📚 Unit Test Mock Paper Analysis: Memory | 单元测试模拟卷解析：记忆
This article provides a detailed walk-through of a mock unit test for the Edexcel Year 10 Psychology topic on Memory. Each question type is broken down with full explanations, model answers, and commentary on how marks are awarded. By studying this analysis, you will consolidate key theories, strengthen your research methods skills, and learn how to structure high‑scoring responses.
本文详细解析一份针对爱德思 Year 10 心理学“记忆”单元的模拟测试卷。每一类题型都配有完整讲解、参考答案以及评分注释。通过学习本解析，你将巩固核心理论、提升研究方法技能，并掌握高分答题的结构。
1. Introduction to the Mock Paper | 模拟试卷简介
The mock test is divided into four sections. Section A contains five multiple‑choice questions worth 1 mark each, covering basic terminology and model components. Section B includes three short‑answer questions (4 marks each) that require description and brief application of memory theories. Section C presents a research methods scenario with identification of variables, experimental design, and a simple calculation (6 marks). Section D is an extended writing question (9 marks) asking students to evaluate the multi‑store and working memory models. The total mark is 40, simulating a complete end‑of‑unit assessment.
本模拟卷分为四个部分。A 部分包含五道单选题，每题 1 分，考查基本术语与模型成分。B 部分有三道简答题（各 4 分），要求描述并简单应用记忆理论。C 部分是一道研究方法情景题，涉及变量识别、实验设计识别和简单计算（6 分）。D 部分为论述题（9 分），要求学生评价多存储模型与工作记忆模型。总分 40 分，模拟完整的单元测评。
2. Multiple‑Choice Questions: Core Concepts in Memory | 选择题：记忆核心概念
Question 1 asks which component of the multi‑store model holds information for less than a second. The correct answer is Sensory Memory. Atkinson and Shiffrin (1968) proposed that environmental stimuli are first registered in sensory stores (iconic and echoic), where they last only fractions of a second before they decay or are attended to.
第 1 题问多存储模型中哪个成分保存信息不到一秒。正确答案是感觉记忆。阿特金森和希夫林（1968）提出，环境刺激首先被感觉存储器（图像和声像）登记，在那里只持续几分之一秒，随后消退或被注意。
Question 2 tests knowledge of the working memory model. It asks which component is responsible for coordinating the slave systems. The answer is the Central Executive. This supervisory system directs attention, allocates data to the phonological loop and visuospatial sketchpad, and has limited capacity.
第 2 题考查工作记忆模型的知识，问哪个组件负责协调从属系统。答案是中央执行器。这个监控系统主导注意力分配，将数据分配给语音环路和视空间画板，并且容量有限。
Question 3 presents a scenario: a student recalls old physics formulas while trying to learn new chemistry equations, causing confusion. This is an example of Proactive Interference, where older information disrupts the recall of newer learning. Retroactive interference works in the opposite direction.
第 3 题给出情景：一名学生在试图学习新的化学方程式时回忆起旧的物理公式，导致混淆。这是前摄干扰的例子，即旧信息干扰对新内容的提取。倒摄干扰则方向相反。
Question 4 requires identifying the study that demonstrated reconstructive memory. The classic research by Bartlett (1932), using the ‘War of the Ghosts’ story, showed that participants distorted details to fit their own cultural schemas. The answer is Bartlett’s ‘War of the Ghosts’.
第 4 题需要识别展示重构记忆的研究。巴特利特（1932）利用“幽灵战争”故事进行的经典研究表明，参与者会歪曲细节以符合自己的文化图式。答案是巴特利特的“幽灵战争”。
Question 5 asks about duration of short‑term memory without rehearsal. The typical estimate from Peterson & Peterson (1959) is 18–30 seconds. In their experiment, recall of trigrams dropped rapidly when rehearsal was prevented by a Brown‑Peterson task.
