📚 Congress Approval Rating: Statistical Analysis | 国会支持率：统计分析
In the United States, Congress is a central institution that drafts, debates, and passes laws. Public opinion on Congress is frequently measured through approval ratings, which give the percentage of people who approve of the way Congress is handling its job. For an A-Level Mathematics student, these data provide a real-world context to explore binomial distributions, hypothesis testing, confidence intervals, and common pitfalls like sampling bias. This article unpacks the statistical reasoning behind Congress approval polls, with clear Edexcel-style notation and step-by-step calculations, so you can apply the same methods to any proportional data.
美国国会是起草、辩论和通过法律的核心机构。公众对国会的看法常常通过支持率来衡量，即表示认可国会工作方式的人所占的百分比。对于 A-Level 数学学生来说，这些数据为二项分布、假设检验、置信区间以及抽样偏差等常见问题提供了真实的探究情境。本文用清晰的 Edexcel 风格符号和逐步计算，拆解国会支持率调查背后的统计推理，让你能将同样的方法应用于任何比例数据。
1. Understanding Congress Approval Data | 理解国会支持率数据
A typical Gallup or Pew Research Center poll might ask 1,500 adults: “Do you approve of the way Congress is handling its job?” The result is a sample proportion, often denoted as p̂. For instance, a recent poll might report a 25 % approval rating, so p̂ = pprove or not), each respondent can be modelled as a Bernoulli trial with probability p of “success”.
典型的盖洛普或皮尤研究中心民调可能会询问 1,500 名成年人：“您是否认可国会当前的工作方式？” 结果是一个样本比例，通常记为 p̂。例如，近期民调报告支持率为 25 %，即 p̂ = 0.25。真实总体比例 p 是未知的，是统计学家试图估计的对象。由于我们处理的是二元结果（认可或不认可），每位受访者都可以被建模为成功概率为 p 的伯努利试验。
For the sample size n = 1,500, the number of people who approve, X, follows a binomial distribution B(n, n conditions, we can approximate the binomial with a normal distribution. The condition np > 5 and n(1−p) > 5 ensures the approximation is valid. With a reported p̂ of 0.25, we have np̂ = 375 and n(1−p̂) = 1125, both well above 5.
对于样本量 n = 1,500，表示认可的人数 X 服从二项分布 B(n, p)。在特定条件下，我们可以用正态分布来近似二项分布。条件 np > 5 和 n(1−p) > 5 保证近似的有效性。当 p̂ 为 0.25 时，np̂ = 375，n(1−p̂) = 1125，两者远大于 5。
2. Binomial Distribution Recap | 二项分布回顾
If X ~ B(n, p), the probability of obtaining exactly k successes is given by P(X = = ⁿCₖ pᵏ (1−p)ⁿ⁻ᵏ. The mean is np and the variance is np(1−p). For a Congress poll with n = 1,500 and p = 0.25, the expected number of approvers is 375, with a standard deviation √375 × 0.75 ≈ 16.77.
若 X ~ B(n, p)，则恰好获得 k 次成功的概率为 P(X = k) = ⁿCₖ pᵏ (1−p)ⁿ⁻ᵏ。均值为 np，方差为 np(1−p)。对于 n = 1,500、p = 0.25 的国会民调，认可人数的期望值为 375，标准差 √375 × 0.75 ≈ 16.77。
We do not usually calculate individual binomial probabilities for large n by hand; rather, we use the normal approximation with continuity correction. The norm-based calculations allow us to construct confidence intervals and conduct hypothesis tests on the proportion p.
对于大样本 n，我们通常不会手工计算单个二项概率，而是采用带连续性校正的正态近似。基于正态的计算使我们能够对比例 p 构建置信区间并进行假设检验。
3. Sampling Distribution of the Proportion | 样本比例的抽样分布
When n is large, the sampling distribution of p̂ is approximately normal: p̂ ~ central limit theorem for proportions. In the context of a Congress approval survey with n = 1,500 and assumed p = 0.25, the standard error of p̂ is SE = √0.25 × 0.75 ÷ 1500 ≈ 0.01118.
当 n 很大时，p̂ 的抽样分布近似正态：p̂ ~ N(p, √[p(1−p)/n])。这源于比例的中心极限定理。在 n = 1,500、假定 p = 0.25 的国会支持率调查中，p̂ 的标准误为 SE = √0.25 × 0.75 ÷ 1500 ≈ 0.01118。
This small standard error means the poll is fairly precise: margin of error (at 95 % confidence) is about 1.96 × SE ≈ 2.19 percentage points. Thus, the true approval could reasonably be between 22.8 % and 27.2 %. Confidence intervals will formalise this.
