̂₁ − p̂₂) / √(p̂(1−p̂)(1/n₁ + 1/n₂))，其中p̂₁ = 0.7, p̂₂ = 0.6333。计算得z ≈ 1.73。单侧检验5%临界值1.645。因为1.73 > 1.645，拒绝H₀。有充分证据表明新药康复率更高。
2. Psychology: Paired Memory Test | 心理学：配对记忆测试
A psychologist investigates whether a memory training programme improves recall. Eight participants are tested before and after the programme, yielding the following scores: differences (after − before): 3, −1, 4, 2, 0, 5, 3, 1. Assuming differences are normally distributed, test at the 1% significance level whether the programme is effective. Report the paired t-test result.
一位心理学家研究记忆训练项目是否提高回忆能力。八名参与者分别在训练前后测试，差值（后减前）为：3, −1, 4, 2, 0, 5, 3, 1。假设差值服从正态分布，在1%显著性水平下检验训练是否有效。给出配对t检验结果。
H₀: μ_d = 0, H₁: μ_d > 0. Mean difference d̄ = (3−1+4+2+0+5+3+1)/8 = 17/8 = 2.125. Sample standard deviation s_d = √((Σ(d−d̄)²)/(n−1)) = √(( (0.875)²+(−3.125)²+(1.875)²+(−0.125)²+(−2.125)²+(2.875)²+(0.875)²+(−1.125)² )/7) ≈ √(27.875/7) ≈ 1.996. Test statistic t = d̄/(s_d/√n) = 2.125/(1.996/√8) ≈ 3.01. Degrees of freedom = 7, critical value t₀.₀₁,₇ (one-tailed) ≈ 2.998. Since 3.01 > 2.998, reject H₀. The programme significantly improves recall.
H₀: μ_d = 0，H₁: μ_d > 0。均值差d̄ = 2.125。样本标准差s_d ≈ 1.996。检验统计量t = 2.125/(1.996/√8) ≈ 3.01。自由度7，单侧1%临界值约2.998。因3.01 > 2.998，拒绝H₀。训练显著提高回忆能力。
3. Economics: Interest Rate and Investment | 经济学：利率与投资
An economist collects annual data on interest rate (x) and investment volume in billion pounds (y) for 10 years. Summary statistics: Σx = 45, Σy = 520, Σx² = 285, Σy² = 29840, Σxy = 2040. (a) Find the equation of the regression line of y on x. (b) Test whether the slope is significantly less than zero at the 5% level, using a t-test for the slope coefficient.
一位经济学家收集了10年间利率(x)与投资额（十亿英镑）(y)的年数据。汇总统计量：Σx = 45, Σy = 520, Σx² = 285, Σy² = 29840, Σxy = 2040。(a) 求y对x的回归直线方程。(b) 在5%显著性水平下检验斜率是否显著小于零，使用斜率系数的t检验。
(a) n = 10, S_xx = Σx² − (Σx)²/n = 285 − 45²/10 = 285 − 202.5 = 82.5. S_xy = Σxy − (Σx)(Σy)/n = 2040 − (45×520)/10 = 2040 − 2340 = −300. Slope b = S_xy / S_xx = −300/82.5 ≈ −3.636. Intercept a = ȳ − b x̄ = (520/10) − (−3.636)(45/10) = 52 + 16.364 = 68.364. Regression line: y = 68.4 − 3.64x (3 s.f.).
(a) S_xx = 82.5，S_xy = −300。斜率b = −3.636，截距a = 68.364。回归线：y = 68.4 − 3.64x。
(b) H₀: β = 0, H₁: β < 0. Residual sum of squares: S_yy = Σy² − (Σy)²/n = 29840 − 520²/10 = 29840 − 27040 = 2800. SS_res = S_yy − b² S_xx = 2800 − (−3.636)²×82.5 ≈ 2800 − 1090.9 = 1709.1. Estimate of error variance s² = SS_res/(n−2) = 1709.1/8 = 213.64. Standard error of b: SE(b) = √(s²/S_xx) = √(213.64/82.5) ≈ √2.589 = 1.609. t = b/SE(b) = −3.636/1.609 ≈ −2.26. Degrees of freedom 8, critical value t₀.₀₅,₈ (one-tailed, lower) = −1.860. Since −2.26 < −1.860, reject H₀. Slope is significantly negative; higher interest rates tend to reduce investment.
