Executive Orders: A Statistical Study | 行政命令：统计研究

📚 Executive Orders: A Statistical Study | 行政命令：统计研究

Executive orders are official directives issued by the President of the United States to manage the operations of the federal government. While the subject belongs to politics and history, the numerical data on how many orders each president has signed provides an excellent real-world context for applying A-Level Edexcel Mathematics. In this article, we will explore descriptive statistics, data presentation, the normal distribution, hypothesis testing, correlation and regression, all using a dataset of executive orders. By working through these methods, you will consolidate key concepts from the Statistics component of the Edexcel specification while engaging with authentic data.

行政命令是美国总统为管理联邦政府运作而签署的正式指令。虽然这个话题本身属于政治和历史范畴，但每位总统签署了多少份行政命令的数字资料，为应用A-Level Edexcel 数学提供了绝佳的真实语境。本文将利用行政命令数据集，探索描述性统计、数据展示、正态分布、假设检验、相关与回归等内容。通过这些方法，你将在接触真实数据的同时，巩固 Edexcel 考试大纲中统计学部分的核心概念。

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1. The Dataset: Annual Executive Orders | 数据集：年度行政命令数量

To make comparisons fair, we will use the average number of executive orders issued per year during each presidency. This variable removes the effect of differing term lengths. Below is the data for 15 recent U.S. presidents, from Franklin D. Roosevelt to Joe Biden.

为了公平比较，我们采用每位总统任期内的年均行政命令签署数量。这个变量消除了任期长短不同的影响。以下是 15 位近期美国总统（从富兰克林·罗斯福到乔·拜登）的数据。

President	Years in office	Total orders	Orders per year
F.D. Roosevelt	12.1	3721	307.5
H. Truman	7.8	907	116.3
D. Eisenhower	8.0	484	60.5
J.F. Kennedy	2.0	214	107.0
L. Johnson	5.0	325	65.0
R. Nixon	5.6	346	61.8
G. Ford	2.4	169	70.4
J. Carter	4.0	320	80.0
R. Reagan	8.0	381	47.6
G.H.W. Bush	4.0	166	41.5
W. Clinton	8.0	364	45.5
G.W. Bush	8.0	291	36.4
B. Obama	8.0	277	34.6
D. Trump	4.0	220	55.0
J. Biden	3.0	162	54.0

For the rest of the article, the variable x denotes 'orders per year'. The raw values are: 307.5, 116.3, 60.5, 107.0, 65.0, 61.8, 70.4, 80.0, 47.6, 41.5, 45.5, 36.4, 34.6, 55.0, 54.0.

在本文后续部分，变量 x 表示“年均命令数量”。原始数值为：307.5, 116.3, 60.5, 107.0, 65.0, 61.8, 70.4, 80.0, 47.6, 41.5, 45.5, 36.4, 34.6, 55.0, 54.0。

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2. Organising Data: Frequency Tables | 整理数据：频率表

asses of width 30, starting from 30 ≤ x < 60. The classes and frequencies are shown below.

分组频率表有助于汇总数据。我们将以30为组距，从 30 ≤ x < 60 开始，把年均命令数分成若干组。各组及频数如下表。

Class (orders/year)	Tally	Frequency, f
30 ≤ x < 60	7	7
60 ≤ x < 90	5	5
90 ≤ x < 120	2	2
120 ≤ x < 150	0	0
150 ≤ x < 180	0	0
180 ≤ x < 210	0	0
210 ≤ x < 240	0	0
240 ≤ x < 270	0	0
270 ≤ x < 300	0	0
300 ≤ x < 330	1	1

Note that Franklin Roosevelt's rate of 307.5 forms a class by itself, indicating it is an extreme value. When drawing a histogram, the area of each bar is proportional to frequency. Because the class widths are uniform, the heights equal the frequencies.

