📚 A-Level Further Mathematics: Large-Sample Inference for Unknown Populations | A-Level 进阶数学：未知总体的大样本推断
In statistics, we often need to make inferences about a population whose underlying distribution is unknown. When the sample size is large, the Central Limit Theorem allows us to use the normal distribution as an approximation, even when the population is not normal. This article explains the key ideas, formulas, and exam-style techniques for large-sample inference in Edexcel A-Level Further Statistics.
在统计学中，我们经常需要对分布形式未知的总体进行推断。当样本量很大时，中心极限定理使我们即使面对非正态总体，也可以使用正态分布作为近似。本文将讲解 Edexcel A-Level 进阶统计中关于大样本推断的核心思想、公式和考试技巧。
1. The Problem of Unknown Populations | 未知总体的问题
Real-world populations rarely come labelled with their exact distribution. A population might be skewed, discrete, bounded, or otherwise non-normal, and its parameters such as the mean μ and variance σ² are usually unknown.
现实中的总体很少会直接标明其精确分布。一个总体可能偏斜、离散、有界，或者根本不是正态分布，而其参数如均值 μ 和方差 σ² 通常是未知的。
To infer information about such an unknown population, we take a random sample of size n and use statistics such as the sample mean x̄ or sample proportion p̂. For large n, these statistics have sampling distributions that are approximately normal.
为了对这样的未知总体进行推断，我们随机抽取容量为 n 的样本，并使用样本均值 x̄ 或样本比例 p̂ 等统计量。当 n 很大时，这些统计量的抽样分布近似正态。
The main condition for these approximations is that the sample is random and the observations are independent. The larger the sample, the better the normal approximation becomes.
这些近似的主要条件是样本为随机样本且观测值相互独立。样本越大，正态近似通常就越精确。
2. The Central Limit Theorem | 中心极限定理
The Central Limit Theorem states that if X₁, X₂, … , Xₙ is a random sample from any distribution with mean μ and finite variance σ², then for sufficiently large n, the sample mean x̄ has an approximately normal distribution.
中心极限定理指出：如果 X₁, X₂, … , Xₙ 是来自均值为 μ、方差为 σ² 的任意分布的随机样本，那么当 n 足够大时，样本均值 x̄ 近似服从正态分布。
x̄ ≈ N( μ , σ² / n )
Equivalently, the standardised sample mean follows an approximate standard normal distribution:
等价地，标准化后的样本均值近似服从标准正态分布：
Z = ( x̄ − μ ) / ( σ / √n ) ≈ N(0, 1)
Notice that the mean of the sampling distribution is still μ, but the variance is reduced by a factor of n. This explains why sample means become more concentrated around μ as n increases.
注意，抽样分布的均值仍然是 μ，但方差缩小了 n 倍。这解释了为什么随着 n 增大，样本均值会越来越集中在 μ 附近。
3. Sampling Distribution of the Sample Mean | 样本均值的抽样分布
If the original population is exactly normal, then x̄ has exactly the normal distribution N(μ, σ²/n) for any sample size. If the population is not normal, the distribution of x̄ is only approximately normal, but the approximation is good when n is large.
如果原始总体恰好服从正态分布，那么对于任意样本量，x̄ 都精确服从 N(μ, σ²/n)。如果总体不是正态分布，那么 x̄ 的分布只是近似正态，但当 n 很大时近似效果很好。
The standard deviation of the sample mean is called the standard error:
样本均值的标准差称为标准误：
Standard Error = σ / √n
In many practical situations σ is unknown. For large samples, we can replace σ with the sample standard deviation s, giving the estimated standard error s / √n.
在许多实际问题中 σ 未知。对于大样本，我们可以用样本标准差 s 代替 σ，得到估计标准误 s / √n。
It is essential to remember that x̄ is an unbiased estimator of μ, meaning E(x̄) = μ. This property holds for any population distribution, provided the sample is random.
