Interdisciplinary Integrated Question Practice for Year 12 OCR Statistics

📚 Interdisciplinary Integrated Question Practice for Year 12 OCR Statistics | 跨学科综合题型训练

Year 12 OCR Statistics is not just about learning mathematical techniques in isolation. Many exam questions are set in real‑world, cross‑disciplinary contexts – from testing new medical treatments to analysing climate data or modelling consumer behaviour. This article provides a structured, bilingual revision resource to help you master these integrated problems. Each section pairs an English explanation with its Chinese equivalent, then presents a typical exam‑style question and a worked solution. The aim is to build your confidence in transferring statistical skills seamlessly across subjects.

OCR 统计学 Year 12 并不仅仅是孤立地学习数学技巧。许多考题被置于现实世界、跨学科的背景中——从检验新药疗效到分析气候数据，或建模消费者行为。本文提供一份结构化的双语复习资源，帮助你掌握这类综合题型。每个小节先呈现英文解释，再提供对应的中文解释，随后给出典型考题及详细解答示范。目标是让你能够自信地将统计技能自如地迁移到不同学科中。

1. Hypothesis Testing in Psychology | 心理学中的假设检验

Psychologists often use hypothesis tests to determine whether a therapy has a significant effect. For example, they might test whether a new mindfulness programme reduces anxiety scores compared with a control. The data are usually small samples from normally distributed populations, so a t‑test for two independent means is appropriate. You must state the null and alternative hypotheses, calculate the test statistic using the formula with pooled variance, find the p‑value from the t‑distribution with the correct degrees of freedom, and compare it with the significance level (commonly 5%). If the p‑value is below 0.05, the result is statistically significant and the psychologist can reject H₀.

心理学家经常使用假设检验来判断某种疗法是否具有显著效果。例如，他们可能测试一项新的正念计划是否能降低焦虑分数，并与对照组进行比较。数据通常是来自正态分布总体的小样本，因此适用两个独立样本均值的t检验。你必须陈述零假设与备择假设，使用合并方差公式计算检验统计量，根据正确的自由度查出 t 分布的 p 值，并将其与显著性水平（通常取 5%）比较。如果 p 值低于 0.05，则结果具有统计显著性，心理学研究者可以拒绝零假设。

Example Question: A psychologist measures anxiety scores for 10 patients who received mindfulness therapy and 10 who did not. The mindfulness group has mean 24.2 and standard deviation 5.1, while the control group has mean 30.6 and standard deviation 6.3. Test at the 5% level whether mindfulness reduces anxiety, assuming equal population variances.

例题： 一位心理学家测量了 10 名接受正念治疗与 10 名未接受治疗患者的焦虑得分。正念组均值为 24.2、标准差 5.1，对照组均值为 30.6、标准差 6.3。假定总体方差相等，在 5% 显著性水平下检验正念是否能降低焦虑。

Worked Solution Outline: H₀: μ₁ = μ₂, H₁: μ₁ < μ₂. Pooled variance sₚ² = (9×5.1² + 9×6.3²) / (10+10−2) ≈ 32.85. Test statistic t = (24.2 − 30.6) / (√(32.85×(1/10+1/10))) ≈ −2.50. Degrees of freedom = 18. One‑tailed critical value from t₍₁₈₎ at 5% is −1.734. Since −2.50 < −1.734, reject H₀; there is sufficient evidence that mindfulness reduces anxiety.

解答要点： H₀: μ₁ = μ₂，H₁: μ₁ < μ₂。合并方差 sₚ² = (9×5.1² + 9×6.3²) / (10+10−2) ≈ 32.85。检验统计量 t = (24.2 − 30.6) / (√(32.85×(1/10+1/10))) ≈ −2.50。自由度 = 18。t₍₁₈₎ 在 5% 显著性水平下单尾临界值为 −1.734。由于 −2.50 < −1.734，拒绝 H₀；有充分证据表明正念能够降低焦虑。

2. Chi‑squared Tests in Biology | 生物学中的卡方检验

Biologists frequently use chi‑squared tests to analyse categorical data, especially in genetics (to see if observed offspring ratios match Mendelian expectations) or in ecology (to check for association between species and habitat type). For a goodness‑of‑fit test, you compare observed frequencies with expected frequencies under the null hypothesis using the statistic χ² = Σ (O−E)² / E. The degrees of freedom are (number of categories − 1) for a simple ratio, or (rows−1)×(columns−1) for a test of association. A high χ² value relative to the critical value from the chi‑squared distribution leads to rejection of the null hypothesis.

