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

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

This article provides a detailed, question-by-question walkthrough of a representative Year 13 WJEC Statistics unit test mock paper. Covering regression, correlation, probability distributions, hypothesis tests and confidence intervals, it aims to reinforce core A2 techniques. Worked solutions, examiner-style commentary and common pitfalls are included to sharpen your exam performance.

本文对一份典型的 WJEC Year 13 统计单元测试模拟卷进行逐题详解，涵盖回归、相关、概率分布、假设检验与置信区间等核心 A2 内容。通过完整演算、阅卷视角点拨与常见失分点提醒，帮助你提升应试精度与速度。

1. Mock Paper Structure and Key Topics | 模拟试卷结构与核心考点

The mock paper contains eight compulsory questions totalling 75 marks, mirroring the style of WJEC Statistics Unit 3 or Unit 4. Questions span Pearson's and Spearman's correlation, linear regression, discrete and continuous distributions, goodness-of-fit, contingency tables, confidence intervals and hypothesis testing for a binomial proportion. Data sets are concise, and exact critical values from the WJEC formula booklet must be used.

模拟卷共设八道必答题，满分 75 分，贴近 WJEC 统计单元 3 或单元 4 的命题风格。知识点覆盖皮尔逊与斯皮尔曼相关、线性回归、离散与连续分布、拟合优度、列联表、均值的置信区间以及二项比例假设检验。题目数据量适中，必须使用 WJEC 公式手册中的精确临界值。

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2. Question 1: Pearson's PMCC Calculation and Test | 皮尔逊积矩相关系数计算与检验

ations: x = 1, 2, 3, 4, 5, 6, 7 and y = 2.1, 2.9, 4.2, 5.0, 5.8, 6.9, 8.1. First compute x̄ = 4, ȳ = 5.0, Sₓₓ = Σ(xᵢ − x̄)² = 28, S_yy = Σ(yᵢ − ȳ)² ≈ 27.68 and S_xy = Σ(xᵢ − x̄)(yᵢ − ȳ) ≈ 27.6. The product‑moment correlation coefficient r = S_xy / √Sₓₓ S_yy ≈ 27.6 / √28 × 27.68 ≈ 0.991.

研究者记录 7 对观测值：x = 1, 2, 3, 4, 5, 6, 7 和 y = 2.1, 2.9, 4.2, 5.0, 5.8, 6.9, 8.1。先计算 x̄ = 4，ȳ = 5.0，Sₓₓ = Σ(xᵢ − x̄)² = 28，S_yy ≈ 27.68，S_xy ≈ 27.6。积矩相关系数 r = S_xy / √Sₓₓ S_yy ≈ 0.991。

To test H₀: ρ = 0 against H₁: ρ ≠ 0 at the 5% significance level, compare |r| with the critical value from the table for n = 7. The two‑ 1 > 0.7545, we reject H₀ and conclude there is significant evidence of linear correlation.

在 5% 显著性水平下检验 H₀: ρ = 0，H₁: ρ ≠ 0，将 |r| 与 n = 7 的临界值比较。双尾临界值为 0.7545。因为 0.991 > 0.7545，拒绝 H₀，即有显著证据表明存在线性相关。

r = S_xy / √Sₓₓ S_yy = 27.6 / √28 × 27.68 ≈ 0.991

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3. Question 2: Regression Line and Forecasting | 回归直线与预测

Using the same data, find the least‑squares regression line of y on x. The gradient b = S_xy / Sₓₓ = 27.6 / 28 ≈ 0.9857. The intercept a = ȳ − b x̄ = 5.0 − 0.9857×4 ≈ 1.057. Hence the equation is y = 1.057 + 0.986x (to 3 s.f.).

使用相同数据，求 y 对 x 的最小二乘回归线。斜率 b = S_xy / Sₓₓ = 27.6 / 28 ≈ 0.9857，截距 a = ȳ − b x̄ = 5.0 − 0.9857×4 ≈ 1.057。因此回归方程为 y = 1.057 + 0.986x（保留三位有效数字）。

943. The prediction is reasonably reliable because the correlation is very strong, but the x‑value lies just outside the original data range (extrapolation), so caution is advisable.