如此小的标准误意味着民调相当精确：95 % 置信水平的误差幅度约为 1.96 × SE ≈ 2.19 个百分点。因此，真实支持率很可能在 22.8 % 到 27.2 % 之间。置信区间将对此进行规范。
4. Building a Confidence Interval for p | 构建 p 的置信区间
The most common interval is the Wald interval: p̂ ± z* √[p̂(1−p̂)/n]. For a 95 % confidence level, z* = 1.96. Using p̂ = 0.25 and n = 1,500, we get 0.25 ± 1.96 √0.25 × 0.75/1500 = (0.2281, 0.2719). This means we are 95 % confident that the true Congress approval rating lies between 22.8 % and 27.2 %.
最常用的区间是 Wald 区间：p̂ ± z* √[p̂(1−p̂)/n]。对于 95 % 置信水平，z* = 1.96。代入 p̂ = 0.25、n = 1,500，得到 0.25 ± 1.96 √0.25 × 0.75/1500 = (0.2281, 0.2719)。这意味着我们有 95 % 的信心认为国会真实支持率介于 22.8 % 和 27.2 % 之间。
The Wald interval can behave poorly when p̂ is near 0 or 1. An alternative is the Wilson score interval, which solves a quadratic and gives better coverage. For Edexcel, you will typically use the Wald interval as a first approach, provided the success-failure condition is met.
当 p̂ 接近 0 或 1 时，Wald 区间可能表现不佳。替代方法是 Wilson 得分区间，它求解二次方程并提供更好的覆盖率。在 Edexcel 考试中，只要满足成功-失败条件，通常首先使用 Wald 区间。
5. Hypothesis Test for a Single Proportion | 单一比例的假设检验
Suppose a political commentator claims that Congress’s true approval rating is still 30 % despite recent scandals. We can test this with a two-tailed hypothesis test at α = 0.05. Let H₀: p = 0.30, H₁: p ≠ 0.30. Using the poll’s p̂ = 0.25 and n = 1,500, the test statistic is z = (p̂ − p₀) / √[p₀(1−p₀)/n].
假设某政治评论员声称，尽管最近发生丑闻，国会的真实支持率仍为 30 %。我们可以在 α = 0.05 显著性水平下进行双尾假设检验。设 H₀: p = 0.30，H₁: p ≠ 0.30。利用民调的 p̂ = 0.25 和 n = 1,500，检验统计量为 z = (p̂ − p₀) / √[p₀(1−p₀)/n]。
Under H₀, SE = √0.30 × 0.70/1500 . Then z = (0.25 − 0.30) / 0.01183 ≈ −4.23. The corresponding p-value is extremely small (≪0.0001). Since ∣z∣ > 1.96, we reject H₀. There is strong evidence that the true approval rating differs from 30 %.
在 H₀ 下，SE = √0.30 × 0.70/1500 ≈ 0.01183。于是 z = (0.25 − 0.30) / 0.01183 ≈ −4.23。对应的 p 值极小 (≪0.0001)。由于 ∣z∣ > 1.96，我们拒绝 H₀。有有力证据表明真实支持率不同于 30 %。
6. Using Continuity Correction | 使用连续性校正
When performing the test with the normal approximation, the continuity correction improves accuracy. The corrected z is zc = (|p̂ − p₀| 2n) = 1/3000 ≈ 0.000333. The adjusted numerator becomes 0.05 − 0.000333 = 0.049667. Then zc .20, still significant. Edexcel papers sometimes ask for continuity correction explicitly.
当使用正态近似进行检验时，连续性校正可提高准确性。校正后的 z 为 zc = (|p̂ − p₀| − 1/(2n)) / SE。对于我们的数据，1/(2n) = 1/3000 ≈ 0.000333。调整后的分子为 0.05 − 0.000333 = 0.049667。于是 zc ≈ −4.20，仍然显著。Edexcel 试卷有时会明确要求连续性校正。
For a one-tailed test, say H₁: p < 0.30, the critical region changes. The critical z-value at α = rch question. When a commentator firmly predicts a drop, a one-tailed test is justifiable.
对于单尾检验，例如 H₁: p < 0.30，临界域会改变。α = 0.05 时的临界 z 值为 −1.645。我们的 z = −4.23 很容易落入拒绝域。单尾还是双尾的选择取决于研究问题。当评论员坚定地预测下降时，采用单尾检验是合理的。
7. Comparing Two Congress Polls | 比较两次国会民调
Imagine a second poll taken one month later, surveying n₂ = 1,200 adults, with p̂₂ = 0.22. We want to test if approval has significantly fallen. H₀: p₁ = p₂ versus H₁: p₁ > p₂. The pooled proportion is p_combined = (x₁ + x₂)/(n₁ + n₂) = (375 + 264)/(1500 + 1200) = 639/2700 ≈ 0.2367.
