Cross-Disciplinary Integrated Question Practice for Year 12 AQA Statistics

📚 Cross-Disciplinary Integrated Question Practice for Year 12 AQA Statistics | 跨学科综合题型训练

In AQA Statistics, mastering the ability to apply statistical techniques to real-world, cross-disciplinary problems is essential for top marks. This article will guide you through integrated question practice, covering biology, psychology, geography, economics, and more.

在AQA统计中，掌握将统计技术应用于现实世界跨学科问题的能力对于获得高分至关重要。本文将指导你进行综合题型训练，涵盖生物学、心理学、地理学、经济学等多个领域。

1. Understanding the Interdisciplinary Approach | 理解跨学科方法

Statistics transcends subject boundaries — the same hypothesis test might appear in a medical trial or a psychological study. You must interpret contexts and extract the relevant statistical model.

统计超越学科界限——同一假设检验可能出现在医学试验或心理学研究中。你必须解读情境并提取相关的统计模型。

Key skills include defining variables, selecting the correct test, checking assumptions, and writing conclusions that refer back to the original problem. In AQA exams, marks are often awarded for context-specific justifications.

关键技能包括定义变量、选择正确的检验方法、检查假设条件，以及撰写回扣原问题的结论。在AQA考试中，给出与情境相关的理由往往能够得分。

2. Biology: Binomial Hypothesis Testing for Proportions | 生物学：比例的二项分布假设检验

In genetics or drug trials, you often test a proportion. For example, a biologist claims that a new fertiliser increases the germination rate above the standard 60%. She plants 25 seeds and observes 19 germinate. Test at the 5% significance level.

在遗传学或药物试验中，你经常需要检验一个比例。例如，一位生物学家声称新肥料可以使发芽率超过标准的60%。她种下25粒种子，观察到19粒发芽。在5%显著性水平下进行检验。

Let X ~ B(25, 0.6). H₀: p = 0.6, H₁: p > 0.6 (one-tailed). Under H₀, P(X ≥ 19) = 1 – P(X ≤ 18). Using binomial tables, P(X ≤ 18) = 0.9536, so p-value = 0.0464. Since 0.0464 < 0.05, reject H₀. There is sufficient evidence, at the 5% level, that the new fertiliser increases the germination rate.

设X ~ B(25, 0.6)。H₀: p = 0.6，H₁: p > 0.6（单尾检验）。在H₀下，P(X ≥ 19) = 1 – P(X ≤ 18)。查二项分布表，P(X ≤ 18) = 0.9536，因此p值为0.0464。由于0.0464 < 0.05，拒绝H₀。在5%显著性水平下，有充分证据表明新肥料提高了发芽率。

Always check that the binomial model is valid: fixed number of trials, independence, two outcomes, and constant probability. In biology questions, the context often justifies these assumptions.

务必检查二项模型的适用性：固定试验次数、独立性、两种结果以及恒定概率。在生物学问题中，情境通常能够为这些假设提供依据。

3. Psychology: Normal Distribution and IQ Scores | 心理学：正态分布与智商分数

IQ scores are modelled by a normal distribution N(100, 15²). A psychologist wants to find the proportion of adults with an IQ exceeding 130. Standardise: Z = (130 – 100)/15 = 2. P(Z > 2) = 1 – 0.9772 = 0.0228. So about 2.3% of the population.

智商分数服从正态分布 N(100, 15²)。一位心理学家想求智商超过130的成年人比例。标准化：Z = (130 – 100)/15 = 2。P(Z > 2) = 1 – 0.9772 = 0.0228。因此大约有2.3%的人口。

In a study, a psychologist tests whether the mean IQ of a sample of 36 children from a special programme differs from 100. Their mean is 104. The test statistic is z = (x̄ – μ) / (σ/√n) = (104 – 100) / (15/6) = 4/2.5 = 1.6. For a two-tailed test at 5% significance, the critical value is 1.96. Since 1.6 < 1.96, do not reject H₀. There is insufficient evidence that the programme affects IQ.

在一项研究中，心理学家检验某特殊项目中36名儿童的平均智商是否不同于100。他们的样本均值为104。检验统计量为 z = (x̄ – μ) / (σ/√n) = (104 – 100) / (15/6) = 4/2.5 = 1.6。进行5%显著性水平的双尾检验，临界值为1.96。由于1.6 < 1.96，不拒绝H₀。没有足够证据表明该项目会影响智商。

Remember to state the assumption that the sample mean is normally distributed by the Central Limit Theorem, and that the population variance is known. In psychology, measurement tools often provide a known standard deviation.

记住要说明根据中心极限定理，样本均值服从正态分布，且总体方差已知。在心理学中，测量工具通常会提供已知的标准差。

4. Geography: Spearman’s Rank Correlation and Environmental Variables | 地理学：斯皮尔曼等级相关与环境变量

A geography student investigates the relationship between river depth and flow velocity at 10 sites. The data are ranked, and the Spearman’s rank correlation coefficient r₝ = 1 – (6Σd²)/(n(n²-1)) is calculated as 0.86. The null hypothesis is ρ = 0 (no correlation).

