Prime Minister and Executive: Statistical Analysis of Approval Ratings | 首相与行政机关：支持率的统计分析

📚 Prime Minister and Executive: Statistical Analysis of Approval Ratings | 首相与行政机关：支持率的统计分析

In Edexcel A-Level Mathematics, applied statistics often uses real-world scenarios to test core techniques. The relationship between a Prime Minister and the Executive can be turned into a numerical problem: we can treat approval ratings, policy decisions, and survey results as data. This article focuses on binomial models, normal approximations, hypothesis tests, confidence intervals, correlation, and regression, all built around the theme of the Prime Minister and the executive.

在 Edexcel A-Level 数学中，应用统计常用真实情境考查核心方法。首相与行政机关之间的关系可以转化为数值问题：我们可以把支持率、政策决定和调查结果看作数据。本文重点讨论二项分布模型、正态近似、假设检验、置信区间、相关性和回归分析，全部围绕首相与行政机关这一主题展开。

1. Modelling Approval Ratings with the Binomial Distribution | 用二项分布建模支持率

Suppose a polling company asks n = 200 randomly chosen voters whether they approve of the Prime Minister’s executive decisions. Each response is independent, and the national approval proportion is p. Let X be the number of voters who approve. Then X ~ B(n, p), and P(X = r) = ⁿCᵣ pʳ (1 – p)ⁿ⁻ʳ. For example, if p = 0.45, the expected number of approvals is np = 200 × 0.45 = 90, with variance np(1 – p) = 200 × 0.45 × 0.55 = 49.5.

假设某民调公司随机询问 n = 200 名选民是否认可首相的行政决策。每个回答相互独立，全国认可比例为 p。设 X 为认可的选民人数，则 X ~ B(n, p)，且 P(X = r) = ⁿCᵣ pʳ (1 – p)ⁿ⁻ʳ。例如，若 p = 0.45，则认可的期望人数为 np = 200 × 0.45 = 90，方差为 np(1 – p) = 200 × 0.45 × 0.55 = 49.5。

• Conditions: fixed number of trials, independent trials, two outcomes, constant p.
• 条件：固定试验次数、独立试验、两种结果、p 恒定。

2. Sampling Methods for Opinion Polls | 民意调查的抽样方法

A valid survey of the Prime Minister’s executive performance requires a random sample. Simple random sampling gives every voter an equal chance; stratified sampling ensures regions or age bands are represented. Quota sampling is often used by pollsters but can introduce bias. In Edexcel questions, you must identify advantages and disadvantages of each method.

对首相行政表现的有效调查需要随机样本。简单随机抽样让每位选民有平等机会；分层抽样确保地区或年龄段得到代表。配额抽样常被民调机构使用，但可能引入偏差。在 Edexcel 考题中，你必须指出每种方法的优缺点。

3. Normal Approximation to the Binomial | 二项分布的正态近似

When n is large and p is close to 0.5, we can approximate X ~ B(n, p) by Y ~ N(np, np(1 – p)). A continuity correction is needed. For X ≤ r, use Y < r + 0.5; for X ≥ r, use Y > r – 0.5. If n = 200 and p = 0.45, then Y ~ N(90, 49.5), so σ = √49.5 ≈ 7.04.

当 n 较大且 p 接近 0.5 时，可用 Y ~ N(np, np(1 – p)) 近似 X ~ B(n, p)。需要进行连续性校正。对于 X ≤ r，使用 Y < r + 0.5；对于 X ≥ r，使用 Y > r – 0.5。若 n = 200 且 p = 0.45，则 Y ~ N(90, 49.5)，因此 σ = √49.5 ≈ 7.04。

Z = (X – μ) / σ

4. Hypothesis Testing for a Proportion | 比例假设检验

A political analyst claims that the Prime Minister’s approval rate has fallen below 40%. We test H₀: p = 0.40 against H₁: p < 0.40. In a sample of 150 voters, 51 approve, so p̂ = 51/150 = 0.34. The test statistic is z = (p̂ - p) / √[p(1 - p)/n]. Calculate z = (0.34 - 0.40) / √(0.4 × 0.6 / 150) ≈ -1.50. At the 5% significance level, the critical value is -1.645, so we do not reject H₀. There is insufficient evidence to support the claim at 5%.