(b) H₀: β = 0, H₁: β < 0。误差方差估计s² = 213.64，SE(b) ≈ 1.609。t = −2.26，自由度8，单侧临界值−1.860。因−2.26 < −1.860，拒绝H₀。斜率为显著负值，利率越高投资越少。
4. Physics: Measurement Error and Confidence Interval | 物理学：测量误差与置信区间
A physics student measures the acceleration due to gravity g by timing a pendulum. Ten independent measurements give a sample mean of 9.82 m/s² and a sample standard deviation of 0.15 m/s². Assuming the measurements are normally distributed, construct a 95% confidence interval for the true g. Additionally, if another student claims that true g is 9.80 m/s², does the interval support this claim?
一位物理系学生通过单摆计时测量重力加速度g。十次独立测量得到样本均值9.82 m/s²，样本标准差0.15 m/s²。假设测量值服从正态分布，构造真实g的95%置信区间。若另一位学生声称真实g为9.80 m/s²，该区间是否支持此说法？
The 95% CI for the population mean is x̄ ± t_α/2, n−1 × (s/√n). Here n = 10, degrees of freedom 9, t₀.₀₂₅,₉ = 2.262. Standard error = 0.15/√10 ≈ 0.0474. Margin of error = 2.262 × 0.0474 ≈ 0.107. CI: (9.82 − 0.107, 9.82 + 0.107) = (9.713, 9.927). Because 9.80 lies within this interval, the data do not contradict the claim at the 95% confidence level.
总体均值的95%置信区间为x̄ ± t_α/2, n−1 × (s/√n)。n=10，自由度9，t₀.₀₂₅,₉ = 2.262。标准误 = 0.0474，边际误差 ≈ 0.107。区间：(9.713, 9.927)。9.80落在此区间内，故数据在95%置信水平下不反驳该声称。
5. Environmental Science: Poisson Fit for Rare Species | 环境科学：稀有物种的泊松拟合
An ecologist counts the number of a rare plant in 40 quadrats of equal size. Observed frequencies: 0 plants in 12 quadrats, 1 plant in 15, 2 plants in 8, 3 plants in 4, and 4 or more in 1 quadrat. Test at the 5% level whether the number per quadrat follows a Poisson distribution. Use mean from the data.
一位生态学家在40个等面积样方中计数某种稀有植物。观测频数：0株（12个样方），1株（15个），2株（8个），3株（4个），4株及以上（1个）。在5%显著性水平下检验每个样方的株数是否服从泊松分布。使用从数据估计的均值。
Total plants = 0×12 + 1×15 + 2×8 + 3×4 + 4×1 = 0+15+16+12+4 = 47. Mean λ̂ = 47/40 = 1.175. Poisson probabilities: P(X=0) = e⁻¹·¹⁷⁵ = 0.3088, P(X=1)= 1.175×0.3088 = 0.3628, P(X=2)= (1.175²/2)×0.3088 = 0.2132, P(X=3)= (1.175³/6)×0.3088 = 0.0835, P(X≥4) = 1 − (0.3088+0.3628+0.2132+0.0835) = 0.0317. Expected frequencies: 0:40×0.3088=12.35, 1:14.51, 2:8.53, 3:3.34, ≥4:1.27. Combine last two groups (expected <5): 3 or more: expected 3.34+1.27=4.61, observed 5. Chi-squared contributions: (12−12.35)²/12.35 = 0.0099, (15−14.51)²/14.51 = 0.0165, (8−8.53)²/8.53 = 0.0330, (5−4.61)²/4.61 = 0.0330. Total X² = 0.0924. Degrees of freedom: 4 groups − 1 − 1 (estimated λ) = 2. Critical value χ²₀.₀₅,₂ = 5.991. 0.0924 < 5.991, do not reject H₀. Data consistent with Poisson distribution.