预测 x = 8 时的 y 值。代入得 y ≈ 1.057 + 0.9857×8 = 8.943。由于相关性极强，预测较为可信，但 x 值已在原始数据范围之外（外推），因此解读时应谨慎。

y = 1.057 + 0.986x, 预测值 (x=8) ≈ 8.94

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4. Question 3: Poisson Goodness-of-Fit Test | 泊松分布拟合优度检验

downs → 50 days, 1 → 55, 2 → 30, 3 → 10, 4 or more → 5. Estimate λ from the sample mean: λ̂ λ = 1.2, calculate expected frequencies for 0, 1, 2, 3 and ≥4 from Poisson probabilities.

某工厂记录 150 天中每天机器故障次数。频数如下：0 次故障 → 50 天，1 → 55，2 → 30，3 → 10，4 次及以上 → 5。用样本均值估计 λ：λ̂ = (0×50+1×55+2×30+3×10+4×5)/150 = 1.2。取 λ = 1.2，根据泊松概率计算 0,1,2,3 和 ≥4 的期望频数。

x	Observed O	P(X=x)	Expected E
0	50	0.3012	45.18
1	55	0.3614	54.21
2	30	0.2169	32.54
3	10	0.0867	13.01
≥4	5	0.0338	5.07

d; here all E ≥ 5, so no merging is required. Compute χ² = Σ(O−E)²/E = (50−45.18)²/45.18 + … 1.23. Degrees of freedom ν = 5 − 1 − 1 = 3 (classes minus 1 minus 1 parameter estimated).

必要时合并期望频数低于 5 的类别；本例所有 E ≥ 5，无需合并。计算 χ² = Σ(O−E)²/E ≈ 1.23。自由度 ν = 5 − 1 − 1 = 3（类别数减 1 再减估计参数个数）。

we do not reject H₀. The Poisson(1.2) distribution provides an adequate fit for the data.

取 α = 0.05，临界值 χ²₃ = 7.815。因 1.23 < 7.815，不拒绝 H₀，即泊松分布（λ=1.2）能充分拟合该数据。

χ² = 1.23, ν = 3, 临界值 = 7.815 → 不拒绝 H₀

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5. Question 4: Chi-Squared Test for Association | 卡方独立性检验

= 45; Female preferring A = 60, Female preferring B = 40. Set up H₀: gender and preference are independent. Compute expected frequencies under independence: e.g. E(Male, A) = (100×115)/200 = 57.5.

一项 200 位顾客的调查按性别与产品偏好（A 或 B）分类。观察频数为：男性偏好 A = 55，偏好 B = 45；女性偏好 A = 60，偏好 B = 40。设 H₀：性别与偏好独立。在独立性假设下计算期望频数，例如 E(男,A) = (100×115)/200 = 57.5。

²/42.5 ≈ 0.109 + 0.147 + 0.109 + 0.147 = 0.512. Degrees of freedom = (2−1)(2−1) = 1. The 5% from χ²₁ is 3.841. Since 0.512 < 3.841, we retain H₀; there is no significant association.

检验统计量 χ² = Σ(O−E)²/E 计算得 0.512。自由度 = (2−1)(2−1) = 1。χ²₁ 在 5% 水平的临界值为 3.841。因 0.512 < 3.841，保留 H₀，即没有显著关联。

χ² = 0.512, ν = 1, p > 0.05 → 独立性成立

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6. Question 5: Confidence Interval for Population Mean (σ known) | 总体均值的置信区间（σ已知）

A sample of 50 tubes yields a mean lifetime of x̄ = 72 hours. The population standard deviation is known to be σ = 8 hours. Construct a 95% confidence interval for the true mean μ. The formula is x̄ ± z₀.₀₂₅ × σ/√n. Here z₀.₀₂₅ = 1.960 from Normal tables.

随机抽取 50 根管子，得平均寿命 x̄ = 72 小时。已知总体标准差 σ = 8 小时。求总体均值 μ 的 95% 置信区间。公式为 x̄ ± z₀.₀₂₅ × σ/√n，其中 z₀.₀₂₅ = 1.960。

Standard error = σ/√n = 8/√50 .96 × 1.1314 ≈ 2.218. Hence the interval is 72 ± 2.218, i.e. (69.8, 74.2) hours. We are 95% confident that the true mean lifetime lies between 69.8 and 74.2 hours.

