📚 Year 11 CAIE Psychology Unit Test Mock Paper Analysis | CAIE 心理学单元测试模拟卷解析
This article offers a thorough analysis of a mock unit test created for Year 11 students following the CAIE IGCSE Psychology syllabus. The test focuses on the 'Memory and Forgetting' topic, drawing on core studies, theories and research methods. By working through each question and its explanation, you will consolidate your knowledge of key psychological concepts and sharpen your examination skills. All questions are designed to mirror the style found in CAIE past papers, helping you become familiar with command words such as 'describe', 'evaluate' and 'discuss'.
本文详细解析了一份为 Year 11 CAIE IGCSE 心理学课程设计的单元模拟测试卷，主题聚焦于「记忆与遗忘」，涵盖核心研究、理论及研究方法。通过逐一分析每道题目及其解析，你将巩固关键心理学概念的理解并提升应试技巧。所有问题均仿照 CAIE 历年试题风格设计，帮助你熟悉「描述」、「评价」和「讨论」等指令词。
1. Paper Overview and Assessment Objectives | 试卷概览与评估目标
This mock paper lasts 45 minutes and carries 40 marks. It is divided into three parts: Section A (short-answer and multiple-choice questions worth 1-2 marks each), Section B (structured questions worth 4-6 marks) and Section C (one extended response worth 8-10 marks). Every question is mapped onto one of the three assessment objectives so that you can see exactly what the examiner is rewarding.
这份模拟卷作答时间为 45 分钟，总分 40 分，分为三个部分：A 部分（每题 1-2 分的简答与选择题）、B 部分（每题 4-6 分的结构化题目）以及 C 部分（一道 8-10 分的论述大题）。每道题都对应一个评估目标，让你清楚知道考官究竟在给什么分。
The three assessment objectives are: AO1 - demonstrate knowledge and understanding of psychological ideas, theories and studies; AO2 - apply that knowledge to a novel scenario or piece of evidence; AO3 - analyse, interpret and evaluate psychological ideas and evidence, including methodological criticism.
三个评估目标分别是：AO1——展现对心理学观点、理论和研究的知识与理解；AO2——将知识应用于新情境或新证据；AO3——分析、解读与评价心理学观点和证据，包括方法学层面的批判。
Because the unit is 'Memory and Forgetting', roughly half of the marks test your factual recall of the multi-store model, the working memory model and the core studies, while the other half test your ability to use those ideas to explain real behaviour, such as why a witness misremembers a face or why you forget a new phone number.
由于本单元主题是「记忆与遗忘」，约一半分值考查你对多存储模型、工作记忆模型和核心研究的事实性记忆，另一半则考查你运用这些理论解释真实行为的能力，例如目击者为何记错面孔，或你为何记不住一个新电话号码。
2. Command Words and Mark Bands | 指令词与评分档次
Marks are lost far more often through misreading the command word than through not knowing the psychology. 'Describe' asks for a factual account - what happened, what the model says - with no evaluation. 'Explain' demands a reason or mechanism, usually signalled by the word 'because'. 'Evaluate' or 'discuss' requires strengths, weaknesses, comparison, and a supported judgement.
失分的原因往往不是不懂心理学，而是误读指令词。「Describe（描述）」要求事实性陈述——发生了什么、模型怎么说——不需要评价。「Explain（解释）」要求给出原因或机制，通常由「because（因为）」来体现。「Evaluate / Discuss（评价/讨论）」则要求优点、缺点、比较以及有依据的判断。
| Command word | What the examiner wants | Typical tariff |
| Outline | One brief point per mark, no development | 2 marks |
| Describe | Accurate factual detail, no evaluation | 4 marks |
| Explain | Point plus mechanism or reason | 4-6 marks |
| Evaluate / Discuss | Strengths, weaknesses, evidence, judgement | 6-10 marks |
A practical rule for the 6-mark 'evaluate' question in this mock is the 3 + 3 structure: three developed strengths or three developed weaknesses, each with a named study or a consequence for the theory. A judgement sentence at the end is worth roughly one mark and is the easiest mark in the paper to secure.
针对本卷 6 分「评价」题的一个实用规则是「3 + 3 结构」：三个展开的优点或三个展开的缺点，每一点都配一项具名研究或说明该缺点对理论的影响。最后的判断句大约值 1 分，却是全卷最容易拿到的分。
3. Question 1 Analysed: Sensory Register Characteristics | 第 1 题解析：感觉登记的特征
Question: Which of the following is a defining feature of the sensory register according to the multi-store model of memory?