某政治分析师声称首相的支持率已跌至 40% 以下。我们检验 H₀: p = 0.40 对 H₁: p < 0.40。在 150 名选民的样本中，51 人认可，因此 p̂ = 51/150 = 0.34。检验统计量为 z = (p̂ - p) / √[p(1 - p)/n]。计算得 z = (0.34 - 0.40) / √(0.4 × 0.6 / 150) ≈ -1.50。在 5% 显著性水平下，临界值为 -1.645，因此我们不拒绝 H₀。在 5% 水平下没有足够证据支持该说法。

5. Confidence Intervals for Approval Ratings | 支持率的置信区间

A 95% confidence interval for the true approval proportion p is p̂ ± z* × √[p̂(1 – p̂)/n]. If p̂ = 0.34 and n = 150, the standard error is √(0.34 × 0.66 / 150) ≈ 0.0387. Using z* = 1.96, the interval is 0.34 ± 1.96 × 0.0387 = (0.264, 0.416). We are 95% confident that the true approval rate lies between 26.4% and 41.6%.

真实支持率 p 的 95% 置信区间为 p̂ ± z* × √[p̂(1 – p̂)/n]。若 p̂ = 0.34 且 n = 150，标准误为 √(0.34 × 0.66 / 150) ≈ 0.0387。使用 z* = 1.96，区间为 0.34 ± 1.96 × 0.0387 = (0.264, 0.416)。我们有 95% 的置信度认为真实支持率介于 26.4% 和 41.6% 之间。

6. Comparing Two Prime Ministers’ Approval Ratings | 比较两位首相的支持率

To compare two independent approval proportions, the standard error of the difference is √[p̂₁(1 – p̂₁)/n₁ + p̂₂(1 – p̂₂)/n₂]. Suppose PM A has p̂₁ = 0.52 in n₁ = 300 and PM B has p̂₂ = 0.48 in n₂ = 250. The difference is 0.04, with SE ≈ √(0.52 × 0.48 / 300 + 0.48 × 0.52 / 250) ≈ 0.0423. A 95% CI is 0.04 ± 1.96 × 0.0423 = (-0.043, 0.123), so the difference is not significant.

比较两个独立支持率时，差异的标准误为 √[p̂₁(1 – p̂₁)/n₁ + p̂₂(1 – p̂₂)/n₂]。假设首相 A 在 n₁ = 300 中 p̂₁ = 0.52，首相 B 在 n₂ = 250 中 p̂₂ = 0.48。差异为 0.04，标准误约为 √(0.52 × 0.48 / 300 + 0.48 × 0.52 / 250) ≈ 0.0423。95% 置信区间为 0.04 ± 1.96 × 0.0423 = (-0.043, 0.123)，因此差异不显著。

7. Correlation Between Executive Decisions and Public Confidence | 行政决策与公众信心的相关性

Suppose we record the number of major executive orders issued per month (x) and the Prime Minister’s approval rating (y) over 10 months. The product moment correlation coefficient r measures linear association. If r = 0.78, there is a strong positive correlation: as executive activity increases, approval tends to rise. However, correlation does not imply causation; other variables may be involved.

假设我们记录每月发布的重大行政命令数量 (x) 和首相支持率 (y)，共 10 个月。积矩相关系数 r 衡量线性相关程度。若 r = 0.78，则存在较强正相关：行政活动增加时，支持率往往上升。但相关不等于因果；可能涉及其他变量。

r = Sxy / √(Sxx × Syy)

8. Regression Analysis for Approval Trends | 支持率趋势的回归分析

The least squares regression line is y = a + bx, where b = Sxy / Sxx and a = ȳ – bx̄. If b = 1.5, then each additional executive order is associated with a 1.5 percentage point increase in approval rating. Use the line to predict y for a given x, but avoid extrapolating far beyond the data range.

最小二乘回归直线为 y = a + bx，其中 b = Sxy / Sxx，a = ȳ – bx̄。若 b = 1.5，则每增加一项行政命令

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