Cross-Disciplinary Exam Practice for AQA Statistics | AQA统计跨学科综合题型训练

📚 Cross-Disciplinary Exam Practice for AQA Statistics | AQA统计跨学科综合题型训练

Welcome to this targeted cross-disciplinary revision resource for Year 13 AQA Statistics. In your examination, you will be expected to apply a wide range of statistical techniques to authentic contexts drawn from biology, psychology, economics, physics, geography, and beyond. The worked examples that follow will guide you through hypothesis tests, probability distributions, correlation, regression, and decision-making, always with a focus on the precise reasoning and interpretation demanded by AQA mark schemes.

欢迎使用这份专为13年级AQA统计设计的跨学科综合题型训练。在考试中，你需要将多种统计方法应用于生物学、心理学、经济学、物理学、地理学等真实情境。下面的例题将带你一步步完成假设检验、概率分布、相关与回归以及决策分析，始终紧扣AQA评分方案所要求的严谨推理与解释。

1. Genetics: Chi-Squared Goodness-of-Fit Test | 遗传学：卡方拟合优度检验

Gregor Mendel’s classic pea experiments predicted that the phenotypes of dihybrid crosses should appear in a 9:3:3:1 ratio. A modern replication records the following counts: round-yellow 315, round-green 108, wrinkled-yellow 101, wrinkled-green 32. Use a 5% significance level to test whether the data are consistent with the theoretical ratio.

孟德尔的经典豌豆实验预测双因子杂交的表型应按9:3:3:1的比例出现。一项现代重复实验记录了以下数据：圆黄315，圆绿108，皱黄101，皱绿32。使用5%的显著性水平检验数据是否符合理论比例。

Step 1: State the hypotheses. H₀: The data follow a 9:3:3:1 ratio; H₁: The data do not follow this ratio.

步骤1：提出假设。H₀：数据符合9:3:3:1比例；H₁：数据不符合该比例。

Step 2: Calculate expected frequencies. Total observations n = 556. Expected counts are 9/16 × 556 = 312.75, 3/16 × 556 = 104.25, 3/16 × 556 = 104.25, 1/16 × 556 = 34.75.

步骤2：计算期望频数。总观测数 n = 556。期望频数为：9/16 × 556 = 312.75，3/16 × 556 = 104.25，3/16 × 556 = 104.25，1/16 × 556 = 34.75。

Step 3: Compute the test statistic.

步骤3：计算检验统计量。

χ² = Σ (O_i – E_i)² / E_i

Phenotype	Oi	Ei	(Oi-Ei)2/Ei
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

χ² = 0.016 + 0.135 + 0.101 + 0.218 = 0.470

χ² = 0.016 + 0.135 + 0.101 + 0.218 = 0.470

Step 4: Degrees of freedom = number of categories – 1 = 3. Critical value from chi-squared table at 5% (3 d.f.) is 7.815.

步骤4：自由度 = 类别数 – 1 = 3。查卡方分布表，在5%显著性水平、自由度为3时，临界值为7.815。

Step 5: Since 0.470 < 7.815, we do not reject H₀. There is insufficient evidence to suggest the data deviate from the 9:3:3:1 ratio.

步骤5：由于0.470 < 7.815，我们不拒绝H₀。没有足够证据表明数据偏离9:3:3:1比例。

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2. Psychology: Two-Sample t-Test for Independent Groups | 心理学：独立样本t检验

A psychologist measures response times (ms) for a cognitive task under two conditions: a control group (n₁=10) and a sleep-deprived group (n₂=10). The summary statistics are: control mean x¯₁ = 245 ms, standard deviation s₁ = 18 ms; sleep-deprived mean x¯₂ = 267 ms, s₂ = 22 ms. Assume equal population variances and test whether sleep deprivation increases response time at the 5% significance level.

