# 0x03 两个总体参数的检验

## 两个总体均值之差的检验

### $$\sigma^2\_1$$, $$\sigma^2\_2$$已知

$$\bar{x}\_1-\bar{x}\_2$$的抽样分布服从正态分布, 可以使用$$z$$统计量进行检验, 如下转换:

$$z=\frac{(\bar{x}\_1-\bar{x}\_2)-(\mu\_1-\mu\_2)}{\sqrt{\frac{\sigma^2\_1}{n\_1}+\frac{\sigma^2\_2}{n\_2}}}$$

### $$\sigma^2\_1$$, $$\sigma^2\_2$$未知, 且样本量$$n$$较小, $$\sigma^2\_1=\sigma^2\_2$$

对$$\hat{\sigma}\_{\bar{x}\_1-\bar{x}\_2}$$的估计为:

$$\hat{\sigma}\_{\bar{x}\_1-\bar{x}\_2}=s\_p\sqrt{\frac{1}{n\_1}+\frac{1}{n\_2}}$$

其中:

$$s^2\_p=\frac{(n\_1-1)s^2\_1+(n\_2-1)s^2\_2}{n\_1+n\_2-2}$$

使用$$t$$检验统计量, 自由度为$$n\_1+n\_2-2$$:

$$t=\frac{(\bar{x}\_1-\bar{x}\_2)-(\mu\_1-\mu\_2)}{s\_p\sqrt{\frac{1}{n\_1}+\frac{1}{n\_2}}}$$

### $$\sigma^2\_1$$, $$\sigma^2\_2$$未知, 且样本量$$n$$较小, $$\sigma^2\_1 \ne \sigma^2\_2$$

对$$\hat{\sigma}\_{\bar{x}\_1-\bar{x}\_2}$$的估计为:

$$\hat{\sigma}\_{\bar{x}\_1-\bar{x}\_2}=\sqrt{\frac{s^2\_1}{n\_1}+\frac{s^2\_2}{n\_2}}$$

使用$$t$$检验统计量, 但此时的自由度为$$f$$:

$$\begin{aligned}f=\frac{(\frac{s^2\_1}{n\_1}+\frac{s^2\_2}{n\_2})^2}{\frac{(s^2\_1/n\_1)^2}{n\_1-1}\frac{(s^2\_2/n\_2)^2}{n\_2-1}}\end{aligned}$$

$$t=\frac{(\bar{x}\_1-\bar{x}\_2)-(\mu\_1-\mu\_2)}{\sqrt{\frac{s^2\_1}{n\_1}+\frac{s^2\_2}{n\_2}}}$$

## 两个总体比例之差的检验

### 检验两个总体比例相等的假设

$$H\_0:\pi\_1=\pi\_2$$

在原假设成立的条件下, 首先计算两个样本合并后得到的**比例估计量**, 即:

$$p=\frac{x\_1+x\_2}{n\_1+n\_2}=\frac{p\_1n\_1+p\_2n\_2}{n\_1+n\_2}$$

则此时样本比例抽样分布的方差为$$p(1-p)$$, 使用$$z$$统计量:

$$z=\frac{p\_1-p\_2}{\sqrt{p(1-p)(\frac{1}{n\_1}+\frac{1}{n\_2})}}$$

### 检验两个总体比例之差不为零的假设

$$H\_0: \pi\_1-\pi\_2=d\_0$$

此时两个样本比例之差$$p\_1-p\_2$$服从以$$\pi\_1-\pi\_2=d\_0$$为期望的正态分布, 使用$$z$$统计量:

$$z=\frac{(p\_1-p\_2)-d\_0}{\sqrt{\frac{p\_1(1-p\_1)}{n\_1}+\frac{p\_2(1-p\_2)}{n\_2}}}$$

## 两个总体方差比的检验

检验两个总体方差是否相等, 通过方差比是否等于1来进行. 如果$$s^2\_1/s^2\_2$$接近于1, 说明两个总体方差$$\sigma^2\_1$$和$$\sigma^2\_2$$很接近.

两个方差之比服从自由度为$$(n\_1-1, n\_2-1)$$的$$F$$分布:

$$F=\frac{s^2\_1/\sigma^2\_1}{s^2\_2/\sigma^2\_2}$$

在单侧检验中, 一般把较大的$$s^2$$放在分子的位置, 此时$$F>1$$, 拒绝域在$$F$$分布的右侧, 对应的原假设为$$H\_0: \sigma^2\_1 \le \sigma^2\_2$$, 临界点为$$F\_{\alpha}(n\_1-1,n\_2-1)$$.

双侧检验中, 拒绝域在$$F$$分布的两侧, 两个临界点分别为: $$F\_{\alpha/2}(n\_1-1,n\_2-1)$$, $$F\_{1-\alpha/2}(n\_1-1,n\_2-1)$$.


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