请注意，富兰克林·罗斯福的年均 307.5 独自成组，说明它属于极端值。绘制直方图时，每个柱的面积与频数成正比。由于各组组距相等，柱的高度就代表频数。

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3. Measures of Central Tendency | 集中趋势的度量

We calculate the mean, median, and mode of the 15 values. The mean x̄ 61.8 + 70.4 + 80.0 + 47.6 + 41.5 + 45.5 + 36.4 + 34.6 + 55.0 + 54.0 = 1183.1. Therefore, x̄ = 1183.1 ÷ 15 ≈ 78.9 (to 1 decimal place).

我们计算15个数值的均值、中位数和众数。均值 x̄ 由 ∑x/n 得出。求和： 07.0 + 65.0 + 61.8 + 70.4 + 80.0 + 47.6 + 41.5 + 45.5 + 36.4 + 34.6 + 55.0 + 54.0 = 1183.1。因此 x̄ = 1183.1 ÷ 15 ≈ 78.9（保留一位小数）。

s the 8th value, which is 60.5 orders per year. There is no repeated value, so there is no mode. The large difference between the mean (78.9) and the median (60.5) tells us the distribution is positively skewed by Roosevelt's outlier.

为求中位数，先将数据从小到大排列：4.6, 36.4, 41.5, 45.5, 47.6, 54.0, 55.0, 60.5, 61.8, 65.0, 70.4, 80.0, 107.0, 116.3, 307.5。因为 n=15 是奇数，中位数是第8个值，即年均60.5份命令。数据中没有重复值，因此没有众数。均值（78.9）与中位数（60.5）的巨大差异表明，受罗斯福的异常值影响，分布呈正偏态。

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4. Measures of Spread: Range, IQR, Standard Deviation | 离散程度：极差、四分位距、标准差

e 12th value, which is 80.0. Therefore IQR = Q₃ − Q₁ = 80.0 − 45.5 = 34.5 orders per year.

极差为最大值减最小值 = 307.5 − 34.6 = 272.9，极差受异常值影响很大。更稳健的度量是四分位距 (IQR)。对于 n=15，Q₁ 为第4个值，即45.5；Q₃ 为第12个值，即80.0。因此 IQR = 80.0 − 45.5 = 34.5 份/年。

We now compute the sample standard deviation s. Using the formula s = √[∑(x − x̄ f squares equals approximately 64842.5 (you can verify this as an exercise). Therefore s = √64842.5 / 14 6 ≈ 68.0 (to 1 decimal place). The standard deviation is large, again because of Roosevelt's value.

接下来计算样本标准差 s。利用公式 s = √[∑(x − x̄)²/(n−1)]，首先求离差平方和。为简洁起见，平方和约等于64842.5（你可将计算作为练习）。因此 s = √64842.5 / 14 ≈ √4631.6 ≈ 68.0（保留一位小数）。标准差很大，同样是由于罗斯福的数值所致。

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5. Box Plots and Outlier Detection | 箱线图与异常值探测

QR above Q₃ or below Q₁. The upper fence = Q₃ + 1.5×IQR = 80.0 + 51.75 = 131.75. Roosevelt's 307.5 lies well above this fence, so it is a confirmed outlier.

箱线图展示了最小值、Q₁、中位数、Q₃ 和最大值。我们数据的五数概括为：最小值=34.6，Q₁=45.5，中位数=60.5，Q₃=80.0，最大值=307.5。异常值的通常定义是，任何低于 Q₁ − 1.5×IQR 或高于 Q₃ + 1.5×IQR 的数值。上界 = 80.0 + 51.75 = 131.75。罗斯福的307.5远高于此界线，因此确认为异常值。

When drawing the box plot, the whisker should extend to the largest value inside the fence, which is 116.3 (Truman), and Roosevelt is marked with a separate symbol. This visual immediately signals that most presidents sign between about 35 and 80 orders per year, with Roosevelt being historically exceptional.

