Chapter 16: Introduction to Quality 329
Chapter 16: Introduction to Quality
16-1 thru 5. Various answers.
16-6 a.
b. CL = = 192.6
LCL = = 192.6 – 1.18(5.42) = 186.2044
UCL = = 192.6 + 1.18(5.42) = 198.9956
c. CL = 5.42
LCL = .12(5.42) = .6504
UCL = 1.88(5.42) = 10.1896
16-7 a.
b. CL = 93.2
LCL = 93.2 – 1.29(3.67) = 88.4657
UCL = 93.2 + 1.29(3.67) = 97.9343
c. CL = 3.67
LCL = .03(3.67) = .1101
UCL = 1.97(3.67) = 7.2299
16-8 a.
b. CL = 19.86
LCL = 19.86 – 1.1(1.23) = 18.507
UCL = 19.86 + 1.1(1.23) = 21.213
c. CL = 1.23
LCL = .18(1.23) = .2214
UCL = 1.82(1.23) = 2.2386
16-9 a.
b. CL = 230.5
LCL = 230.5 – 1.43(1.8617) = 227.8378
UCL = 230.5 + 1.43(1.8617) = 233.1622
c. CL = 1.75
LCL = 0(1.75) = 0.0
UCL = 2.09(1.75) = 3.6575
16-10 a.
b.
c. = 6.2062
d. CL = 149.98
LCL = 149.98 – 1.1(5.989) = 143.3921
UCL = 149.98 + 1.1(5.989) = 156.5679
e. The chart shows a process where the level is under statistical control
f. CL = 5.989
LCL = .18(5.989) = 1.078
UCL = 1.82(5.989) = 10.9
g. S-chart
16-11 a.
b.
c. = 2.0903
d. CL = 19.84
LCL = 19.84 – 1.29(1.99) = 17.2729
UCL = 19.84 + 1.29(1.99) = 22.4071
e.
f. CL = 1.99, LCL = .03(199) = .0597, UCL = 1.97(1.99) = 3.9203
g. S Chart
16-12 a. 192.6 ± 3(5.6517): (175.6449, 209.5551)
These values lie within the tolerance limits
b. The process is not capable since
c. The process is not capable since
16-13 a. 93.2 ± 3(3.855): (81.635, 104.765)
These limits are beyond the tolerances set by management
b. The process is not capable since
c. The process is not capable since
16-14 a. 19.86 ± 3(1.2746): (16.0362, 23.6838)
These limits are beyond the tolerances set by management
b. The process is not capable since
c. The process is not capable since
16-15 a. 149.98 ± 3(6.2062): (131.3614, 168.5986)
These values lie within the tolerance limits
b. The process is not capable since
c. The process is not capable since
16-16 a. 19.84 ± 3(2.0903): (13.5691, 26.1109)
These limits are beyond the tolerances set by management
b. The process is not capable since
c. The process is not capable since
16-17 CL = .056, LCL = .056 – 3 = .0124
UCL = .056 + 3 = .0996
16-18 CL = .016, LCL = 0, UCL = .016 + 3 = .0328
16-19 a. , CL = .1247,
LCL = .1247 – 3 = .0546
UCL= .1247 + 3 = .1948
b. p-chart
The p-chart shows a process that is in statistical control
16-20 a. , CL = .051, LCL = .051 – 3 = .0215
UCL = .051 + 3 = .0805
b. p-chart
No evidence that the process is out of statistical control
16-21 a.
b. CL = .08, LCL = .08 – 3 = .0285, UCL = .08 + 3 =.1315
c. p-chart
The p-chart shows an observation at sample number 9 that is beyond the upper limit.
16-22 a. b. CL = 5.6, LCL = 0, UCL = 5.6 + 3 = 12.70
c. C chart
No evidence that the process is out of statistical control
16-23 a.
b. CL = 13.55, LCL = 13.55 – 3 = 2.507, UCL=13.55+3=24.59
c.
All points are in statistical control
16-24 a.
b. CL = 13.8667, LCL = 13.8667 – 3 = 2.6953
UCL = 13.8667 + 3 = 25.0381
c. c-chart
No evidence of a process out of control; however, keep monitoring the process.
16-25 X bar and s charts for Bolts data using Minitab:
16-26 a. Capability process (normal) is meaningless since the process of manufacturing precision bolts is not stable.
b. Capability sixpack (normal) is meaningless since the process of manufacturing precision bolts is not stable.
16-27 p-chart for Exercise 16-21:
Process has an assignable cause of variation that is present and should be investigated as a special cause of variation
16-28 a. c-chart for Exercise 16-22
b. c-chart for Exercise 16-23
c. c-chart for Exercise 16-24
16-29 a. A process is ‘in control’ when the sample values are randomly distributed around the center line and within the control limits with no discernable data pattern. In contrast, a process that is ‘capable’ of meeting specifications are when the observations lie within the upper and lower specification limits.
b. All processes will generate variations in the data. These variations may simply be due to chance (natural variability) or due to assignable causes, particular circumstances that create errors or data patterns such as incorrect data entry, machines wearing out, operator errors, etc.
