设总体 X 的分布律为 P ( X = 1 ) = theta , P ( X = 2 ) = 2 theta , P ( X = 3 ) = 1 - 3 theta 作检验 H _ 0 : theta = 0.1 , H _ 1 : theta = 0.2 现抽取 容量 3 的样本 X _ 1 , X _ 2 , X _ 3 若拒绝域 ( X _ 1 = 1 , X _ 2 = 2 , X _ 3 = 3 ) 为则犯第一类错误及第二类错误的概率是多少

答案分别为0.014，0.968。由犯第一类错误计算公式则有第一类错误： P \{ X_{1}=1,X_{2}=2,X_{3}=3 \mid \theta =0.1 \} =P \{ X_{1}=1 \mid \theta =0.1 \} \times P \{ X_{2}=2 \mid \theta =0.1 \} \times P \{ X_{3}=3 \mid \theta =0.1 \} =0.1 \times 0.2 \times 0.7 =0.014 由犯第二类错误计算公式则有第二类错误 P \{ X_{1} \neq 1 \cup X_{2} \neq 2 \cup X_{3} \neq 3 \mid \theta =0.2 \} =P \{ X_{1} \neq 1 \mid \theta =0.2 \} +P \{ X_{2} \neq 2 \mid \theta =0.2 \} +P \{ X_{3} \neq 3 \mid \theta =0.2 \} -P \{ X_{1} \neq 1 \mid \theta =0.2 \} P \{ X_{2} \neq 2 \mid \theta =0.2 \} -P \{ X_{1} \neq 1 \mid \theta =0.2 \} P \{ X_{3} \neq 3 \mid \theta =0.2 \} -P \{ X_{2} \neq 2 \mid \theta =0.2 \} P \{ X_{3} \neq 3 \mid \theta =0.2 \} +P \{ X_{1} \neq 1 \mid \theta =0.2 \} P \{ X_{2} \neq 2 \mid \theta =0.2 \} \times P \{ X_{3} \neq 3 \mid \theta =0.2 \} 887^{\circ}0+98^{\circ}0-8秒0-9'0+9'0+9'0= =0.968 。