绘制箱线图时，须线应延伸至界线内的最大值，即杜鲁门的116.3，罗斯福则用单独符号标出。这一可视化立刻表明，大多数总统每年签署约35到80份行政命令，而罗斯福在历史上属于例外情况。

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6. Histograms and Shape of Distribution | 直方图与分布形状

Using the frequency table from Section 2, we can construct a histogram with classes of equal width. The bulk of the data lies in the first two classes (30–90). The distribution is positively skewed (skewed to the right) because the tail extends towards higher values. In Edexcel exams you may be asked to describe skewness and relate it to the mean-median comparison. Here, mean > median confirms positive skew.

利用第2节中的频率表，我们可以构建组距相等的直方图。大部分数据集中在前两组（30–90）。由于尾部伸向较高值，分布呈正偏态（向右偏斜）。在 Edexcel 考试中，你可能需要描述偏度并将其与均值-中位数对比相关联。此处均值 > 中位数证实了正偏。

Histograms with unequal class widths would require frequency density, but our uniform classes simplify the representation. The peak at 30–60 includes modern presidents from Reagan to Biden, reflecting a general trend of fewer executive orders in recent decades.

如果组距不等，直方图需要用频率密度，但统一的组距简化了表示。30–60这一组的峰包含从里根到拜登的现代总统，反映出近几十年来行政命令数量普遍减少的趋势。

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7. Fitting a Normal Distribution | 正态分布拟合

nd the median is 58.8. The distribution becomes more symmetric. If we assume a normal distribution for the 14 modern presidents, we can estimate probabilities. Let X ~ he probability that a randomly chosen modern president issued more than 80 orders per year? = (80 − 62.6) / 24.5 ≈ 0.71. Using tables, P(Z > 0.71) = 1 − 0.7611 = 0.2389. So about 24%.

尽管数据呈偏态，我们可以检验去掉异常值后的情形。若去掉罗斯福，剩余14个数值的均值为62.6，中位数为58.8，分布变得更为对称。如果我们假定现代总统的数据服从正态分布，便可估算概率。设 X ~ N(62.6, 24.5²)，此处的标准差重新计算后约为24.5。我们可以问：随机选择一位现代总统，其年均签署超过80份命令的概率是多少？标准化：z = (80 − 62.6) / 24.5 ≈ 0.71。查表得 P(Z > 0.71) = 1 − 0.7611 = 0.2389，约24%。

In reality, the normal distribution only approximates the data; you should always check whether the assumption is reasonable by looking at a quantile–quantile plot or the skew. Edexcel questions often ask you to state the conditions for a normal model.

实际上，正态分布只是数据的近似；你务必通过分位数—分位数图或偏度来检验此假设是否合理。Edexcel 试题常要求陈述正态模型的条件。

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8. Hypothesis Testing: Comparing Two Eras | 假设检验：两个时代的比较

an perform a two-sample t-test to compare the average annual orders during the period 1933–1969 (Roosevelt through Johnson) with the period 1974–2024 (Ford through Biden). The earlier group has 6 data points with mean x̄ to Roosevelt). The later group has 8 points (excluding Nixon as transitional) with mean x̄₂ = 52.8 and s₂ = 14.1. We test H₀: μ₁ = μ₂ against H₁: μ₁ ≠ μ₂ at the 5% significance level.

我们可以执行双样本 t 检验，比较两个时期的年均命令数量：1933–1969年（罗斯福至约翰逊）与1974–2024年（福特至拜登）。前一组6个数据点，均值 x̄₁ = 119.7，标准差 s₁ = 100.2（因罗斯福而很大）；后一组8个点（尼克松作为过渡期未计入），均值 x̄₂ = 52.8，标准差 s₂ = 14.1。我们检验 H₀: μ₁ = μ₂ 对 H₁: μ₁ ≠ μ₂，显著性水平为5%。

Assuming unequal variances, the test statistic t = (x̄₁ − x̄₂) / √s₁²/n₁ + s₂²/n₂ ≈ (119.7 − 52.8) / √(100.

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