Webster chemical company produces mastics and

Webster Chemical Company produces mastics and caulking for the construction
industry. The product is blended in large mixers and then pumped into tubes and
capped.

6-8 Chapter 6: Process Performance and Quality
Webster is concerned whether the filling process for tubes of caulking is in statistical
control. The process should be centered on 8 ounces per tube. Several samples of
eight tubes are taken and each tube is weighed in ounces.
Assuming that taking only 6 samples is sufficient, is the process in statistical control?
Conclusion on process variability:
708 . 0 ) 38 . 0 ( 864 . 1
4
= = = R D UCL
R
052 . 0 ) 38 . 0 ( 136 . 0
3
= = = R D LCL
R
Since the range is out of control, the x calculation is moot.
Consider dropping sample 6 because of an inoperative scale, causing inaccurate
measures.
What is the conclusion on process variability and process average?
( 29 ( 29
061 . 0 45 . 0 136 . 0 839 . 0 45 . 0 864 . 1
3 4
= = = = = = R D LCL R D UCL
R R
( 29 ( 29
832 . 7 45 . 0 373 . 0 034 . 8 202 . 8 45 . 0 373 . 0 034 . 8
2 2
= - = - = = + = + = R A x LCL R A x UCL
x x
The resulting control charts indicate that the process is actually in control.
Tube Number
Sample 1 2 3 4 5 6 7 8 Avg Range
1 7.98 8.34 8.02 7.94 8.44 7.68 7.81 8.11 8.040 0.76
2 8.23 8.12 7.98 8.41 8.31 8.18 7.99 8.06 8.160 0.43
3 7.89 7.77 7.91 8.04 8.00 7.89 7.93 8.09 7.940 0.32
4 8.24 8.18 7.83 8.05 7.90 8.16 7.97 8.07 8.050 0.41
5 7.87 8.13 7.92 7.99 8.10 7.81 8.14 7.88 7.980 0.33
6 8.13 8.14 8.11 8.13 8.14 8.12 8.13 8.14 8.130 0.03
Avgs 8.050 0.38
Tube Number
Sample 1 2 3 4 5 6 7 8 Avg Range
1 7.98 8.34 8.02 7.94 8.44 7.68 7.81 8.11 8.040 0.76
2 8.23 8.12 7.98 8.41 8.31 8.18 7.99 8.06 8.160 0.43
3 7.89 7.77 7.91 8.04 8.00 7.89 7.93 8.09 7.940 0.32
4 8.24 8.18 7.83 8.05 7.90 8.16 7.97 8.07 8.050 0.41
5 7.87 8.13 7.92 7.99 8.10 7.81 8.14 7.88 7.980 0.33
Avgs 8.034 0.45

Chapter 6: Process Performance and Quality 6-
9
2. Control charts for attributes
TEACHING TIP
Mention examples from banking and manufacturing. Emphasize that defects can be
counted.
a. p -chart — population proportion defective
• Sampling for a p -chart involves a yes or no decision, based on the binomial
distribution
• Take a random sample of n units.
• Count the number of defectives.
• Proportion defective = number of defectives ÷ sample size
• Plot sample proportion defective on a chart. If it is outside the range between
the upper and lower control limits, search for an assignable cause. If cause is
found do not use these data to determine the control limits.
UCL
p
= p + z σ
p
LCL
p
= p – z σ
p
where
σ
p
=
p 1 – p
( 29
n
• Two things to note:
⇒ The lower control limit cannot be negative
⇒ When the number of defects is less than the LCL , then the system is out
of control in a good way. We want to find the assignable cause. Find
what was unique about this event that caused things to work out so well.
• Use Application 6.2: p -Chart for Attributes for an example of a p -chart
problem. The short answers are:
p = 0.025; UCL
p
= 0.064; LCL
p
= 0; IN CONTROL
A sticky scale brings Webster’s attention to whether caulking tubes are being
properly capped. If a significant proportion of the tubes aren’t being sealed,
Webster is placing their customers in a messy situation. Tubes are packaged in
large boxes of 144. Several boxes are inspected and the following number of
leaking tubes are found:
Sample Tubes Sample Tubes Sample Tubes
1 3 8 6 15 5
2 5 9 4 16 0
3 3 10 9 17 2
4 4 11 2 18 6
5 2 12 6 19 2
6 4 13 5 20 1
7 2 14 1 Total = 72

6-10 Chapter 6: Process Performance and Quality
Calculate the p