Question

Answer the following questions 1. A company manufacturing oil seals wants to establish -R control charts on the process. There are 25 subgroups and number of observations are four on the internal diameter of the seal. The summary data (in mm) are as follows ŽE=123.Ša=1406 a) Find the control limits that should be used on the I-R control charts. b) Assume that the 25 subgroups samples plot in control on both charts. Estimate the process mean() and standard deviation ) c) If the specifications are at 50.1 +0.5mm, calculate both , and pk- d) What conclusion would you draw about process capability?

Answer 1

we have 25 given, 25 : R =25 ; = 1253 = 14.08 , n = 4 - a) control limits for ñ chartare, = { di = 1253 = 50.12 & R = 14.08 =0.5632 25 25 LCL y = x - A₂ Ř = 49.7094 CLx = = 50.12 UCLY = + A₂ R = 50.5306 where Az = 0.729 for n=4 control limits to R chart are, LCLR = D₂ Ř = 0 CLR = ł = 0.5632 UcLe =D4 R = 1.2852 where Dz = 0 & D4 = 2.282 for n = 4 b) The process mean n ) = 50.12 & standard deviation cô) = ł Here = da = 2.059 for res. h = 0.5632 = 0.2735 2.059 son to.smm. - c) The specifications are ie (49.6 50.6) The upper specification limit - + 3 3 50.12 + 3 x0.2735 = 50.9405 Scanned with CamScanner

The lower specification limit = x-30 = 50.12 - 3x02735 = 49.2995 Now, So = USL LSL - 50.9405-49. 2995 - 1 6x0,2735 min 2 USL LSL2 2 38 3 = min 250.9405-50112 50112-49.29952 2 3x0.2735 3x0,2735 pk = 1 d) p = 1, i.e. the process variability just meets the specifications. The process is capable & CK=1 lie Cp = CK = M = M where M = USLELSL 2 The process is good. es Cartner