📚 Campaign Finance and Party Funding: An A-Level Mathematical Modelling Approach | 竞选资金与政党筹款：A-Level 数学建模视角
Campaign finance and party funding are usually discussed in politics, but the underlying data can be analysed using A-Level Edexcel Mathematics. This article shows how percentages, index numbers, exponential growth, logarithms, probability, correlation, regression, sampling, hypothesis testing and calculus help us understand donations, spending and electoral outcomes.
竞选资金与政党筹款通常属于政治学讨论范畴，但其背后的数据可以用 A-Level Edexcel 数学进行分析。本文展示如何运用百分比、指数、指数增长、对数、概率、相关性、回归、抽样、假设检验和微积分来理解捐款、支出及选举结果。
1. Interpreting Campaign Finance Data | 解读竞选资金数据
Raw funding data are often presented as totals in millions of pounds across several parties. Before any advanced modelling, we must convert these figures into shares, percentage changes and per-vote measures so that parties of different sizes can be compared fairly.
原始筹款数据通常以百万英镑为单位，列示多个政党。在进行任何高级建模之前，我们必须将这些数字转换为份额、百分比变化以及每票平均金额，以便公平比较不同规模的政党。
For example, if Party A raises £8.4 million and Party B raises £5.6 million from a grand total of £20 million, the share of Party A is (8.4 ÷ 20) × 100% = 42%. A year later, if Party A raises £9.2 million, the percentage change is ((9.2 − 8.4) ÷ 8.4) × 100% ≈ 9.52%.
例如，如果政党 A 筹得 840 万英镑，政党 B 筹得 560 万英镑，总筹款额为 2000 万英镑，则政党 A 的份额为 (8.4 ÷ 20) × 100% = 42%。一年后，若政党 A 筹得 920 万英镑，则百分比变化为 ((9.2 − 8.4) ÷ 8.4) × 100% ≈ 9.52%。
Percentage change = ((New − Old) ÷ Old) × 100%
2. Percentages and Index Numbers | 百分比与指数
Index numbers are useful for comparing funding levels over time. A base year is selected and assigned an index of 100; all other years are expressed relative to that base year.
指数用于比较不同时期的筹款水平。选择一个基年并赋予其指数 100；所有其他年份都相对于该基年表示。
Suppose a party raised £10 million in 2016 and £14 million in 2021. Taking 2016 as the base year, the 2021 funding index is (14 ÷ 10) × 100 = 140. This indicates a 40% increase in money raised, which is clearer than simply stating the difference of £4 million when comparing many parties.
假设某政党 2016 年筹得 1000 万英镑，2021 年筹得 1400 万英镑。以 2016 年为基年，2021 年的筹款指数为 (14 ÷ 10) × 100 = 140。这表明筹款额增长了 40%，在比较多个政党时比仅仅说明相差 400 万英镑更加清晰。
Funding index = (Current year total ÷ Base year total) × 100
3. Modelling Donation Growth with Exponential Functions | 用指数函数建模捐款增长
Party funding often grows by a similar percentage each year, so an exponential model is appropriate. The general form is D(t) = D₀ × (1 + r)t, where D₀ is the initial funding, r is the annual growth rate as a decimal, and t is the number of years.
政党筹款通常每年以相近的百分比增长，因此适合使用指数模型。一般形式为 D(t) = D₀ × (1 + r)t，其中 D₀ 为初始筹款额，r 为年增长率（以小数表示），t 为年数。
For example, a party starts with £2 million and funding grows at 8% per year. After 5 years, D = 2 × 1.085 ≈ 2 × 1.4693 ≈ 2.94 million pounds. The table below shows the projected growth.
例如，某政党初始筹款 200 万英镑，每年增长 8%。5 年后，D = 2 × 1.085 ≈ 2 × 1.4693 ≈ 294 万英镑。下表展示了预测增长情况。
| Year t | 0 | 1 | 2 | 3 | 4 | 5 |
| Funding £m | 2.00 | 2.16 | 2.33 | 2.52 | 2.72 | 2.94 |
In Edexcel questions, you may be asked to find the growth rate from two data points. For D(0) = 3 and D(4) = 4.5, solving 4.5 = 3(1+r)4 gives r = (4.5 ÷ 3)1/4 − 1 ≈ 0.1067, or about 10.7% per year.
在 Edexcel 考题中，你可能需要根据两个数据点求增长率。若 D(0) = 3，D(4) = 4.5，解 4.5 = 3(1+r)4 可得 r = (4.5 ÷ 3)1/4 − 1 ≈ 0.1067，即每年约 10.7%。
4. Logarithms in Funding Comparisons | 对数在筹款比较中的应用
Logarithms allow us to solve for the time needed to reach a target funding level. If a party grows from £1.5 million at 10% annually, the time to reach £3 million is found by solving 3 = 1.5 × 1.1t.