必须记住，x̄ 是 μ 的无偏估计量，即 E(x̄) = μ。只要样本是随机的，这一性质对任何总体分布都成立。
4. Confidence Interval for μ (Known σ) | 已知 σ 时 μ 的置信区间
A confidence interval gives a range of plausible values for an unknown parameter. For a large-sample estimate of μ, the confidence interval has the form:
置信区间给出了未知参数的一组可能取值范围。对于 μ 的大样本估计，置信区间具有如下形式：
x̄ ± z* × σ / √n
Here z* is the critical value from the standard normal distribution. It is chosen so that the required percentage of the distribution lies between −z* and +z*.
其中 z* 是来自标准正态分布的临界值。选择 z* 使得所需百分比的分布落在 −z* 与 +z* 之间。
| Confidence Level | 置信水平 | 90% | 95% | 99% |
| z* | z* 值 | 1.645 | 1.960 | 2.576 |
Example: A sample of n = 64 observations has mean x̄ = 150 and known population standard deviation σ = 20. A 95% confidence interval for μ is:
示例：一个容量为 n = 64 的样本均值为 x̄ = 150，已知总体标准差 σ = 20。μ 的 95% 置信区间为：
150 ± 1.96 × 20 / √64 = 150 ± 4.9
So the interval is (145.1, 154.9). We say we are 95% confident that μ lies in this interval.
因此区间为 (145.1, 154.9)。我们称有 95% 的把握认为 μ 落在这个区间内。
5. Using Sample Standard Deviation | 使用样本标准差
When σ is unknown, as is almost always true in practice, we can use the sample standard deviation s to estimate it. For large n, the confidence interval becomes:
当 σ 未知时（现实中几乎总是如此），我们可以使用样本标准差 s 来估计它。对于大样本，置信区间变为：
x̄ ± z* × s / √n
This approximation is justified by the fact that, for a sufficiently large sample, s estimates σ very precisely.
这一近似是合理的，因为当样本量足够大时，s 能够非常精确地估计 σ。
Example: Suppose n = 100, x̄ = 42.5, and s = 12. The 90% confidence interval for μ is:
示例：假设 n = 100，x̄ = 42.5，s = 12。μ 的 90% 置信区间为：
42.5 ± 1.645 × 12 / √100 = 42.5 ± 1.974
Thus the interval is approximately (40.53, 44.47).
因此区间约为 (40.53, 44.47)。
If the sample size were small, we would need the t-distribution instead. In this A-Level Further Statistics context, the large-sample normal approximation is the central focus.
如果样本量较小，则需要使用 t 分布。在本课程的进阶统计背景下，大样本正态近似是核心要点。
6. Sampling Distribution of the Sample Proportion | 样本比例的抽样分布
For categorical data, we often want to estimate the population proportion p of items with a particular characteristic. If X is the number of successes in n independent trials, then the sample proportion is:
对于分类数据，我们常常希望估计具有某种特征的总体比例 p。如果 X 是 n 次独立试验中的成功次数，则样本比例为：
p̂ = X / n
For large n, the sampling distribution of p̂ is approximately normal:
当 n 很大时，p̂ 的抽样分布近似正态：
p̂ ≈ N( p , p(1 − p) / n )
The standard error of the sample proportion is therefore:
因此样本比例的标准误为：
√( p(1 − p) / n )
A common rule is that the approximation is reliable when np and n(1 − p) are both greater than 5. Some textbooks require 10; always check the question context and formula booklet.
常用的规则是：当 np 和 n(1 − p) 都大于 5 时，近似是可靠的。有些教材要求大于 10；务必结合题目背景和公式册进行判断。
7. Confidence Interval for a Proportion | 比例的置信区间
To construct a confidence interval for an unknown population proportion p, we replace p by its estimate p̂ in the standard error. The interval is:
为了构造未知总体比例 p 的置信区间，我们在标准误中用估计值 p̂ 代替 p。区间为：
p̂ ± z* × √( p̂(1 − p̂) / n )
Example: In a survey of 200 students, 60 said they study further mathematics. Here n = 200 and p̂ = 60/200 = 0.30.
