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26 December, 11:45

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?

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  1. 26 December, 13:26
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    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
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