Campaign Finance: Statistical Analysis for Edexcel A-Level Maths | 竞选资金：Edexcel A-Level 数学统计应用

📚 Campaign Finance: Statistical Analysis for Edexcel A-Level Maths | 竞选资金：Edexcel A-Level 数学统计应用

Campaign finance provides a rich real-world context for A-Level Mathematics. In Edexcel Statistics, you are expected to summarise data, calculate measures of central tendency and spread, work with probability distributions, and carry out hypothesis tests. Using political donations and spending figures allows these techniques to be applied to meaningful data sets.

竞选资金为 A-Level 数学提供了丰富的现实背景。在 Edexcel 统计学部分，你需要总结数据、计算集中趋势和离散程度的度量、使用概率分布并进行假设检验。运用政治捐款和开支数据可以把这些方法应用到有意义的数据集上。

1. Introduction to Campaign Finance Data | 竞选资金数据简介

Campaign finance data typically includes the amount each donor gives, the total spending of each candidate, the number of donors, and the proportion of small donations. These variables can be discrete or continuous, so it is important to classify them correctly before choosing a statistical technique.

竞选资金数据通常包括每个捐款人的捐款金额、每位候选人的总开支、捐款人数以及小额捐款所占比例。这些变量可以是离散型或连续型，因此在选择统计方法之前，正确分类变量非常重要。

For example, the number of individual donations is a discrete variable, while the total amount spent in pounds is a continuous variable. Edexcel exam questions often ask you to identify the type of data and suggest an appropriate diagram.

例如，个人捐款的次数是离散变量，而以英镑计算的总开支是连续变量。Edexcel 考试题常常要求你识别数据类型并建议合适的图表。

2. Sampling Methods in Political Donations | 政治捐款中的抽样方法

When studying campaign finance, we cannot always record every donation. Instead, we take a sample. A simple random sample gives every donation an equal chance of being selected, while a stratified sample divides donors into groups, such as small, medium and large donors, and samples proportionally from each.

在研究竞选资金时，我们不可能记录每一笔捐款，因此需要抽样。简单随机抽样使每笔捐款被选中的机会相等，而分层抽样将捐款人分为不同组，例如小额、中额和大额捐款人，然后按比例从每组中抽样。

A systematic sample might select every 20th donation from a list. However, if donations are listed in date order and large donations cluster near election day, systematic sampling may introduce bias.

系统抽样可以每 20 笔捐款选择一笔。但是，如果捐款按日期顺序排列，而大额捐款集中在选举日前后，系统抽样就可能引入偏差。

Simple random sampling: unbiased but may miss small subgroups.
Stratified sampling: improves representation of key donor groups.
Systematic sampling: quick to use but can be biased by hidden patterns.

简单随机抽样：无偏但可能遗漏小子群体。分层抽样：改善关键捐款人群体的代表性。系统抽样：使用快捷但可能因隐藏规律而产生偏差。

3. Descriptive Statistics: Mean, Median and Mode | 描述性统计：平均数、中位数和众数

Suppose five candidates spend the following amounts in thousands of pounds: 40, 55, 60, 65, 180. The mean is (40 + 55 + 60 + 65 + 180) ÷ 5 = 80. The median is 60 because it is the middle value in an ordered list. The mode is the most frequent value, but here no value repeats.

假设五位候选人的开支（单位：千英镑）为 40、55、60、65、180。平均数为 (40 + 55 + 60 + 65 + 180) ÷ 5 = 80。中位数为 60，因为它是排序后位于中间的值。众数是最常见的值，但这里没有重复值。

The large value of 180 pulls the mean upwards, so the median gives a better summary of a typical candidate. In finance data, outliers such as a billionaire self-funded campaign are common, so always compare the mean and median.

