Question

A bicycle manufacturer randomly selects 9 frames each day and tests for defects. The number of defective frames found over the last 14 days is 1, 1, 2, 2, 1, 2, 2, 0, 1, 0, 4, 6, 0, 4.

  Click here for the Excel Data File

Determine the control limits for the process. (Round your answers to 1 decimal place.)

Comment on whether the process is "in control."

Answer 1

D = p = Total number of defectives Total number of items Sampled number of frames sampled per day p - 3 BVPlace = LCU UCL p+ 3 Velees

A E F G 1 Days 1 N B С D Number of frames sampled Number of defective frames Proportion defective 9 1 0.1 9 1 0.1 9 2 0.2 9 2 0.2 3 2 4 3 5 4 6 5 9 1 0.1 7 6 9 2 0.2 co 7 9 2 0.2 9 8 9 0 0.0 10 9 9 1 0.1 11 10 0 0.0 9 9 11 4 0.4 12 13 14 12 6 0 13 14 9 9 9 126 0.7 0.0 0.4 4 TOTAL 26 17 0.206349206 19 n 9 20 21 LCL -0.2 LCL should be equal to zero as negative proportion of defective is not possible. 22 UCL 0.6 23 24 a) The control limit for the process is 0 to 0.6 25 b) Proportion defective on day 12 is above the UCL and therefore the process is not in control on day 12.

А. Days B Number of frames sampled C Number of defective frames D Proportion defective 9 9 9 1 2 1 3 2 4 3 5 4 6 5 7 6 8 7 1 1 2 2 1 2 19 19 9 9 9 2 =C2/B2 =C3/B3 =C4/B4 =C5/B5 =C6/B6 =C7/B7 =C8/B8 =C9/B9 =C10/10 =C11/B11 =C12/B12 =C13/B13 =C14/B14 =C15/B15 98 19 0 10 9 19 1 9 0 11 10 12 11 9 4 19 6 13 12 14 13 19 0 15 14 19 =SUM(B2:B15) 4 =SUM(C2:C15) 16 TOTAL 17 18 p =C16/B16 9 19 n 20 21 LCL 22 UCL LCL should be equal to zero as negative proportion of defective is not possible. =B18-3*((B18*(1-B18)/B19)^0.5) =B18+3*((B18*(1-B18)/B19)^0.5) 23 24 a) The control limit for the process is 0 to 0.6 25 b) Proportion defective on day 12 is above the UCL and therefore the process is not in control on day 12.

0.7 0.6 UCL 0.5 0.4 PROPORTION DEFECTIE 0.3 0.2 0.1 LCL 0.0 0 2 4 10 12 14 6 8 DAYS