第 5 题问无复述时短时记忆的持续时间。彼得森夫妇（1959）的典型估计是18–30 秒。在他们的实验中，当布朗‑彼得森任务阻止了复述后，三字母组的回忆成绩迅速下降。
3. Short‑Answer Question: Multi‑Store Model | 简答题：多存储模型
The question is: ‘Outline the multi‑store model of memory. In your answer, refer to the capacity and duration of two of its stores.’ (4 marks). A strong response identifies the three unitary stores: sensory memory, short‑term memory (STM), and long‑term memory (LTM). It states that information flows linearly via attention and rehearsal. For capacity and duration, choose STM (capacity 7 ± 2 items, duration 18–30 seconds) and LTM (capacity potentially unlimited, duration up to a lifetime). One mark is awarded for each accurate description and each correctly stated feature.
题目是：“概述记忆的多存储模型。在你的回答中，要提及其中两个存储器的容量和持续时间。”（4 分）。一份优秀回答会指明三个单一存储器：感觉记忆、短时记忆和长时记忆，并说明信息通过注意和复述线性流动。关于容量和持续时间，可选择短时记忆（容量 7 ± 2 个项目，持续时间 18–30 秒）和长时记忆（容量可能无限，持续时间可达终身）。每个准确描述和正确陈述的特征各得一分。
A common mistake is confusing STM with sensory memory or failing to mention rehearsal. Also, simply saying ‘STM duration is short’ without a time range loses the mark. Reference to Atkinson & Shiffrin (1968) strengthens the answer but is not compulsory for full credit.
常见错误是把短时记忆与感觉记忆混淆，或者未提到复述。同样，只说“短时记忆持续时间短”而不给出时间范围会失分。提及阿特金森和希夫林（1968）可以增强答案，但满分并不强制要求。
4. Short‑Answer Question: Working Memory Model | 简答题：工作记忆模型
‘Describe two components of Baddeley and Hitch’s (1974) working memory model.’ (4 marks). The phonological loop can be described as a temporary storage system for verbal and auditory information, subdivided into the phonological store and the articulatory loop. The visuospatial sketchpad handles visual and spatial data, allowing mental imagery and navigation. A description of the central executive as an attention controller that coordinates the slave systems is also acceptable. Two marks per component: one for naming and one for an accurate description with an example.
“描述巴德利和希奇（1974）工作记忆模型中的两个成分。”（4 分）。语音环路可被描述为一个临时储存言语和听觉信息的系统，细分为语音存储器和发音环路。视空间画板则处理视觉和空间数据，支持心理意象和导航。中央执行器作为协调从属系统的注意力控制器也是可接受的描述。每个成分两分：一分给命名，一分给准确的描述并举例。
To earn all four marks, avoid vague phrases. Instead of ‘it processes visual stuff’, write ‘the visuospatial sketchpad enables you to picture the layout of your classroom and mentally rotate objects.’ The episodic buffer was added in 2000 and is not required here, but including it accurately may show depth.
要拿到满分四分，应避免模糊表述。不要写“它处理视觉东西”，而应写“视空间画板使你能够想象教室的布局并在头脑中旋转物体”。虽然情景缓冲区于 2000 年被加入，在此不必提及，但准确加入可以展示深度。
5. Short‑Answer Question: Forgetting Theories | 简答题：遗忘理论
‘Explain one reason for forgetting, using a study to support your answer.’ (4 marks). The most effective route is to choose retrieval failure (cue‑dependent forgetting) or interference theory. For retrieval failure, describe Tulving’s encoding specificity principle: recall is best when the context at retrieval matches the context at encoding. Godden & Baddeley (1975) showed that divers who learned words underwater recalled more underwater than on land, demonstrating context‑dependent forgetting.