一位地理学学生调查研究10个地点的河流深度与流速之间的关系。对数据排序后计算斯皮尔曼等级相关系数 r₝ = 1 – (6Σd²)/(n(n²-1))，得到0.86。原假设为ρ = 0（无相关）。

Using the critical value table for n = 10, the 5% two-tailed critical value is 0.648. Since 0.86 > 0.648, reject H₀. There is significant evidence of a positive rank correlation between depth and velocity.

查 n = 10 的临界值表，5%双尾临界值为0.648。因为0.86 > 0.648，拒绝H₀。有显著证据表明深度和流速之间存在正等级相关。

Spearman’s rank is appropriate when data are not necessarily linear or when outliers may be present. In fieldwork, it is a robust tool for ordinal or continuously distributed environmental variables.

当数据不一定呈线性关系或可能存在离群值时，斯皮尔曼等级相关系数是合适的。在实地考察中，对于定序或连续分布的环境变量，它是一种稳健的工具。

5. Economics: Least Squares Regression for Cost Prediction | 经济学：用于成本预测的最小二乘回归

An economist models the relationship between units produced (x) and total cost y (pound;). Given n = 8 data points, the following summations are computed: Σx = 440, Σy = 3360, Σxy = 189800, Σx² = 25200, Σy² = 1446000.

一位经济学家对产量（x）和总成本y（英镑）之间的关系建模。给定8个数据点，计算出以下求和值：Σx = 440, Σy = 3360, Σxy = 189800, Σx² = 25200, Σy² = 1446000。

Calculate S₝₠ = Σxy – (Σx)(Σy)/n = 189800 – (440 times;3360)/8 = 189800 – 184800 = 5000. S₝₝ = Σx² – (Σx)²/n = 25200 – 193600/8 = 25200 – 24200 = 1000. Hence b = S₝₠/S₝₝ = 5000/1000 = 5. a = ȳ – b x̄ = (3360/8) – 5 times;(440/8) = 420 – 5 times;55 = 420 – 275 = 145. The regression line is y = 145 + 5x.

计算 S₝₠ = Σxy – (Σx)(Σy)/n = 189800 – (440 times;3360)/8 = 189800 – 184800 = 5000。S₝₝ = Σx² – (Σx)²/n = 25200 – 193600/8 = 25200 – 24200 = 1000。因此 b = S₝₠/S₝₝ = 5000/1000 = 5。a = ȳ – b x̄ = (3360/8) – 5 times;(440/8) = 420 – 5 times;55 = 420 – 275 = 145。回归直线为 y = 145 + 5x。

Predict the cost for 70 units: y = 145 + 5 times;70 = pound;495. However, if 70 is outside the original x-range, this is extrapolation and may be unreliable. Always comment on the validity in economic contexts.

预测70单位产量的成本：y = 145 + 5 times;70 = 495英镑。但如果70在原始x范围之外，则属于外推，可能不可靠。在经济学情境中必须对有效性加以评论。

6. Sociology: Chi-Squared Test of Independence | 社会学：卡方独立性检验

A sociologist surveys voting intention (Yes/No) and gender (Male/Female). The observed frequencies are: Male-Yes 30, Male-No 20, Female-Yes 45, Female-No 35. Test at 1% significance whether gender and intention are independent.

一位社会学家调查投票意向（是/否）和性别（男/女）。观察频数为：男-是30，男-否20，女-是45，女-否35。在1%显著性水平下检验性别和意向是否独立。

Expected frequencies: for Male-Yes: row total times;col total / grand total = (50 times;75)/130 = 3750/130 asymp; 28.85. Complete the table: Male-No expected = 50 times;55/130 asymp; 21.15, Female-Yes = 80 times;75/130 asymp; 46.15, Female-No = 80 times;55/130 asymp; 33.85.

期望频数：男-是：行合计 times;列合计 / 总合计 = (50 times;75)/130 = 3750/130 asymp; 28.85。完成表格：男-否期望 = 50 times;55/130 asymp; 21.15，女-是期望 = 80 times;75/130 asymp; 46.15，女-否期望 = 80 times;55/130 asymp; 33.85。

χ² = Σ (O – E)²/E = (30-28.85)²/28.85 + (20-21.15)²/21.15 + (45-46.15)²/46.15 + (35-33.85)²/33.85 asymp; 0.046 + 0.062 + 0.029 + 0.039 = 0.176. Degrees of freedom = (2-1)(2-1) = 1. Critical value at 1% is 6.635. Since 0.176 < 6.635, do not reject H₀. There is no significant association.

