Year 13 OCR Statistics: Unit Test Mock Paper Analysis | 单元测试模拟卷解析

📚 Year 13 OCR Statistics: Unit Test Mock Paper Analysis | 单元测试模拟卷解析

This mock paper is designed to mirror the OCR A Level Mathematics specification for Statistics in Year 13, covering core topics such as the Binomial, Poisson and Normal distributions, hypothesis testing, chi-squared tests, correlation and regression, and sampling distributions. Each question has been crafted to test both computational fluency and conceptual understanding under timed conditions. The worked solutions and commentary below highlight the reasoning required to achieve full marks and avoid common pitfalls.

本模拟卷紧密贴合 OCR A Level 数学中 Year 13 统计学的考纲要求，涵盖了二项分布、泊松分布、正态分布、假设检验、卡方检验、相关与回归以及抽样分布等核心考点。试卷中的每一道题都旨在限时条件下同时检验计算熟练度和概念理解。以下逐题解析与点评将重点展示获取满分所需的推理过程，并指出需要规避的常见错误。

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1. Mock Paper Overview | 模拟卷概览

The paper contains six questions to be completed in 75 minutes, with a total of 60 marks. A question typically combines several Assessment Objectives: recall of probability models, selection of appropriate distributions, carrying out hypothesis tests, and interpreting results in context. Tables for the standard normal, Poisson, chi-squared and the product moment correlation coefficient are provided or assumed as given in the examination.

试卷包含六道大题，要求在 75 分钟内完成，总分 60 分。每道题通常综合考查多个评价目标：概率模型的记忆、选择恰当的分布、实施假设检验，以及结合情境解释结果。标准正态分布表、泊松分布表、卡方分布表以及积差相关系数临界值表均由试卷提供或假定已给。

Question	Topic	Marks
1	Binomial & Normal approximation	10
2	Poisson hypothesis test	9
3	Chi-squared goodness of fit	12
4	Product moment correlation	8
5	Regression line & prediction	10
6	Sampling distribution of the mean	11

Careful time management is essential; allocate roughly 1.2 minutes per mark and leave at least 10 minutes for checking calculations and interpretations.

合理的时间分配至关重要：按约 1.2 分钟/分的节奏推进，并至少预留 10 分钟检查计算步骤与文字解释。

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2. Question 1: Binomial Distribution and Normal Approximation | 题目1：二项分布与正态近似

In this question, a random variable X ~ B(200, 0.35) models the number of defective items. Students are asked to estimate P(X ≥ 80) using a normal approximation with a continuity correction. The first step is to verify that the approximation is valid: both np = 70 and nq = 130 exceed 5, so the normal distribution is appropriate.

此题中随机变量 X ~ B(200, 0.35) 表示次品数量。要求使用正态近似并配合连续性校正来估计 P(X ≥ 80)。第一步应验证近似的合理性：np = 70 与 nq = 130 均大于 5，故使用正态分布是合适的。

The approximating normal distribution has mean μ = np = 70 and variance σ² = npq = 200 × 0.35 × 0.65 = 45.5, so X ≈ N(70, 45.5). To find P(X ≥ 80), the continuity correction converts the discrete boundary to 79.5, giving P(X ≥ 80) ≈ P(X > 79.5).

近似正态分布具有均值 μ = np = 70，方差 σ² = npq = 200 × 0.35 × 0.65 = 45.5，故 X ≈ N(70, 45.5)。求 P(X ≥ 80) 时，连续性校正将离散边界调整为 79.5，从而得到 P(X ≥ 80) ≈ P(X > 79.5)。

z = (79.5 – 70) / √45.5 ≈ 1.408

Using standard normal tables, P(Z < 1.41) ≈ 0.9207, so the required probability is 1 – 0.9207 = 0.0793. The final answer should be stated as approximately 0.0793, and a brief comment linking this to the practical chance of observing 80 or more defectives would earn the interpretation mark.

查标准正态分布表得 P(Z < 1.41) ≈ 0.9207，故所求概率为 1 – 0.9207 = 0.0793。最终答案应写出约为 0.0793，并简要说明在实际中观测到 80 件或以上次品的可能性，以获得解释分数。

A common error is to omit the continuity correction and use 80 directly, which yields z = 1.48 and a probability of 0.0694 — an underestimate that loses accuracy marks. Always remember to subtract 0.5 when approximating a ‘≥’ inequality.