“解释遗忘的一个原因，并引用一项研究支持你的答案。”（4 分）。最有效的路线是选择提取失败（线索依赖遗忘）或干扰理论。对于提取失败，描述塔尔文的编码特异性原则：提取时的情境与编码时匹配时，回忆效果最佳。戈登和巴德利（1975）表明，在水下学习单词的潜水员在水下回忆得比在陆地上更多，展示了情境依赖遗忘。
If using interference, you might outline proactive interference with Underwood (1957), who found that participants who learned multiple lists recalled fewer items from the final list. Marks are awarded for naming the theory (1), describing it (1), linking to the study (1), and a brief conclusion (1).
若使用干扰理论，你可以概述前摄干扰并引用安德伍德（1957），他发现学习了多组词表的参与者从最后一组词表中回忆的项目更少。分数分配为：命名理论（1 分），描述理论（1 分），关联研究（1 分），简短结论（1 分）。
6. Research Methods: Experimental Design | 研究方法：实验设计
This scenario describes a study where two groups of Year 10 students were tested on memory for a 20‑word list. Group A listened to the words in silence; Group B listened while loud pop music played. The task was to write down as many words as possible in order. Students were randomly allocated to groups. Identify the experimental design and name the independent variable (IV) and dependent variable (DV).
该情景描述了一项研究：两组 Year 10 学生接受对一份 20 个单词的词表进行记忆测试。A 组在安静环境中听单词；B 组在响亮的流行音乐中听单词。任务是在听完后按顺序写下尽可能多的单词。学生被随机分配到各组。请识别实验设计，并写出自变量和因变量。
This is an independent groups design because different participants are used in each condition. The IV is the presence or absence of background music (or sound condition: silence vs. music). The DV is the number of words correctly recalled in the correct position (or proportion of words recalled). Mentioning order accuracy shows deeper understanding.
这是独立组设计，因为每种条件下使用的是不同的参与者。自变量是背景音乐的存在与否（或声音条件：安静 vs. 音乐）。因变量是在正确位置上正确回忆出的单词数量（或单词回忆比例）。提及顺序准确性能展现更深的理解。
Explain one strength of this design: there are no order effects, and participants are less likely to guess the aim. One limitation: participant variables (e.g. music preference influence) may confound results, though random allocation helps to control this.
解释该设计的一个优点：没有顺序效应，参与者也不容易猜出研究目的。一个局限：参与者变量（如对音乐的偏好）可能会混淆结果，尽管随机分配有助于控制这一点。
7. Data Analysis: Measures of Central Tendency | 数据分析：集中趋势量数
The question provides a data set: the number of words recalled in the correct position by the silence group were 8, 12, 7, 9, 13, 15, 12, 10. Calculate the mean, median, and mode. Show your working. (3 marks). The mean is (8+12+7+9+13+15+12+10)/8 = 86/8 = 10.75. The mode is 12, as it appears twice. For the median, order the scores: 7, 8, 9, 10, 12, 12, 13, 15. With 8 values, median is (10+12)/2 = 11. Always present the answers to one decimal place if required.
题目给出一个数据集：安静组在正确位置上回忆的单词数为 8, 12, 7, 9, 13, 15, 12, 10。计算均值、中位数和众数，并展示计算过程。（3 分）。均值为 (8+12+7+9+13+15+12+10)/8 = 86/8 = 10.75。众数为 12，因为出现两次。中位数：将分数排序为 7, 8, 9, 10, 12, 12, 13, 15。有 8 个值，中位数是 (10+12)/2 = 11。若要保留小数位，始终根据要求保留一位小数。
When interpreting, note that the mean (10.75) is slightly lower than the median (11) because of the low outlier ‘7’. This makes the median a better measure of central tendency for this skewed data set. Examiners reward such commentary in evaluation questions.
