21.4 An inspector’s accuracy has been assessed as follows: p1 = 0.96 and p2 = 0.60. The inspector is given the task of inspecting a batch of 500 parts and sorting out the defects from good units. If the actual defect rate in the batch is 0.04, determine (a) the expected number of Type I and (b) Type II errors the inspector will make. (c) Also, what is the expected fraction defect rate that the inspector will report at the end of the inspection task?

a) 19.2

b) 8

c) 0.0624

Explanation:

a) for type I error:

(1 - p1)(1 - q)(1 - p1)(1 - q) =

= (1 - 0.94)×(1 - 0.04)

= (0.04)×(0.96) = 0.0384

For 500 pts:

E × (number of Type I errors) =

=0.0384 × 500= 19.2 pc

(b) For Type II error, we use:

(1 - p2)q(1 - p2)q

E = (1 - 0.60)×(0.04)

= 0.4×0.04 = 0.016

For 500 pts:

E × (number of Type II errors) = 0.016 ×500 = 8pc

(c) number of defects reported:

1 - p1 - q(1 - p1 - p2)

= 1 - 0.96 - 0.04(1 - 0.96 - 0.60)

= 0.0624

For 500 pts:

E×number of defects reported

= 0.0624 × 500

=31.2pc

Therefore, the expected fraction defect rate reported will be:

= 31.2 / 500 = 0.0624