Year 12 WJEC Statistics: Cross-disciplinary Integrated Problem Practice

📚 Year 12 WJEC Statistics: Cross-disciplinary Integrated Problem Practice | 跨学科综合题型训练

Welcome to an integrated revision session designed to sharpen your WJEC Year 12 Statistics skills through cross-disciplinary problem-solving. Real-world applications in biology, economics, business, psychology, and beyond provide the perfect testing ground for the statistical tools you have studied—probability distributions, hypothesis testing, confidence intervals, correlation, and regression. This article presents a series of interdisciplinary worked examples to help you bridge theory and practice, exactly as examiners expect.

欢迎来到为提升 WJEC Year 12 统计技能而设计的综合复习课程。通过生物学、经济学、商业、心理学等领域的跨学科问题解决，你将把所学的概率分布、假设检验、置信区间、相关与回归等统计工具付诸实践。本文提供一系列跨学科范例题，帮助你在理论与应用之间搭建桥梁，贴近考试要求。

1. Mendelian Genetics & Chi-Squared Goodness-of-Fit | 孟德尔遗传与卡方拟合度检验

In a dihybrid cross of pea plants, a genetic theory predicts a 9:3:3:1 phenotypic ratio for round yellow, round green, wrinkled yellow, and wrinkled green. An experiment yields observed counts: 315, 108, 101, and 32. Use a chi-squared goodness-of-fit test at the 5% significance level to assess whether the data support the Mendelian model.

在豌豆的双因子杂交实验中，遗传理论预测圆黄、圆绿、皱黄、皱绿的表型比例为9:3:3:1。实验结果观测到以下数量：315, 108, 101, 32。请使用卡方拟合度检验，在5%显著性水平下评估数据是否支持孟德尔模型。

The null hypothesis H₀ states that the observed frequencies follow the 9:3:3:1 ratio; the alternative H₁ states they do not. The total sample size is 556. Expected frequencies are calculated as 556 × (9/16) = 312.75, 556 × (3/16) = 104.25, 556 × (3/16) = 104.25, and 556 × (1/16) = 34.75.

原假设 H₀ 为观测频数符合9:3:3:1比例；备择假设 H₁ 为不符合。样本总数为556。期望频数计算为 556 × (9/16) = 312.75，556 × (3/16) = 104.25，556 × (3/16) = 104.25，556 × (1/16) = 34.75。

Phenotype	Observed (O)	Expected (E)	(O – E)² / E
Round Yellow	315	312.75	0.016
Round Green	108	104.25	0.135
Wrinkled Yellow	101	104.25	0.101
Wrinkled Green	32	34.75	0.218

The test statistic is computed using the formula below. Summing the contributions gives χ² ≈ 0.016 + 0.135 + 0.101 + 0.218 = 0.47. Degrees of freedom = number of categories − 1 = 3.

检验统计量利用下方公式计算。将各贡献值相加得到 χ² ≈ 0.016 + 0.135 + 0.101 + 0.218 = 0.47。自由度 = 类别数 − 1 = 3。

χ² = Σ (Oᵢ – Eᵢ)² / Eᵢ

The critical value from the χ²(3) distribution at the 5% level is 7.815. Since 0.47 < 7.815, we do not reject H₀. The data are consistent with the predicted Mendelian ratio, and there is no evidence of genetic linkage or deviation.

在5%显著性水平下，χ²(3)分布的临界值为7.815。由于0.47 < 7.815，我们不拒绝原假设。数据与预期的孟德尔比例一致，没有证据表明存在遗传连锁或偏差。

2. Economics: Correlation between Price and Demand | 经济学：价格与需求的相关性

A small business records the unit price (£) and weekly demand (units sold) for a new product over 6 weeks. The data are: price 5, 6, 7, 8, 9, 10; demand 180, 168, 155, 140, 132, 120. Investigate the linear relationship by calculating the product moment correlation coefficient and testing its significance.

一家小企业记录了新产品6周内的单价（英镑）和每周需求（销售量）。数据如下：价格 5, 6, 7, 8, 9, 10；需求 180, 168, 155, 140, 132, 120。通过计算积矩相关系数并检验其显著性，研究两者之间的线性关系。

Let x represent price and y represent demand. The summary statistics are: n = 6, Σx = 45, Σy = 895, Σx² = 355, Σy² = 135449, Σxy = 6515. The formula for r is centered below.

设 x 表示价格，y 表示需求。汇总统计量：n = 6, Σx = 45, Σy = 895, Σx² = 355, Σy² = 135449, Σxy = 6515。r 的计算公式居中如下。

r = [n Σxᵢ yᵢ – (Σxᵢ)(Σyᵢ)] / √[ (n Σxᵢ² – (Σxᵢ)²) (n Σyᵢ² – (Σyᵢ)²) ]

Substituting the values yields r = [6×6515 – 45×895] / √[(6×355 – 45²)×(6×135449 – 895²)] = (39090 – 40275) / √[(2130 – 2025)×(812694 – 801025)] = -1185 / √[105 × 11669] ≈ -1185 / 1107.2 ≈ -1.07. The correlation is effectively −1, indicating a perfect negative linear association—higher price strongly predicts lower demand.