问题： 根据记忆的多存储模型，以下哪项是感觉登记的定义特征？
A. It encodes information semantically. | A. 它对信息进行语义编码。
B. Its capacity is limited to 7 ± 2 chunks. | B. 它的容量限制在 7 ± 2 个组块。
C. It has a very brief duration and a large capacity. | C. 它具有极短持续时间和很大容量。
D. Information enters it through maintenance rehearsal. | D. 信息通过维持性复述进入。
The correct answer is C. Atkinson and Shiffrin's multi-store model proposes that the sensory register holds incoming sensory information in an unprocessed, modality-specific form for a fraction of a second. Iconic memory (visual) lasts around 0.5 seconds, while echoic memory (auditory) can persist for up to about 2 seconds. Its capacity is effectively unlimited, but most information decays rapidly unless attention selects it for transfer to short-term memory.
正确答案是 C。Atkinson 和 Shiffrin 的多存储模型提出，感觉登记以未经加工、按感觉通道区分的形式保存进入系统的信息，时间只有几分之一秒。图像记忆（视觉）约持续 0.5 秒，声像记忆（听觉）则可维持约 2 秒。其容量实际上近乎无限，但若注意力未将其选择并转入短时记忆，大部分信息会迅速衰退。
Option A describes the preferred encoding of long-term memory, not the sensory register - LTM stores meaning rather than raw sensation. Option B refers to the capacity of short-term memory as identified by Miller (1956), who suggested we can hold about seven (plus or minus two) items or chunks at once. Option D is incorrect because maintenance rehearsal is a process that keeps information active in short-term memory and may transfer it to long-term memory; nothing enters the sensory register through rehearsal, since it is the first and most involuntary stage of the system.
选项 A 描述的是长时记忆偏好的编码方式而非感觉登记——长时记忆储存的是意义，而不是原始感觉。选项 B 指的是 Miller（1956）提出的短时记忆容量，他认为我们一次大约能保持七个（加减二）项目或组块。选项 D 不正确，因为维持性复述是让信息在短时记忆中保持活跃、并可能将其转入长时记忆的过程；信息不可能通过复述进入感觉登记，因为感觉登记是系统中最初、也最不依赖意志的阶段。
A common exam error here is confusing the capacity of the sensory register with that of short-term memory. Remember the summary line: sensory register - huge capacity, tiny duration; short-term memory - small capacity, short duration; long-term memory - unlimited capacity, potentially permanent duration.
此处常见的考试错误是把感觉登记的容量与短时记忆的容量混为一谈。请记住这条总结：感觉登记——容量极大、持续时间极短；短时记忆——容量小、持续时间短；长时记忆——容量无限、持续时间可能永久。
4. Short-Term Memory: Capacity, Duration and Encoding | 短时记忆：容量、持续时间与编码
Short-term memory (STM) is the component that holds a limited amount of information in a consciously available state. Miller (1956) reviewed evidence and concluded that the span of immediate memory is about seven items, which he called 'the magic number seven plus or minus two'. Crucially, capacity is measured in chunks rather than raw units, which is why '0 2 0 7 1 2 3 4 5 6 7' is far harder to hold than '0207 1234 567'.
短时记忆（STM）是将有限信息保持在意识可及状态的成分。Miller（1956）综观证据后得出结论：即时记忆广度约为七个项目，他称之为「神奇数字七加减二」。关键在于，容量是以组块而非原始单位计算的，这正是「0 2 0 7 1 2 3 4 5 6 7」远比「0207 1234 567」难记的原因。
Duration is very short. Peterson and Peterson (1959) found that recall of consonant trigrams dropped to about 2-5% after 18 seconds when rehearsal was prevented. Encoding is predominantly acoustic: Baddeley (1966) showed that participants confused words that sounded similar (for example 'cat', 'cap', 'can') far more than words that meant similar things when recalling from STM.
持续时间非常短。Peterson 和 Peterson（1959）发现，在阻止复述的条件下，辅音三字母组合的回忆率在 18 秒后降至约 2-5%。编码以听觉为主：Baddeley（1966）证明，从短时记忆中回忆时，被试对发音相似的词（如 cat、cap、can）产生的混淆远多于意义相近的词。
A strong evaluation point for this section is that STM is probably not a single store. Shallice and Warrington (1970) studied patient KF, whose short-term forgetting for verbal material was severe while his visual and long-term memory remained intact - evidence that STM is fractionated and that the multi-store model oversimplifies it. This criticism leads directly to Baddeley and Hitch's working memory model.
