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Table A Areas under the normal curve, 0 to z 03 Z 00 0080 0120 01 0478 0517 15 16 19 o00 DI 02 E#
e.What alpha risk would control limits of .0470 and .0136 provide? (Round your intermediate calculations to 4 decimal places.


Problem 10-5 Using samples of 198 credit card statements, an auditor found the following: Use Table-A 3 4 7 B Sanple Nunber v
00303 Mean ao122 Standard deviation d.What control limits would give an alpha risk of 03 for this process? (Round your interm
Table A Areas under the normal curve, O to z 01 04 07 08 10 8 3 4 6 7 8 R T Y 5 R NN LL
Table A Areas under the normal curve, 0 to z 03 Z 00 0080 0120 01 0478 0517 15 16 19 o00 DI 02 E#
e.What alpha risk would control limits of .0470 and .0136 provide? (Round your intermediate calculations to 4 decimal places. Round your "z" value to 2 decimal places and "alpha risk" value to 4 decimal places.) 1, alpha risk z= fUsing control limits of .0470 and .0136, is the process in control? ook Oyes rint Ono rences g.Suppose that the long-term fraction defective of the process is known to be 2 percent. What are the values of the mean and standard deviation of the sampling distribution? (Round your intermediate calculations and final answers to 2 decimal places.) Mean Standard deviation h.Construct a control chart for the process, assuming a fraction defective of 2 percent, using two-sigma control limits. Is the process in control? Yes ONo 1 of 1
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Answer #1

Arsuwov Sample Fvaction defecttue O-o052 = o-oo hqg -o-o3 53 4 O-0404 (b) stimate fue faactton defectue of the Pocess o-035 oOo4 10 -o-0302 = C) o.o12 o- ol 36 OO30L -38 Oo12 = o.166 the Pro Yes, Control as all hefactfon cess 5 dafective Valeues lra

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