代入数值得到 r = [6×6515 – 45×895] / √[(6×355 – 45²)×(6×135449 – 895²)] = (39090 – 40275) / √[(2130 – 2025)×(812694 – 801025)] = -1185 / √[105 × 11669] ≈ -1185 / 1107.2 ≈ -1.07。相关系数接近−1，表明完全负线性相关——更高的价格强烈预示着更低的需求。

To test H₀: ρ = 0 against H₁: ρ < 0 at the 5% level, we use the t-test with 4 degrees of freedom. The critical value for a one-tailed test is t₄ = −2.132. The observed test statistic t = r√(n-2)/√(1-r²) is extremely large in magnitude; thus we reject H₀ and conclude that a significant negative correlation exists. This insight helps the business set optimal pricing.

在5%水平下检验 H₀: ρ = 0 对 H₁: ρ < 0，使用自由度为4的 t 检验。单尾检验的临界值为 t₄ = −2.132。观测到的检验统计量 t = r√(n-2)/√(1-r²) 绝对值极大；因此我们拒绝 H₀，得出结论：存在显著的负相关。这一洞察有助于企业制定最优定价。

3. Business Quality Control: Confidence Interval for a Proportion | 商业质量控制：比例的置信区间

A factory randomly inspects 200 items from a production line and finds 12 defective items. Management wants a 95% confidence interval for the true proportion of defective items to decide whether the process meets the target of at most 5% defectives.

工厂从生产线上随机抽取200个产品进行检验，发现12个不合格品。管理层希望得到真实不合格率的95%置信区间，以决定该工艺是否满足不合格率至多5%的目标。

The sample proportion is p̂ = 12/200 = 0.06. For a 95% confidence interval, the z-value is 1.96. The standard error is estimated by √(p̂(1 – p̂)/n) = √(0.06 × 0.94 / 200) ≈ 0.0168.

样本比例为 p̂ = 12/200 = 0.06。对于95%置信区间，z 值为1.96。标准误差的估计为 √(p̂(1 – p̂)/n) = √(0.06 × 0.94 / 200) ≈ 0.0168。

CI: p̂ ± z* √(p̂(1 – p̂)/n)

The confidence limits are 0.06 ± 1.96 × 0.0168, giving (0.027, 0.093). Since the entire interval is below 0.10 but the lower bound is above 0, and notably the lower bound is above 0.05 for some interpretation? The interval includes 0.05, so we cannot be certain the true proportion is less than 5%. However, the upper confidence limit of 9.3% suggests the defect rate could be higher than the target. Management might consider process adjustments.

置信界限为 0.06 ± 1.96 × 0.0168，得出 (0.027, 0.093)。由于整个区间低于0.10，但下界高于0，且值得注意的是下界高于0.05的某种解释？区间包含了0.05，因此我们不能确定真实比例低于5%。然而，置信上限9.3%表明不合格率可能高于目标。管理层可能会考虑工艺调整。

A one-sided test could also be performed: H₀: p = 0.05 vs H₁: p > 0.05. The test statistic z = (0.06 – 0.05) / √(0.05×0.95/200) ≈ 0.65, with a p-value of about 0.258. There is insufficient evidence to reject H₀, so the process might still be acceptable, but the confidence interval provides richer information for decision-making.

也可执行单侧检验：H₀: p = 0.05 vs H₁: p > 0.05。检验统计量 z = (0.06 – 0.05) / √(0.05×0.95/200) ≈ 0.65，p值约为0.258。没有充分证据拒绝原假设，因此工艺可能仍可接受，但置信区间为决策提供了更丰富的信息。

4. Psychology Experiment: Two-Sample t-Test | 心理学实验：双样本 t 检验

A psychologist compares the effect of two study techniques on memory recall. Group A (n=10) uses visual mnemonics and scores a mean recall of 78 with a standard deviation of 10. Group B (n=12) uses repetition and scores a mean of 70 with a standard deviation of 9. Assuming equal population variances, test whether the mnemonics technique leads to significantly higher scores at the 1% level.

一位心理学家比较两种学习方法对记忆回忆的影响。A组（n=10）使用视觉记忆法，平均回忆得分为78，标准差为10。B组（n=12）使用重复记忆法，平均得分为70，标准差为9。假设总体方差相等，检验在1%显著性水平下视觉记忆法是否导致显著更高的得分。

We use a two-sample t-test with pooled variance. The pooled estimate of the common variance is sₚ² = [(n₁-1)s₁² + (n₂-1)s₂²] / (n₁ + n₂ – 2) = [9×100 + 11×81] / 20 = (900 + 891)/20 = 89.55. Hence sₚ ≈ 9.46.

我们使用合并方差的双样本 t 检验。共同方差的合并估计为 sₚ² = [(n₁-1)s₁² + (n₂-1)s₂²] / (n₁ + n₂ – 2) = [9×100 + 11×81] / 20 = (900 + 891)/20 = 89.55。因此 sₚ ≈ 9.46。

t = (x̄₁ – x̄₂) / [sₚ √(1/n₁ + 1/n₂)]

Substituting, t = (78 – 70) / [9.46 × √(1/10 + 1/12)] = 8 / [9.46 × √(0.1833)] ≈ 8 / [9.46 × 0.428] ≈ 8 / 4.

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Year 12 WJEC Statistics: Cross-disciplinary Integrated Problem Practice | 跨学科综合题型训练