对数可用于求解达到目标筹款水平所需的时间。若某政党从 150 万英镑开始，每年增长 10%，则达到 300 万英镑所需时间可通过解 3 = 1.5 × 1.1t 求得。
Dividing both sides by 1.5 gives 2 = 1.1t. Taking natural logarithms: ln 2 = t ln 1.1, so t = ln 2 ÷ ln 1.1 ≈ 0.6931 ÷ 0.0953 ≈ 7.27 years. This type of calculation often appears in Edexcel Pure Mathematics, but it has a direct application in campaign finance planning.
两边除以 1.5 得 2 = 1.1t。取自然对数：ln 2 = t ln 1.1，因此 t = ln 2 ÷ ln 1.1 ≈ 0.6931 ÷ 0.0953 ≈ 7.27 年。这类计算常见于 Edexcel 纯数学，但在竞选筹款规划中有直接应用。
t = ln(D ÷ D₀) ÷ ln(1 + r)
5. Probability and Donation Distributions | 概率与捐款分布
Individual donations often approximately follow a normal distribution. If the mean donation is £30 with a standard deviation of £8, we can calculate the probability that a randomly selected donation exceeds £45.
个人捐款金额通常近似服从正态分布。若平均捐款为 30 英镑，标准差为 8 英镑，我们可以计算随机选取的一笔捐款超过 45 英镑的概率。
First calculate the z-score: z = (45 − 30) ÷ 8 = 1.875. Using the standard normal distribution, P(Z > 1.875) = 1 − Φ(1.875) ≈ 1 − 0.9696 = 0.0304. Therefore, about 3.04% of donations exceed £45.
首先计算 z 分数：z = (45 − 30) ÷ 8 = 1.875。使用标准正态分布，P(Z > 1.875) = 1 − Φ(1.875) ≈ 1 − 0.9696 = 0.0304。因此，约 3.04% 的捐款超过 45 英镑。
This probabilistic thinking helps party treasurers understand how common large donations are and whether an unusually large donation should be treated as an outlier.
这种概率思维有助于政党财务主管了解大额捐款的普遍程度，以及某笔异常大的捐款是否应被视为离群值。
6. Measures of Central Tendency and Spread | 集中趋势与离散程度
Summarising donation data requires measures of location and spread. Consider ten recorded donations in pounds: 10, 12, 15, 20, 22, 25, 28, 35, 40, 200.
汇总捐款数据需要使用集中趋势和离散程度指标。考虑十笔记录的捐款（英镑）：10、12、15、20、22、25、28、35、40、200。
The mean is 407 ÷ 10 = 40.7 pounds, but the median is (22 + 25) ÷ 2 = 23.5 pounds. The very large donation of £200 pulls the mean upward, so the median is a better measure of central tendency in this skewed data set.
平均数为 407 ÷ 10 = 40.7 英镑，但中位数为 (22 + 25) ÷ 2 = 23.5 英镑。200 英镑这笔非常大的捐款拉高了平均数，因此在这组偏斜数据中，中位数是更好的集中趋势度量。
The lower quartile Q₁ is 15, the upper quartile Q₃ is 35, so the interquartile range is 20 pounds. This provides a robust measure of spread that is not distorted by the outlier.
下四分位数 Q₁ 为 15，上四分位数 Q₃ 为 35，因此四分位距为 20 英镑。这提供了一个不受离群值扭曲的稳健离散程度度量。
7. Correlation between Spending and Vote Share | 支出与得票率的相关性
A key question in campaign finance is whether higher spending is associated with a higher vote share. The product moment correlation coefficient r measures the strength and direction of a linear relationship between two variables.
竞选资金中的一个关键问题是，更高的支出是否与更高的得票率相关。积矩相关系数 r 衡量两个变量之间线性关系的强度和方向。
Suppose data from 12 constituencies give r = 0.78 between campaign spending per voter and vote share. This indicates a strong positive correlation, but it does not prove that spending causes a higher vote share because other variables, such as party popularity and local demographics, may also be important.
假设来自 12 个选区的数据显示，每位选民的竞选支出与得票率之间的 r = 0.78。这表明存在较强的正相关，但并不能证明支出导致得票率提高，因为政党受欢迎程度和当地人口结构等其他变量也可能很重要。
r = Sxy ÷ √(Sxx × Syy)
8. Regression Lines and Predictions | 回归直线与预测
When a linear relationship is present, a least squares regression line can be used to predict vote share from spending. The equation is y = a + bx, where b = Sxy ÷ Sxx and a = ȳ − bx̄.
当存在线性关系时，可以使用最小二乘回归直线根据支出来预测得票率。方程为 y = a + bx，其中 b = Sxy ÷ Sxx，a = ȳ − bx̄。
For instance, a model gives vote share = 18 + 0.4 × spending, where spending is measured in pounds per voter. If a party spends £5 per voter, the predicted vote share is 18 + 0.4 × 5 = 20%.
例如，某模型给出得票率 = 18 + 0.4 × 支出，其中支出以每位选民的英镑数计量。若某政党在每位选民身上支出 5 英镑，则预测得票率为 18 + 0.4 × 5 = 20%。
Extrapolation beyond the observed spending range can be unreliable, a point often examined in Edexcel Statistics questions.
超出观测支出范围的外推可能不可靠，这一点在 Edexcel 统计题中经常考查。
9. Sampling Methods in Party Funding Surveys | 政党筹款调查中的抽样方法
Surveys about party funding need careful sampling. A simple random sample gives every donor an equal chance of selection, but a stratified sample ensures that different donor types are represented proportionally.
关于政党筹款的调查需要谨慎抽样。简单随机抽样使每位捐款者都有均等的入选机会，而分层抽样则确保不同捐款者类型按比例被代表。
If a party has donors in three regions making up 50%, 30% and 20% of the membership, a stratified sample of 200 donors should include 100, 60 and 40 donors from those regions respectively. This reduces bias compared with a voluntary response sample, which may over-represent highly motivated donors.
如果某政党的捐款者分布在三个地区，分别占党员人数的 50%、30% 和 20%，则一个容量为 200 的分层样本应分别从这些地区抽取 100 名、60 名和 40 名捐款者。与自愿回应样本相比，这可以减少偏差，因为自愿回应样本可能过度代表积极性很高的捐款者。
10. Hypothesis Testing for Funding Differences | 资金差异的假设检验
We can test whether two parties have significantly different mean donation amounts. Let the null hypothesis be H₀: μ₁ = μ₂ and the alternative be H₁: μ₁ ≠ μ₂, where μ₁ and μ₂ are the true mean donations for Party A and Party B.
我们可以检验两个政党的平均捐款金额是否存在显著差异。设原假设为 H₀: μ₁ = μ₂，备择假设为 H₁: μ₁ ≠ μ₂，其中 μ₁ 和 μ₂ 分别是政党 A 和政党 B 的真实平均捐款额。
Suppose sample means are £40 and £35, both standard deviations are £10, and sample sizes are 50. The test statistic is z = (40 − 35) ÷ √(10² ÷ 50 + 10² ÷ 50) = 5 ÷ √(2 + 2) = 5 ÷ 2 = 2.5.
假设样本均值分别为 40 英镑和 35 英镑，两者的标准差均为 10 英镑，样本容量均为 50。检验统计量为 z = (40 − 35) ÷ √(10² ÷ 50 + 10² ÷ 50) = 5 ÷ √(2 + 2) = 5 ÷ 2 = 2.5。
At the 5% significance level, the critical value is 1.96, so we reject H₀ and conclude the mean donations differ significantly. At the 1% level, the critical value is 2.576, so we would not reject H₀.
在 5% 显著性水平下，临界值为 1.96，因此我们拒绝 H₀，得出平均捐款额显著不同的结论。在 1% 显著性水平下，临界值为 2.576，因此我们不会拒绝 H₀。
11. Calculus: Marginal Cost of Fundraising | 微积分：筹款的边际成本
Calculus can be used to model the cost of fundraising. Suppose the total cost in pounds of contacting x hundred potential donors is C(x) = 1200 + 5x + 0.02x².
微积分可用于建立筹款成本模型。假设联系 x 百位潜在捐款者的总成本（英镑）为 C(x) = 1200 + 5x + 0.02x²。
The marginal cost function is the derivative C'(x) = 5 + 0.04x. When x = 100, the marginal cost is 5 + 0.04 × 100 = £9 per additional hundred donors contacted. In other words, contacting one more hundred donors at that level adds about £9 to the total cost.
边际成本函数是导数 C'(x) = 5 + 0.04x。当 x = 100 时，边际成本为 5 + 0.04 × 100 = 9 英镑，即每多联系一百位捐款者所增加的成本。换句话说，在该水平上多联系一百位捐款者会使总成本增加约 9 英镑。
C'(x) = dC/dx = 5 + 0.04x
12. Exam-Style Integrated Problem | 考试风格综合题
A party’s donation rate, measured in £ million per month during a six-month campaign, is modelled by r(t) = 0.2t + 0.3 for 0 ≤ t ≤ 6. Find the total funds raised over the six-month period.
在为期六个月的竞选活动中，某政党的捐款速率（以每月百万英镑计）由 r(t) = 0.2t + 0.3 给出，其中 0 ≤ t ≤ 6。求这六个月期间筹集的资金总额。
We integrate the rate function: ∫₀⁶ (0.2t + 0.3) dt = [0.1t² + 0.3t]₀⁶ = (0.1 × 36 + 0.3 × 6)
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