Hypothesis Testing for the Variance of a Normal Distribution | 正态分布方差的假设检验

📚 Hypothesis Testing for the Variance of a Normal Distribution | 正态分布方差的假设检验

In A-Level statistics, hypothesis tests are often introduced for means and proportions. However, Edexcel also expects you to test whether the variance of a normal population has a particular value. This uses the chi-squared distribution and is a key skill in further statistics.

在 A-Level 统计中，假设检验通常从均值和比例开始。但 Edexcel 也要求你检验正态总体的方差是否等于某个特定值。这需要使用卡方分布，是进阶统计中的关键技能。

---

1. Why Test a Variance? | 为什么要检验方差？

Many real-world quality control problems are about variability rather than the average. A production process might have a correct mean but too much spread, so individual items can still fail a specification.

许多现实中的质量控制问题关注的是波动性而不是平均值。一个生产过程可能均值正确，但离散程度过大，导致个别产品仍然不符合规格。

Testing the population variance σ² lets us decide whether observed sample variability is consistent with a claimed value, or whether the process has become more or less variable.

检验总体方差 σ² 可以帮助我们判断观察到的样本波动是否与声称的值一致，或者过程是否变得更大或更小的波动。

---

2. The Chi-Squared Distribution | 卡方分布

If X is normally distributed with variance σ², and a random sample of size n has sample variance s², then the statistic below follows a chi-squared distribution with ν = n − 1 degrees of freedom:

如果 X 服从方差为 σ² 的正态分布，且容量为 n 的随机样本的样本方差为 s²，那么下面的统计量服从自由度为 ν = n − 1 的卡方分布：

χ² = (n − 1)s² / σ²

The chi-squared distribution is not symmetric. It is skewed to the right and only takes positive values, because variances and squared quantities cannot be negative.

卡方分布不是对称的。它右偏，并且只取正值，因为方差和平方量不可能是负数。

---

3. Degrees of Freedom | 自由度

The degrees of freedom for a single-sample variance test are ν = n − 1. We lose one degree of freedom because the sample variance uses the sample mean in its calculation.

单样本方差检验的自由度为 ν = n − 1。我们损失一个自由度，因为样本方差在计算中使用了样本均值。

Degrees of freedom determine the shape of the chi-squared distribution. A small ν gives a highly skewed curve, while a large ν makes the distribution more symmetric and closer to a normal shape.

自由度决定卡方分布的形状。较小的 ν 会得到高度偏斜的曲线，而较大的 ν 会使分布更对称，更接近正态形状。

---

4. Conditions for the Test | 检验的条件

This variance test is valid only when the underlying population is normally distributed. The test is not robust to strong departures from normality, especially with small samples.

该方差检验仅在总体服从正态分布时有效。该检验对严重偏离正态的情况不稳健，尤其是在小样本时。

You also need a random sample and the sample variance s² calculated using n − 1 in the denominator. Edexcel questions will normally state that the population is normal.

你还需要一个随机样本，并且样本方差 s² 的分母使用 n − 1。Edexcel 的题目通常会说明总体是正态分布。

---

5. Stating the Hypotheses | 提出原假设与备择假设

The null hypothesis always fixes the population variance at a claimed value. The alternative hypothesis can be two-tailed, upper-tailed, or lower-tailed.

原假设总是将总体方差固定在一个声称的值上。备择假设可以是双尾、上尾或下尾。

For a claim that the variance has changed, use H₁: σ² ≠ σ₀². For a claim that it has increased, use H₁: σ² > σ₀². For a claim that it has decreased, use H₁: σ² < σ₀².

如果声称方差发生了变化，使用 H₁: σ² ≠ σ₀²。如果声称方差增大，使用 H₁: σ² > σ₀²。如果声称方差减小，使用 H₁: σ² < σ₀²。

The significance level α is usually 5% or 1%, and the degrees of freedom are ν = n − 1.

显著性水平 α 通常为 5% 或 1%，自由度为 ν = n − 1。

---

6. Test Statistic Formula | 检验统计量公式

Once the sample variance s² and the hypothesised variance σ₀² are known, compute the test statistic:

一旦知道样本方差 s² 和假设的方差 σ₀²，计算检验统计量：

χ² = (n − 1)s² / σ₀²

This observed value is then compared with critical values from the chi-squared distribution with ν = n − 1 degrees of freedom.

然后将这个观测值与自由度为 ν = n − 1 的卡方分布的临界值进行比较。

---

7. Critical Value Method: Two-Tailed Test | 临界值法：双尾检验

For a two-tailed test at significance level α, split α equally between the two tails. The lower critical value is the α/2 quantile and the upper critical value is the 1 − α/2 quantile.

对于显著性水平为 α 的双尾检验，将 α 平均分配到两个尾部。下临界值是 α/2 分位数，上临界值是 1 − α/2 分位数。

Reject H₀ if the observed χ² is less than the lower critical value or greater than the upper critical value. Otherwise, do not reject H₀.

如果观测到的 χ² 小于下临界值或大于上临界值，则拒绝 H₀。否则，不拒绝 H₀。

---

8. Critical Value Method: One-Tailed Tests | 临界值法：单尾检验

For H₁: σ² > σ₀², use the upper α critical value. Reject H₀ if the observed χ² is greater than this value.