生物学家经常使用卡方检验来分析分类数据，尤其是在遗传学中（检验观察到的子代表型比是否符合孟德尔预期）或生态学中（检验物种与栖息地类型之间的关联）。对于拟合优度检验，将观察频数与零假设下的期望频数进行比较，计算统计量 χ² = Σ (O−E)² / E。检验简单比例时自由度为（分类数 − 1），独立性检验的自由度为（行数−1）×（列数−1）。若 χ² 值相对卡方分布的临界值较大，则拒绝零假设。

Example Question: In a genetic cross, the expected ratio of round yellow, round green, wrinkled yellow, and wrinkled green seeds is 9:3:3:1. Observed counts among 160 seeds are 86, 35, 26, and 13. Test at the 5% level whether the data fit the expected ratio.

例题： 在一次遗传杂交中，预期圆黄、圆绿、皱黄、皱绿种子的比例为 9:3:3:1。160 粒种子中观察数量分别为 86、35、26、13。在 5% 水平下检验数据是否符合预期比例。

Worked Solution: Expected counts: 160×(9/16)=90, 160×(3/16)=30, 30, 10. χ² = (86−90)²/90 + (35−30)²/30 + (26−30)²/30 + (13−10)²/10 ≈ 0.178 + 0.833 + 0.533 + 0.9 = 2.444. Degrees of freedom = 3; critical value at 5% = 7.815. 2.444 < 7.815, do not reject H₀. The observed frequencies are consistent with the 9:3:3:1 ratio.

解答： 期望频数：160×(9/16)=90，160×(3/16)=30、30、10。χ² = (86−90)²/90 + (35−30)²/30 + (26−30)²/30 + (13−10)²/10 ≈ 0.178 + 0.833 + 0.533 + 0.9 = 2.444。自由度 = 3；5% 水平临界值 = 7.815。2.444 < 7.815，不拒绝 H₀。观察频数与 9:3:3:1 比例一致。

3. Regression Analysis in Economics | 经济学中的回归分析

Economists model relationships between variables such as income and consumption, or interest rates and investment. Simple linear regression estimates the line of best fit y = a + bx through least squares. The slope b is interpreted as the change in the response variable for a one‑unit increase in the explanatory variable. Hypothesis tests for the slope (H₀: β = 0 vs. H₁: β ≠ 0) tell us whether the relationship is statistically significant. The coefficient of determination r² measures how much of the variation in y is explained by x. In exam questions, you may be given summary statistics or computer output and asked to interpret these quantities in the economic context.

经济学家通过建立模型来研究变量间的关系，如收入与消费、利率与投资。简单线性回归用最小二乘法估计最佳拟合直线 y = a + bx。斜率 b 被解释为解释变量每增加一个单位，响应变量的变化量。对斜率进行假设检验（H₀: β = 0 对比 H₁: β ≠ 0），可以判断这种关系是否统计显著。决定系数 r² 衡量了 y 的变异中有多大比例可以由 x 解释。在考题中，你可能会得到汇总统计量或计算机输出，并要求在经济语境下解释这些数量。

Example Question: Annual data for 10 years on advertising spend x (₤1000s) and sales revenue y (₤10 000s) yield: Σx=420, Σy=850, Sxx=2850, Syy=19 600, Sxy=6800. Find the regression equation of y on x and interpret the slope. Test at the 1% level whether the slope is significantly different from zero.

例题： 10 年的广告支出 x（千英镑）与销售收入 y（万英镑）数据：Σx=420，Σy=850，Sxx=2850，Syy=19 600，Sxy=6800。求 y 对 x 的回归方程并解释斜率。在 1% 水平下检验斜率是否显著不等于零。

Solution: b = Sxy / Sxx = 6800/2850 ≈ 2.386. a = (850/10) − 2.386×(420/10) = 85 − 100.21 = −15.21. Equation: y = −15.21 + 2.386x. Slope: each additional ₤1000 spent on advertising is associated with an increase of about ₤23 860 in sales. Test H₀: β = 0. s² = (Syy − b·Sxy)/(n−2) = (19600−2.386×6800)/8 ≈ (19600−16224.8)/8 = 421.9. SE(b) = √(s²/Sxx) = √(421.9/2850) ≈ 0.3848. t = 2.386/0.3848 ≈ 6.20. Critical t₍₈₎(two‑tailed, 1%) = 3.355. |t| > 3.355, reject H₀. There is strong evidence of a linear relationship.