总植株数47，均值λ̂ = 1.175。泊松概率和期望频数如表，合并后四组。卡方统计量X² ≈ 0.0924。自由度 = 4−1−1 = 2。临界值5.991。0.0924 < 5.991，不拒绝H₀。数据与泊松分布一致。
6. Sociology: Gender and Voting Preference | 社会学：性别与投票偏好
A survey on voting preference (Party A, B, C) and gender yields the following table. Test at the 1% significance level whether gender and voting preference are independent.
| Party A | Party B | Party C | Total | |
|---|---|---|---|---|
| Male | 50 | 70 | 30 | 150 |
| Female | 60 | 50 | 40 | 150 |
| Total | 110 | 120 | 70 | 300 |
一份关于投票偏好（党派A、B、C）与性别的调查数据如上表。在1%显著性水平下检验性别与投票偏好是否独立。
H₀: gender and party preference independent; H₁: not independent. Expected frequency for each cell = (row total × column total)/grand total. E.g., Male-Party A: 150×110/300 = 55. Similarly, Male-B: 150×120/300 = 60, Male-C: 150×70/300 = 35. Female-A: 150×110/300 = 55, Female-B: 60, Female-C: 35. Chi-squared contributions: (50−55)²/55=0.455, (70−60)²/60=1.667, (30−35)²/35=0.714, (60−55)²/55=0.455, (50−60)²/60=1.667, (40−35)²/35=0.714. Sum X² = 5.672. Degrees of freedom = (2−1)×(3−1)=2. Critical value χ²₀.₀₁,₂ = 9.210. Since 5.672 < 9.210, do not reject H₀. Insufficient evidence to suggest association.
H₀：性别与偏好独立。期望频数如上。卡方统计量X² = 5.672，自由度2，1%临界值9.210。5.672 < 9.210，不拒绝H₀。没有足够证据表明存在关联。
7. Medicine: Diagnostic Test Evaluation | 医学：诊断测试评估
A new diagnostic test for a disease has a sensitivity of 95% and specificity of 90%. The disease prevalence in the population is 2%. If a randomly selected person tests positive, what is the probability that they actually have the disease? Use Bayes’ theorem and interpret the result.
一种新诊断测试的灵敏度为95%，特异度为90%。该病在人群中的患病率为2%。若随机一人检测呈阳性，其真实患病的概率是多少？用贝叶斯定理计算并解释结果。
Let D be having the disease, + be positive test. P(D)=0.02, P(+|D)=0.95, P(−|D’)=0.90 → P(+|D’)=0.10. P(+) = P(+|D)P(D) + P(+|D’)P(D’) = 0.95×0.02 + 0.10×0.98 = 0.019 + 0.098 = 0.117. Then P(D|+) = P(+|D)P(D) / P(+) = 0.019 / 0.117 ≈ 0.1624. Even with a positive test, the probability of having the disease is only about 16.2%, due to low prevalence – an important consideration in medical decision-making.
设D为患病，+为阳性。P(D)=0.02，P(+|D)=0.95，P(+|D’)=0.10。P(+) = 0.019 + 0.098 = 0.117。P(D|+) = 0.019/0.117 ≈ 0.1624。即使检测阳性，患病概率仅约16.2%，原因在于患病率低——这是医学决策的重要考量。
8. Engineering: Lifetime of Components | 工程学：元件寿命
The lifetime (in thousands of hours) of a certain electronic component follows an exponential distribution with unknown mean μ. A sample of 20 components is tested, and the total lifetime sum is 68.2 thousand hours. Test at the 5% level whether the mean lifetime is less than 4 thousand hours. (For an exponential distribution, the sum is Gamma; use the chi-squared distribution: 2n X̄ / μ ~ χ²(2n) under H₀).