常见错误是忽略连续性校正直接使用 80，此时 z = 1.48，概率为 0.0694，属于低估，导致失分。务必记住，当近似 “≥” 型不等式时要减去 0.5。

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3. Question 2: Poisson Hypothesis Test | 题目2：泊松分布假设检验

A call centre receives an average of 2.5 emergency calls per hour. After a system upgrade, six calls are recorded in the first hour. The task is to test, at the 5% significance level, whether there is evidence of an increase in the underlying rate λ.

某呼叫中心每小时平均接到 2.5 通紧急电话。系统升级后的第一个小时内记录到 6 通电话。题目要求在 5% 显著性水平下检验是否有证据表明底层发生率 λ 有所上升。

Let X denote the number of calls in one hour. The hypotheses are H₀: λ = 2.5 against H₁: λ > 2.5. Under H₀, X ~ Po(2.5). The test statistic is the observed value x = 6, and we compute P(X ≥ 6) using the Poisson cumulative formula or tables.

设 X 表示一小时内接到的电话数。假设为 H₀: λ = 2.5，H₁: λ > 2.5。在 H₀ 下，X ~ Po(2.5)。检验统计量为观测值 x = 6，并利用泊松累积公式或表格计算 P(X ≥ 6)。

P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – 0.9580 = 0.0420

Since 0.042 < 0.05, the result is significant. We reject H₀ and conclude that there is sufficient evidence at the 5% level to suggest the rate of emergency calls has increased. It is good practice to explicitly state the comparison: 0.042 < 0.05.

由于 0.042 < 0.05，结果具有显著性。拒绝 H₀，结论为在 5% 显著性水平下有足够证据表明紧急电话发生率有所增加。良好习惯是明确写出比较过程：0.042 < 0.05。

If the test were two-tailed, the p-value would need to be compared against 0.025, but careful reading of the context indicates a one-sided upper-tail test. Misinterpreting the direction of the test is a frequent mistake, so highlight the phrase ‘increase’ when formulating hypotheses.

若为双尾检验，则需将 p 值与 0.025 比较，但仔细阅读上下文可知此处为单侧上尾检验。错误判断检验方向是常见失误，因此在构建假设时应圈出“上升”这类关键词。

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4. Question 3: Chi-Squared Goodness-of-Fit Test | 题目3：卡方拟合优度检验

A six-sided die is rolled 120 times, and the observed frequencies are: 16, 24, 18, 22, 20, 20. The question requires a χ² goodness-of-fit test to determine whether the die is fair at the 5% level.

一颗六面骰子投掷了 120 次，观察频数依次为：16, 24, 18, 22, 20, 20。要求进行 χ² 拟合优度检验，在 5% 水平下判断骰子是否均匀。

Under the null hypothesis of fairness, each face has an expected frequency of 120/6 = 20, which is greater than 5, so no pooling is needed. The test statistic is:

在均匀性的原假设下，每个面的期望频数均为 120/6 = 20，均大于 5，因此无需合并。检验统计量为：

χ² = Σ (Oᵢ – Eᵢ)² / Eᵢ

Face	Observed	Expected	(O – E)²/E
1	16	20	0.8
2	24	20	0.8
3	18	20	0.2
4	22	20	0.2
5	20	20	0.0
6	20	20	0.0
Total	120	120	2.0

The calculated χ² value is 2.0. The number of degrees of freedom is (6 – 1) = 5. From tables, the 5% critical value for χ²₅ is 11.07. Since 2.0 < 11.07, we do not reject H₀. There is insufficient evidence to suggest the die is biased.

计算得 χ² 值为 2.0。自由度为 (6 – 1) = 5。查表得 χ²₅ 的 5% 临界值为 11.07。由于 2.0 < 11.07，不拒绝 H₀。没有充分证据表明骰子不均匀。

Always check expected frequencies before calculating; if any expected value falls below 5, adjacent categories must be combined. In this case all are 20, so the approximation is reliable. Present the calculations in a neat table to secure method marks even if arithmetic slips.

