📚 Year 13 OCR Statistics: Mock Unit Test Analysis | Year 13 OCR 统计：单元测试模拟卷解析
This detailed walkthrough breaks down a full OCR Year 13 Statistics mock unit test, covering key topics such as hypothesis testing, Poisson and normal approximations, confidence intervals, chi‑squared tests, correlation, and non‑parametric methods. Each question is explained step‑by‑step, highlighting common mistakes and exam technique to help you secure top marks.
这份详细解析拆解了一套完整的 OCR Year 13 统计单元测试模拟卷，涵盖假设检验、泊松与正态近似、置信区间、卡方检验、相关分析以及非参数方法等重点内容。每个问题都逐步讲解，突出常见错误与应试技巧，帮助你稳拿高分。
1. Binomial Hypothesis Test | 二项分布假设检验
A manufacturer claims that at most 20% st this, an inspector randomly selects 25 components and finds 8 defectives. Test at the 5% significance level whether the proportion of defectives is greater than 20%. Let X ~ e smallest r such that P(X ≥ r) ≤ 0.05. Using the binomial cumulative tables, P(X ≥ 9) = 1 – P(X ≤ 8) = 1 – 0.9532 = 0.0468, and P(X ≥ 8) = 1 – P(X ≤ 7) = 1 – not reject H₀. There is insufficient evidence to suggest the proportion is greater than 20%.
某制造商声称其元件的缺陷率最多为 20%。为检验该声明，检验员随机抽取 25 个元件，发现 8 个有缺陷。在 5% 显著性水平下检验缺陷比例是否高于 20%。设 X ~ B(25, 0.20)，假设为 H₀: p = 0.20，H₁: p > 0.20。我们需要满足 P(X ≥ r) ≤ 0.05 的最小 r 值。查二项累积表，P(X ≥ 9) = 1 – P(X ≤ 8) = 0.0468，P(X ≥ 8) = 1 – P(X ≤ 7) = 0.1091，因此拒绝域为 X ≥ 9。观测值 8 未落入拒绝域，故不拒绝 H₀。没有充分证据表明缺陷率高于 20%。
Always define the random variable and state both hypotheses clearly. If you use the alternative ‘less than’ or ‘two‑tailed’, make sure the critical region matches the direction. Many students forget to use the correct tail probability when reading tables.
务必定义随机变量并清晰写出原假设和备择假设。若使用‘小于’或‘双尾’备择假设，需确保拒绝域与方向一致。很多学生查表时会忘记使用正确的尾部概率。
2. Poisson Approximation to Binomial | 二项分布的泊松近似
proximate probability that more than 5 resistors fail. Since n is large (1500) and p is small (0.002), we can approximate X ~ B(1500, 0.002) with Y ~ Poisson(λ) where λ = np = 1500 × 0.002 = 3. Then P(X > 5) ≈ P(Y > 5) = 1 – P(Y ≤ 5) = 1 – 0.9161 = 0.0839 (from Poisson cumulative tables with λ = 3).
一大批电阻器中，单个电阻器未能通过质控的概率为 0.002。随机抽取 1500 个电阻器。求超过 5 个电阻器失效的近似概率。由于 n 大 (1500) 且 p 小 (0.002)，可用 Y ~ Poisson(λ) 近似 X ~ B(1500, 0.002)，其中 λ = np = 3。则 P(X > 5) ≈ P(Y > 5) = 1 – P(Y ≤ 5) = 0.0839（查 λ = 3 的泊松累积表）。
The Poisson approximation is valid when n is large and p is small (typically n > 50 and np < 10). Always check that the mean and variance have been calculated correctly, and remember that a continuity correction is not needed for the Poisson approximation.
