AS Eduqas Statistics: Interdisciplinary Problem-Solving Training | AS Eduqas 统计：跨学科综合题型训练

📚 AS Eduqas Statistics: Interdisciplinary Problem-Solving Training | AS Eduqas 统计：跨学科综合题型训练

In today’s data-driven world, statistical reasoning is no longer confined to the mathematics classroom—it under pins research in biology, psychology, economics, geography and many other fields. The AS Eduqas Statistics specification expects you to apply core methods such as probability, distributions, hypothesis testing and correlation to real‑world contexts that blend several subject areas. This revision guide offers a series of integrated problem‑solving exercises designed to build your confidence in transferring your statistical tool kit across disciplinary boundaries.

在当今数据驱动的世界里，统计推理早已不再局限于数学课堂——它支撑着生物学、心理学、经济学、地理学以及其他众多领域的研究。AS Eduqas 统计大纲要求你将概率、分布、假设检验和相关分析等核心方法运用于融合了多个学科的真实情境中。本复习指南提供了一系列综合性的问题解决练习，旨在帮助你增强自信，能够在不同学科之间灵活迁移你的统计工具。

1. Genetics and Binomial Probability | 遗传学与二项概率

In Mendelian genetics, the inheritance of a recessive disorder often follows a binomial model. Consider a couple who are both carriers of cystic fibrosis (heterozygous). The probability that a child of theirs inherits the disease is 0.25. If they plan to have five children, we can find the probability that exactly two of them are affected. Let X be the number of affected children, so X ~ B(5, 0.25). P(X = 2) is calculated from the binomial formula: P(X = 2) = C(5,2) × (0.25)² × (0.75)³.

在孟德尔遗传学中，隐性疾病的遗传常常符合二项模型。假设一对夫妇都是囊性纤维化的携带者（杂合子），他们的任一孩子患病的概率为 0.25。如果他们计划生育五个孩子，我们可以计算恰好有两个孩子患病的概率。记 X 为患病的子女数，则 X ~ B(5, 0.25)。从二项公式可得：P(X = 2) = C(5,2) × (0.25)² × (0.75)³。

P(X = 2) = 10 × 0.0625 × 0.421875 = 0.2637 (4 d.p.)

Now suppose a geneticist suspects that in a specific population the disorder rate is higher than 0.25. A random sample of 20 children from carrier parents reveals 8 affected cases. Test this claim at the 5% significance level. Define H₀: p = 0.25 and H₁: p > 0.25. Under H₀, the test statistic X ~ B(20, 0.25). The observed value is 8, so the p‑value is P(X ≥ 8) = 1 − P(X ≤ 7). Using binomial cumulative tables, P(X ≤ 7) = 0.8982, giving a p‑value of 0.1018. Since 0.1018 > 0.05, we do not reject H₀; there is insufficient evidence that the proportion has increased.

现在假设一位遗传学家怀疑在某个特定人群中该病发生率高于 0.25。从携带者父母的 20 个孩子中随机抽样，发现有 8 例患病。在 5% 的显著性水平下检验这个说法。定义原假设 H₀: p = 0.25，备择假设 H₁: p > 0.25。在 H₀ 下，检验统计量 X ~ B(20, 0.25)。观测值为 8，因此 p 值 = P(X ≥ 8) = 1 − P(X ≤ 7)。查二项累积分布表得 P(X ≤ 7) = 0.8982，故 p 值 = 0.1018。由于 0.1018 > 0.05，我们不拒绝 H₀；没有足够证据表明该比例确实上升了。

2. Psychology: Sign Test as a Binomial Test | 心理学：符号检验作为二项检验

A psychologist measures the anxiety levels of 12 participants before and after a mindfulness intervention. The intention is to test whether the intervention reduces anxiety. The data are summarised as follows: 9 participants showed a decrease, 2 showed an increase, and 1 showed no change. After discarding the tie, we apply the sign test with n = 11. Let θ be the probability of a decrease. Set H₀: θ = 0.5 and H₁: θ > 0.5. Under the null hypothesis, the number of decreases X ~ B(11, 0.5). The observed number is 9.