计算前务必核查期望频数；若任一期望值低于 5，则需合并相邻类别。此题中期望值均为 20，因此近似可靠。将计算过程整理为清晰的表格，即使在算术上出现小失误，也能保证得到方法分。

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5. Question 4: Product Moment Correlation Coefficient | 题目4：积差相关系数

Given the data (x, y): (1,4), (2,7), (3,8), (4,13), (5,16), (6,18), (7,19), (8,24), (9,27), (10,30), calculate Pearson’s product moment correlation coefficient r and test for evidence of correlation at the 5% level.

给定数据 (x, y)：(1,4), (2,7), (3,8), (4,13), (5,16), (6,18), (7,19), (8,24), (9,27), (10,30)，要求计算皮尔逊积差相关系数 r，并在 5% 显著性水平下检验是否存在相关关系。

Summary statistics: n = 10, Σx = 55, Σy = 166, Σx² = 385, Σy² = 3300, Σxy = 1168. Then S_{xx} = Σx² – (Σx)²/n = 385 – 3025/10 = 82.5; S_{yy} = Σy² – (Σy)²/n = 3300 – 27556/10 = 544.4; S_{xy} = Σxy – (Σx)(Σy)/n = 1168 – 9130/10 = 255.

汇总统计量：n = 10, Σx = 55, Σy = 166, Σx² = 385, Σy² = 3300, Σxy = 1168。从而 S_{xx} = 385 – 3025/10 = 82.5；S_{yy} = 3300 – 27556/10 = 544.4；S_{xy} = 1168 – 9130/10 = 255。

r = S_{xy} / √(S_{xx} × S_{yy}) = 255 / √(82.5 × 544.4) ≈ 0.970

This indicates a very strong positive linear correlation. To test H₀: ρ = 0 vs H₁: ρ ≠ 0, use the t-statistic:

该值表明存在很强的正线性相关。为检验 H₀: ρ = 0 与 H₁: ρ ≠ 0，使用 t 统计量：

t = r √(n – 2) / √(1 – r²) = 0.970 × √8 / √(1 – 0.9409) ≈ 11.2

With 8 degrees of freedom, the two-tail critical value at 5% is about 2.306. Since 11.2 > 2.306, we reject H₀ and conclude there is significant evidence of correlation. Alternatively, comparing r = 0.970 with the Pearson table critical value for n=10 (0.632) leads to the same decision.

自由度为 8，双尾 5% 临界值约为 2.306。由于 11.2 > 2.306，拒绝 H₀，结论为有显著证据表明存在相关关系。或者，将 r = 0.970 与 n=10 时的皮尔逊表临界值（0.632）比较，也会得到相同结论。

When computing r, ensure you use the full precision of intermediate sums. Early rounding can drastically change the final value, often causing it to fall below a critical threshold.

在计算 r 时，必须使用中间求和项的精确值。过早舍入会显著改变最终结果，常使其低于临界阈值。

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6. Question 5: Regression Line and Prediction | 题目5：回归线与预测

Using the same data set, find the equation of the regression line of y on x and use it to predict y when x = 12. Explain why this prediction may be unreliable.

利用同一组数据，求 y 对 x 的回归线方程，并由此预测 x = 12 时 y 的值。解释该预测可能为何不可靠。

The gradient b = S_{xy} / S_{xx} = 255 / 82.5 ≈ 3.091. The intercept a = ȳ – b x̄ = 16.6 – 3.091×5.5 ≈ –0.40. Thus the regression equation is:

斜率 b = S_{xy} / S_{xx} = 255 / 82.5 ≈ 3.091。截距 a = ȳ – b x̄ = 16.6 – 3.091 × 5.5 ≈ –0.40。因此回归方程为：

y = –0.40 + 3.09x

When x = 12, predicted y = –0.40 + 3.09 × 12 = 36.68 ≈ 36.7. The prediction is extrapolation because x = 12 lies outside the observed range [1,10]. The linear relationship observed within the data may not hold beyond this interval, so the forecast is unreliable.

