WJEC Year 13 Statistics Unit Test Mock Paper Walkthrough | WJEC 高三统计单元测试模拟卷解析

📚 WJEC Year 13 Statistics Unit Test Mock Paper Walkthrough | WJEC 高三统计单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for the WJEC Year 13 Statistics syllabus. We dissect each question, highlight the essential statistical techniques, and point out frequent errors. Whether you are revising probability distributions, confidence intervals, hypothesis tests, or chi-squared methods, this analysis will reinforce your understanding and exam technique.

本文详细解析一份针对 WJEC 高三统计课程的单元测试模拟卷。我们逐一分析每道题目，强调核心统计方法，并指出常见错误。无论你是在复习概率分布、置信区间、假设检验还是卡方检验，这份解析都将巩固你的理解并提升应试技巧。

1. Overview of the Mock Paper | 模拟卷概述

The mock paper consists of six compulsory questions, each worth 12 marks, making a total of 72 marks. The questions are designed to test the core topics of the WJEC A2 Statistics unit: discrete distributions, continuous approximations, statistical inference, and bivariate data analysis. Time allowed: 1 hour 15 minutes. You are expected to show full working, state any assumptions, and interpret your results in context.

模拟卷包含六道必答题，每题12分，共计72分。题目旨在覆盖 WJEC A2 统计单元的核心主题：离散分布、连续近似、统计推断以及双变量数据分析。考试时间：1小时15分钟。要求写出完整计算过程，说明所作假设，并结合实际情境解释结果。

A formula booklet is provided, so memorising every distribution function is unnecessary, but you must know when and how to apply them correctly. The mock paper rewards clear logical steps, correct use of notation, and linking conclusions back to the original problem.

考试会提供公式手册，因此无需记住每一个分布函数，但必须清楚何时以及如何正确应用它们。模拟卷考查清晰的逻辑步骤、正确的符号用法，以及能否将结论与实际情境联系起来。

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2. Question 1: Poisson Distribution Calculations | 题目1：泊松分布计算

A call centre receives calls at an average rate of 3 per minute. For a randomly chosen 2-minute interval, find (i) the probability of receiving exactly 5 calls, and (ii) the probability of receiving at least 1 call. State one assumption required for your model.

某呼叫中心平均每分钟接到3通电话。对于随机选取的2分钟时段，求 (i) 恰好接到5通电话的概率，以及 (ii) 至少接到1通电话的概率。说明该模型所需的一条假设。

Let X represent the number of calls in 2 minutes. Since calls occur randomly and independently at a constant average rate, X ~ Po(λ) with λ = 3 × 2 = 6. The probability mass function is:

设 X 为2分钟内的电话数。由于通话随机、独立且以恒定平均速率发生，X ~ Po(λ)，其中 λ = 3 × 2 = 6。概率质量函数为：

P(X = x) = e⁻λ λˣ / x!

For exactly 5 calls, P(X = 5) = e⁻⁶ × 6⁵ ÷ 5! Using a calculator, e⁻⁶ ≈ 0.00248, 6⁵ = 7776, 5! = 120. Hence P(X = 5) ≈ 0.00248 × 7776 / 120 ≈ 0.1606.

对于恰好5通电话，P(X = 5) = e⁻⁶ × 6⁵ ÷ 5！。通过计算器，e⁻⁶ ≈ 0.00248，6⁵ = 7776，5！= 120。因此 P(X = 5) ≈ 0.00248 × 7776 / 120 ≈ 0.1606。

For at least one call, P(X ≥ 1) = 1 – P(X = 0) = 1 – e⁻⁶ ≈ 1 – 0.00248 = 0.9975. This answers part (ii).

对于至少一通电话，P(X ≥ 1) = 1 – P(X = 0) = 1 – e⁻⁶ ≈ 1 – 0.00248 = 0.9975。这回答了 (ii) 问。

The key assumption is that calls arrive independently and at a constant average rate over time. Also, two calls cannot occur at exactly the same instant. Many students forget to multiply the rate by 2 for the 2-minute interval – always align λ with the time base.

关键假设是通话独立且以恒定平均速率随时间到达。此外，两个通话不能在同一瞬间发生。许多同学忘记将2分钟的速率乘以2——务必使 λ 与时间基数一致。

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3. Question 2: Normal Approximation to Binomial | 题目2：二项分布的正态近似

A new medication is effective for 70% of patients. A random sample of 200 patients receives the treatment. Using a normal approximation with a continuity correction, estimate the probability that at least 150 patients benefit from the medication. Comment on the validity of the approximation.