解答： b = Sxy / Sxx = 6800/2850 ≈ 2.386。a = (850/10) − 2.386×(420/10) = 85 − 100.21 = −15.21。回归方程：y = −15.21 + 2.386x。斜率意义：广告支出每增加 1000 英镑，销售收入平均增加约 23 860 英镑。检验 H₀: β = 0。s² = (Syy − b·Sxy)/(n−2) = (19600−2.386×6800)/8 ≈ 421.9。SE(b) = √(421.9/2850) ≈ 0.3848。t = 6.20。临界值 t₍₈₎（双尾，1%）= 3.355。|t| > 3.355，拒绝 H₀，线性关系极显著。

4. Sampling Methods in Geography | 地理学中的抽样方法

Geography fieldwork often requires sampling from a large area or population, for example when measuring pebble sizes along a beach or collecting questionnaire responses from households. A simple random sample is theoretically ideal but often impractical, so systematic sampling (every kth unit) or stratified sampling (dividing into groups and sampling proportionally) are commonly used. The choice of method affects the precision of estimates and the validity of statistical inference. Questions might ask you to critique a given sampling plan, suggest improvements, or calculate estimates and standard errors appropriate to the design.

地理实地考察通常需要从大片区域或总体中进行抽样，例如测量海滩上的卵石大小或收集住户问卷。简单随机抽样在理论上是最理想的，但往往不切实际，因此常采用系统抽样（每隔 k 个单位取样）或分层抽样（先划分群组再按比例抽样）。抽样方法的选择会影响到估计的精度以及统计推断的有效性。考题可能会要求你评判给定的抽样方案、提出改进建议，或计算与设计相匹配的估计值和标准误。

Example Question: A geographer wants to estimate the mean length of pebbles on a 2 km beach. She divides the beach into 20 sections of 100 m and randomly selects one 10 m transect within each section, measuring all pebbles along that transect. Identify the sampling method and explain one advantage. If the sample mean is 6.3 cm with standard deviation 1.8 cm from 200 pebbles, calculate a 95% confidence interval for the population mean.

例题： 一位地理学家想估计一条 2 km 海滩上卵石的平均长度。她将海滩分成 20 个 100 m 区段，在每个区段内随机选取一条 10 m 样带，测量样带内的所有卵石。试识别该抽样方法并说明一个优点。若从 200 块卵石中得到样本均值 6.3 cm、标准差 1.8 cm，计算总体均值的 95% 置信区间。

Solution: The method is stratified random sampling (each section acts as a stratum). Advantage: it ensures representation from all parts of the beach, improving accuracy. Confidence interval: x̄ ± z* × (s/√n). For 95%, z* = 1.96. Interval: 6.3 ± 1.96×(1.8/√200) = 6.3 ± 1.96×0.1273 ≈ 6.3 ± 0.249, i.e. (6.05 cm, 6.55 cm).

解答： 该方法是分层随机抽样（每个区段为一层）。优点：确保海滩各部分均有代表，提高准确性。置信区间：x̄ ± z* × (s/√n)。95% 时 z* = 1.96。区间：6.3 ± 1.96×(1.8/√200) = 6.3 ± 0.249，即 (6.05 cm, 6.55 cm)。

5. Error Analysis in Physics | 物理学中的误差分析

Physicists must quantify uncertainty in repeated measurements. The standard deviation of the mean (also called the standard error) describes how well the sample mean estimates the true value. When several quantities are combined, errors propagate. For sums or differences, absolute uncertainties add in quadrature; for products or quotients, relative (percentage) uncertainties add in quadrature. In an OCR Statistics paper, a physics‑themed questions might give you a set of measurements and ask you to compute the mean, standard deviation, and standard error, then combine uncertainties in a derived quantity.

物理学家必须量化重复测量中的不确定度。均值的标准差（也称标准误）反映了样本均值估计真值的好坏。当多个量被组合时，误差会传递。对于和或差，绝对不确定度以平方和开方的方式合成；对于积或商，相对（百分比）不确定度以平方和开方的方式合成。在 OCR 统计试卷中，物理学主题的题目可能提供一组测量值，要求你计算均值、标准差和标准误，然后对导出量进行不确定度合成。

Example Question: A student measures the diameter d of a wire five times: 0.512, 0.508, 0.515, 0.510, 0.513 mm. Calculate the mean, standard deviation, and standard error. The cross‑sectional area A = πd²/4. Given the mean d = 0.5116 mm, SE(d) = 0.0012 mm, find the percentage uncertainty in A.