设想一个月后进行了第二次民调，调查了 n₂ = 1,200 名成年人，p̂₂ = 0.22。我们想检验支持率是否显著下降。H₀: p₁ = p₂ vs H₁: p₁ > p₂。合并比例为 p_combined = (x₁ + x₂)/(n₁ + n₂) = (375 + 264)/(1500 + 1200) = 639/2700 ≈ 0.2367。
The standard error of the difference under H₀ is SE_diff = √[p_combined(1−p_combined)(1/n₁ + 1/n₂)] = √[0.2367 × 0.7633 × (1/1500 + 1/1200)] ≈ √0.1807 × 0.0015 ≈ 0.01647. The test statistic z = (0.25−0.22)/0.01647 ≈ 1.82. With α = 0.05 (one-tailed critical value 1.645), we reject H₀; the drop appears significant.
在 H₀ 下，差异的标准误为 SE_diff = √[p_combined(1−p_combined)(1/n₁ + 1/n₂)] = √[0.2367 × 0.7633 × (1/1500 + 1/1200)] ≈ √0.1807 × 0.0015 ≈ 0.01647。检验统计量 z = (0.25−0.22)/0.01647 ≈ 1.82。在 α = 0.05（单尾临界值 1.645）下，我们拒绝 H₀；这一下降似乎具有显著性。
However, a confidence interval for the difference (p̂₁ − p̂₂) ± 1.96 × SE_unpooled may also be informative. Using unpooled SE = √[p̂₁(1−p̂₁)/n₁ + p̂₂(1−p̂₂)/n₂] ≈ 0.0176, the 95 % .0645). Because it contains 0, the difference is borderline; always check both approaches.
然而，差值 (p̂₁ − p̂₂) ± 1.96 × SE_unpooled 的置信区间也可能提供信息。使用非合并标准误 SE_unpooled = √[p̂₁(1−p̂₁)/n₁ + p̂₂(1−p̂₂)/n₂] ≈ 0.0176，95 % 置信区间为 (0.03 ± 0.0345) = (−0.0045, 0.0645)。由于包含 0，差异处于边缘状态；务必同时检查两种方法。
8. Sample Size Determination | 样本量的确定
rge must the sample be to estimate the approval rating within ±3 percentage points with 95 % confidence? Using the conservative p = zes p(1−p)), we set margin of error E = z* √[p(1−p)/n]. Solving for n: n = (z*/E)² p(1−p).
在开展国会民调之前，研究人员可能会问：需要多大的样本量，才能在 95 % 置信度下以 ±3 个百分点的误差估计支持率？采用保守的 p = 0.5（最大化 p(1−p)），设误差幅度 E = z* √[p(1−p)/n]。求解 n：n = (z*/E)² p(1−p)。
With z* = 1.96, E = 0.03, p = respondents. If prior polls suggest p closer to 0.25, we can reduce n: n ≈ (65.33)² × 0.25 × 0.75? ≈ 4268.44 × 0.1875 ≈ 800.3, so about 801 respondents. Using prior knowledge reduces cost.
取 z* = 1.96，E = 0.03，p = 0.5，得到 n ≈ (1.96/0.03)² × 0.25 ≈ (65.33)² × 0.25 ≈ 4268.44 × 0.25 ≈ 1067.1。向上取整，至少需要 1,068 名受访者。如果以往民调表明 p 接近 0.25，我们可以减少样本量：p(1−p)=0.1875，因此 n ≈ 4268.44 × 0.1875 ≈ 800.3，约 801 名受访者。利用先验知识可降低成本。
9. Sources of Bias in Congress Polls | 国会民调中的偏倚来源
Even with perfect sampling techniques, biases can creep in. Non-response bias occurs when people with strong negative views about Congress are less likely to answer surveys. Selection bias is introduced if the poll relies solely on landline phones, excluding younger demographics. Weighting adjustments attempt to correct for such biases by aligning the sample demographics with known population proportions from census data.
即便抽样技术完美，偏倚仍可能产生。无回答偏倚发生在对国会持强烈负面看法的人不太可能接受调查时。如果民调仅依赖固定电话，就会引入选择偏倚，排除年轻群体。加权调整通过将样本人口统计特征与人口普查数据中的已知总体比例对齐，试图纠正这类偏倚。
For an A-Level exam, you may be asked to critique a sampling method. For example, a telephone poll of 1,500 randomly dialled numbers may underrepresent Hispanic voters if they are less likely to answer unfamiliar numbers. Always consider the frame, the method of contact, and the weighting scheme.