一位心理学家测量了 12 名参与者在正念干预前后的焦虑水平，想检验该干预是否降低了焦虑。数据概括如下：9 人焦虑下降，2 人上升，1 人无变化。去掉持平的那一个，我们用符号检验，样本量 n = 11。记 θ 为焦虑下降的概率。设 H₀: θ = 0.5，H₁: θ > 0.5。在原假设下，下降的人数 X ~ B(11, 0.5)，实际观测值为 9。

Compute the p‑value: P(X ≥ 9) = P(X=9) + P(X=10) + P(X=11). Using the binomial distribution: P(X=9) = C(11,9)×(0.5)¹¹ = 55 × 0.00048828 = 0.02686, P(X=10) = 11 × 0.00048828 = 0.005371, P(X=11) = 0.0004883. Summing gives approximately 0.0327. Since 0.0327 < 0.05, we reject H₀ and conclude that the mindfulness session significantly reduces anxiety at the 5% level.

计算 p 值：P(X ≥ 9) = P(X=9) + P(X=10) + P(X=11)。利用二项分布：P(X=9) = C(11,9)×(0.5)¹¹ = 55 × 0.00048828 = 0.02686，P(X=10) = 11 × 0.00048828 = 0.005371，P(X=11) = 0.0004883。三者相加约等于 0.0327。由于 0.0327 < 0.05，我们拒绝 H₀，并得出结论：在 5% 的显著性水平上，正念课程显著降低了焦虑。

3. Economics: Correlation between Interest Rates and Inflation | 经济学：利率与通胀的相关性

An economist collects annual data for a country over eight years: nominal interest rate (x%) and inflation rate (y%). The pairs are (4.0, 2.1), (4.5, 2.5), (5.0, 3.0), (5.5, 3.2), (6.0, 3.8), (6.5, 4.0), (7.0, 4.5), (7.5, 4.8). Use a calculator or formula to find the product‑moment correlation coefficient r. The necessary sums are ∑x = 46.0, ∑y = 27.9, ∑xy = 167.65, ∑x² = 279.5, ∑y² = 103.83. Then r = [8×167.65 − 46.0×27.9] / √{[8×279.5 − 46.0²] × [8×103.83 − 27.9²]} = 0.992 (3 s.f.). This suggests a very strong positive linear correlation between nominal interest rate and inflation.

一位经济学家收集了某国八年间的年度数据：名义利率 (x%) 与通货膨胀率 (y%)。数据对为 (4.0, 2.1), (4.5, 2.5), (5.0, 3.0), (5.5, 3.2), (6.0, 3.8), (6.5, 4.0), (7.0, 4.5), (7.5, 4.8)。用公式或计算器求积差相关系数 r。所需的和为 ∑x = 46.0, ∑y = 27.9, ∑xy = 167.65, ∑x² = 279.5, ∑y² = 103.83。由此 r = [8×167.65 − 46.0×27.9] / √{[8×279.5 − 46.0²] × [8×103.83 − 27.9²]} = 0.992（3 位有效数字）。这表明名义利率与通货膨胀之间存在极强的正线性相关。

To test whether the population correlation coefficient ρ is zero, compare r = 0.992 with the critical value from the Pearson correlation table for n = 8 at the 1% level (two‑tailed test). The critical value is 0.834. Since 0.992 > 0.834, we reject H₀: ρ = 0 and conclude that there is significant evidence of linear association. This correlation might help economists forecast inflation when interest rates change.

