Mathematical Case Studies of Three Key General Elections | 三次关键大选的数学案例研究

📚 Mathematical Case Studies of Three Key General Elections | 三次关键大选的数学案例研究

This article uses three landmark UK general elections – 1945, 1979 and 1997 – as data-rich case studies for Edexcel A-Level Mathematics. We apply percentages, weighted averages, swing, probability, hypothesis testing and regression to real electoral data. These examples link pure statistical techniques to a familiar political context, helping you revise data handling, probability distributions and statistical inference.

本文以三次具有里程碑意义的英国大选——1945年、1979年和1997年——作为数据丰富的案例研究，适用于爱德思A-Level数学。我们将百分比、加权平均、摆动、概率、假设检验和回归应用于真实选举数据。这些例子将纯统计技术与熟悉的政治背景联系起来，帮助你复习数据处理、概率分布和统计推断。

1. Context and Data Collection | 背景与数据收集

Each election provides a two-party swing dataset. In 1945 Labour won 393 seats with 47.7% of the vote; in 1979 the Conservatives won 339 seats with 43.9%; in 1997 Labour won 418 seats with 43.2%. Turnout also varied from 71.4% to 76.0%, which affects the interpretation of percentage shares.

每次大选都提供一个两党摆动数据集。1945年工党以47.7%的得票率赢得393席；1979年保守党以43.9%的得票率赢得339席；1997年工党以43.2%的得票率赢得418席。投票率也从71.4%到76.0%不等，这会影响对百分比份额的解释。

2. Percentage Share and Proportionality | 百分比份额与比例性

Percentage share is the number of votes for a party divided by total valid votes, multiplied by 100. Seat share is seats won divided by total seats, multiplied by 100. Comparing the two measures reveals how proportional the electoral outcome is under a first-past-the-post system.

百分比份额是某党获得的选票数除以总有效票数，再乘以100。席位份额是赢得的席位数除以总席位数，再乘以100。比较这两个指标可以揭示在简单多数制下选举结果的比例性如何。

Seat share = (Seats won ÷ Total seats) × 100

In 1945 Labour’s vote share was 47.7%, but its seat share was 393 ÷ 640 × 100 ≈ 61.4%. The difference of 13.7 percentage points shows how a plurality of votes can be converted into a large seat majority. In 1997 the gap was even larger: Labour won 63.4% of the seats on 43.2% of the vote.

1945年工党的得票率为47.7%，但其席位份额为393 ÷ 640 × 100 ≈ 61.4%。13.7个百分点的差距表明，相对多数选票可以转化为巨大的席位多数。1997年的差距更大：工党以43.2%的得票率赢得了63.4%的席位。

3. Swing Calculation | 摆动计算

Swing measures the transfer of support between the two main parties from one election to the next. The standard two-party swing formula compares changes in Conservative and Labour vote shares. A positive swing to one party means its vote share rose while the other party’s share fell.

摆动衡量两次大选之间两大主要政党支持率的转移。标准两党摆动公式比较保守党和工党得票率的变化。对某党而言，正摆动意味着其得票率上升，而另一党的得票率下降。

Swing = ½ × ((C₂ – C₁) + (L₁ – L₂))

Here C₁ and L₁ are Conservative and Labour vote shares at the first election, and C₂ and L₂ are the shares at the second election. Using 1945 as election 1 and 1979 as election 2: C₁ = 36.2%, L₁ = 47.7%, C₂ = 43.9%, L₂ = 36.9%. The swing to the Conservatives is ½ × ((43.9 – 36.2) + (47.7 – 36.9)) = ½ × (7.7 + 10.8) = 9.25 percentage points.

这里C₁和L₁是第一次大选中保守党和工党的得票率，C₂和L₂是第二次大选中的得票率。以1945年为第一次大选、1979年为第二次大选：C₁ = 36.2%，L₁ = 47.7%，C₂ = 43.9%，L₂ = 36.9%。对保守党的摆动为 ½ × ((43.9 – 36.2) + (47.7 – 36.9)) = ½ × (7.7 + 10.8) = 9.25 个百分点。

4. Seat-Vote Elasticity and Cube Law | 席位-选票弹性与立方定律

A simple mathematical model connecting votes to seats is the cube law. If the vote ratio of two parties is V₁ : V₂, the seat ratio is approximately V₁³ : V₂³. This explains why a modest lead in votes can produce a much larger lead in seats.

连接选票与席位的一个简单数学模型是立方定律。如果两党的得票比为 V₁ : V₂，那么席位比约为 V₁³ : V₂³。这解释了为什么适度的选票领先会转化为大得多的席位领先。

Seat ratio ≈ (V₁ / V₂)³

In 1997 Labour and Conservative vote shares were 43.2% and 30.7%, giving a vote ratio of 1.407. The cube law predicts a seat ratio of 1.407³ ≈ 2.79. The actual seat ratio was 418 : 165 ≈ 2.53. The model over-predicts Labour’s dominance, but it still captures the exaggerative effect of first-past-the-post.

1997年工党和保守党的得票率分别为43.2%和30.7%，得票比为1.407。立方定律预测席位比为1.407³ ≈ 2.79。实际席位比为418 : 165 ≈ 2.53。该模型高估了工党的优势，但仍捕捉到了简单多数制的放大效应。

5. Weighted Averages and Turnout | 加权平均与投票率

National swing is not a simple average of constituency swings. It is a weighted average, where each constituency is weighted by its electorate size or total votes cast. The formula for a weighted mean is central to A-Level data

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

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