在 A-Level 考试中，你可能需要评述某种抽样方法。例如，通过随机拨号开展的 1,500 人电话调查，如果西班牙裔选民不太愿意接听陌生号码，就可能对其代表性不足。务必考虑抽样框、联系方式以及加权方案。
10. Type I and Type II Errors in Context | 情景中的第一类错误和第二类错误
Rejecting a true H₀ (Type I error) in the Congress approval context could mean concluding the rating has fallen when it actually has not, potentially causing a political party to change strategy unnecessarily. Failing to reject a false H₀ (Type II error) means missing a real shift in public sentiment. The probability of Type I error is α, while the probability of Type II error is β, related to the power of the test (1−β).
在国会支持率的语境中，拒绝一个为真的 H₀（第一类错误）可能意味着得出支持率下降的结论而实际并未下降，可能导致一个政党不必要地改变策略。未拒绝一个为假的 H₀（第二类错误）则意味着遗漏了公众情绪的真实变化。第一类错误的概率为 α，第二类错误的概率为 β，并与检验功效 (1−β) 相关。
We can calculate β for a specific alternative, say p = 0.26. Using n = 1,500 and testing H₀: p = 0.30, we find the critical value under H₀ (one-tailed, left side). The reject region is z < −1.645. The corresponding sample proportion threshold is p̂_crit = p₀ − 1.645 SE₀ = 0.30 − 1.645 × 0.01183 ≈ 0.2805. If true p = 0.26, the probability of p̂ < 0.2805 is calculated using the normal distribution with mean 0.26 and SE from p = 0.26. That yields power and β, which you could be asked to compute.
我们可以针对一个特定备择假设，例如 p = 0.26，计算 β。使用 n = 1,500 并检验 H₀: p = 0.30，找到 H₀ 下的临界值（左单尾）。拒绝域为 z < −1.645。相应的样本比例阈值为 p̂_crit = p₀ − 1.645 SE₀ = 0.30 − 1.645 × 0.01183 ≈ 0.2805。如果真实 p = 0.26，则 p̂ < 0.2805 的概率可利用均值为 0.26、标准误基于 p = 0.26 的正态分布计算。这便会得出功效和 β，考试中可能有所要求。
11. Interpreting p-Values Correctly | 正确解读 p 值
A common mistake is to say “a p-value of 0.03 means there is a 3 % chance that H₀ is true.” This is incorrect. The p-value is the probability of obtaining a sample result at least as extreme as the one observed, assuming H₀ is true. In the Congress poll example, a p-value < 0.05 indicates strong evidence against the commentator’s claim of 30 % approval.
一个常见错误是说“p 值为 0.03 表示 H₀ 为真的概率是 3 %”。这是不正确的。p 值是在假定 H₀ 为真时，获得至少与观测结果一样极端的样本结果的概率。在国会民调例子中，p 值 < 0.05 表明有强有力的证据反对评论员关于支持率为 30 % 的说法。
₀; otherwise, do not reject. For Edexcel, you will often be given a significance level like 5 % and asked to state your conclusion clearly in context: “There is sufficient evidence at the 5 % level to suggest that the true Congress approval rating is not 30 %.”
显著性水平 α 在检验前选定。如果 p 值 ≤ α，则拒绝 H₀；否则不拒绝。在 Edexcel 考试中，通常会给定 5 % 的显著性水平，并要求你在语境中清楚陈述结论：“在 5 % 显著性水平下，有足够证据表明国会真实支持率并非 30 %。”
12. Real Data and Further Exploration | 真实数据与拓展探索
You can access historical Congress approval ratings from sources like Gallup. Plotting the monthly approval rate over time reveals dramatic fluctuations: it was above 80 % after 9/11, dropped below 15 % during government shutdowns. Analysing such time series involves moving averages and control charts, linking A-Level Statistics with real politics.
你可以从盖洛普等来源获取国会支持率的历史数据。绘制月支持率随时间变化的图表会揭示剧烈的波动：9/11 事件后支持率曾超过 80 %，而在政府停摆期间则跌至 15 % 以下。分析这类时间序列会涉及移动平均和控制图，将 A-Level 统计与现实政治联系起来。
| Year | Approval (%) | Significant Event |
|---|---|---|
| 2001 | 84 | Post 9/11 unity |
| 2013 | 14 | Government shutdown |
| 2023 | 21 | Post midterm average |
As an extension, you could test whether the mean approval in crisis years differs from non-crisis years using a two-sample t-test on the arcsine-transformed proportions. This type of investigation strengthens statistical reasoning and gives you a powerful set of skills for university.
作为拓展，你可以对该比例进行反正弦变换后，使用双样本 t 检验，来检验危机年份与非危机年份的平均支持率是否存在差异。这类探究能强化统计推理能力，并为大学阶段的学习提供一套强有力的技能。
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