某种电子元件的寿命（千小时）服从指数分布，均值μ未知。测试20个元件，总寿命和为68.2千小时。在5%显著性水平下检验平均寿命是否小于4千小时。（指数分布下，2n X̄ / μ 在H₀下服从χ²(2n)分布。）
H₀: μ = 4, H₁: μ < 4. X̄ = 68.2/20 = 3.41. Test statistic: 2n X̄ / μ₀ = (2×20×3.41)/4 = 136.4/4 = 34.1. Under H₀, this statistic ~ χ²(40). For a one-tailed (lower) test, we check if statistic is too small. Reject H₀ if 2n X̄/μ₀ < χ²_lower. Critical value χ²₀.₀₅,₄₀ (lower tail) = 26.509 (since P(χ² > 55.758) = 0.05 for upper, the lower 5% point is 26.509 from tables). Alternatively, compute p-value. Since 34.1 > 26.509, do not reject H₀. No significant evidence that mean is less than 4.
H₀: μ = 4, H₁: μ < 4。X̄ = 3.41，检验统计量 = 2n X̄ / μ₀ = 34.1。H₀下服从χ²(40)。拒绝域为统计量 < χ²下侧临界值。χ²(40)的下侧5%点为26.509。34.1 > 26.509，不拒绝H₀。无显著证据支持均值低于4千小时。
9. Geography: River Flow and Rainfall | 地理学：河流流量与降雨量
A geographer studies the relationship between monthly rainfall (x mm) and river flow rate (y m³/s) at 12 locations. The Pearson correlation coefficient r is 0.68. Test at the 5% level whether there is positive correlation. Assume bivariate normality.
一位地理学家研究12个地点月降雨量(x mm)与河流流量(y m³/s)的关系。皮尔逊相关系数r为0.68。在5%显著性水平下检验是否存在正相关。假设二元正态分布。
H₀: ρ = 0, H₁: ρ > 0. Test statistic t = r √(n−2) / √(1−r²) = 0.68 × √10 / √(1−0.4624) = 0.68×3.1623 / √0.5376 ≈ 2.150 / 0.733 = 2.934. Degrees of freedom 10. Critical value t₀.₀₅,₁₀ (one-tailed) = 1.812. Since 2.934 > 1.812, reject H₀. There is significant positive correlation.
H₀: ρ = 0, H₁: ρ > 0。检验统计量t = r√(n−2)/√(1−r²) ≈ 2.934，自由度10。单侧临界值1.812。2.934 > 1.812，拒绝H₀。存在显著正相关。
10. Sports Science: Non-parametric Comparison | 运动科学：非参数比较
A coach compares two training methods by recording the number of push-ups performed after 4 weeks. Group A (n₁=8): 25, 30, 28, 35, 32, 27, 31, 29. Group B (n₂=7): 22, 24, 20, 26, 23, 21, 19. Use the Mann–Whitney U test at the 5% significance level to determine if method A yields higher results. State hypotheses and conclusion.
一位教练通过记录4周后俯卧撑完成次数比较两种训练方法。A组（n₁=8）：25, 30, 28, 35, 32, 27, 31, 29。B组（n₂=7）：22, 24, 20, 26, 23, 21, 19。使用Mann–Whitney U检验在5%显著性水平下判断方法A是否产生更高结果。陈述假设和结论。
H₀: the two population distributions are identical; H₁: A tends to have larger values (one-tailed). Combine and rank all 15 observations: 19(1), 20(2), 21(3), 22(4), 23(5), 24(6), 25(7), 26(8), 27(9), 28(10), 29(11), 30(12), 31(13), 32(14), 35(15). Sum of ranks for A: 7+12+10+15+14+9+13+11 = 91. For B: 1+2+3+4+5+6+8 = 29. U₁ = n₁n₂ + n₁(n₁+1)/2 − R₁ = 8×7 + 8×9/2 − 91 = 56 + 36 − 91 = 1. U₂ = 56 + 7×8/2 − 29 = 56+28−29 = 55. Smaller U = 1. For n₁=8, n₂=7, the critical value from tables (one-tailed 5%) is 7 (for U ≤ 7 reject). Since 1 < 7, reject H₀. Method A significantly outperforms B.
H₀：两总体分布相同；H₁：A组倾向于更大值（单侧）。混合排序，A组秩和91，B组29。U₁ = 1, U₂ = 55，较小U = 1。查表临界值（n₁=8, n₂=7, 单侧5%）为7。因1 < 7，拒绝H₀。方法A显著优于B。
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