当 n 很大而 p 很小（通常 n > 50 且 np < 10）时，泊松近似是有效的。务必检查均值与方差是否正确，并牢记泊松近似无须连续性校正。
3. Normal Approximation with Continuity Correction | 正态近似与连续性校正
roximation to find the probability of obtaining between 185 and 210 heads inclusive. Let X ~ B(400, 0.5). Then μ = np = 200 and σ² = np(1 – p) = 100. Applying continuity correction, P(185 ≤ X ≤ 210) ≈ P(184.5 < Y < 210.5) where Y ~ N(200, 100). Standardising: z₁ = (184.5 – 200) / 10 = –1.55, z₂ = (210.5 – 200) / 10 = 1.05. Then P = Φ(1.05) – Φ(–1.55) = Φ(1.05) – [1 – Φ(1.55)] = 0.8531 – (1 – 0.9394) = 0.7925.
抛掷一枚硬币 400 次。用正态近似求正面朝上次数在 185 至 210 之间（含）的概率。设 X ~ B(400, 0.5)，则 μ = np = 200，σ² = 100。应用连续性校正，P(185 ≤ X ≤ 210) ≈ P(184.5 < Y < 210.5)，其中 Y ~ N(200, 100)。标准化：z₁ = (184.5 – 200)/10 = –1.55，z₂ = (210.5 – 200)/10 = 1.05。则概率 P = Φ(1.05) – Φ(–1.55) = 0.8531 – (1 – 0.9394) = 0.7925。
The continuity correction adjusts the discrete interval to a continuous interval by 0.5 on each boundary. Many candidates lose accuracy by omitting the half‑unit correction or by incorrectly applying it to open vs closed boundaries. Practice converting inequalities like ‘less than 210’ to ‘less than 209.5’ when using the normal approximation.
连续性校正在两个边界上各加减 0.5，将离散区间转为连续区间。许多考生因遗漏半单位校正或对开闭区间校正不当而失分。如‘小于 210’在用正态近似时需转为‘小于 209.5’。
4. Continuous Random Variable – PDF and CDF | 连续随机变量与概率密度函数
The continuous random variable X has probability density function f(x) = k(4 – constant k, the cumulative distribution function F(x), and the median. Integrate: ∫₀² k(4 – x²) dx = k[4x – x³/3]₀² = k(8 – 8/3) = k(16/3) = 1, so k = 3/16. For the CDF, for 0 ≤ x ≤ 2, F(x) = ∫₀ˣ (3/16)(4 – t²) dt = (3/16)[4t – t³/3]₀ˣ = (3/16)(4x – x³/3). The median m satisfies F(m) = 0.5, so (3/16)(4m – m³/3) = 0.5. Multiply by 16: 12m – m³ = 8, giving m³ – 12m + 8 = 0. Solve numerically: m ≈ 0.70 (exactly m ≈ 0.695).
连续随机变量 X 的概率密度函数为 f(x) = k(4 – x²)，0 ≤ x ≤ 2，其余为 0。求常数 k、累积分布函数 F(x) 以及中位数。积分：∫₀² k(4 – x²) dx = k[4x – x³/3]₀² = k(16/3) = 1，故 k = 3/16。对于 CDF，当 0 ≤ x ≤ 2 时，F(x) = ∫₀ˣ (3/16)(4 – t²) dt = (3/16)(4x – x³/3)。中位数 m 满足 F(m) = 0.5，即 (3/16)(4m – m³/3) = 0.5，乘以 16 得 12m – m³ = 8，整理为 m³ – 12m + 8 = 0。数值求解得 m ≈ 0.70（精确约 0.695）。
Always verify that the total area under the PDF equals 1. When finding the median, you may need to solve a cubic equation – typically a simple numerical trial‑and‑improvement is accepted in OCR exams. Remember to state the range for which the CDF is valid.
务必验证 PDF 下方面积等于 1。求中位数时可能需要解三次方程——OCR 考试中通常接受简单的试凑法。别忘了给出 CDF 的有效区间。
5. Confidence Interval for a Population Mean | 总体均值的置信区间
tributed population has sample mean 52.3 and sample standard deviation 4.8. Construct a 95% interval for the population mean μ. Since the population variance is unknown, we use the t‑ 39 degrees of freedom. The critical value t₀.₀₂₅,₃₉ ≈ 2.023. The confidence interval is x̄ ± t × s/√n = 52.3 ± 2.023 × (4.8/√40) = 52.3 ± 1.536, i.e. (50.76, 53.84). We are 95% confident that the true population mean lies between 50.76 and 53.84.