一种新药对70%的患者有效。随机抽取200名患者接受该治疗。使用带连续性校正的正态近似，估计至少150名患者获益的概率。并对近似的有效性进行评述。

Define Y as the number of patients who benefit. Y ~ B(200, 0.7). The mean μ = np = 200 × 0.7 = 140, and the variance σ² = npq = 200 × 0.7 × 0.3 = 42. We approximate Y by a normal distribution N(140, 42).

定义 Y 为获益患者人数。Y ~ B(200, 0.7)。均值 μ = np = 200 × 0.7 = 140，方差 σ² = npq = 200 × 0.7 × 0.3 = 42。我们用正态分布 N(140, 42) 来近似 Y。

For P(Y ≥ 150), apply the continuity correction: P(Y > 149.5). Standardising: z = (149.5 – 140) / √42 = 9.5 / 6.4807 ≈ 1.466. Using standard normal tables, Φ(1.466) ≈ 0.9285. Thus P(Z > 1.466) = 1 – 0.9285 = 0.0715.

对于 P(Y ≥ 150)，使用连续性校正：P(Y > 149.5)。标准化：z = (149.5 – 140) / √42 = 9.5 / 6.4807 ≈ 1.466。查标准正态表，Φ(1.466) ≈ 0.9285。因此 P(Z > 1.466) = 1 – 0.9285 = 0.0715。

The normal approximation is valid because np = 140 > 5 and nq = 60 > 5, satisfying the usual rule of thumb. The continuity correction improves accuracy, as we are approximating a discrete distribution with a continuous one.

正态近似是有效的，因为 np = 140 > 5 且 nq = 60 > 5，满足常用经验法则。连续性校正提高了精度，因为我们是用连续分布来近似离散分布。

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4. Question 3: Confidence Interval for a Mean | 题目3：均值的置信区间

A random sample of 25 students’ test scores has a mean of 72 marks and a standard deviation of 8 marks. Assuming the scores are normally distributed, calculate a 95% confidence interval for the population mean. Explain why the t-distribution rather than the normal distribution is used.

一个随机样本包含25名学生的测验成绩，平均分为72分，标准差为8分。假设成绩服从正态分布，计算总体均值的95%置信区间。解释为何使用 t 分布而非正态分布。

Since the population variance is unknown and the sample size is small (n = 25), the appropriate interval uses the t-distribution with ν = 24 degrees of freedom. The 95% critical value t₀.₀₂₅,₂₄ = 2.064.

由于总体方差未知且样本量较小（n = 25），应使用自由度为 ν = 24 的 t 分布求解置信区间。95%临界值 t₀.₀₂₅,₂₄ = 2.064。

Confidence interval = x̄ ± t × (s / √n)

The standard error = 8 / √25 = 1.6. So the margin of error = 2.064 × 1.6 ≈ 3.30. Hence the 95% CI is 72 ± 3.30, i.e. (68.7, 75.3).

标准误 = 8 / √25 = 1.6。因此误差界限 = 2.064 × 1.6 ≈ 3.30。所以95%置信区间为 72 ± 3.30，即 (68.7, 75.3)。

The t-distribution is used because we have estimated the population standard deviation by the sample standard deviation, introducing additional uncertainty. If σ were known, a z-interval would be appropriate, but here the t-interval correctly reflects the extra variability.

之所以使用 t 分布，是因为我们用样本标准差估计了总体标准差，引入了额外的不确定性。如果 σ 已知，则应使用 z 区间；但此处 t 区间能正确反映这种额外波动。

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5. Question 4: Hypothesis Test for a Binomial Proportion | 题目4：二项比例假设检验

A manufacturer claims that the proportion of defective items in a batch is at most 5%. To investigate, a random sample of 100 items is inspected, and 8 are found to be defective. Test the manufacturer’s claim at the 5% significance level, stating your hypotheses, p-value and conclusion clearly.

某制造商声称其一批产品中次品率不超过5%。为检验该声称，随机抽取100件产品进行检查，发现8件次品。在5%显著性水平下检验该声称，清楚地陈述原假设、备择假设、p 值及结论。

Let p be the true proportion of defective items. H₀: p = 0.05, H₁: p > 0.05. Under H₀, the number of defectives X ~ B(100, 0.05). The test statistic is X = 8, and we compute the p-value as P(X ≥ 8 | p = 0.05).

