已知某产品的质量指标服从正态分布，但由于原料、设备条件、操作人员不同或工艺过程的改变等原因，都会引起总体的均值或方差有所改变，我们需要知道这种改变有多大？这就需要考察两个正态总体的均值差或方差比的区间估计问题．下面我们给出正态总体参数区间估计的几种常见类型．对于非正态总体情形，即总体不服从正态分布或者不知道总体服从什么分布，一般采用大容量的样本，根据中心极限定理，按照正态分布近似处理．
两个正态总体的参数的区间估计
设总体 X ∼ N ( μ 1 , σ 1 2 ) X \sim N\left(\mu_1, \sigma_1^2\right) X ∼ N ( μ 1 , σ 1 2 ) Y ∼ N ( μ 2 , σ 2 2 ) , X Y \sim N\left(\mu_2, \sigma_2^2\right), X Y ∼ N ( μ 2 , σ 2 2 ) , X Y Y Y X 1 , X 2 , ⋯ , X n 1 X_1, X_2, \cdots, X_{n_1} X 1 , X 2 , ⋯ , X n 1 X X X Y 1 , Y 2 , ⋯ , Y n 2 Y_1, Y_2, \cdots, Y_{n_2} Y 1 , Y 2 , ⋯ , Y n 2 Y Y Y
（1）σ 1 2 \sigma_1^2 σ 1 2 σ 2 2 \sigma_2^2 σ 2 2 μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2 σ 1 2 , σ 2 2 \sigma_1^2, \sigma_2^2 σ 1 2 , σ 2 2 X ˉ , Y ˉ \bar{X}, \bar{Y} X ˉ , Y ˉ μ 1 , μ 2 \mu_1, \mu_2 μ 1 , μ 2 X ˉ − Y ˉ \bar{X}-\bar{Y} X ˉ − Y ˉ μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2 X ˉ ∼ N ( μ 1 , σ 1 2 n 1 ) , Y ˉ ∼ N ( μ 2 , σ 2 2 n 2 ) \bar{X} \sim N\left(\mu_1, \frac{\sigma_1^2}{n_1}\right), \bar{Y} \sim N\left(\mu_2, \frac{\sigma_2^2}{n_2}\right) X ˉ ∼ N ( μ 1 , n 1 σ 1 2 ) , Y ˉ ∼ N ( μ 2 , n 2 σ 2 2 ) X ˉ \bar{X} X ˉ Y ˉ \bar{Y} Y ˉ
X ˉ − Y ˉ ∼ N ( μ 1 − μ 2 , σ 1 2 n 1 + σ 2 2 n 2 ) \bar{X}-\bar{Y} \sim N\left(\mu_1-\mu_2, \frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}\right) X ˉ − Y ˉ ∼ N ( μ 1 − μ 2 , n 1 σ 1 2 + n 2 σ 2 2 ) 取枢轴量为
U = X ˉ − Y ˉ − ( μ 1 − μ 2 ) σ 1 2 n 1 + σ 2 2 n 2 ∼ N ( 0 , 1 ) , U=\frac{\bar{X}-\bar{Y}-\left(\mu_1-\mu_2\right)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}} \sim N(0,1), U = n 1 σ 1 2 + n 2 σ 2 2 X ˉ − Y ˉ − ( μ 1 − μ 2 ) ∼ N ( 0 , 1 ) , 可得 μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2 1 − α 1-\alpha 1 − α
[ ( X ˉ − Y ˉ ) − u α 2 σ 1 2 n 1 + σ 2 2 n 2 , ( X ˉ − Y ˉ ) + u α 2 σ 1 2 n 1 + σ 2 2 n 2 ] \boxed{
\left[(\bar{X}-\bar{Y})-u_{\frac{\alpha}{2}} \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}},(\bar{X}-\bar{Y})+u_{\frac{\alpha}{2}} \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\right]
} ( X ˉ − Y ˉ ) − u 2 α n 1 σ 1 2 + n 2 σ 2 2 , ( X ˉ − Y ˉ ) + u 2 α n 1 σ 1 2 + n 2 σ 2 2 （2）σ 1 2 \sigma_1^2 σ 1 2 σ 2 2 \sigma_2^2 σ 2 2 σ 1 2 = σ 2 2 \sigma_1^2=\sigma_2^2 σ 1 2 = σ 2 2 μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2 由于总体方差 σ 1 2 \sigma_1^2 σ 1 2 σ 2 2 \sigma_2^2 σ 2 2 σ 1 2 = σ 2 2 \sigma_1^2=\sigma_2^2 σ 1 2 = σ 2 2 正态总体的抽样分布 , 故取枢轴量
T = X ˉ − Y ˉ − ( μ 1 − μ 2 ) S w 1 n 1 + 1 n 2 ∼ t ( n 1 + n 2 − 2 ) . . . ( 2 ) \boxed{
T=\frac{\bar{X}-\bar{Y}-\left(\mu_1-\mu_2\right)}{S_w \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \sim t\left(n_1+n_2-2\right) ...(2)
} T = S w n 1 1 + n 2 1 X ˉ − Y ˉ − ( μ 1 − μ 2 ) ∼ t ( n 1 + n 2 − 2 ) ... ( 2 ) 且
S w 2 = ( n 1 − 1 ) S 1 2 + ( n 2 − 1 ) S 2 2 n 1 + n 2 − 2 S_w^2=\frac{\left(n_1-1\right) S_1^2+\left(n_2-1\right) S_2^2}{n_1+n_2-2} S w 2 = n 1 + n 2 − 2 ( n 1 − 1 ) S 1 2 + ( n 2 − 1 ) S 2 2 可得 μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2 1 − α 1-\alpha 1 − α
[ X ˉ − Y ˉ − t α 2 ( n 1 + n 2 − 2 ) S w 1 n 1 + 1 n 2 , X ˉ − Y ˉ + t α 2 ( n 1 + n 2 − 2 ) S w 1 n 1 + 1 n 2 ] \boxed{
\left[\bar{X}-\bar{Y}-t_{\frac{\alpha}{2}}\left(n_1+n_2-2\right) S_w \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}, \bar{X}-\bar{Y}+t_{\frac{\alpha}{2}}\left(n_1+n_2-2\right) S_w \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\right]
} [ X ˉ − Y ˉ − t 2 α ( n 1 + n 2 − 2 ) S w n 1 1 + n 2 1 , X ˉ − Y ˉ + t 2 α ( n 1 + n 2 − 2 ) S w n 1 1 + n 2 1 ] 两个正态分布参数估计主要用于两条生产线生产同一批产品的对比，请结合下面例题理解。
例 为比较 I，II 两种型号步枪子弹的枪口速度，随机地取 I 型子弹 10发，得到枪口速度的平均值为 x ˉ 1 = 500 m / s \bar{x}_1=500 \mathrm{~m} / \mathrm{s} x ˉ 1 = 500 m / s s 1 = 1.10 m / s s_1=1.10 \mathrm{~m} / \mathrm{s} s 1 = 1.10 m / s x ˉ 2 = 496 m / s \bar{x}_2=496 \mathrm{~m} / \mathrm{s} x ˉ 2 = 496 m / s s 2 = 1.20 m / s s_2=1.20 \mathrm{~m} / \mathrm{s} s 2 = 1.20 m / s μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2
解 按实际情况，可认为分别来自两个总体的样本是相互独立的．又因由假设两总体的方差相等，但数值未知，故可用（2）式求均值差的置信区间．
由于 1 − α = 0.95 , α / 2 = 0.025 , n 1 = 10 , n 2 = 20 , n 1 + n 2 − 2 = 28 , t 0.025 ( 28 ) = 2.0484 1-\alpha=0.95, \alpha / 2=0.025, n_1=10, n_2=20, n_1+n_2-2=28, t_{0.025}(28)=2.0484 1 − α = 0.95 , α /2 = 0.025 , n 1 = 10 , n 2 = 20 , n 1 + n 2 − 2 = 28 , t 0.025 ( 28 ) = 2.0484 s w 2 = ( 9 × 1.10 2 + 19 × 1.20 2 ) / 28 , s w = s w 2 = 1.1688 s_w^2=\left(9 \times 1.10^2+19 \times 1.20^2\right) / 28, s_w=\sqrt{s_w^2}=1.1688 s w 2 = ( 9 × 1.1 0 2 + 19 × 1.2 0 2 ) /28 , s w = s w 2 = 1.1688 μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2
( x ˉ 1 − x ˉ 2 ± s w × t 0.025 ( 28 ) 1 10 + 1 20 ) = ( 4 ± 0.93 ) \left(\bar{x}_1-\bar{x}_2 \pm s_w \times t_{0.025}(28) \sqrt{\frac{1}{10}+\frac{1}{20}}\right)=(4 \pm 0.93) ( x ˉ 1 − x ˉ 2 ± s w × t 0.025 ( 28 ) 10 1 + 20 1 ) = ( 4 ± 0.93 ) 即
( 3.07 , 4.93 ) . (3.07,4.93) . ( 3.07 , 4.93 ) . 本题中得到的置信区间的下限大于零，在实际中我们就认为 μ 1 \mu_1 μ 1 μ 2 \mu_2 μ 2
例为提高某一化学生产过程的得率，试图采用一种新的催化剂．为慎重起见，在实验工厂先进行试验．设采用原来的催化剂进行了 n 1 = 8 n_1=8 n 1 = 8 x ˉ 1 = 91.73 \bar{x}_1=91.73 x ˉ 1 = 91.73 s 1 2 = 3.89 s_1^2=3.89 s 1 2 = 3.89 n 2 = 8 n_2=8 n 2 = 8 x ˉ 2 = 93.75 \bar{x}_2=93.75 x ˉ 2 = 93.75 s 2 2 = 4.02 s_2^2=4.02 s 2 2 = 4.02 μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2
解 现在
s w 2 = ( n 1 − 1 ) s 1 2 + ( n 2 − 1 ) s 2 2 n 1 + n 2 − 2 = 3.96 , s w = 3.96 s_w^2=\frac{\left(n_1-1\right) s_1^2+\left(n_2-1\right) s_2^2}{n_1+n_2-2}=3.96, s_w=\sqrt{3.96} s w 2 = n 1 + n 2 − 2 ( n 1 − 1 ) s 1 2 + ( n 2 − 1 ) s 2 2 = 3.96 , s w = 3.96 由（5．12）式得所求的置信区间为
( x ˉ 1 − x ˉ 2 ± t 0.025 ( 14 ) s w 1 8 + 1 8 ) = ( − 2.02 ± 2.13 ) \left(\bar{x}_1-\bar{x}_2 \pm t_{0.025}(14) s_w \sqrt{\frac{1}{8}+\frac{1}{8}}\right)=(-2.02 \pm 2.13) ( x ˉ 1 − x ˉ 2 ± t 0.025 ( 14 ) s w 8 1 + 8 1 ) = ( − 2.02 ± 2.13 ) 即 ( − 4.15 , 0.11 ) (-4.15,0.11) ( − 4.15 , 0.11 )
（3）μ 1 \mu_1 μ 1 μ 2 \mu_2 μ 2 σ 1 2 σ 2 2 \frac{\sigma_1^2}{\sigma_2^2} σ 2 2 σ 1 2 由于 μ 1 \mu_1 μ 1 μ 2 \mu_2 μ 2 正态分布参数估计 故取枢轴量为
F = S 1 2 σ 1 2 S 2 2 σ 2 2 ∼ F ( n 1 − 1 , n 2 − 1 ) , F=\dfrac{\frac{S_1^2}{\sigma_1^2}}{\frac{S_2^2}{\sigma_2^2}} \sim F\left(n_1-1, n_2-1\right), F = σ 2 2 S 2 2 σ 1 2 S 1 2 ∼ F ( n 1 − 1 , n 2 − 1 ) , 可得 σ 1 2 σ 2 2 \frac{\sigma_1^2}{\sigma_2^2} σ 2 2 σ 1 2 1 − α 1-\alpha 1 − α
[ S 1 2 S 2 2 F 1 − α 2 ( n 2 − 1 , n 1 − 1 ) , S 1 2 S 2 2 F α 2 ( n 2 − 1 , n 1 − 1 ) ] . \left[\frac{S_1^2}{S_2^2} F_{1-\frac{\alpha}{2}}\left(n_2-1, n_1-1\right), \frac{S_1^2}{S_2^2} F_{\frac{\alpha}{2}}\left(n_2-1, n_1-1\right)\right] . [ S 2 2 S 1 2 F 1 − 2 α ( n 2 − 1 , n 1 − 1 ) , S 2 2 S 1 2 F 2 α ( n 2 − 1 , n 1 − 1 ) ] . 例 某厂利用两条自动化流水线罐装辣椒酱，现分别从两条流水线上抽取了容量分别为 13 与 17 的两个相互独立的样本，其中
x ˉ = 10.6 g , y ˉ = 9.5 g , s 1 2 = 2.4 g 2 , s 2 2 = 4.7 g 2 . \bar{x}=10.6 g, \bar{y}=9.5 g, s_1^2=2.4 g^2, s_2^2=4.7 g^2 . x ˉ = 10.6 g , y ˉ = 9.5 g , s 1 2 = 2.4 g 2 , s 2 2 = 4.7 g 2 . 假设两条流水线上罐装的辣椒酱质量分别服从正态分布 N ( μ 1 , σ 1 2 ) N\left(\mu_1, \sigma_1^2\right) N ( μ 1 , σ 1 2 ) N ( μ 2 , σ 2 2 ) N\left(\mu_2, \sigma_2^2\right) N ( μ 2 , σ 2 2 )
（1）求它们的方差比 σ 1 2 σ 2 2 \frac{\sigma_1^2}{\sigma_2^2} σ 2 2 σ 1 2 σ 1 2 = σ 2 2 = σ 2 \sigma_1^2=\sigma_2^2=\sigma^2 σ 1 2 = σ 2 2 = σ 2 μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2
解（1）由于 μ 1 \mu_1 μ 1 μ 2 \mu_2 μ 2 σ 1 2 σ 2 2 \frac{\sigma_1^2}{\sigma_2^2} σ 2 2 σ 1 2 1 − α 1-\alpha 1 − α
[ S 1 2 S 2 2 F 1 − α 2 ( n 2 − 1 , n 1 − 1 ) , S 1 2 S 2 2 F α 2 ( n 2 − 1 , n 1 − 1 ) ] . \left[\frac{S_1^2}{S_2^2} F_{1-\frac{\alpha}{2}}\left(n_2-1, n_1-1\right), \frac{S_1^2}{S_2^2} F_{\frac{\alpha}{2}}\left(n_2-1, n_1-1\right)\right] . [ S 2 2 S 1 2 F 1 − 2 α ( n 2 − 1 , n 1 − 1 ) , S 2 2 S 1 2 F 2 α ( n 2 − 1 , n 1 − 1 ) ] . 由 n 1 = 13 , n 2 = 17 , α = 0.05 n_1=13, n_2=17, \alpha=0.05 n 1 = 13 , n 2 = 17 , α = 0.05 F F F 附表 ）得
F 0.025 ( 16 , 12 ) = 3.16 , F 0.975 ( 16 , 12 ) = 1 F 0.025 ( 12 , 16 ) ≈ 1 2.89 = 0.346 F_{0.025}(16,12)=3.16, \quad F_{0.975}(16,12)=\frac{1}{F_{0.025}(12,16)} \approx \frac{1}{2.89}=0.346 F 0.025 ( 16 , 12 ) = 3.16 , F 0.975 ( 16 , 12 ) = F 0.025 ( 12 , 16 ) 1 ≈ 2.89 1 = 0.346 由公式得方差比 σ 1 2 σ 2 2 \frac{\sigma_1^2}{\sigma_2^2} σ 2 2 σ 1 2
[ 2.4 4.7 × 0.346 , 2.4 4.7 × 3.16 ] = [ 0.1767 , 1.6136 ] \left[\frac{2.4}{4.7} \times 0.346, \frac{2.4}{4.7} \times 3.16\right]=[0.1767,1.6136] [ 4.7 2.4 × 0.346 , 4.7 2.4 × 3.16 ] = [ 0.1767 , 1.6136 ] （2）由于方差未知，但 σ 1 2 = σ 2 2 = σ 2 \sigma_1^2=\sigma_2^2=\sigma^2 σ 1 2 = σ 2 2 = σ 2 μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2 1 − α 1-\alpha 1 − α
[ ( X ˉ − Y ˉ ) − t α 2 ( n 1 + n 2 − 2 ) 1 n 1 + 1 n 2 ( n 1 − 1 ) S 1 2 + ( n 2 − 1 ) S 2 2 n 1 + n 2 − 2 ( X ˉ − Y ˉ ) + t α 2 ( n 1 + n 2 − 2 ) 1 n 1 + 1 n 2 ( n 1 − 1 ) S 1 2 + ( n 2 − 1 ) S 2 2 n 1 + n 2 − 2 ] \begin{aligned}
& {\left[(\bar{X}-\bar{Y})-t_{\frac{\alpha}{2}}\left(n_1+n_2-2\right) \sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \sqrt{\frac{\left(n_1-1\right) S_1^2+\left(n_2-1\right) S_2^2}{n_1+n_2-2}}\right.} \\
& \left.(\bar{X}-\bar{Y})+t_{\frac{\alpha}{2}}\left(n_1+n_2-2\right) \sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \sqrt{\frac{\left(n_1-1\right) S_1^2+\left(n_2-1\right) S_2^2}{n_1+n_2-2}}\right]
\end{aligned} ( X ˉ − Y ˉ ) − t 2 α ( n 1 + n 2 − 2 ) n 1 1 + n 2 1 n 1 + n 2 − 2 ( n 1 − 1 ) S 1 2 + ( n 2 − 1 ) S 2 2 ( X ˉ − Y ˉ ) + t 2 α ( n 1 + n 2 − 2 ) n 1 1 + n 2 1 n 1 + n 2 − 2 ( n 1 − 1 ) S 1 2 + ( n 2 − 1 ) S 2 2 由 n 1 = 13 , n 2 = 17 , α = 0.05 n_1=13, n_2=17, \alpha=0.05 n 1 = 13 , n 2 = 17 , α = 0.05 t t t t 0.025 ( 28 ) = 2.0484 t_{0.025}(28)=2.0484 t 0.025 ( 28 ) = 2.0484 μ 1 − μ 2 \mu_1-\mu_2 μ 1 − μ 2 [ 0.3545 , 2.5545 ] [ 0.3545,2.5545] [ 0.3545 , 2.5545 ]
例 研究由机器 A A A B B B A A A s 1 2 = 0.34 s_1^2=0.34 s 1 2 = 0.34 B B B s 2 2 = 0.29 s_2^2=0.29 s 2 2 = 0.29 A A A B B B N ( μ 1 , σ 1 2 ) , N ( μ 2 , σ 2 2 ) N\left(\mu_1, \sigma_1^2\right), N\left(\mu_2, \sigma_2^2\right) N ( μ 1 , σ 1 2 ) , N ( μ 2 , σ 2 2 ) μ i , σ i 2 ( i = 1 , 2 ) \mu_i, \sigma_i^2(i=1,2) μ i , σ i 2 ( i = 1 , 2 ) σ 1 2 / σ 2 2 \sigma_1^2 / \sigma_2^2 σ 1 2 / σ 2 2
解 现在 n 1 = 18 , s 1 2 = 0.34 , n 2 = 13 , s 2 2 = 0.29 , α = 0.10 , F α / 2 ( n 1 − 1 , n 2 − 1 ) = F 0.05 ( 17 , 12 ) = 2.59 , F 1 − α / 2 ( 17 , 12 ) = F 0.95 ( 17 , 12 ) = 1 F 0.05 ( 12 , 17 ) = 1 2.38 n_1=18, s_1^2=0.34, n_2=13, s_2^2=0.29, \alpha=0.10, F_{\alpha / 2}\left(n_1-1, n_2-1\right) =F_{0.05}(17,12)=2.59, F_{1-\alpha / 2}(17,12)=F_{0.95}(17,12)=\frac{1}{F_{0.05}(12,17)}=\frac{1}{2.38} n 1 = 18 , s 1 2 = 0.34 , n 2 = 13 , s 2 2 = 0.29 , α = 0.10 , F α /2 ( n 1 − 1 , n 2 − 1 ) = F 0.05 ( 17 , 12 ) = 2.59 , F 1 − α /2 ( 17 , 12 ) = F 0.95 ( 17 , 12 ) = F 0.05 ( 12 , 17 ) 1 = 2.38 1 σ 1 2 / σ 2 2 \sigma_1^2 / \sigma_2^2 σ 1 2 / σ 2 2
( 0.34 0.29 × 1 2.59 , 0.34 0.29 × 2.38 ) , \left(\frac{0.34}{0.29} \times \frac{1}{2.59}, \quad \frac{0.34}{0.29} \times 2.38\right), ( 0.29 0.34 × 2.59 1 , 0.29 0.34 × 2.38 ) , ＝
( 0.45 , 2.79 ) . (0.45,2.79) . ( 0.45 , 2.79 ) . 由于 σ 1 2 / σ 2 2 \sigma_1^2 / \sigma_2^2 σ 1 2 / σ 2 2 σ 1 2 , σ 2 2 \sigma_1^2, \sigma_2^2 σ 1 2 , σ 2 2
对以上关于正态总体参数的区间估计的讨论进行总结，单个正态总体参数的区间估计如表附表
正态分布参数估计表