当 x = 12 时，预测值 y = –0.40 + 3.09 × 12 = 36.68 ≈ 36.7。该预测属于外推，因为 x = 12 超出了观测区间 [1,10]。数据中观测到的线性关系可能在此区间外不成立，因此预测不可靠。

In the exam, always interpret the intercept in context if asked, but here the negative intercept has no physical meaning since x cannot be zero. Good practice requires stating the danger of extrapolation explicitly.

考试中若要求解释截距，应结合具体情境，但此处负截距在 x 不可能为零的情况下没有实际意义。良好习惯是明确指出外推的危险性。

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7. Question 6: Sampling Distribution of the Mean | 题目6：样本均值的抽样分布

Question: The weight of apples from an orchard is normally distributed with mean 150 g and standard deviation 18 g. A random sample of 25 apples is selected. Find the probability that the sample mean weight is less than 145 g.

题目：某果园苹果重量服从正态分布，均值为 150 克，标准差为 18 克。随机抽取 25 个苹果，求样本均值小于 145 克的概率。

If X ~ N(150, 18²), then for a sample of size n = 25, the sample mean X̄ follows a normal distribution with μ_{X̄} = 150 and σ_{X̄} = σ/√n = 18/5 = 3.6. Thus X̄ ~ N(150, 3.6²).

设 X ~ N(150, 18²)，则对于样本容量 n = 25，样本均值 X̄ 服从正态分布，其均值 μ_{X̄} = 150，标准差 σ_{X̄} = σ/√n = 18/5 = 3.6。即 X̄ ~ N(150, 3.6²)。

P(X̄ < 145) = P(Z < (145 – 150) / 3.6) = P(Z < –1.389) ≈ 0.0823

Thus, there is about an 8.23% chance of obtaining a sample mean below 145 g. Many students confuse the standard deviation of the sample mean with the population standard deviation; the division by √n is essential.

因此，样本均值低于 145 克的概率约为 8.23%。许多学生容易混淆样本均值的标准差与总体标准差，除以 √n 这一步至关重要。

If the population were not normal, the Central Limit Theorem would still justify normality of X̄ for n ≥ 30, but here n = 25 and the population is normal, so no approximation is needed. This question emphasizes the exact sampling distribution.

若总体非正态，对于 n ≥ 30 的大样本中心极限定理可保证 X̄ 近似正态，但本题 n = 25 且总体正态，故无需近似。该题强调了精确的抽样分布。

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8. Question 7: Contingency Table and Chi-Squared Independence Test | 题目7：列联表与卡方独立性检验

In a survey, 200 people were classified by gender and whether they prefer online shopping. The observed frequencies are: Male–Yes: 55, Male–No: 45; Female–Yes: 80, Female–No: 20. Test at the 5% level whether preference is independent of gender, using Yates’ correction.

在一项调查中，200 人按性别及是否偏好网购进行分类。观察频数为：男–是：55，男–否：45；女–是：80，女–否：20。要求在 5% 水平下检验偏好是否独立于性别，并使用耶茨连续性校正。

Expected frequencies under independence: E = (row total × column total) / grand total. For Male–Yes: (100×135)/200 = 67.5; Male–No: (100×65)/200 = 32.5; Female–Yes: (100×135)/200 = 67.5; Female–No: (100×65)/200 = 32.5. All exceed 5, so the test is valid.

独立性假设下的期望频数：E = （行合计 × 列合计）/ 总合计。男–是：(100×135)/200 = 67.5；男–否：32.5；女–是：67.5；女–否：32.5。所有期望值均大于 5，检验有效。

With Yates’ correction, the test statistic is:

χ² = Σ (|O – E| – 0.5)² / E

For Male–Yes: (|55 – 67.5| – 0.5)² / 67.5 = (12.5 – 0.5)² / 67.5 = 144/67.5 ≈ 2.133. Repeat for others: Male–No: (|45 – 32.5| – 0.5)² / 32.5 = (12.5 – 0.5)²/32.5 = 144/32.5 ≈ 4.431; Female–Yes: same as Male–Yes = 2.133; Female–No: same as Male–No = 4.431. Sum = 13.128 ≈ 13.13.

使用耶茨校正后

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