标准误 = 8/√50 ≈ 1.1314，误差幅度 = 1.96 × 1.1314 ≈ 2.218。置信区间为 72 ± 2.218，即 (69.8, 74.2) 小时。我们有 95% 的把握认为真实平均寿命介于 69.8 至 74.2 小时之间。

95% CI: 72 ± 1.96 × 8/√50 = (69.8, 74.2)

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7. Question 6: Binomial Hypothesis Test (One-Tailed) | 二项分布假设检验（单尾）

A manufacturer claims that at least 90% of its components are fault‑ inspected, revealing 2 defectives. Test at α = 0.05 whether the true proportion p of fault‑free items is less than 0.9. H₀: p = 0.9, H₁: p < 0.9. Under H₀, the number of fault‑free items X ~ B(20, 0.9).

某厂商声称至少 90% 的元件无故障。抽取 20 件检验，发现 2 件有缺陷。在 α = 0.05 下检验无故障比例 p 是否小于 0.9。H₀: p = 0.9，H₁: p < 0.9。在 H₀ 下，无故障元件数 X ∼ B(20, 0.9)。

The observed fault‑free count is 18. The p‑value = P(X ≤ 18 | ual exact cumulative gives around 0.677). Since 0.677 > 0.05, we do not reject H₀. The evidence is insufficient to claim the proportion is below 90%.

观察到的无故障数为 18。p 值 = P(X ≤ 18 | p=0.9)。查表或计算：P(X=20)≈0.1216，P(X=19)≈0.2702，P(X=18)≈0.2852，故 P(X ≤ 18) ≈ 0.677。因为 0.677 > 0.05，不拒绝 H₀，即证据不足以声称比例低于 90%。

X ~ B(20, 0.9), P(X ≤ 18) ≈ 0.677 > 0.05 → 不拒绝 H₀

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8. Question 7: Spearman's Rank Correlation | 斯皮尔曼等级相关系数

alculate the difference d in ranks, square: d² values: 1, 1, 1, 1, 0, 0. Σd² = 4. Spearman's rank coefficient r_s = 1 − [6Σd² / n(n²−1)] = 1 − (6×4) / (6×35) = 1 − 24/210 = 1 − 0.1143 = 0.8857.

六位学生由两位评委就其表达技能排序（1 为最优）。配对秩为 (1,2), (2,1), (3,4), (4,3), (5,5), (6,6)。计算秩差 d 及 d²：1,1,1,1,0,0，Σd² = 4。斯皮尔曼等级相关系数 r_s = 1 − [6Σd² / n(n²−1)] = 1 − 24/210 ≈ 0.886。

For n = 6, the two‑tailed 5% critical value for r_s is 0.886 (exact from table). This is a borderline result. Since r_s tly we do not reject H₀ of no correlation; the agreement is not quite significant at the 5% level. In an exam, state the comparison clearly.

对于 n=6，双尾 5% 临界值通常为 0.886。r_s = 0.8857 略低于临界值，严格来说不能拒绝无相关的原假设；在 5% 水平下一致性尚不显著。答题时需明确说明这一比较。

r_s = 0.886 (approx), 临界值 = 0.886 → 临界边缘，不拒绝 H₀

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9. Question 8: Continuous Random Variable and PDF | 连续随机变量与概率密度函数

er the domain to equal 1. ∫₀² kx(2−x) dx = k ∫₀² (2x − x²) dx = k [x² − x³/3]₀² = k (4 − 8/3) = (4/3)k = 1 ⇒ k = 3/4.

连续随机变量 X 的概率密度函数为 f(x) = kx(2 − x)，0 ≤ x ≤ 2，其余为零。由全域积分等于 1 可得 k。∫₀² kx(2−x) dx = k [x² − x³/3]₀² = k(4 − 8/3) = (4/3)k = 1，故 k = 3/4。

)x³ − x⁴/4]₀² = (3/4)[(16/3) − 4] = (3/4)(16/3 − 12/3) = (3/4)(4/3) = 1. So the mean is 1.

接着求 ) dx = (3/4)∫₀² (2x² − x³) dx = (3/4)[(2/3)x³ − x⁴/4]₀² = (3/4)[(16/3)−4] = (3/4)(4/3) = 1。均值为 1。

32/5] = (3/4)[8 − 6.4] = (3/4)(1.6) = 1.2. Then Var(X) = E(X²) − [E(X)]² = 1.2 − 1 = 0.2.

为求方差，先算 E(X²) = (3/4)∫₀² (2x³ − x⁴) dx = (3/4)[(1/2)x⁴ − x⁵/5]₀² = (3/4)[8 − 6.4] = 1.2。方差 Var(X) = E(X²) − [E(X)]² = 1.2 − 1 = 0.

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