Year 12 CCEA Psychology: Mock Exam Paper Walkthrough | CCEA Year 12 心理学模拟卷解析

📚 Year 12 CCEA Psychology: Mock Exam Paper Walkthrough | CCEA Year 12 心理学模拟卷解析

This article provides a step-by-step walkthrough of a mock examination for Year 12 CCEA Psychology. It covers the structure of AS Unit 1 (Research Methods) and Unit 2 (The Individual and Social Psychology), analysing typical questions, model responses and common pitfalls across experimental design, descriptive statistics, ethical issues, cognitive models, biological explanations and social influence. Each section is written in exam-style format with examiner commentary to support effective revision.

本文为 CCEA Year 12 心理学模拟试卷提供逐步解析，涵盖 AS 单元 1（研究方法）和单元 2（个体与社会心理学）的试卷结构，分析实验设计、描述统计、伦理议题、认知模型、生物解释及社会影响等典型题目、参考答案与常见失分点。每节采用考题形式并配有考官式评注，帮助高效复习。

---

1. Mock Exam Overview and Deconstructing Command Words | 模拟卷概览与指令词解析

Success in CCEA psychology papers begins with mastering command words. ‘Identify’ requires a brief, accurate statement; ‘describe’ calls for detailed factual knowledge; ‘explain’ demands causal links or mechanisms; and ‘evaluate’ expects a balanced discussion of strengths, weaknesses and a final judgement. In the mock paper, a 2-mark ‘Identify the experimental design’ question only needs the design name and one justification, whereas a 6-mark ‘Evaluate the multi-store model’ question requires supporting evidence, limitations and a concluding comment on the model’s usefulness.

在 CCEA 心理学试卷中取得高分首先要掌握指令词。’Identify’（识别）要求简短准确的陈述；’describe’（描述）需要详细的事实性知识；’explain’（解释）要求说明因果联系或机制；’evaluate’（评价）则期望对优点、缺点进行平衡讨论并给出最终判断。在模拟卷中，一道 2 分的 ‘Identify the experimental design’ 题只需写出设计名称和一条理由，而一道 6 分的 ‘Evaluate the multi-store model’ 题则需要提供支持证据、局限性以及对模型实用性的总结。

For instance, a question may read: ‘Identify the sampling method used.’ Answering ‘opportunity sampling’ and adding that the researcher approached available classmates fully satisfies the command word. Contrast this with ‘Evaluate the use of opportunity sampling’, which would require discussing the risk of bias, lack of representativeness and the practical advantage of speed, finishing with a justified verdict.

例如，题目可能问：’Identify the sampling method used.’（识别所使用的取样方法）。回答’opportunity sampling’（机会取样）并补充研究者招募了在场的同学，即可完全满足指令词要求。相比之下，’Evaluate the use of opportunity sampling’（评价机会取样的使用）则需要讨论偏见风险、缺乏代表性以及速度上的实用优势，并以合理的结论收尾。

---

2. Research Methods: Experimental Design and Variables | 研究方法：实验设计与变量

A typical mock question presents a scenario: ‘A researcher investigated the effect of caffeine on reaction time. Group A drank caffeinated coffee; Group B drank decaffeinated coffee. Reaction times were recorded in milliseconds.’ The experimental design is independent groups because different participants took part in each condition. The independent variable (IV) is the presence or absence of caffeine, and the dependent variable (DV) is reaction time. The independent groups design avoids order effects but may be affected by participant variables, so researchers often use random allocation to control for individual differences.

模拟卷中常见的情景题：’一位研究者调查了咖啡因对反应时间的影响。A 组饮用含咖啡因的咖啡，B 组饮用无咖啡因的咖啡，记录反应时间（毫秒）。’实验设计是独立组设计，因为不同参与者参与了不同条件。自变量（IV）是咖啡因的有无，因变量（DV）是反应时间。独立组设计避免了顺序效应，但可能受到参与者变量的影响，因此研究者常采用随机分配来控制个体差异。

Another question might ask: ‘Identify one possible extraneous variable and explain how it could be controlled.’ A strong answer would highlight room temperature; if the caffeinated group were tested in a warmer room, reaction times could slow independently of caffeine. This can be controlled by keeping temperature constant across conditions – a standardised procedure. Always link the control to the potential confound to demonstrate understanding.