在进行解释时，注意均值（10.75）略低于中位数（11），因为有一个低分异常值“7”。这使得中位数成为这组偏态数据更好的集中趋势量数。评分者在评价类问题中会青睐这类评论。
8. Extended Writing: Evaluating Memory Models | 论述题：评价记忆模型
‘Compare the multi‑store model and the working memory model. Evaluate both models using evidence. (9 marks)’ A top‑band answer will structure a balanced discussion. Start by briefly describing each model. Then identify two similarities: both are structural models that distinguish between temporary and permanent storage, and both recognise the limited capacity of short‑term/working memory.
“比较多存储模型与工作记忆模型。运用证据评价两个模型。（9 分）”高分答案会构建一个平衡的讨论。首先简要描述每个模型。然后指出两个相似点：两者都是区分临时和永久存储的结构模型，且两者都承认短期/工作记忆的容量有限。
Key differences should include: the multi‑store model views STM as a single unitary store, whereas working memory presents multiple components. The multi‑store model emphasises role of rehearsal in transfer, but working memory focuses on processing. Evidence for multi‑store: the serial position effect (Glanzer & Cunitz, 1966) supports separate STM and LTM. Evidence for working memory: dual task studies (Baddeley & Hitch, 1976) demonstrate the existence of separate verbal and visual subsystems. Critical evaluation: the multi‑store model is oversimplified; it cannot explain why some patients with damaged STM still form LTM (e.g. KF case study). The working memory model says little about LTM or how it links. Conclude that while both are valuable, the working memory model offers a more detailed account of immediate memory processes.
关键差异应包括：多存储模型将短时记忆视为一个单一存储器，而工作记忆则提出多个成分。多存储模型强调复述在信息传输中的作用，但工作记忆则侧重加工处理。支持多存储模型的证据：系列位置效应（格兰茨和库尼茨，1966）支持独立的短时和长时记忆。支持工作记忆的证据：双任务研究（巴德利和希奇，1976）证明存在分离的言语和视觉子系统。批判性评价：多存储模型过于简化；它无法解释为何有些短时记忆受损的患者仍能形成长时记忆（如 KF 案例研究）。工作记忆模型对长时记忆及其联系着墨甚少。结论可以是：虽然两者都很有价值，但工作记忆模型对即时记忆过程提供了更详尽的阐释。
9. Applying Memory Research to Study Skills | 记忆研究在学习技巧中的应用
One application question in the mock asks: ‘Using your knowledge of memory research, suggest two strategies a student could use to improve revision. Justify each strategy.’ (4 marks). Valid strategies include using chunking, which breaks information into smaller meaningful groups, thereby reducing the load on working memory and making use of LTM patterns. This links to Miller’s magic number 7 ± 2.
模拟卷中有一道应用题问：“运用你的记忆研究知识，提出学生可以用于提升复习效果的两个策略，并为每个策略提供依据。”（4 分）。有效的策略包括使用组块化，将信息分解成更小的、有意义的组合，从而减轻工作记忆的负荷并利用长时记忆模式。这与米勒的神奇数字 7 ± 2 有关。
Another strategy is elaborative rehearsal, which involves connecting new information to existing knowledge and giving it meaning, facilitating transfer to LTM. This can be supported by Craik & Lockhart’s Levels of Processing theory: deeper, semantic processing leads to more durable memories. Simply rereading notes involves shallow processing and is less effective.
另一个策略是精细复述，即将新信息与已有知识建立联系并赋予意义，促进向长时记忆的转移。这可以用克雷克和洛克哈特的加工层次理论来支持：更深层的语义加工会产生更持久的记忆。仅仅重读笔记涉及浅层加工，效果较差。
Using retrieval practice, such as self‑testing or flashcards, strengthens memory by taking advantage of the testing effect. Each time information is successfully recaptured, the memory trace becomes stronger. This combats storage and retrieval failures. Always justify with a psychological concept.
使用提取练习，如自测或闪卡，利用测试效应来强化记忆。信息每被成功提取一次，记忆痕迹就变得更强。这对抗了存储失败和提取失败。必须用一个心理学概念来提供依据。
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