Question

Application 13.2A A metal fabricator produces connecting rods with an outer diameter that has a 1±0.02 inch specification. A machine operator takes several sample measurements over time and determines the sample mean outer to be 1.002 inches with a standard deviation of 0.008 inch. Calculate the process capability index for this process.

Application 13.2B A metal fabricator produces connecting rods with an outer diameter that has a 1±0.04 inch specification. A machine operator takes several sample measurements over time and determines the sample mean outer to be 1.002 inches with a standard deviation of 0.009 inch. Calculate the process capability index for this process.

Answer 1

Answer 13.2A

Given that, Mean = m = 1.002

Standard Deviation = SD = 0.008

Upper Specificaton limit = 1 + 0.02 = 1.02

Lower Specificaton limit = 1 - 0.02 = 0.98

Cp = (USL - LSL) / 6SD

= (1.02 - 0.98) / (6 x 0.008)

= 0.8333

CpU = (USL - m) / 3SD

= (1.02 - 1.002) / (3 x 0.008)

= 0.7500

CpL = (m - LSL) / 3SD

= (1.002 - 0.98) / (3 x 0.008)

= 0.9167

CpK = Process Capability Index = Min (CpU, CpL)

= Min (0.7500, 0.9167)

= 0.7500

Answer 13.2B

Given that, Mean = m = 1.002

Standard Deviation = SD = 0.009

Upper Specificaton limit = 1 + 0.04 = 1.04

Lower Specificaton limit = 1 - 0.04 = 0.96

Cp = (USL - LSL) / 6SD

= (1.04 - 0.96) / (6 x 0.009)

= 1.4815

CpU = (USL - m) / 3SD

= (1.04 - 1.002) / (3 x 0.009)

= 1.4074

CpL = (m - LSL) / 3SD

= (1.002 - 0.96) / (3 x 0.009)

= 1.5556

CpK = Process Capability Index = Min (CpU, CpL)

= Min (1.4074, 1.5556)

= 1.4074