χ² = Σ (O – E)²/E = (30-28.85)²/28.85 + (20-21.15)²/21.15 + (45-46.15)²/46.15 + (35-33.85)²/33.85 asymp; 0.046 + 0.062 + 0.029 + 0.039 = 0.176。自由度为 (2-1)(2-1) = 1。1%显著性水平下的临界值为6.635。因为0.176 < 6.635，不拒绝H₀。两者之间没有显著关联。

The chi-squared test is widely used in social sciences to examine relationships between categorical variables. Ensure all expected frequencies are at least 5; if not, combine categories.

卡方检验在社会科学中被广泛用于检验分类变量之间的关系。确保所有期望频数至少为5；否则需要合并类别。

7. Environmental Science: Sampling Techniques and Confidence Intervals | 环境科学：抽样技术与置信区间

An environmental scientist collects a random sample of 50 soil specimens to estimate the mean lead concentration. The sample mean is 22.6 ppm, and from a previous large-scale study, the population standard deviation is known to be 4.8 ppm.

一位环境科学家随机抽取了50份土壤样本以估计平均铅浓度。样本均值为22.6 ppm，根据之前的大规模研究，已知总体标准差为4.8 ppm。

A 95% confidence interval for the population mean μ is given by x̄ plusmn; z* times; (σ/√n) = 22.6 plusmn; 1.96 times; (4.8/√50) = 22.6 plusmn; 1.96 times; 0.679 = 22.6 plusmn; 1.33. Hence the interval is (21.27, 23.93) ppm. We are 95% confident that the true mean lead concentration lies within this range.

总体均值μ的95%置信区间为 x̄ plusmn; z* times; (σ/√n) = 22.6 plusmn; 1.96 times; (4.8/√50) = 22.6 plusmn; 1.96 times; 0.679 = 22.6 plusmn; 1.33。因此该区间为 (21.27, 23.93) ppm。我们有95%的把握认为真实的平均铅浓度落在此范围内。

In environmental sampling, it is crucial to use a random sampling method to avoid bias. If the population variance is unknown, you might use the sample standard deviation and a t-distribution (covered later), but for large samples, the normal approximation is adequate.

在环境抽样中，使用随机抽样方法以避免偏差至关重要。如果总体方差未知，你可能需要使用样本标准差和t分布（后续学习），但对于大样本，正态近似是可行的。

8. Physics: Propagation of Errors in Measurement | 物理学：测量中的误差传播

In a physics experiment, you measure the length L = 2.50 m with uncertainty plusmn;0.02 m, and width W = 1.20 m with uncertainty plusmn;0.01 m. The area A = L times; W. The uncertainties are independent random errors expressed as standard deviations.

在一个物理实验中，你测得长度 L = 2.50 m，不确定度为 plusmn;0.02 m；宽度 W = 1.20 m，不确定度为 plusmn;0.01 m。面积 A = L times; W。这些不确定度是以标准差表示的独立随机误差。

Using the formula for the variance of a product of independent variables: Var(A) asymp; (W²)Var(L) + (L²)Var(W). So Var(A) asymp; (1.20²)(0.02²) + (2.50²)(0.01²) = (1.44)(0.0004) + (6.25)(0.0001) = 0.000576 + 0.000625 = 0.001201. Therefore σᵧ = √0.001201 asymp; 0.0346 m². The area is A = 3.00 plusmn; 0.03 m² (to 1 s.f. for uncertainty).

使用独立变量乘积的方差公式：Var(A) asymp; (W²)Var(L) + (L²)Var(W)。故 Var(A) asymp; (1.20²)(0.02²) + (2.50²)(0.01²) = (1.44)(0.0004) + (6.25)(0.0001) = 0.000576 + 0.000625 = 0.001201。因此 σᵧ = √0.001201 asymp; 0.0346 m²。面积 A = 3.00 plusmn; 0.03 m²（不确定度保留一位有效数字）。

This technique, rooted in statistical theory, is essential in experimental sciences. AQA questions may ask you to combine standard deviations and interpret the final uncertainty in context.

这种基于统计理论的技术在实验科学中至关重要。AQA题目可能会要求你合并标准差，并在情境中解释最终的不确定度。

9. Exam Strategy: Identifying the Correct Statistical Test | 考试策略：识别正确的统计检验

Many students lose marks by applying an inappropriate test. Develop a systematic approach based on the data type and the aim of the investigation. The following decision table can guide you.

许多学生因应用了不合适的检验而失分。可根据数据类型和研究目的建立一套系统方法。下面的决策表可以为你提供指导。

Data Type	Goal	Test
Single categorical (binary)	Test proportion p	Binomial hypothesis test
Single continuous (known sigma)	Test mean mu	One-sample z-test (normal)
Two continuous variables	Association	PMCC (Pearson) or Spearman’s rank
Two categorical variable	Independence	Chi-squared test

数据类型

目标

检验

单一分类（二值）

检验比例p

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Cross-Disciplinary Integrated Question Practice for Year 12 AQA Statistics | 跨学科综合题型训练