对于 H₁: σ² > σ₀²，使用上侧 α 临界值。如果观测到的 χ² 大于该值，则拒绝 H₀。

For H₁: σ² < σ₀², use the lower α critical value. Reject H₀ if the observed χ² is less than this value.

对于 H₁: σ² < σ₀²，使用下侧 α 临界值。如果观测到的 χ² 小于该值，则拒绝 H₀。

Always draw a small diagram and shade the rejection region. This reduces sign errors in exam answers.

始终画一个小图并涂出拒绝区域。这可以减少考试答案中的方向错误。

---

9. p-Value Method | p 值法

The p-value is the probability of obtaining a test statistic at least as extreme as the observed value, assuming H₀ is true. For an upper-tailed test, p = P(χ² ≥ observed χ²).

p 值是在 H₀ 为真的前提下，得到至少与观测值一样极端的检验统计量的概率。对于上尾检验，p = P(χ² ≥ 观测 χ²)。

For a two-tailed test, double the smaller tail probability: p = 2 × min(P(χ² ≥ observed), P(χ² ≤ observed)).

对于双尾检验，将较小的尾部概率乘以 2：p = 2 × min(P(χ² ≥ 观测值), P(χ² ≤ 观测值))。

Reject H₀ if the p-value is less than α. Otherwise, do not reject H₀.

如果 p 值小于 α，则拒绝 H₀。否则，不拒绝 H₀。

---

10. Worked Example: Two-Tailed Test | 例题：双尾检验

A machine fills bottles with a volume that is normally distributed. The manufacturer claims the variance is 10 ml². A sample of 10 bottles gives s² = 18.5 ml². Test at the 5% significance level whether the variance has changed.

一台机器灌装瓶子，容量服从正态分布。制造商声称方差为 10 ml²。一个容量为 10 的样本给出 s² = 18.5 ml²。在 5% 显著性水平下检验方差是否改变。

Hypotheses: H₀: σ² = 10, H₁: σ² ≠ 10. Degrees of freedom: ν = 10 − 1 = 9.

假设：H₀: σ² = 10，H₁: σ² ≠ 10。自由度：ν = 10 − 1 = 9。

χ² = (9 × 18.5) / 10 = 16.65

At 5% two-tailed with 9 degrees of freedom, the lower 2.5% critical value is 2.700 and the upper 2.5% critical value is 19.023. Since 2.700 < 16.65 < 19.023, we do not reject H₀.

在 5% 双尾、自由度 9 下，下 2.5% 临界值为 2.700，上 2.5% 临界值为 19.023。因为 2.700 < 16.65 < 19.023，我们不拒绝 H₀。

Conclusion: there is insufficient evidence at the 5% level to suggest that the variance has changed from 10 ml².

结论：在 5% 水平下，没有足够证据表明方差已从 10 ml² 改变。

---

11. Worked Example: One-Tailed Test | 例题：单尾检验

An engineer claims that the variance of the diameters of a metal rod is at most 0.03 mm². A sample of 15 rods gives s² = 0.045 mm². Test at the 5% level whether the variance exceeds 0.03 mm².

一位工程师声称金属棒直径的方差至多为 0.03 mm²。一个容量为 15 的样本给出 s² = 0.045 mm²。在 5% 水平下检验方差是否超过 0.03 mm²。

Hypotheses: H₀: σ² = 0.03, H₁: σ² > 0.03. Degrees of freedom: ν = 15 − 1 = 14.

假设：H₀: σ² = 0.03，H₁: σ² > 0.03。自由度：ν = 15 − 1 = 14。

χ² = (14 × 0.045) / 0.03 = 21.0

The upper 5% critical value with 14 degrees of freedom is 23.685. Since 21.0 < 23.685, we do not reject H₀.

自由度为 14 的上 5% 临界值为 23.685。因为 21.0 < 23.685，我们不拒绝 H₀。

Conclusion: there is insufficient evidence at the 5% level that the variance exceeds 0.03 mm².

结论：在 5% 水平下，没有足够证据表明方差超过 0.03 mm²。

---

12. Exam Tips and Common Errors | 考试技巧与常见错误

Always state hypotheses, degrees of freedom, the test statistic, the critical value or p-value, and a conclusion in context. A bare numeric answer is not enough for Edexcel.

务必写出假设、自由度、检验统计量、临界值或 p 值，以及在情境中的结论。仅有数字答案对 Edexcel 来说是不够的。

Do not use the normal distribution critical values. This test uses χ² critical values with n − 1 degrees of freedom.

不要使用正态分布的临界值。该检验使用自由度为 n − 1 的 χ² 临界值。

Check the direction of the rejection region. For an upper-tailed test, the rejection region is on the right; for a lower-tailed test, it is on the left.

检查拒绝区域的方向。对于上尾检验，拒绝区域在右边；对于下尾检验，拒绝区域在左边。

Remember that s² must be calculated using the unbiased formula with denominator n − 1. Edexcel usually provides Σx, Σx², or s² directly, but you should be prepared to compute it.

记住 s² 必须使用分母为 n − 1 的无偏公式计算。Edexcel 通常会直接给出 Σx、Σx² 或 s²，但你应准备好自己计算。

---

Published by TutorHao | A-Level Further Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）