例题： 一位学生五次测量导线直径 d：0.512、0.508、0.515、0.510、0.513 mm。计算均值、标准差和标准误。截面积 A = πd²/4。已知平均 d = 0.5116 mm，SE(d) = 0.0012 mm，求 A 的百分比不确定度。

Solution: Mean = 0.5116 mm. Sample s.d. ≈ 0.00270 mm. Standard error = s.d./√5 ≈ 0.00121 mm. For A proportional to d², percentage uncertainty in A = 2 × (percentage uncertainty in d) = 2 × (0.0012/0.5116 × 100%) ≈ 2 × 0.234% ≈ 0.47%.

解答： 均值 = 0.5116 mm。样本标准差 ≈ 0.00270 mm。标准误 = 标准差/√5 ≈ 0.00121 mm。A 正比于 d²，故 A 的百分比不确定度 = 2 × (d 的百分比不确定度) = 2 × (0.0012/0.5116 × 100%) ≈ 2 × 0.234% ≈ 0.47%。

6. Correlation and Causation in Sociology | 社会学中的相关与因果

Sociologists collect data on variables such as educational attainment, income, and health. They often compute the Pearson product‑moment correlation coefficient r to quantify linear association. However, a strong correlation does not imply causation; there may be confounding variables or reverse causality. Exam questions may present a scenario and ask you to interpret r, test its significance using t = r√(n−2)/√(1−r²), and then critically discuss the limitations of correlational evidence.

社会学家收集关于教育程度、收入与健康等变量的数据。他们常常计算皮尔逊积矩相关系数 r 来量化线性关联的强度。然而，强相关并不意味着因果关系；可能存在混杂变量或反向因果。考题可能会给出一个情境，要求你解释 r，用 t = r√(n−2)/√(1−r²) 对其进行显著性检验，然后批判性地讨论相关证据的局限性。

Example Question: A study of 15 regions finds a correlation of 0.72 between average household income and life expectancy. Test whether the correlation is significant at the 1% level. Suggest a reason why this correlation might not indicate that increasing income causes longer life.

例题： 一项针对 15 个地区的研究发现，家庭平均收入与预期寿命之间的相关系数为 0.72。检验该相关系数在 1% 水平下是否显著。提出一个理由，说明为什么这种相关性未必表明增加收入会导致寿命延长。

Solution: H₀: ρ = 0, H₁: ρ ≠ 0. t = 0.72√(15−2)/√(1−0.72²) = 0.72√13/√0.4816 ≈ 2.595/0.694 = 3.74. df = 13; two‑tailed critical t at 1% = 3.012. 3.74 > 3.012, reject H₀; significant correlation. Limitation: Wealthier regions may also have better healthcare, nutrition, and education—confounding variables that could explain the correlation.

解答： H₀: ρ = 0，H₁: ρ ≠ 0。t = 0.72√13/√(1−0.72²) ≈ 3.74。自由度 13，双尾 1% 临界值 t = 3.012。3.74 > 3.012，拒绝 H₀，相关显著。局限性：较富裕地区可能也拥有更好的医疗、营养和教育条件——这些混杂变量可以解释该相关，而不是直接的因果关系。

7. Probability Distributions in Environmental Science | 环境科学中的概率分布

Environmental scientists often model rare events, such as floods, extreme temperatures, or the occurrence of a pollutant exceeding a threshold. The Poisson distribution is a natural choice for counts of independent events occurring at a fixed rate in time or space. Alternatively, the binomial distribution is used when sampling from a population with a fixed proportion affected by pollution. Questions may ask you to calculate probabilities of specific events, find the mean and variance, or use the distribution to test whether the observed frequency exceeds what would be expected by chance.

环境科学家经常对稀有事件建模，例如洪水、极端温度或污染物超过阈值的发生次数。泊松分布是为在固定时间或空间内独立事件计数而设的自然选择。另外，当从一个具有固定受污染比例的总体中抽样时，可使用二项分布。考题可能要求你计算特定事件的概率、求均值与方差，或利用分布检验观察到的频率是否超出了偶然预期。

Example Question: The number of days per summer with an air quality index above 100 in a city follows a Poisson distribution with mean 3.2. Calculate the probability that next summer there will be (i) exactly 2 days, (ii) at most 1 day above 100. Use the formula P(X=k) = e⁻ᵏⁱ³·²(3.2)ᵏⁱ/k!.