从正态总体中抽取 40 个观测值，样本均值 52.3，样本标准差 4.8。构建总体均值 μ 的 95% 置信区间。由于总体方差未知，使用自由度为 39 的 t 分布，临界值 t₀.₀₂₅,₃₉ ≈ 2.023。置信区间为 x̄ ± t × s/√n = 52.3 ± 1.536，即 (50.76, 53.84)。我们有 95% 的把握认为真实总体均值介于 50.76 与 53.84 之间。
If the population standard deviation σ were known, you would use the z‑ , or the sample size must be large enough (n ≥ 30) for the Central Limit Theorem to apply.
若总体标准差 σ 已知，则改用 z 分布。同时需检查条件：数据为随机样本且近似正态，或样本量足够大 (n ≥ 30) 适用中心极限定理。
6. Chi‑squared Test for Independence | 独立性卡方检验
A survey investigates whether there is an association between age group and preference for three brands of cereal. The observed frequencies are shown in the table below. Test at the 1% significance level.
| Brand A | Brand B | Brand C | Total | |
|---|---|---|---|---|
| Under 30 | 21 | 35 | 24 | 80 |
| 30 – 50 | 37 | 42 | 31 | 110 |
| Over 50 | 32 | 28 | 20 | 80 |
| Total | 90 | 105 | 75 | 270 |
total) / grand total. For example, E₁₁ = (80×90)/270 = 26.67. The test statistic χ² = Σ (O – fter computing all cells, χ² ≈ 5.44 (details omitted for brevity). Degrees of freedom = (3–1)×(3– 3.277. Since 5.44 < 13.277, do not reject H₀. There is no significant association at the 1% level.
H₀：年龄组与品牌偏好独立；H₁：不独立。期望频数按 (行合计 × 列合计) / 总计 计算，例如 E₁₁ = (80×90)/270 = 26.67。检验统计量 χ² = Σ (O – E)² / E，计算所有单元格后得 χ² ≈ 5.44。自由度 = (3–1)×(3–1) = 4，临界值 χ²₀.₀₁,₄ = 13.277。由于 5.44 < 13.277，不拒绝 H₀。在 1% 显著性水平下无显著关联。
Remember to state the degrees of freedom correctly and check that all expected frequencies are at least 5 to meet the chi‑squared approximation conditions. If any expected value is below 5, you may need to combine categories. Many candidates lose marks by writing the hypotheses incorrectly – they must mention independence or association, not differences in means.
务必正确给出自由度并检查所有期望频数至少为 5，以满足卡方近似的条件。若任一期望值低于 5，可能需要合并类别。许多考生因假设陈述错误而失分——假设必须提及独立性或关联性，而非均值差异。
7. Product Moment Correlation Test | 积矩相关系数检验
(x, y), the calculated product moment correlation coefficient is r = 0.632. Test at the 5% e level whether there is positive correlation in the population. H₀: ρ = 0, H₁: ρ > 0 (one‑tailed). Critical value for n = 12 at 5% one‑ is sufficient evidence to suggest a positive linear correlation. Alternatively, using a t‑test: t = r√n–2/√1–r² = 0.632×√10/√1–0.399 = 2.000/0.775 = 2.58, which exceeds t₀.₀₅,₁₀ = 1.812, leading to the same conclusion.
从 12 对 (x, y) 的样本中算得积矩相关系数 r = 0.632。在 5% 显著性水平下检验总体是否存在正相关。H₀: ρ = 0，H₁: ρ > 0（单尾）。查皮尔逊表，n = 12 时 5% 单尾临界值为 0.497。因 r = 0.632 > 0.497，拒绝 H₀。有充分证据表明存在正线性相关。也可用 t 检验：t = r√n–2/√1–r² = 2.58，超过 t₀.₀₅,₁₀ = 1.812，结论相同。
Be careful to use the correct critical value for one‑tailed or two‑tailed tests. If r is positive but the test is two‑tailed, the positive critical value for |r| is used. When data is given, you may need to compute Sxx, Syy, Sxy from sums of squares to find r before testing.