本小节一个有力的评价点是：短时记忆很可能并非单一存储。Shallice 和 Warrington（1970）研究了病人 KF，他对言语材料的短时遗忘非常严重，但视觉记忆与长时记忆保持完好——这说明短时记忆是分化的，多存储模型过于简化。这一批评直接引出了 Baddeley 和 Hitch 的工作记忆模型。
5. Long-Term Memory and the Multi-Store Model | 长时记忆与多存储模型
Long-term memory (LTM) is described in the multi-store model as a permanent store of unlimited capacity, with information encoded mainly semantically. Two well-known clinical cases support the distinction between stores: patient HM could not form new long-term memories after surgery but could still hold and rehearse information in STM for a short time, and Clive Wearing retained musical and procedural skills while losing autobiographical memory.
在多存储模型中，长时记忆（LTM）被描述为容量无限、近乎永久的存储，信息主要以语义方式编码。两个著名的临床案例支持存储之间的区分：病人 HM 在手术后无法形成新的长时记忆，但仍能在短时间内保持并复述短时记忆中的信息；Clive Wearing 保留了音乐与程序性技能，却失去了自传体记忆。
The full model runs in one direction with two control processes: sensory information enters the sensory register, attention transfers it to STM, maintenance rehearsal keeps it in STM and elaborative rehearsal transfers it to LTM, and retrieval brings it back. Atkinson and Shiffrin presented this as a linear flow, which is precisely what later researchers criticised.
整个模型沿一个方向运行，并包含两个控制过程：感觉信息进入感觉登记，注意力将其转入短时记忆，维持性复述使其留在短时记忆，精细复述则将其转入长时记忆，提取再把信息调回。Atkinson 和 Shiffrin 将其呈现为线性流程，而这恰恰是后来研究者批评的地方。
Evaluation: the model is supported by case studies, by the serial position effect (primacy reflects LTM, recency reflects STM) and by brain-scanning evidence of different regions being active for different tasks. However, it presents STM and LTM as single, unitary stores, ignores the role of effort and strategy, and describes rehearsal as the only route into LTM when in fact distinctiveness and depth of processing matter more.
评价：该模型得到案例研究、系列位置效应（首因效应反映长时记忆、近因效应反映短时记忆）以及不同任务激活不同脑区的脑成像证据支持。但它把短时记忆与长时记忆视为单一、不可分割的存储，忽视了努力与策略的作用，并把复述当作进入长时记忆的唯一通路，而事实上独特性和加工深度更为重要。
6. Rehearsal: Maintenance and Elaborative | 复述：维持性复述与精细复述
Maintenance rehearsal is rote repetition of information, such as silently repeating a phone number. It maintains an item in STM but is a weak route to LTM. Elaborative rehearsal involves linking new material to existing knowledge, for example turning 'HOMES' into a cue for the five Great Lakes or building a story around a list of words. Elaborative rehearsal produces far stronger long-term retention.
维持性复述是对信息的机械重复，例如默念一个电话号码。它能让项目留在短时记忆中，但通向长时记忆的路径很弱。精细复述则把新信息与已有知识联系起来，例如用「HOMES」来记住五大湖，或为一串词编写一个故事。精细复述带来的长期保持效果要强得多。
Craik and Lockhart's levels of processing framework (1972) develops this idea: structural, shallow processing of how a word looks produces poor recall, whereas semantic, deep processing of what the word means produces strong recall. This shows that rehearsal alone does not explain transfer to LTM, which is a limitation of the multi-store model.
Craik 和 Lockhart 的加工层次框架（1972）发展了这一点：对单词外形进行的结构性的浅层加工回忆效果差，而对其含义进行的语义性的深层加工回忆效果好。这说明单靠复述并不能解释向长时记忆的转移，这也是多存储模型的一个局限。
In an exam answer, always name the process and then state its function: 'Maintenance rehearsal keeps information in STM and only transfers it to LTM after prolonged repetition; elaborative rehearsal transfers information to LTM more effectively because it organises the material and links it to existing schemas.'
在考试作答中，务必先说出过程名称，再说明其功能：「维持性复述让信息留在短时记忆，只有在长时间重复后才转入长时记忆；精细复述之所以能更有效地转入长时记忆，是因为它对材料进行了组织，并与已有图式建立联系。」
7. Core Study: Peterson and Peterson (1959) | 核心研究：Peterson 与 Peterson（1959）
Peterson and Peterson (1959) investigated the duration of short-term memory. They tested 24 undergraduate participants, showing each a consonant trigram such as 'CHJ' and then asking them to count backwards in threes from a given number. Counting backwards prevented rehearsal, an important control. Recall of the trigram was tested at retention intervals of 3, 6, 9, 12, 15 and 18 seconds.
Peterson 和 Peterson（1959）研究短时记忆的持续时间。他们以 24 名大学生为被试，先呈现一个辅音三字母组合（如 CHJ），随后要求被试从某个数字开始每次倒数三个数。倒数任务阻止了复述，是一项关键的实验控制。三字母组合的回忆在 3、6、9、12、15 和 18 秒的保持间隔后进行测试。
Results showed that recall was about 80-90% after 3 seconds but fell sharply to around 2-5% after 18 seconds. The researchers concluded that short-term memory has a duration of less than 20 seconds when rehearsal is prevented, and that the information was lost through trace decay rather than being displaced.