数值 180 较大，会将平均数向上拉，因此中位数能更好地概括典型候选人的开支。在金融数据中，像亿万富翁自筹竞选经费这样的异常值很常见，所以要始终比较平均数和中位数。

4. Measures of Spread: Range, Variance and Standard Deviation | 离散程度：极差、方差与标准差

The range is the difference between the largest and smallest values: 180 – 40 = 140. The variance measures the average squared distance from the mean. For the five spending values, the variance is calculated using the formula:

极差是最大值与最小值之差：180 – 40 = 140。方差衡量各数据点与平均数之间距离平方的平均值。对于这五个开支值，方差用以下公式计算：

σ² = Σ(xᵢ – μ)² / n

Using μ = 80, the squared deviations are (40-80)² = 1600, (55-80)² = 625, (60-80)² = 400, (65-80)² = 225 and (180-80)² = 10000. The variance is (1600+625+400+225+10000) ÷ 5 = 12850 ÷ 5 = 2570. The standard deviation is √2570 ≈ 50.7.

使用 μ = 80，偏差平方分别为 (40-80)² = 1600, (55-80)² = 625, (60-80)² = 400, (65-80)² = 225 和 (180-80)² = 10000。方差为 (1600+625+400+225+10000) ÷ 5 = 12850 ÷ 5 = 2570。标准差为 √2570 ≈ 50.7。

Standard deviation is preferred because it is in the same units as the original data. It tells you that spending values typically deviate from the mean by about 50.7 thousand pounds.

标准差更常用，因为它与原始数据的单位相同。它告诉你开支值通常偏离平均数约 50.7 千英镑。

5. Data Presentation: Histograms and Box Plots | 数据展示：直方图与箱线图

A histogram displays continuous data using area to represent frequency. Because the class widths may differ, you must calculate frequency density = frequency ÷ class width. In a campaign spending histogram, small class intervals can reveal a spike in small donations.

直方图用面积表示频数来展示连续数据。由于组距可能不同，必须计算频率密度 = 频数 ÷ 组距。在竞选开支直方图中，较小的组距可以揭示小额捐款的峰值。

A box plot shows the minimum, lower quartile Q₁, median Q₂, upper quartile Q₃ and maximum. It is very useful for comparing spending distributions between two candidates. Outliers are often defined as values more than 1.5 × IQR above Q₃ or below Q₁.

箱线图显示最小值、下四分位数 Q₁、中位数 Q₂、上四分位数 Q₃ 和最大值。它对于比较两位候选人的开支分布非常有用。异常值通常定义为高于 Q₃ 超过 1.5 × IQR 或低于 Q₁ 超过 1.5 × IQR 的值。

Candidate A: Min=20, Q₁=35, Median=50, Q₃=70, Max=120

Candidate B: Min=10, Q₁=25, Median=40, Q₃=55, Max=200

Candidate A has higher typical spending but a smaller spread, while Candidate B has a more right-skewed distribution caused by a few very large expenses.

候选人 A 的典型开支较高但分布较窄，而候选人 B 的分布更右偏，原因是少数几笔非常大的开支。

6. Probability Models for Donation Amounts | 捐款金额的概率模型

A random variable X can represent the amount of a single donation. If X takes a finite number of values, it is discrete; if it can take any value in an interval, it is continuous. The sum of all probabilities in a discrete distribution must equal 1.

随机变量 X 可以表示单笔捐款的金额。如果 X 取有限个值，它就是离散的；如果它在某个区间内可取任意值，它就是连续的。离散分布中所有概率之和必须等于 1。

For example, a charity fundraiser models small donations as: P(X = 10) = 0.4, P(X = 20) = 0.35, P(X = 50) = 0.25. The expected value E(X) is 10(0.4) + 20(0.35) + 50(0.25) = 4 + 7 + 12.5 = 23.5 pounds.

例如，某慈善筹款活动将小额捐款建模为：P(X = 10) = 0.4，P(X = 20) = 0.35，P(X = 50) = 0.25。期望值 E(X) = 10(0.4) + 20(0.35) + 50(0.25) = 4 + 7 + 12.5 = 23.5 英镑。

The expected value is a long-run average, so it is extremely useful when forecasting total funds raised from a large number of donors.

期望值是长期平均值，因此在预测大量捐款人筹集的总资金时非常有用。

7. Binomial Distribution: Small-Donor Success | 二项分布：小额捐款者成功概率

A binomial distribution arises when there are a fixed number of independent trials, each with the same probability of success. In campaign finance, we might define success as a donor giving more than 50 pounds.