注意区分单尾与双尾检验的临界值。若 r 为正但检验为双尾，应使用 |r| 的正临界值。当给出原始数据时，你可能需要先通过平方和计算 Sxx、Syy、Sxy 再求得 r，然后进行检验。
8. Wilcoxon Signed‑Rank Test | 威尔科克森符号秩检验
ht participants perform a task before and after training. The improvement scores are: 2.3, –0.5, 1.8, 4.1, 0.2, –1.2, 3.0, 0.6. Test at the 5% sitive median improvement. Exclude zero differences, none here. Rank absolute differences: |–1.2|→ rank 3, |–0.5|→ rank 2, |0.2|→ rank 1, |0.6|→ rank 4, |1.8|→ rank 5, |2.3|→ rank 6, |3.0|→ rank 7, |4.1|→ rank 8. Sum of ranks for positive differences T⁺ = 1+4+5+6+7+8 = 31 (note: –0.5 and – 3 go to T⁻). The test statistic is T = min(T⁺, T⁻) = min(31, 2+3=5) = 5. For n = 8, the 5% two‑ There is insufficient evidence to conclude the training increases the median improvement.
八名参与者分别在训练前后完成一项任务。改善分数为：2.3, –0.5, 1.8, 4.1, 0.2, –1.2, 3.0, 0.6。在 5% 水平下检验训练是否带来正的中位数改善。排除零差值（此处无）。对绝对值之差排序：|–1.2|→ 秩 3，|–0.5|→ 秩 2，|0.2|→ 秩 1，等。正差值的秩和 T⁺ = 1+4+5+6+7+8 = 31（负差值秩和 T⁻ = 2+3=5）。检验统计量 T = min(31,5) = 5。n=8 时 5% 双尾临界值为 4。因 T=5 > 4，不拒绝 H₀。没有充分证据表明训练提高了中位数改善。
When using the Wilcoxon signed‑rank test, always list the paired differences, remove any zeros, and rank the absolute values, assigning average ranks for ties. State the test statistic clearly and compare to the critical value according to the sample size and hypothesis direction. If using a one‑tailed test, check the appropriate table.
使用威尔科克森符号秩检验时，应列出成对差值，剔除零值，对绝对值排序，出现相等值时分配平均秩。清晰写出检验统计量，并按样本量和假设方向与临界值比较。若为单尾检验，需查相应表格。
9. Common Pitfalls and Exam Tips | 常见错误与应试技巧
ng the random variable, misstating hypotheses, or using the wrong tail for critical regions. In hypothesis tests, always finish with a conclusion in context – 'reject H₀' or 'do not reject H₀' – and never say 'accept H₀' fidence intervals, check whether σ is known or estimated to decide between z and t. In chi‑ rrect degrees of freedom. In regression, confirm the direction of correlation matches the hypothesis before choosing the critical value.
在所有 OCR 统计试卷中，学生常因未定义随机变量、假设陈述错误或临界域方向用错而失分。假设检验最后务必将结论置于情境中——“拒绝 H₀”或“不拒绝 H₀”，切勿写“接受 H₀”。置信区间中，需判断 σ 已知还是估计，以选择 z 或 t 分布。卡方检验确保期望频数 ≥ 5 且自由度正确。在相关分析里，选择临界值前先确认相关方向是否与假设一致。
Practice reading tables accurately: binomial, Poisson, normal, t, and chi‑squared tables each have their own layout. For the normal distribution, sketch a bell curve and shade the required region to avoid sign errors. When using approximate methods, always state the approximation and justify its use (e.g., np and nq > 5 for normal approximation). Manage your time by allocating roughly one mark per minute and leave time to check arithmetic.
熟练读表：二项、泊松、正态、t 和卡方分布表各有不同布局。对于正态分布，画钟形曲线并标记所需区域以避免符号错误。使用近似方法时，务必写明近似方法并说明理由（如 np 和 nq > 5 方可用正态近似）。按每分钟一分的节奏分配时间，留出时间检查计算。
Published by TutorHao | Statistics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图