医学实验室精密度和准确度的确认ep15-a2_user_verifica

Deviation )5 25 24 23 22 21 2, ?????????? run run run run run average run sd sd sd sd sd sd = ( 0 + 0.3334 + 0.3334 + 0.3334 + 1 )÷ 5 = 0.40004 1、 2、 2, average run r sd s?= 0.632 < 1 mg/dL 计算扩展σ r v :自由度( Degrees Of Freedom ) C :自由度为 10 置信度为 97.5% 卡平方分布值( ) 扩展σ r = v c r??( Verification value )σ r = 1 ( mg/dL ); v = 10 ; C = 20.48 2? = 1.431 ∵s r = 0.632 < 1.431 ; ∴可以认为待验重复性( s r)与标示的重复性( claimed )σ r保持一致, 待验系统重复性( s r)被证实。(结论) (S l) 总不精密度( With-laboratory Standard Deviation ) = ( ( 140-141.33) 2 + (138.33-141.33) 2 +(143.67-141.33) 2 + (142.67-141.33) 2 + (142-141.33) 2)÷ 4 = 4.62225 1、 4 )( 51 2 2???? d d bxxss r = 0.632 ; (S l) 总不精密度( With-laboratory Standard Deviation ) 2、= SQRT ( 2 ÷3× 0.40004 + 4.62225 ) ; 221 brlssn ns????= 2.21 > 2 mg/dL