16-30 There are two types of errors that can be made – 1) identifying a special or assignable cause of variation when there is none and 2) ignoring a special cause by assuming that it is due to natural variability. Control limits that are ‘too narrow’ will incorrectly flag natural causes of variability as a special cause and the researcher is sent off to find a special cause of variability that does not exist. Control limits that are ‘too wide’ imply that the researcher is not going off to correct processes that are out of control. The use of the three sigma limits as control limits was set by Shewhart as an appropriate balance between the two types of errors.
16-31 a.
b.
c.
d. CL = 119.825
LCL = 119.825 – 1.43(1.985) = 116.9864
UCL = 119.825 + 1.43(1.985) = 122.6635
e.
No discernable data patterns – process is in control
f. CL = 1.985
LCL = 0; UCL = 2.09(1.985) = 4.1486
g. s-chart
No discernable data patterns – process is in control
h. i) 119.825 ± 3(2.117): (113.4899, 126.1601)
These limits are beyond the tolerances set by management
ii) Therefore, the process is not capable
iii) Therefore, the process is not capable
16-32 a.
b.
c.
d. CL = 350.416
LCL = 350.416 – 1.1(5.602) = 344.2538
UCL = 350.416 + 1.1(5.602) = 356.5782
e. The X-bar chart provides no evidence that the process is out of control.
f. CL=5.602
LCL = .18(5.602) = 1.0084
UCL = 1.82(5.602) = 10.1956
g. s-chart
No pattern analysis rules are violated. Process is in control.
h. i) 350.416 ± 3(5.8052): (333.0004, 367.8316)
ii)
Therefore, the process is capable
iii)
Therefore, the process is capable
16-33 a.
b. CL = .0723
LCL = .0723 – 3 = .0375
LCL = .0723 + 3 = .1070
c., d. p-chart
Process is in control
16-34 a.
b. CL = 15
LCL = 15 – 3 = 3.381
UCL = 15 + 3 = 26.619
c., d. C chart shows no evidence of a process that is out of statistical control:
16-35 a. Means and standard deviations
x1 x2 x3 x4 XBar s
340 346 351 338 343.75 5.9090
332 348 330 344 338.50 8.8506
339 343 339 347 342.00 3.8297
342 338 346 338 341.00 3.8297
350 340 345 347 345.50 4.2032
344 332 347 351 343.50 8.1854
336 348 362 331 344.25 13.8173
345 342 349 330 341.50 8.1854
356 342 348 329 343.75 11.3835
337 361 332 344 343.50 12.6623
353 329 323 360 341.25 18.0069
348 367 323 320 339.50 22.2186
370 354 358 340 355.50 12.3693
368 328 339 347 345.50 16.9017
366 328 343 351 347.00 15.8535
330 323 364 339 339.00 17.9072
b.
c.
d.
e. CL = 343.44
LCL = 343.44 – 1.63(11.51) = 324.6787
UCL = 343.44 + 1.63(11.51) = 362.2013
f., h., i.
g. CL = 11.51
LCL = 0
UCL = 2.274(11.51) = 26.1277
16-36 a. Common cause – affects all workers within the process
b. Common cause – affects all workers within the process
c. Assignable cause
d. Assignable cause
e. Assignable cause
16-37 a. X bar and s chart for the Bottles data:
b. The mean for sample 5 is less than the LCL in the Xbar chart. Hence the process as it stands it out of control. In the ‘real world’ one would first check to see if there were any data entry errors for the five values recorded at 9:00 AM. If so, adjust and check again to see if the process is in control. If there were no data entry errors, then one could remove subgroup 5 (the values at 9:00 AM) and test the remaining nineteen subgroups. You will then find that the process is in control, but by checking capability analysis, the bottling process is not capable of meeting the specifications indicated since Cp = 1.00 and Cpk = 0.98.
c. Process capability:
16-38 a. Machine 1 – X bar – s chart:
Machine 1 shows no evidence of being ‘out of statistical control’
b. Machine 2 – X bar – s chart:
Likewise, Machine 2 shows no evidence of being ‘out of statistical control’
c. Machine 1 – capability analysis shows that Cp =0.44 and Cpk=0.14. Machine 1 is not capable of meeting specifications.
d. Machine 2 – capability analysis shows that Cp =0.80 and Cpk=0.72. Machine 2 is not capable of meeting specifications.
e. Neither machine is capable of meeting specifications. Both of the machines produce a product with greater variability than the specification limits call for. Note that Machine 1 has greater variability than Machine 2.
16-39 a. X bar – s chart for Granola:
The process does not show any evidence of being ‘out of statistical control’
b. Capability analysis of Granola:
The process is capable of meeting specifications. Note that the Cp index is 1.50 which is greater than 1.33.
16-40 X bar chart for TOC data:
All data points are within the control limits. No pattern analysis rule has been violated.