例题： 某城市夏季空气质量指数超过 100 的天数服从均值为 3.2 的泊松分布。计算下一年夏季 (i) 恰好有 2 天，(ii) 最多有 1 天指数超过 100 的概率。使用公式 P(X=k) = e⁻³·²(3.2)ᵏⁱ/k!。

Solution: (i) P(X=2) = e⁻³·² × (3.2²)/2 = e⁻³·² × 5.12 ≈ 0.0408×5.12 = 0.209. (ii) P(X≤1) = P(0)+P(1) = e⁻³·² + e⁻³·² ×3.2 ≈ 0.0408 + 0.1306 = 0.1714.

解答： (i) P(X=2) = e⁻³·² × (3.2²)/2 = e⁻³·² × 5.12 ≈ 0.0408×5.12 = 0.209。(ii) P(X≤1) = P(0)+P(1) = e⁻³·² + e⁻³·²×3.2 ≈ 0.0408 + 0.1306 = 0.1714。

8. Decision Trees and Expected Values in Business | 商业中的决策树与期望值

Business decisions involve uncertainty. A decision tree diagrams out alternatives and their possible payoffs, each with a probability. The expected monetary value (EMV) of each option is calculated by summing (payoff × probability). The option with the highest EMV (or lowest expected cost) is chosen. This falls under probability and discrete random variables in the OCR specification. A typical question might provide a business scenario (launch a new product or not) and ask you to draw and evaluate the tree.

商业决策涉及不确定性。决策树图示化了各种方案及其可能的收益，每个收益都附有概率。每个选项的期望货币价值 (EMV) 通过求和 (收益 × 概率) 来计算，并选择 EMV 最高（或预期成本最低）的方案。这属于 OCR 考试大纲中概率与离散随机变量的内容。典型的考题可能会给出一个商业情景（是否推出新产品）并要求你绘制并评估决策树。

Example Question: A company can launch a new product or stay with the old one. Launching costs £200 000. If the market is favourable (probability 0.6), profit is £800 000; if unfavourable (0.4), profit is £100 000. The old product guarantees a profit of £300 000. Draw a decision tree and advise the company based on EMV.

例题： 一家公司可以推出新产品或继续使用旧产品。推出新产品需花费 20 万英镑。若市场有利（概率 0.6），利润为 80 万英镑；若不利（0.4），利润为 10 万英镑。旧产品可保证 30 万英镑利润。绘制决策树并根据 EMV 提出建议。

Solution: EMV(new) = (0.6 × £800k + 0.4 × £100k) − £200k = £480k + £40k − £200k = £320k. EMV(old) = £300k. Since £320k > £300k, launch the new product. (Tree would show decision node, two branches, chance nodes, and payoffs.)

解答： EMV(新) = (0.6 × 80 万 + 0.4 × 10 万) − 20 万 = 48 万 + 4 万 − 20 万 = 32 万英镑。EMV(旧) = 30 万。因 32 万 > 30 万，应推出新产品。（树图应包含决策节点、两个方案分支、机会节点及收益值。）

9. Confidence Intervals in Medicine | 医学中的置信区间

In clinical trials, confidence intervals for the difference between two means or two proportions are vital. If the interval for the difference in recovery rates between a new drug and a placebo does not contain zero, the difference is statistically significant at the chosen confidence level. The formula involves the observed difference ± (critical value × standard error). Medical questions often require interpretation in context: a confidence interval provides a range of plausible values for the true effect size and is more informative than a simple reject/do-not-reject decision.

在临床试验中，两个均值或两个比例之差的置信区间至关重要。如果新药与安慰剂的康复率差异的置信区间不包含零，则表明在选定的置信水平下差异具有统计显著性。公式为观察到的差异 ± (临界值 × 标准误)。医学题目通常要求结合语境进行解释：置信区间给出了真实效应大小的合理范围，比简单的拒绝/不拒绝结论更能提供信息。

Example Question: In a trial, 120 of 200 patients on a new drug recovered, compared with 90 of 190 on a placebo. Construct a 95% confidence interval for the difference in population recovery rates. Interpret the interval.