设 p 为真实的次品率。H₀: p = 0.05，H₁: p > 0.05。在 H₀ 下，次品个数 X ~ B(100, 0.05)。检验统计量为 X = 8，计算 p 值 = P(X ≥ 8 | p = 0.05)。

Using binomial tables or a calculator, P(X ≤ 7) ≈ 0.8720, so P(X ≥ 8) = 1 – 0.8720 = 0.1280. Since p-value = 0.1280 > 0.05, there is insufficient evidence to reject H₀.

使用二项分布表或计算器，P(X ≤ 7) ≈ 0.8720，因此 P(X ≥ 8) = 1 – 0.8720 = 0.1280。由于 p 值 = 0.1280 > 0.05，没有足够的证据拒绝 H₀。

We conclude that the sample does not provide significant evidence to dispute the manufacturer’s claim. However, failing to reject H₀ does not prove the claim is true – it simply means the observed 8 defectives could reasonably occur by chance under a 5% defect rate.

我们的结论是，该样本并未提供显著证据来质疑制造商的说法。然而，未能拒绝 H₀ 并不能证明该说法为真——只是说明在5%次品率下，观测到8件次品有可能自然地偶然发生。

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6. Question 5: Chi-Squared Test for Independence | 题目5：卡方独立性检验

A survey asks 100 people about their gender and whether they prefer a new drink. The observed frequencies are shown below. Test, at the 5% level, whether gender and drink preference are independent.

一项调查询问了100人的性别以及他们是否偏爱一款新饮品。实际观测频数如下表所示。在5%显著性水平下检验性别与饮品偏好是否独立。

	Prefer	Not Prefer	Total
Male	28	12	40
Female	42	18	60
Total	70	30	100

First, compute expected frequencies under independence: expected = (row total × column total) / grand total. For Male & Prefer: (40×70)/100 = 28; Male & Not Prefer: (40×30)/100 = 12; Female & Prefer: (60×70)/100 = 42; Female & Not Prefer: (60×30)/100 = 18. Notice that in this constructed example the observed equal the expected, so χ² = 0, but let us work through the method as if there were small differences to illustrate the process.

首先，计算独立性假设下的期望频数：期望值 = (行合计 × 列合计) / 总计。如 男性&偏好：(40×70)/100 = 28；男性&不偏好：12；女性&偏好：42；女性&不偏好：18。注意在此构建的例子中观测值等于期望值，因此 χ² = 0，但我们假定略有差异以展示计算过程。

The test statistic is χ² = Σ (O – E)² / E. With 2 rows and 2 columns, degrees of freedom = (2–1)×(2–1) = 1. The 5% critical value from χ² tables is 3.841. If the computed χ² is less than 3.841, we do not reject independence. In a real scenario, even small deviations might produce a χ² below this threshold, so you must compare correctly.

检验统计量为 χ² = Σ (O – E)² / E。对于2行2列表格，自由度 = (2–1)×(2–1) = 1。查卡方分布表得5%临界值为3.841。若计算 χ² 小于3.841，则不拒绝独立性假设。在实际情况下，即便有微小偏差，χ² 也可能低于此阈值，因此务必正确比较。

In practice, the chi-squared test requires expected frequencies to be at least 5 for the approximation to be valid. Here all expected values are ≥ 12, so the test is appropriate. Always state your conclusion in context, e.g. ‘There is insufficient evidence to suggest an association between gender and drink preference.’

在实际应用中，卡方检验要求期望频数至少为5才能保证近似的有效性。此处所有期望值均 ≥ 12，故检验可行。请务必结合情境陈述结论，例如“没有足够证据表明性别与饮品偏好之间存在关联”。

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7. Question 6: Regression Line and Prediction | 题目6：回归直线与预测

A study records the number of years of education (x) and annual income in thousands of pounds (y) for 10 individuals. Summary statistics: Σx = 140, Σy = 480, Σxy = 7490, Σx² = 2250, n = 10. (i) Find the equation of the regression line of y on x. (ii) Predict the income for someone with 16 years of education and comment on the reliability of your prediction.

一项研究记录了10个人的受教育年限（x）和年收入（千英镑，y）。汇总统计量：Σx = 140，Σy = 480，Σxy = 7490，Σx² = 2250，n = 10。(i) 求 y 对 x 的回归直线方程。(ii) 预测受教育16年者的收入，并对预测的可靠性加以评述。

First compute Sₓₓ = Σx²

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