示例：在一项针对 200 名学生的调查中，60 人表示他们学习进阶数学。这里 n = 200，p̂ = 60/200 = 0.30。
A 95% confidence interval for p is:
p 的 95% 置信区间为：
0.30 ± 1.96 × √( 0.30 × 0.70 / 200 ) = 0.30 ± 0.0635
Hence the interval is approximately (0.2365, 0.3635), or about (23.7%, 36.4%).
因此区间约为 (0.2365, 0.3635)，即约 (23.7%, 36.4%)。
Notice that the width of the interval depends on n and on the value of p̂. The width is maximised when p̂ = 0.5, which is useful when planning sample sizes.
注意，区间宽度取决于 n 和 p̂ 的值。当 p̂ = 0.5 时宽度最大，这在规划样本量时非常有用。
8. Hypothesis Testing: z-test for μ | 假设检验：μ 的 z 检验
In a large-sample hypothesis test for an unknown population mean μ, the null hypothesis is typically H₀: μ = μ₀. The alternative hypothesis may be one-sided or two-sided, for example H₁: μ ≠ μ₀.
在未知总体均值 μ 的大样本假设检验中，原假设通常为 H₀: μ = μ₀。备择假设可以是单侧或双侧的，例如 H₁: μ ≠ μ₀。
The test statistic is:
检验统计量为：
z = ( x̄ − μ₀ ) / ( σ / √n )
If σ is unknown and n is large, we replace σ by s. The test statistic is then compared with critical values from the normal distribution.
如果 σ 未知且 n 很大，我们用 s 代替 σ。然后使用正态分布的临界值与检验统计量进行比较。
Example: Test H₀: μ = 50 against H₁: μ ≠ 50 at the 5% significance level. Given n = 100, x̄ = 52.3, and σ = 8, the test statistic is:
示例：在 5% 显著性水平下检验 H₀: μ = 50 对 H₁: μ ≠ 50。已知 n = 100，x̄ = 52.3，σ = 8，检验统计量为：
z = ( 52.3 − 50 ) / ( 8 / √100 ) = 2.3 / 0.8 = 2.875
For a two-tailed test at 5%, the critical values are ±1.96. Since 2.875 > 1.96, we reject H₀ and conclude that there is significant evidence that the population mean is not 50.
对于 5% 双侧检验，临界值为 ±1.96。因为 2.875 > 1.96，所以我们拒绝 H₀，并得出结论：有显著证据表明总体均值不是 50。
The p-value can also be used. For this z-value, the two-tailed p-value is approximately 0.004, which is less than 0.05.
也可以使用 p 值。对于这个 z 值，双侧 p 值约为 0.004，小于 0.05。
9. Hypothesis Testing for p | 比例的假设检验
For an unknown proportion p, the null hypothesis H₀: p = p₀ is tested using the statistic:
对于未知比例 p，原假设 H₀: p = p₀ 使用以下统计量进行检验：
z = ( p̂ − p₀ ) / √( p₀(1 − p₀) / n )
This uses the value p₀ from the null hypothesis to calculate the standard error, rather than p̂.
这里使用原假设中的 p₀ 来计算标准误，而不是使用 p̂。
Example: A claimed proportion is p₀ = 0.40. In a sample of n = 250, we observe 115 successes, so p̂ = 115/250 = 0.46. If the alternative is H₁: p > 0.40, then:
示例：某声明认为总体比例 p₀ = 0.40。在 n = 250 的样本中观察到 115 次成功，因此 p̂ = 115/250 = 0.46。如果备择假设为 H₁: p > 0.40，则：
z = ( 0.46 − 0.40 ) / √( 0.40 × 0.60 / 250 ) = 0.06 / 0.03098 ≈ 1.936
At the 5% level, the one-tailed critical value is 1.645. Since 1.936 > 1.645, we reject H₀ in favour of H₁, concluding that the proportion is significantly greater than 0.40.
在 5% 水平下，单侧检验的临界值为 1.645。因为 1.936 > 1.645，我们拒绝 H₀，支持 H₁，并得出结论：比例显著大于 0.40。
10. Type I Error, Type II Error, and Power | 第一类错误、第二类错误与检验功效
Every hypothesis test involves the possibility of making a wrong decision. A Type I error occurs when we reject H₀ even though H₀ is true. The probability of a Type I error is the significance level α.