当试验次数固定、各次试验独立且每次成功概率相同时，就产生二项分布。在竞选资金中，我们可以将成功定义为捐款人捐款超过 50 英镑。

If X ~ B(n, p), then

P(X = k) = ⁿCₖ pᵏ (1-p)ⁿ⁻ᵏ

Suppose n = 20 donors and the probability that a donor gives more than 50 pounds is p = 0.3. The probability that exactly 6 donors exceed 50 pounds is ⁿCₖ with n=20, k=6, p=0.3. Using a calculator, ⁿCₖ = 38760, pᵏ = 0.3⁶ ≈ 0.000729 and (1-p)ⁿ⁻ᵏ = 0.7¹⁴ ≈ 0.00678. The product is 38760 × 0.000729 × 0.00678 ≈ 0.1917.

假设 n = 20 位捐款人，某捐款人捐款超过 50 英镑的概率 p = 0.3。恰好有 6 位捐款人超过 50 英镑的概率为 ⁿCₖ，其中 n=20、k=6、p=0.3。使用计算器，ⁿCₖ = 38760，pᵏ = 0.3⁶ ≈ 0.000729，(1-p)ⁿ⁻ᵏ = 0.7¹⁴ ≈ 0.00678。乘积为 38760 × 0.000729 × 0.00678 ≈ 0.1917。

You must be able to use your calculator’s binomial PD function to find such probabilities quickly in the exam.

你必须能够在考试中使用计算器的二项分布 PD 功能快速求出这类概率。

8. Normal Distribution and the 68-95-99.7 Rule | 正态分布与 68-95-99.7 法则

When sample sizes are large, the distribution of total campaign spending across many constituencies can be modelled by a normal distribution with mean μ and standard deviation σ. The standardised z-score is:

当样本量很大时，许多选区的竞选总开支分布可以用均值为 μ、标准差为 σ 的正态分布来建模。标准化的 z 分数为：

Z = (X – μ) / σ

If spending is normally distributed with μ = 80 thousand pounds and σ = 50 thousand pounds, then approximately 68% of constituencies spend between 30 and 130 thousand pounds. About 95% spend between -20 and 180 thousand pounds, but because spending cannot be negative, the normal model is only approximate here.

如果开支呈正态分布，μ = 80 千英镑，σ = 50 千英镑，那么约 68% 的选区开支在 30 到 130 千英镑之间。约 95% 的选区开支在 -20 到 180 千英镑之间，但由于开支不可能为负，正态模型在这里只是近似。

To find the probability that spending is between 100 and 150 thousand pounds, convert each boundary to z-scores and use the normal distribution table or calculator.

要计算开支在 100 到 150 千英镑之间的概率，需要把每个边界值转换为 z 分数，然后使用正态分布表或计算器。

9. Hypothesis Testing on Campaign Spending | 竞选开支的假设检验

Hypothesis testing uses sample data to decide whether a claim about a population parameter is supported. In campaign finance, a regulator may claim that the mean donation is 40 pounds. You take a sample and test this claim at the 5% significance level.

假设检验使用样本数据判断关于总体参数的某个说法是否得到支持。在竞选资金中，监管机构可能声称平均捐款为 40 英镑。你抽取一个样本并在 5% 显著性水平下检验这一说法。

The null hypothesis is H₀: μ = 40 and the alternative is H₁: μ ≠ 40. After calculating the test statistic, you compare it with the critical value, often 1.96 for a two-tailed test at the 5% level when σ is known.

零假设为 H₀: μ = 40，备择假设为 H₁: μ ≠ 40。计算检验统计量后，将其与临界值进行比较，当 σ 已知且在 5% 显著性水平下进行双尾检验时，临界值通常为 1.96。

If the sample mean is 48 pounds, the sample size is 50 and the population standard deviation is 12 pounds, the test statistic is Z = (48 – 40) / (12/√50) ≈ 4.71. Since 4.71 > 1.96, you reject H₀: there is sufficient evidence that the mean donation is not 40 pounds.

如果样本平均数为 48 英镑，样本量为 50，总体标准差为 12 英镑，则检验统计量 Z = (48 – 40) / (12/√50) ≈ 4.71。由于 4.71 大于 1.96，拒绝 H₀：有充分证据表明平均捐款不是 40 英镑。

10. Correlation and Regression: Spending vs Vote Share | 相关与回归：开支与得票率

Scatter diagrams can show the relationship between campaign spending and the percentage of votes received. The product moment correlation coefficient r measures the strength and direction of a linear relationship. The value of r always lies between -1 and 1.

散点图可以显示竞选开支与得票率之间的关系。积矩相关系数 r 衡量线性关系的强度和方向。r 的值始终在 -1 到 1 之间。

The least squares regression line has equation y = a + bx, where b = Sxy / Sxx and a = ȳ – bx̄. You should be able to interpret the gradient b: it is the estimated change in vote share for each additional thousand pounds spent.

最小二乘回归线方程为 y = a + bx，其中 b = Sxy / Sxx，a = ȳ – bx̄。你应该能够解释斜率 b：它表示每多花费一千英镑时得票率的估计变化量。

If r = 0.85, there is a strong positive correlation, but correlation does not imply causation. Other variables such as party popularity and media coverage may also affect vote share.

如果 r = 0.85，则存在强正相关，但相关关系并不意味着因果关系。政党受欢迎程度和媒体报道等其他变量也可能影响得票率。

11. Exam-Style Worked Example | 考试风格示例

A random sample of 8 local campaigns gives total spending in thousands of pounds: 32, 38, 40, 44, 45, 48, 52, 60. Calculate the mean and standard deviation, and construct a 95% confidence interval for the population mean.

从 8 个地方竞选活动中随机抽取样本，总开支（千英镑）为：32、38、40、44、45、48、52、60。计算平均数和标准差，并构建总体平均数的 95% 置信区间。

First find the mean: (32 + 38 + 40 + 44 + 45 + 48 + 52 + 60) ÷ 8 = 359 ÷ 8 = 44.875. Next, calculate the squared deviations from this mean, sum them, and divide by 8 to get the variance. The sum of squared deviations is approximately 546.875, so the variance is 68.36 and the standard deviation is √68.36 ≈ 8.27.

首先求平均数：(32 + 38 + 40 + 44 + 45 + 48 + 52 + 60) ÷ 8 = 359 ÷ 8 = 44.875。接下来，计算每个数据点与平均数之差的平方，求和并除以 8 得到方差。偏差平方和约为 546.875，所以方差为 68.36，标准差为 √68.36 ≈ 8.27。

Since the population standard deviation is unknown and the sample size is small, use the t-distribution with 7 degrees of freedom. The 95% confidence interval is x̄ ± t × (s/√n). The t-value is approximately 2.365, so the interval is 44.875 ± 2.365 × (8.27/√8) ≈ 44.875 ± 6.91. This gives (37.97, 51.79).

由于总体标准差未知且样本量较小，使用自由度为 7 的 t 分布。95% 置信区间为 x̄ ± t × (s/√n)。t 值约为 2.365，因此区间为 44.875 ± 2.365 × (8.27/√8) ≈ 44.875 ± 6.91，得到 (37.97, 51.79)。

This means we are 95% confident that the true mean campaign spending lies between 37.97 and 51.79 thousand pounds.

这意味着我们有 95% 的把握认为总体平均竞选开支在 37.97 到 51.79 千英镑之间。

12. Key Takeaways | 关键总结

Campaign finance data offers a realistic context for nearly every Edexcel A-Level Statistics topic: sampling, descriptive statistics, data presentation, probability distributions, hypothesis testing and correlation. Always classify variables, check for outliers, and interpret your results in the context of the problem.

竞选资金数据为几乎所有 Edexcel A-Level 统计学主题提供了现实背景：抽样、描述性统计、数据展示、概率分布、假设检验和相关分析。始终要对变量进行分类，检查异常值，并结合问题背景解释结果。

In the exam, marks are awarded for method, accuracy and interpretation. Showing your working clearly and using the correct notation is just as important as getting the final numerical answer.

在考试中，方法、准确性和解释都能得分。清楚地展示解题过程并使用正确的符号与得到最终数值答案同样重要。

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Campaign Finance: Statistical Analysis for Edexcel A-Level Maths | 竞选资金：Edexcel A-Level 数学统计应用