📚 The Liberal Democrat Party: An A-Level Mathematics Case Study | 自由民主党：A-Level数学案例研究
In A-Level Mathematics, real-world data can be used to explore core statistical ideas such as percentages, probability distributions, confidence intervals and hypothesis tests. This article uses the Liberal Democrat Party’s electoral performance as a case study. You will see how to turn political data into exam-ready mathematical calculations without taking any political position.
在 A-Level 数学中，真实数据可用于探索核心统计概念，例如百分比、概率分布、置信区间和假设检验。本文以自由民主党的选举表现为案例研究。你将看到如何将政治数据转化为符合考试要求的数学计算，而不采取任何政治立场。
1. Introduction to the Data Set | 数据集简介
Imagine we have collected the Liberal Democrat share of the vote in 10 selected constituencies at a general election. The values are 12.3%, 15.6%, 16.8%, 18.9%, 19.7%, 21.4%, 24.1%, 25.5%, 27.8% and 31.2%. These figures will be used throughout the article.
假设我们在一次大选中收集了自由民主党在 10 个选定选区的得票率。数值为 12.3%、15.6%、16.8%、18.9%、19.7%、21.4%、24.1%、25.5%、27.8% 和 31.2%。这些数据将在本文中反复使用。
| Constituency | A | B | C | D | E | F | G | H | I | J |
|---|---|---|---|---|---|---|---|---|---|---|
| Vote share (%) | 12.3 | 15.6 | 16.8 | 18.9 | 19.7 | 21.4 | 24.1 | 25.5 | 27.8 | 31.2 |
2. Calculating Percentages and Share of Vote | 计算百分比与得票率
If the Liberal Democrats receive 18,345 votes out of a total valid vote of 75,200 in one constituency, their percentage share is (18,345 ÷ 75,200) × 100 = 24.39% (to 2 decimal places). Percentage change is equally important: a rise from 18.2% to 22.6% is a change of (22.6 – 18.2)/18.2 × 100 ≈ 24.18%.
如果在某一选区自由民主党获得 18,345 票，而有效总票数为 75,200，则其得票率为 (18,345 ÷ 75,200) × 100 = 24.39%（保留两位小数）。百分比变化同样重要：从 18.2% 上升到 22.6% 的变化为 (22.6 – 18.2)/18.2 × 100 ≈ 24.18%。
Percentage share = (Votes for party ÷ Total valid votes) × 100
3. Mean, Median and Standard Deviation | 平均数、中位数与标准差
Using the ten constituency shares above, the mean is 21.33% and the median is 20.55%. The sample standard deviation is approximately 5.85 percentage points. This tells us that typical constituency results are spread around 21% with moderate variability.
使用上述十个选区的得票率，平均数为 21.33%，中位数为 20.55%。样本标准差约为 5.85 个百分点。这告诉我们，典型选区的结果围绕 21% 分布，并具有中等程度的离散程度。
s = √[ Σ(x – x̄)² / (n – 1) ]
4. Binomial Distribution: Modelling Seat Wins | 二项分布：模拟席位赢得
Suppose the probability that the Liberal Democrats win a given constituency is p = 0.20. If we examine n = 10 independent constituencies, the number of wins X follows a binomial distribution: X ~ B(10, 0.20). The probability of exactly 3 wins is P(X = 3) = ¹⁰C₃ (0.20)³ (0.80)⁷ ≈ 0.2013.
假设自由民主党在某个选区获胜的概率为 p = 0.20。如果我们考察 n = 10 个独立选区，获胜次数 X 服从二项分布：X ~ B(10, 0.20)。恰好赢得 3 个选区的概率为 P(X = 3) = ¹⁰C₃ (0.20)³ (0.80)⁷ ≈ 0.2013。
P(X = r) = ⁿCᵣ pʳ (1 – p)ⁿ⁻ʳ
5. Normal Approximation to the Binomial | 二项分布的正态近似
When n is large, the binomial distribution can be approximated by a normal distribution. If X ~ B(100, 0.20), then the mean is μ = np = 20 and the variance is σ² = np(1-p) = 16, so σ = 4. Using a continuity correction, P(X ≤ 15) ≈ P(Z ≤ (15.5 – 20)/4) = P(Z ≤ -1.125) ≈ 0.1303.
当 n 较大时，二项分布可以用正态分布近似。若 X ~ B(100, 0.20)，则均值为 μ = np = 20，方差为 σ² = np(1-p) = 16，因此 σ = 4。使用连续性校正，P(X ≤ 15) ≈ P(Z ≤ (15.5 – 20)/4) = P(Z ≤ -1.125) ≈ 0.1303。
z = (x – μ) / σ
6. Confidence Intervals for Opinion Polls | 民意调查的置信区间
A poll of 1,000 voters finds that 180 support the Liberal Democrats, so p̂ = 0.18. The approximate 95% confidence interval is p̂ ± 1.96√(p̂(1-p̂)/n). Substituting gives 0.18 ± 1.96√(0.18×0.82/1000) = 0.18 ± 0.0238, i.e. (0.1562, 0.2038).
一项对 1,000 名选民的调查发现，有 180 人支持自由民主党，因此 p̂ = 0.18。近似 95% 置信区间为 p̂ ± 1.96√(p̂(1-p̂)/n)。代入得 0.18 ± 1.96√(0.18×0.82/1000) = 0.18 ± 0.0238，即 (0.1562, 0.2038)。
p̂ ± z √( p̂(1 – p̂) / n )
7. Hypothesis Testing: Has Support Changed? | 假设检验：支持率是否改变？
We test whether the Liberal Democrats’ national support has changed from a historical level of 20%. Let H₀: p = 0.20 and H₁: p ≠ 0.20. In a sample of 500 voters, 82 support the party, giving p̂ = 0.164. The test statistic is z = (0.164 – 0.20)/√(0.20×0.80/500) ≈ -2.01. For a two-tailed test at the 5% level, the critical values are ±1.96. Since -2.01 < -1.96, we reject H₀.
我们检验自由民主党的全国支持率是否从历史水平 20% 发生了变化。设 H₀: p = 0.20，H₁: p ≠ 0.20。在一个包含 500 名选民的样本中，有 82 人支持该党，因此 p̂ = 0.164。检验统计量为 z = (0.164 – 0.20)/√(0.20×0.80/500) ≈ -2.01。对于 5% 显著性水平下的双尾检验，临界值为 ±1.96。由于 -2.01 < -1.96，我们拒绝 H₀。
8. Chi-Squared Test for Regional Differences | 地区差异的卡方检验Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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