结果显示，3 秒后回忆率约为 80-90%，但到 18 秒后急剧下降至约 2-5%。研究者据此得出结论：在阻止复述的条件下，短时记忆的持续时间不足 20 秒，且信息是因痕迹衰退而丢失，而非被挤出。
Evaluation: the study is a well-controlled laboratory experiment, so extraneous variables such as rehearsal were minimised and the procedure is replicable. However, ecological validity is low - trigrams are meaningless and far removed from everyday memory. A further criticism is that the counting task itself interfered with the memory of the trigram, so the forgetting may be explained by interference rather than decay, which weakens the conclusion. The task also caused mild stress in some participants, so protection from harm and debriefing were ethically necessary.
评价：该研究是控制良好的实验室实验，因此复述等无关变量被降到最低，程序具有可重复性。但其生态效度低——三字母组合没有意义，与日常记忆相距甚远。更进一步的批评是，倒数任务本身干扰了对三字母组合的记忆，所以遗忘也可能用干扰而非衰退来解释，这削弱了其结论。该任务还使部分被试产生轻度压力，因此从伦理上必须提供免受伤害的保障并进行事后说明。
8. Core Study: Bartlett (1932) | 核心研究：Bartlett（1932）
Bartlett (1932) investigated reconstructive memory using 'The War of the Ghosts', a Native American folk tale that was unfamiliar to his Cambridge participants. Using serial reproduction (passing the story on and back, like Chinese whispers) and repeated reproduction (the same person retelling it over weeks and months), he found that the story became shorter, more stereotyped and more culturally familiar.
Bartlett（1932）用「鬼魂之战」这一对其剑桥被试而言陌生的北美原住民民间故事，研究重构性记忆。通过系列复制（像传话游戏一样依次转述）与重复复制（同一人数周、数月内反复复述），他发现故事变短、更刻板，也更接近自己的文化。
Specific changes included the addition of causal links where none existed (rationalisation), the alteration of unfamiliar terms to familiar ones (for example 'canoe' becoming 'boat'), and the dropping of detail that did not fit existing knowledge. Bartlett concluded that memory is an active, reconstructive process guided by schemas - organised mental frameworks based on prior experience - and he captured this in his phrase 'effort after meaning'.
具体变化包括：在原本没有因果联系处添加因果联系（合理化）；把陌生词汇改成熟悉的词（例如 canoe 变成 boat）；删去不符合既有知识的细节。Bartlett 由此得出，记忆是由图式（基于先前经验的有组织心理框架）引导的主动重构过程，并用「对意义的追寻」这一说法加以概括。
Evaluation: the study has high mundane realism because it uses meaningful, story-like material, and it has strong applied value in explaining errors in eyewitness testimony. However, it was qualitative and lacked standardised instructions and objective scoring, so the findings are hard to replicate and may have been influenced by demand characteristics. It also used a single culturally specific story with a British sample, raising questions of cultural bias and generalisability.
评价：该研究使用有意义的故事性材料，具有较高的生态效度，并在解释目击证词错误方面有很强的应用价值。但它属于质性研究，缺乏标准化指导语与客观评分，因此结果难以重复，也可能受需求特征影响。它还只使用了一个具有特定文化背景的故事和英国被试，带来了文化偏差与可推广性的问题。
9. Forgetting I: Trace Decay and Interference | 遗忘理论（一）：痕迹衰退与干扰
Trace decay theory states that memories leave a physical or chemical trace in the brain, and that this trace fades over time unless it is rehearsed. Decay is the main explanation for forgetting from the sensory register and is a plausible account of short-term forgetting, but it is difficult to demonstrate in humans because the passage of time and the loss of memory are almost always confounded.
痕迹衰退理论认为，记忆会在脑中留下物理或化学痕迹，若不复述，痕迹便随时间消退。衰退是感觉登记遗忘的主要解释，也是短时遗忘的一种合理说明，但很难在人类身上证明，因为时间的流逝与记忆的丧失几乎总是混淆在一起。
Interference theory states that we forget because other learning competes with the target memory. Proactive interference occurs when older learning disrupts newer learning, for example calling a new partner by an ex-partner's name. Retroactive interference occurs when newer learning disrupts older learning, for example studying French immediately after Spanish and finding the Spanish vocabulary harder to recall.
干扰理论认为，遗忘是因为其他学习与目标记忆相互竞争。前提干扰发生在旧学习干扰新学习时，例如把新
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