一位心理学家测量了两种条件下的认知任务反应时间（毫秒）：控制组（n₁=10）和睡眠剥夺组（n₂=10）。汇总统计量为：控制组均值 x¯₁ = 245 ms，标准差 s₁ = 18 ms；睡眠剥夺组均值 x¯₂ = 267 ms，s₂ = 22 ms。假设总体方差相等，在5%显著性水平下检验睡眠剥夺是否增加反应时间。

Step 1: Hypotheses. H₀: μ₁ = μ₂; H₁: μ₂ > μ₁ (one-tailed).

步骤1：假设。H₀：μ₁ = μ₂；H₁：μ₂ > μ₁（单侧检验）。

Step 2: Pooled variance estimate.

步骤2：合并方差估计。

s_p² = [(n₁-1)s₁² + (n₂-1)s₂²] / (n₁+n₂-2)

= [9×324 + 9×484] / 18 = (2916 + 4356)/18 = 404.0, so s_p = 20.10 ms.

= [9×324 + 9×484] / 18 = (2916 + 4356)/18 = 404.0，因此 s_p = 20.10 ms。

Step 3: Test statistic.

步骤3：检验统计量。

t = (x¯₂ – x¯₁) / (s_p × √(1/n₁ + 1/n₂))

= (267 – 245) / (20.10 × √(0.1+0.1)) = 22 / (20.10 × 0.4472) = 22 / 8.988 = 2.448

= (267 – 245) / (20.10 × √(0.1+0.1)) = 22 / (20.10 × 0.4472) = 22 / 8.988 = 2.448

Step 4: Degrees of freedom = 18. Critical value for one-tailed t-test at 5% is 1.734. Since 2.448 > 1.734, we reject H₀. There is significant evidence that sleep deprivation increases response time.

步骤4：自由度 = 18。单侧5%显著性水平的t临界值为1.734。因为2.448 > 1.734，我们拒绝H₀。有显著证据表明睡眠剥夺会增加反应时间。

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3. Economics: Linear Regression and Correlation | 经济学：线性回归与相关

An economist collects annual data on household disposable income x (in £1000s) and consumption expenditure y (in £1000s) for five years: (30, 24), (35, 28), (40, 31), (45, 35), (50, 38). Determine the product-moment correlation coefficient, the regression equation of y on x, and predict expenditure when income is £52 000. Test whether the correlation is significant at the 1% level.

一位经济学家收集了五年家庭可支配收入x（千英镑）和消费支出y（千英镑）的数据：(30, 24), (35, 28), (40, 31), (45, 35), (50, 38)。计算积矩相关系数、y对x的回归方程，并预测收入为52000英镑时的支出。在1%显著性水平下检验相关是否显著。

Step 1: Compute summations: n=5, Σx=200, Σy=156, Σx²=8250, Σy²=4942, Σxy=6350.

步骤1：计算总和：n=5，Σx=200，Σy=156，Σx²=8250，Σy²=4942，Σxy=6350。

Step 2: S_xy = 6350 – (200×156)/5 = 6350 – 6240 = 110; S_xx = 8250 – 200²/5 = 8250 – 8000 = 250; S_yy = 4942 – 156²/5 = 4942 – 4867.2 = 74.8.

步骤2：S_xy = 6350 – (200×156)/5 = 110；S_xx = 8250 – 200²/5 = 250；S_yy = 4942 – 156²/5 = 74.8。

Step 3: PMCC r = S_xy / √(S_xxS_yy) = 110 / √(250×74.8) = 110 / √18700 = 110 / 136.75 = 0.804.

步骤3：积矩相关系数 r = 110 / √(250×74.8) = 110 / 136.75 = 0.804。

Step 4: For n=5, critical value for two-tailed test at 1% (3 d.f.) is 0.959. Since 0.804 < 0.959, we cannot reject H₀: ρ=0. The correlation is not significant at the 1% level.

步骤4：n=5，双尾1%显著性水平（自由度3）的临界值为0.959。由于0.804 < 0.959，不能拒绝H₀: ρ=0。在1%水平下相关不显著。

Step 5: Regression line: slope b = S_xy/S_xx = 110/250 = 0.44; intercept a = &ymacr; – b&xmacr; = (156/5) – 0.44×(200

Published by TutorHao | Year 13 统计 Revision Series | aleveler.com

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