每一次假设检验都可能做出错误决策。第一类错误发生在 H₀ 实际上为真时我们却拒绝了 H₀。第一类错误的概率就是显著性水平 α。
A Type II error occurs when we fail to reject H₀ even though H₀ is false. Its probability is denoted by β.
第二类错误发生在 H₀ 实际上为假时我们却没有拒绝 H₀。其概率记为 β。
Because the population distribution is unknown, we can only estimate β and power in a large-sample setting by using the approximate normal sampling distribution of the test statistic.
由于总体分布未知，在大样本情况下我们只能通过检验统计量的近似正态抽样分布来估计 β 和检验功效。
| Decision | 决策 | H₀ True | H₀ 为真 | H₀ False | H₀ 为假 |
| Reject H₀ | 拒绝 H₀ | Type I Error | 第一类错误 | Correct | 正确 |
| Do not reject H₀ | 不拒绝 H₀ | Correct | 正确 | Type II Error | 第二类错误 |
The power of a test is defined as 1 − β. It is the probability of correctly rejecting a false null hypothesis.
检验功效定义为 1 − β。它表示正确拒绝错误原假设的概率。
In practice, increasing the sample size n reduces β and therefore increases power, because the sampling distribution becomes more concentrated.
实际中，增大样本量 n 会减小 β，从而提高检验功效，因为抽样分布变得更加集中。
11. Choosing Sample Size | 样本量的选择
Before collecting data, we may need to determine the sample size required to estimate μ with a given margin of error E and confidence level. Rearranging the confidence interval formula gives:
在收集数据之前，我们可能需要确定样本量，以便在给定的误差幅度 E 和置信水平下估计 μ。对置信区间公式变形可得：
n = ( z* × σ / E )²
Since n must be a whole number, we always round up to the next integer.
由于 n 必须是整数，我们总是向上取整。
Example: We want a 95% confidence interval with a margin of error E = 2. Given σ = 10, the required sample size is:
示例：我们希望构造一个误差幅度 E = 2 的 95% 置信区间。已知 σ = 10，所需样本量为：
n = ( 1.96 × 10 / 2 )² = 9.8² = 96.04
We round up to n = 97.
向上取整得到 n = 97。
For estimating a proportion, the formula is n = z*² p(1 − p) / E². If p is unknown, the conservative choice is p = 0.5, which maximises the required sample size.
对于比例估计，公式为 n = z*² p(1 − p) / E²。如果 p 未知，保守选择 p = 0.5，这会使所需样本量最大化。
12. Summary and Exam Tips | 总结与考试要点
Large-sample inference for unknown populations is built on one powerful result: the Central Limit Theorem. It allows us to use normal approximations for both sample means and sample proportions, even when the original population distribution is unknown.
针对未知总体的大样本推断建立在一个强大的结果之上：中心极限定理。它允许我们对样本均值和样本比例都使用正态近似，即使原始总体分布未知。
You should be prepared to construct confidence intervals and perform z-tests using the correct standard error in each context.
你应该准备好构造置信区间，并在每种情境下使用正确的标准误进行 z 检验。
For the Edexcel exam, keep the following points in mind:
在 Edexcel 考试中，请记住以下几点：
- Check whether σ is known; if not, use s provided n is large. | 检查 σ 是否已知；如果未知，只要 n 很大就使用 s。
- For proportions, always use p₀ under H₀ in the hypothesis test standard error. | 对于比例，检验标准误中始终用 H₀ 下的 p₀。
- For confidence intervals, use p̂ in the standard error. | 对于置信区间，标准误中使用 p̂。
- Match one-tailed or two-tailed alternatives to the correct critical values. | 根据单侧或双侧备择假设选择正确的临界值。
- Round sample sizes up, not down. | 样本量要向上取整，而不是向下。
Always write your conclusions clearly in context, stating whether H₀ is rejected at the chosen significance level.
务必结合题目背景清楚地写出结论，说明在所选显著性水平下是否拒绝 H₀。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply