假设检验之两项参数间的关系

深藏阁楼爱情的钟 2023-06-25 10:06 5阅读 0赞

<table> 
 <thead> 
 <tr> 
 <th>假设情况</th> 
 <th>已知条件</th> 
 <th>何时拒绝 H 0 H_0 H0</th> 
 </tr> 
 </thead> 
 <tbody> 
 <tr> 
 <td> μ 1 − μ 2 = σ \mu_1-\mu_2=\sigma μ1−μ2=σ</td> 
 <td> σ 1 , σ 2 \sigma_1,\sigma_2 σ1,σ2</td> 
 <td> x ‾ 1 − x ‾ 2 − σ σ 1 2 n 1 + σ 2 2 n 2 ≥ Z α 2 \frac{\overline x_1-\overline x_2-\sigma}{\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma_2^2}{n_2}}} \ge Z_{\frac{\alpha}{2}} n1σ12+n2σ22 x1−x2−σ≥Z2α</td> 
 </tr> 
 <tr> 
 <td> μ 1 − μ 2 = σ \mu_1-\mu_2=\sigma μ1−μ2=σ</td> 
 <td> ̸ \not </td> 
 <td> x ‾ 1 − x ‾ 2 − σ S w 1 n 1 + 1 n 2 ≥ t α 2 ( n 1 + n 2 − 2 ) \frac{\overline x_1-\overline x_2-\sigma}{S_w\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \ge t_{\frac{\alpha}{2}(n_1+n_2-2)} Swn11+n21 x1−x2−σ≥t2α(n1+n2−2)</td> 
 </tr> 
 <tr> 
 <td> σ 1 2 = σ 2 2 \sigma_1^2=\sigma_2^2 σ12=σ22</td> 
 <td> ̸ \not </td> 
 <td> S 1 2 S 2 2 ≥ F α 2 ( n 1 − 1 , n 2 − 1 ) 或 者 S 1 2 S 2 2 ≤ F 1 − α 2 ( n 1 − 1 , n 2 − 1 ) \frac{S_1^2}{S_2^2}\ge F_{\frac{\alpha}{2}(n_1-1,n_2-1)}或者\frac{S_1^2}{S_2^2}\le F_{1-\frac{\alpha}{2}(n_1-1,n_2-1)} S22S12≥F2α(n1−1,n2−1)或者S22S12≤F1−2α(n1−1,n2−1)</td> 
 </tr> 
 <tr> 
 <td> μ 1 − μ 2 ≤ σ \mu_1-\mu_2 \le \sigma μ1−μ2≤σ</td> 
 <td> σ 1 , σ 2 \sigma_1,\sigma_2 σ1,σ2</td> 
 <td> x ‾ 1 − x ‾ 2 − σ σ 1 2 n 1 + σ 2 2 n 2 ≥ Z α \frac{\overline x_1 - \overline x_2 - \sigma}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}} \ge Z_{\alpha} n1σ12+n2σ22 x1−x2−σ≥Zα</td> 
 </tr> 
 <tr> 
 <td> μ 1 − μ 2 ≤ σ \mu_1-\mu_2 \le \sigma μ1−μ2≤σ</td> 
 <td> ̸ \not </td> 
 <td> x ‾ 1 − x ‾ 2 − σ S w 1 n 1 + 1 n 2 ≥ t α ( n 1 + n 2 − 2 ) \frac{\overline x_1-\overline x_2-\sigma}{S_w\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \ge t_{\alpha(n_1+n_2-2)} Swn11+n21 x1−x2−σ≥tα(n1+n2−2)</td> 
 </tr> 
 <tr> 
 <td> σ 1 2 ≤ σ 2 2 \sigma_1^2 \le \sigma_2^2 σ12≤σ22</td> 
 <td> ̸ \not </td> 
 <td> S 1 2 S 2 2 ≥ F α ( n 1 − 1 , n 2 − 1 ) \frac{S_1^2}{S_2^2} \ge F_{\alpha}(n_1-1,n_2-1) S22S12≥Fα(n1−1,n2−1)</td> 
 </tr> 
 <tr> 
 <td> μ 1 − μ 2 ≥ σ \mu_1-\mu_2 \ge \sigma μ1−μ2≥σ</td> 
 <td> σ 1 , σ 2 \sigma_1,\sigma_2 σ1,σ2</td> 
 <td> x ‾ 1 − x ‾ 2 − σ σ 1 2 n 1 + σ 2 2 n 2 ≤ − Z α \frac{\overline x_1 - \overline x_2 - \sigma}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}} \le -Z_{\alpha} n1σ12+n2σ22 x1−x2−σ≤−Zα</td> 
 </tr> 
 <tr> 
 <td> μ 1 − μ 2 ≥ σ \mu_1-\mu_2 \ge \sigma μ1−μ2≥σ</td> 
 <td> ̸ \not </td> 
 <td> x ‾ 1 − x ‾ 2 − σ S w 1 n 1 + 1 n 2 ≤ − t α ( n 1 + n 2 − 2 ) \frac{\overline x_1-\overline x_2-\sigma}{S_w\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \le -t_{\alpha(n_1+n_2-2)} Swn11+n21 x1−x2−σ≤−tα(n1+n2−2)</td> 
 </tr> 
 <tr> 
 <td> σ 1 2 ≥ σ 2 2 \sigma_1^2 \ge \sigma_2^2 σ12≥σ22</td> 
 <td> ̸ \not </td> 
 <td> S 1 2 S 2 2 ≤ F 1 − α ( n 1 − 1 , n 2 − 1 ) \frac{S_1^2}{S_2^2} \le F_{1-\alpha}(n_1-1,n_2-1) S22S12≤F1−α(n1−1,n2−1)</td> 
 </tr> 
 </tbody> 
</table>

S w = ( n 1 − 1 ) S 1 2 + ( n 2 − 1 ) S 2 2 n 1 + n 2 − 2 S\_w=\\sqrt\{\\frac\{(n\_1-1)S\_1^2+(n\_2-1)S\_2^2\}\{n\_1+n\_2-2\}\} Sw=n1\+n2−2(n1−1)S12\+(n2−1)S22  
 S = 1 n − 1 \[ ( x 1 − x ‾ ) 2 + ( x 2 − x ‾ ) 2 + . . . + ( x n − x ‾ ) 2 \] S=\\sqrt\{\\frac\{1\}\{n-1\}\[(x\_1-\\overline x)^2+(x\_2-\\overline x)^2+...+(x\_n-\\overline x)^2\]\} S=n−11\[(x1−x)2\+(x2−x)2\+...\+(xn−x)2\]