例题： 在一项试验中，服用新药的 200 名患者中有 120 人康复，服用安慰剂的 190 人中有 90 人康复。构建总体康复率之差的 95% 置信区间并解释。

Solution: p̂₁ = 120/200 = 0.60; p̂₂ = 90/190 ≈ 0.4737. Difference = 0.1263. SE = √[0.6×0.4/200 + 0.4737×0.5263/190] ≈ √[0.0012 + 0.001312] ≈ √0.002512 = 0.05012. z* = 1.96. Interval: 0.1263 ± 1.96×0.05012 = 0.1263 ± 0.0982, i.e. (0.0281, 0.2245). The interval does not include zero, suggesting a significant benefit of the new drug. The true difference in recovery rates is likely between 2.8% and 22.5%.

解答： p̂₁ = 0.60；p̂₂ ≈ 0.4737。差异 = 0.1263。SE = √[0.6×0.4/200 + 0.4737×0.5263/190] ≈ 0.05012。z* = 1.96。区间：0.1263 ± 1.96×0.05012 = (0.0281, 0.2245)。区间不包含零，提示新药具有显著的优势。真实康复率差异很可能在 2.8% 到 22.5% 之间。

10. Integrating Concepts: Multi‑step Problems | 概念综合：多步骤问题

High‑quality OCR questions often combine several topics. For instance, you might be given data in a scientific context, asked to identify the sampling method, calculate summary statistics, construct a confidence interval, and then perform a hypothesis test. The solution demands careful reading of the scenario and methodical application of correct formulas. Always state hypotheses clearly, check assumptions, and relate final conclusions to the real‑world problem.

高质量的 OCR 试题常常会融合多个知识点。例如，你可能被给予一个科学情境中的数据，要求识别抽样方法，计算概括统计量，构建置信区间，然后进行假设检验。解答这类问题需要仔细阅读情景描述，并系统性地应用正确的公式。务必清晰地陈述假设，检查前提条件，并将最终结论联系回现实问题。

Example Question: An ecologist studies the effect of a fertiliser on the growth of seedlings. She randomly allocates 15 seedlings to a treatment group and 15 to a control. After 4 weeks, the heights (cm) are summarised: Treatment: x̄₁=23.4, s₁=4.1; Control: x̄₂=18.7, s₂=3.8. (a) State the experimental design and one advantage. (b) Calculate a 90% confidence interval for the mean difference (assume equal variances). (c) Test at the 5% level whether the fertiliser increases mean height. (d) Comment on the reliability of the test if the data were not normally distributed.

例题： 一位生态学家研究肥料对幼苗生长的影响。他将 15 株幼苗随机分配到处理组和 15 株到对照组。4 周后高度数据汇总如下：处理组 x̄₁=23.4 cm, s₁=4.1；对照组 x̄₂=18.7 cm, s₂=3.8。(a) 说明实验设计及一个优点。(b) 计算均值之差的 90% 置信区间（假设方差相等）。(c) 在 5% 水平下检验肥料是否能增加平均高度。(d) 若数据不服从正态分布，对检验的可靠性发表评论。

Solution Outline: (a) Completely randomised design; advantage: avoids systematic bias. (b) Pooled s² = (14×4.1²+14×3.8²)/28 ≈ 15.625. SE = √(s²/15+s²/15) = √(2×15.625/15) = √2.0833 ≈ 1.443. t₍₂₈₎,0.05 one‑tailed = 1.701? Actually, for 90% CI two‑sided critical is t₀.₀₅,₂₈ = 1.701. CI = (23.4−18.7) ± 1.701×1.443 = 4.7 ± 2.454 → (2.246, 7.154). (c) H₀: μ₁=μ₂ vs H₁: μ₁>μ₂. t = 4.7/1.443 ≈ 3.257. Critical t₀.₀₅,₂₈ one‑tailed = 1.701. Reject H₀; fertiliser increases growth. (d) t‑test is robust to moderate non‑normality with equal sample sizes, but if severely skewed, a non‑parametric alternative like Mann‑Whitney might be needed.

解答要点： (a) 完全随机设计；优点：避免系统偏差。(b) 合并方差 ≈ 15.625，SE ≈ 1.443。90% 置信区间：4.7 ± 1.701×1.443 = (2.25, 7.15)。(c) H₀: μ₁=μ₂ vs H₁: μ₁>μ₂。t = 3.257 > 临界值 1.701，拒绝 H₀，肥料能促进生长。(d) t 检验在样本量相等时对中等程度的非正态性具有稳健性，但若数据严重偏态，可能需要使用 Mann‑Whitney 等非参数方法。

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Interdisciplinary Integrated Question Practice for Year 12 OCR Statistics | 跨学科综合题型训练