另一道题可能是：’识别一个可能的额外变量并解释如何控制。’一个有力的答案会指出室温：如果含咖啡因组在较暖的房间接受测试，反应时间可能独立于咖啡因而变慢。这个问题可以通过在所有条件中保持恒温来控制——即标准化程序。请始终将控制措施与潜在的混淆因素联系起来，以展示理解。

---

3. Research Methods: Descriptive Statistics, Graphs and Z‑Scores | 研究方法：描述统计、图表与Z分数

A data set is given: reaction times (ms) 325, 410, 380, 355, 600, 390. The mean is calculated as (325+410+380+355+600+390) ÷ 6 = 410 ms. Ordering the data – 325, 355, 380, 390, 410, 600 – gives a median of (380+390) ÷ 2 = 385 ms. There is no mode. The range is 600 – 325 = 275 ms. Because the score 600 is an extreme value, the median is a more appropriate measure of central tendency than the mean; it remains stable and closer to the typical performance.

给定数据集：反应时间（毫秒）325, 410, 380, 355, 600, 390。平均数计算为 (325+410+380+355+600+390) ÷ 6 = 410 毫秒。将数据排列——325, 355, 380, 390, 410, 600——得到中位数为 (380+390) ÷ 2 = 385 毫秒。没有众数。全距为 600 – 325 = 275 毫秒。由于数值 600 是极端值，中位数比平均数更适合作为集中趋势的测量指标；中位数保持稳定，更接近典型表现。

A graph question expects a bar chart with fully labelled axes (condition on the x-axis, mean reaction time on the y-axis) and an informative title. Ensure bars are separated, not touching, to represent discrete conditions. Standard deviation may be asked for; state it measures the spread of scores around the mean – a small standard deviation indicates data are clustered closely. For the z‑score of 410 when the population mean μ is 400 and σ is 50:

图表题要求绘制条形图，坐标轴需完整标注（x 轴为条件，y 轴为平均反应时间）并附上明确标题。确保条形之间分开不接触，以代表离散条件。可能会问到标准差；回答时应指出它衡量分数围绕平均数的分散程度——标准差小表明数据紧密聚集。当总体均值 μ 为 400，σ 为 50 时，求数值 410 的 z 分数：

z = (X – μ) ÷ σ = (410 – 400) ÷ 50 = 0.2

A z‑score of 0.2 means the score is 0.2 standard deviations above the mean, a relatively typical result. This interpretation earns higher marks than just quoting the formula.

z 分数为 0.2 意味着该分数高于均值 0.2 个标准差，是一个相对典型的结果。展示这种解释能比仅引用公式获得更高分数。

---

4. Research Methods: Ethical Considerations in Psychological Research | 研究方法：心理学研究中的伦理考量

A scenario states: ‘Participants were told they were taking part in a memory test, but the true aim was to study the effect of aggressive noise on recall. They were not given the full purpose until after all data were collected.’ Two ethical issues arise: deception and lack of fully informed consent. The British Psychological Society (BPS) ethical guidelines require that deception be minimised and that participants know enough to give valid consent. Here, participants were misled about the nature of the noise, which could cause distress.

情景描述：’参与者被告知将参加一项记忆测试，但真实目的是研究高刺激性噪音对回忆的影响。在收集完所有数据之前，他们并不知晓全部目的。’这带来了两个伦理问题：欺骗以及缺乏充分的知情同意。英国心理学会（BPS）伦理准则要求尽量减少欺骗，并确保参与者获得足够信息以给出有效同意。该研究中，参与者在噪音性质上被误导，可能会引起不适。

The recommended resolution is full debriefing immediately after participation, including explaining the true aims, offering the right to withdraw data, and providing access to support if any distress occurred. Researchers should also consider a presumptive consent approach – asking a similar group if they would object to such procedures. These safeguards transform a potentially unethical study into an ethically acceptable one and demonstrate an understanding of cost-benefit analysis.

推荐的解决办法是在参与结束后立即进行全面的事后解释，包括说明真实目的、提供撤回数据的权利，并在出现任何不适时给予支持。研究者还应考虑采用预设同意的方法——询问类似群体是否会反对此类程序。这些保障措施可将原本可能不符合伦理的研究转化为伦理上可接受的研究，并展示了对成本-效益分析的理解。

---

5. Cognitive Psychology: Multi-Store Model and Working Memory Model | 认知心理学：多存储模型与工作记忆模型

The multi-store model (MSM) by Atkinson and Shiffrin proposes three separate stores: sensory memory (large capacity, very brief duration), short-term memory (STM, capacity 7±2 items, duration 18–30 seconds) and long-term memory (LTM, unlimited capacity and duration). Information flows linearly through attention and rehearsal. A simulated question asks: ‘Describe one study that supports the MSM.’ Peterson and Peterson (1959) gave participants trigrams to remember while counting backwards in threes to prevent rehearsal; recall dropped dramatically after 18 seconds, supporting STM’s limited duration without rehearsal.