为了检验总体相关系数 ρ 是否为 0，将 r = 0.992 与 n = 8、双侧检验 1% 显著性水平下的 Pearson 相关系数临界值表进行比较。临界值为 0.834。由于 0.992 > 0.834，我们拒绝 H₀: ρ = 0，并得出结论：有显著证据表明两者存在线性关联。这种相关性有助于经济学家在利率变动时预测通胀。

4. Geography: Is River Depth Normally Distributed? | 地理学：河流深度是否呈正态分布？

A geographer measures the depth (in cm) at 50 random points along a stream. The data are roughly symmetric with sample mean x̄ = 45 cm and sample standard deviation s = 8 cm. Assuming the depth follows a normal distribution, we can estimate the proportion of points where the depth is less than 35 cm. Standardise: z = (35 − 45) / 8 = −1.25. From standard normal tables, Φ(−1.25) = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056. So roughly 10.6% of the stream would be expected to have a depth below 35 cm.

一位地理学家沿一条溪流随机测量了 50 个点的水深（厘米）。数据大致对称，样本均值 x̄ = 45 cm，样本标准差 s = 8 cm。假设水深服从正态分布，我们可以估计水深低于 35 cm 的点所占的比例。标准化：z = (35 − 45) / 8 = −1.25。查标准正态表得，Φ(−1.25) = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056。因此，预计约有 10.6% 的溪流区域水深低于 35 cm。

Imagine a conservation agency claims the mean depth of such streams is 50 cm. Test this at the 5% significance level using the sample (n = 50, x̄ = 45, s = 8). Since σ is unknown and n is large, a z‑test is appropriate. H₀: μ = 50, H₁: μ ≠ 50. The test statistic is z = (45 − 50) / (8/√50) = −5 / 1.1314 = −4.42. The two‑tailed p‑value is extremely small (< 0.0001), so H₀ is rejected. The mean depth is significantly different from 50 cm, highlighting how statistical testing can guide environmental policy.

假设某环保机构声称这类溪流的平均水深为 50 cm。利用样本 (n = 50, x̄ = 45, s = 8) 在 5% 显著性水平下检验该说法。由于 σ 未知而样本量较大，可采用 z 检验。H₀: μ = 50，H₁: μ ≠ 50。检验统计量 z = (45 − 50) / (8/√50) = −5 / 1.1314 = −4.42。双侧 p 值极小（< 0.0001），因此拒绝 H₀。平均水深与 50 cm 存在显著差异，这体现了统计检验如何为环境政策提供依据。

5. Sports Science: Testing Mean Reaction Time | 运动科学：检验平均反应时间

A sports coach claims that a new training programme reduces athletes’ mean reaction time to less than 0.30 seconds. Historical data show that the reaction times are normally distributed with standard deviation σ = 0.04 s. A random sample of 25 athletes who completed the programme gives a mean reaction time of 0.285 s. Perform a hypothesis test at the 5% significance level.

一位运动教练声称，新的训练方案可将运动员的平均反应时间降至 0.30 秒以下。历史数据显示反应时间服从正态分布，标准差 σ = 0.04 秒。从完成该方案的运动员中随机抽取 25 名，得到样本平均反应时间为 0.285 秒。在 5% 显著性水平下进行假设检验。

Set H₀: μ = 0.30 against H₁: μ < 0.30. The test statistic is z = (0.285 − 0.30) / (0.04/√25) = −0.015 / 0.008 = −1.875. From the standard normal table, P(Z < −1.875) = 0.0304. Since 0.0304 < 0.05, we reject H₀. There is sufficient evidence to support the coach’s claim that the programme reduces mean reaction time below 0.30 seconds. This is a classic application of the one‑sample z‑test in sports science.

设 H₀: μ = 0.30，H₁: μ < 0.30。检验统计量 z = (0.285 − 0.30) / (0.04/√25) = −0.015 / 0.008 = −1.875。查标准正态表得 P(Z < −1.875) = 0.0304。由于 0.0304 < 0.05，拒绝 H₀。有充分证据支持教练的说法，即该方案将平均反应时间降至 0.30 秒以下。这是单样本 z 检验在运动科学中的一个典型应用。

6. Biology: Estimating Population Proportion with Confidence Interval | 生物学：用置信区间估计种群比例

更多咨询请联系16621398022（同微信）