Question

You are a quality control inspector for Southeast Texas Chemical Company (SETCCO). SETCCO manufactures lubricants for the auto industry. SETCCO's lubricant production process has three stages, Blending, Refining, and Filling. SETCCO receives all of its chemicals from a single supplier. Once manufactured, SETCCO ships the products to wholesalers throughout the U.S. Raw Materials Blending Refining Filling Distribution At the end of the refining process, samples are taken periodically to test the product for viscosity. Close tolerances are required. After taking eight samples of size four, you have the following data: 1.000 | 0.997 | 0.998 0.999 1.000 1.000 1.000 1.000 0.995 1.000 1.000 1.000 1.000 1.000 1.000 0.997 0.995 1.000 0.995 0.995 1.010 0.996 1.000 0.997 1.000 1.000 0.995 0.995 | 1.000 1.000 0.999 1.000 a) Draw X-bar and R charts for this process. b) Based on your analysis, is the process in control? Justify your answer. c) For this question, assume that the process is in control. The inspector takes a ninth sample with the following measurements: 1.000, 0.996, 1.000, and 0.992. Enter the X-bar and R from the ninth sample on your charts. Is the process in control now? Justify your answer.

Answer 1

For sample size 4, A2 = 0.729, D4 = 2.282 and D3 = 0

a)

WOPRbOOR Views Show 20 om B E G H I 1 8 Mean 7 1 1 C D 2 3 1 0.995 1 1 11 1 1 F 5 0.995 1 0.995 0.995 0.997 0.998 0.999 1.01 0.996 1 0.997 0.999 0.995 0.995 0.997 - 1 7 Mean 8 Range 9 0.9985 0.003' 0 0.99875 0.005 0.99925 0.003' 0.99625 0.005 1.00075 0.014' 0.9975 0.005 0.99975 0.99884375 0.001 0.0045

Formula

Function Library Derihed Names Formula Auditing Calculation F 8 Mean 0.995 0.995 1.01 0.996 1 0.997 0.998 0.999 0.999 0.995 0.995 0.995 0.995 0.997 0.997 7 Mean =AVERAGE(B2:35) AVERAGE(C2:05) SAVERAGE(D2:05) =AVERAGE(E2:E5) SAVERAGE(F2:F5) EAVERAGE(G2:55) FAVERAGE(H2:45) AVERAGE(12:15) EAVERAGE(B7:17) Range =MAX(B2:B5)-MIN(B2:35) =MAX(C2C5)-MIN(C2:05) =MAX(D2:05)-MIN(D2:05) =MAX(E2:E5)-MIN(E2:E5) =MAX(F2:F5)-MIN(F2:F5) =MAX(G2:G5)-MIN(G2:G5) =MAX(H2:H5)-MIN(H2:H5) =MAX(12:15)-MIN(12:15) -AVERAGE(B8:18)

X-bar chart,

Center line = 0.99884375 = x-bar-bar

R-bar = 0.0045  

UCL = x-bar-bar   +A2*R-bar = 0.99884375+0.729*0.0045 = 1.00212425
LCL = x-bar-bar   -A2*R-bar = 0.99884375-0.729*0.0045 = 0.99556325

x-bar chart 1.004 1.002 1 0.998 0.996 0.994 0.992 1 2 3 4 5 6 - Center line — UCL LCL - Sample mean

Graph Data

Center line	UCL	LCL	Sample mean
0.99884375	1.002124	0.995563	0.9985
0.99884375	1.002124	0.995563	1
0.99884375	1.002124	0.995563	0.99875
0.99884375	1.002124	0.995563	0.99925
0.99884375	1.002124	0.995563	0.99625
0.99884375	1.002124	0.995563	1.00075
0.99884375	1.002124	0.995563	0.9975
0.99884375	1.002124	0.995563	0.99975

R-chart

Center line = 0.0045  
UCL = D4*R-bar = 2.282*0.0045 = 0.010269
LCL = D3*R-bar = 0

R-chart 0.016 0.014 0.012 0.01 0.008 0.006 0.004 0.002 - 1 2 3 4 5 6 7 8 - Center line UCL LCL - Sample range

Graph Data

Center line	UCL	LCL	Sample range
0.0045	0.010269	0	0.003
0.0045	0.010269	0	0
0.0045	0.010269	0	0.005
0.0045	0.010269	0	0.003
0.0045	0.010269	0	0.005
0.0045	0.010269	0	0.014
0.0045	0.010269	0	0.005
0.0045	0.010269	0	0.001

b)

No the process is not in control as Sample 6 range breaches upper control limit of Range chart

c)

Ninth sample mean = average(1,0.996,1,0.992) = 0.997 and Range = 1-0.992 = 0.008

With the inclusion of 9th sample,

x-bar chart 1.004 1.002 1 0.998 0.996 0.994 0.992 1 9 2 3 —Center line 4 - UCL 5 — 6 — 7 8 Sample mean LL

R-chart 0.016 0.014 0.012 0.01 0.008 0.006 0.004 0.002 1 9 2 3 Center line 4 UCL 5 —LL 6 — 7 8 Sample range — -

If we assume that the process is already in control then adding sample 9 to the x-bar chart and R-chart do not alter the status.The process remains in control as Sample range and sample mean of the 9th sample are in previously calculated control chart limits