Atkinson 和 Shiffrin 的多存储模型（MSM）提出三个独立的存储库：感觉记忆（容量大、持续时间极短）、短时记忆（STM，容量 7±2 个项目，持续时间 18–30 秒）和长时记忆（LTM，容量和持续时间无限）。信息通过注意和复述线性传递。模拟题可能问：’描述一项支持 MSM 的研究。’Peterson 和 Peterson（1959）让参与者记忆三字母组，同时进行倒减 3 的计算以防止复述；18 秒后回忆率大幅下降，支持了 STM 在无复述时持续时间有限的说法。

However, the MSM is oversimplified. It portrays STM as a single, passive store, yet the working memory model (WMM, Baddeley & Hitch, 1974) replaces it with multiple active components: the central executive, phonological loop and visuospatial sketchpad, later adding the episodic buffer. A typical evaluation question asks for evidence from dual-task studies; for instance, performing a verbal task (shadowing digits) and a visual task (tracking a light) simultaneously causes little disruption, but two verbal tasks interfere strongly, supporting separate slave systems. This challenges the unitary STM of the MSM and carries high AO3 marks.

然而，MSM 过于简化。它将 STM 视作一个单一的、被动的存储库，而工作记忆模型（WMM, Baddeley & Hitch, 1974）用多个主动成分取而代之：中央执行器、语音环路和视空间画板，后来还增加了情景缓冲器。典型的评价题会要求提供双任务研究的证据；例如，同时执行一个言语任务（影子跟读数字）和一个视觉任务（追踪光点）几乎不会造成干扰，但两个言语任务则产生强烈干扰，这支持了独立的从属系统。这一发现挑战了 MSM 的单一 STM，能获得较高的 AO3 分数。

---

6. Biological Psychology: Brain Localisation and Neurotransmitters | 生物心理学：大脑功能定位与神经递质

A question may present: ‘After a stroke, Patient P could understand speech but produced only meaningless, fluent sounds. Name the likely brain area damaged.’ This describes Wernicke’s aphasia, pointing to damage in Wernicke’s area (left temporal lobe). It contrasts with Broca’s aphasia, where language production is halting but comprehension is often intact (left frontal lobe). Both support the localisation of function, though modern neuroimaging suggests language involves wider distributed networks. Phineas Gage’s case further demonstrates localisation, as damage to the frontal lobes changed his personality, linking the prefrontal cortex to impulse control.

考题可能出现：’患者 P 中风后能理解言语，但只能说出无意义的流利声音。请指出最可能受损的脑区。’这描述了韦尼克失语症，指向韦尼克区（左侧颞叶）的损伤。与布洛卡失语症（左侧额叶）形成对比，后者言语生成受阻但理解通常完好。两者都支持功能定位说，但现代神经影像学表明语言涉及更广泛的分布式网络。Phineas Gage 的案例进一步证明功能定位：额叶受损改变了他的性格，将前额叶皮层与冲动控制联系起来。

On neurotransmitters, a scenario studies aggression: ‘One group received a drug that reduces serotonin reuptake, the other a placebo. The drug group showed significantly less aggressive responses.’ Serotonin is thought to inhibit impulsive aggression; low levels correlate with increased irritability. Mann et al. (1990) decreased serotonin in volunteers and observed heightened hostility, supporting a biological explanation. Note that a complete answer describes the synaptic process: serotonin is released into the synapse, binds to postsynaptic receptors, and is removed by reuptake more slowly under the drug, amplifying inhibition.

在神经递质方面，情景可涉及攻击行为：’一组服用减少血清素再摄取的药物，另一组服用安慰剂。药物组表现出显著较少的攻击反应。’血清素被认为能抑制冲动性攻击；低水平与易怒性增加相关。Mann 等人（1990）降低志愿者体内的血清素并观察到敌意升高，支持了生物学解释。注意，完整的答案应描述突

Published by TutorHao | Year 12 Psychology Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）