例 8.2.8 未知均值下检验两种香烟尼古丁含量方差是否相等

数据和总体假设同例 8.2.6 两正态总体均值差的t检验 (方差相等), 并设 $μ_{1}, μ_{2}$ 未知. 试问在显著性水平 $α = 0.05$ 下能否认为两种香烟的尼古丁含量的方差相等?

解

依题意,这是两个正态总体方差比的假设检验问题, 其中 $μ_{1}, μ_{2}$ 未知, 样本容量 $m = n = 6$ .

按题目的要求, 提出

H_{0} : \frac{σ _{2}^{2}}{σ _{1}^{2}} = 1, H_{1} : \frac{σ _{2}^{2}}{σ _{1}^{2}} \neq = 1.

由

表 8.4 两个正态总体方差的假设检验的拒绝域
两个正态总体方差的假设检验的拒绝域 (显著性水平为 $α$ )

序号 $H_{0}$ $H_{1}$ $μ_{1}, μ_{2}$ 已知 $μ_{1}, μ_{2}$ 未知
I $σ_{1}^{2} = σ_{2}^{2}$ $σ_{1}^{2} \neq = σ_{2}^{2}$ $\frac{i = 1 \sum m ( x _{i} - μ _{1} ) ^{2} / m}{j = 1 \sum n ( y _{j} - μ _{2} ) ^{2} / n} \leq F_{\frac{α}{2}} (m, n) 或 \frac{i = 1 \sum m ( x _{i} - μ _{1} ) ^{2} / m}{i = 1 \sum n ( y _{i} - μ _{2} ) ^{2} / n} \geq F_{1 - \frac{α}{2}} (m, n)$ $\frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \leq F_{\frac{α}{2}} (m - 1, n - 1)$ 或
$\frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \geq F_{1 - \frac{α}{2}} (m - 1, n - 1)$
II $σ_{1}^{2} = σ_{2}^{2}$ $σ_{1}^{2} > σ_{2}^{2}$ $\frac{i = 1 \sum m ( x _{i} - μ _{1} ) ^{2} / m}{j = 1 \sum n ( y _{j} - μ _{2} ) ^{2} / n} \geq F_{1 - α} (m, n)$ $\frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \geq F_{1 - α} (m - 1, n - 1)$
III $σ_{1}^{2} \leq σ_{2}^{2}$ $σ_{1}^{2} > σ_{2}^{2}$
IV $σ_{1}^{2} = σ_{2}^{2}$ $σ_{1}^{2} < σ_{2}^{2}$ $\frac{i = 1 \sum m ( x _{i} - μ _{1} ) ^{2} / m}{j = 1 \sum n ( y _{j} - μ _{2} ) ^{2} / n} \leq F_{α} (m, n)$ $\frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \leq F_{α} (m - 1, n - 1)$
V $σ_{1}^{2} \geq σ_{2}^{2}$ $σ_{1}^{2} < σ_{2}^{2}$ $\frac{i = 1 \sum m ( x _{i} - μ _{1} ) ^{2} / m}{j = 1 \sum n ( y _{j} - μ _{2} ) ^{2} / n} \leq F_{α} (m, n)$ $\frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \leq F_{α} (m - 1, n - 1)$
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序号	$H_{0}$	$H_{1}$	$μ_{1}, μ_{2}$ 已知	$μ_{1}, μ_{2}$ 未知
I	$σ_{1}^{2} = σ_{2}^{2}$	$σ_{1}^{2} \neq = σ_{2}^{2}$	$\frac{i = 1 \sum m ( x _{i} - μ _{1} ) ^{2} / m}{j = 1 \sum n ( y _{j} - μ _{2} ) ^{2} / n} \leq F_{\frac{α}{2}} (m, n) 或 \frac{i = 1 \sum m ( x _{i} - μ _{1} ) ^{2} / m}{i = 1 \sum n ( y _{i} - μ _{2} ) ^{2} / n} \geq F_{1 - \frac{α}{2}} (m, n)$	$\frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \leq F_{\frac{α}{2}} (m - 1, n - 1)$ 或
				$\frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \geq F_{1 - \frac{α}{2}} (m - 1, n - 1)$
II	$σ_{1}^{2} = σ_{2}^{2}$	$σ_{1}^{2} > σ_{2}^{2}$	$\frac{i = 1 \sum m ( x _{i} - μ _{1} ) ^{2} / m}{j = 1 \sum n ( y _{j} - μ _{2} ) ^{2} / n} \geq F_{1 - α} (m, n)$	$\frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \geq F_{1 - α} (m - 1, n - 1)$
III	$σ_{1}^{2} \leq σ_{2}^{2}$	$σ_{1}^{2} > σ_{2}^{2}$
IV	$σ_{1}^{2} = σ_{2}^{2}$	$σ_{1}^{2} < σ_{2}^{2}$	$\frac{i = 1 \sum m ( x _{i} - μ _{1} ) ^{2} / m}{j = 1 \sum n ( y _{j} - μ _{2} ) ^{2} / n} \leq F_{α} (m, n)$	$\frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \leq F_{α} (m - 1, n - 1)$
V	$σ_{1}^{2} \geq σ_{2}^{2}$	$σ_{1}^{2} < σ_{2}^{2}$	$\frac{i = 1 \sum m ( x _{i} - μ _{1} ) ^{2} / m}{j = 1 \sum n ( y _{j} - μ _{2} ) ^{2} / n} \leq F_{α} (m, n)$	$\frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \leq F_{α} (m - 1, n - 1)$

的第一行, 选取统计量为

Z = \frac{S _{1 m}^{* 2}}{S _{2 n}^{* 2}} \sim F (6 - 1, 6 - 1) .

令 $P (F_{0.025} (5, 5) \leq Z \leq F_{0.975} (5, 5)) = 0.05$ , 查 $F$ 分布表或用 $R$ 软件计算得 $F_{0.025} (5, 5) = 0.1399$ 和 $F_{0.975} (5, 5) = 7.146$

经计算

Z = \frac{s _{1 m}^{* 2}}{s _{2 n}^{* 2}} = 0.229

由于 $F_{0.025} (5, 5) = 0.1399 < Z = 0.229 < F_{0.975} (5, 5) = 7.146$ , 所以接受 $H_{0}$ , 即在显著性水平 $α = 0.05$ 下认为两种香烟的尼古丁含量的方差无显著差异.

Youliang Zhong

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例 8.2.8 未知均值下检验两种香烟尼古丁含量方差是否相等

解

表 8.4 两个正态总体方差的假设检验的拒绝域