In a chemical plant involved in making paint for household use, the viscosity of
ID: 3336109 • Letter: I
Question
In a chemical plant involved in making paint for household use, the viscosity of the liquid has to be maintained at 15 ± 3 centipoise. The table below shows the viscosity values for 26 samples taken every hour
Construct x-mR charts and conclude on the stability of the chemical process (use k=3 sigma limits)
Estimate the mean and standard deviation of the process
If the mean of the process changes to 17 centipoise due to a special cause, how long -in hours- it would take for the control chart to detect the change? Assume that the process variability has not changed.
Sample>
1
2
3
4
5
6
7
8
9
10
11
12
13
Viscosity>
16.2
13.8
17.0
15.8
13.5
14.7
14.0
14.8
13.2
16.8
14.9
13.0
12.5
Sample>
14
15
16
17
18
19
20
21
22
23
24
25
26
Viscosity>
16.7
15.9
14.6
16.5
18.4
15.2
14.6
17.2
16.1
14.4
17
13.8
15.5
Sample>
1
2
3
4
5
6
7
8
9
10
11
12
13
Viscosity>
16.2
13.8
17.0
15.8
13.5
14.7
14.0
14.8
13.2
16.8
14.9
13.0
12.5
Explanation / Answer
Solution
Back-up Theory
Let xi = viscosity of the ith sample. Then, the control limits are:
Lower Control Limit (LSL) = Xbar – 3s
Upper Control Limit (USL) = Xbar + 3s, where
Xbar = sample mean = (1/n) [i = 1,n]xi;
s = sample standard deviation = sqrt{(1/n-1) [ i = 1,n](xi – xbar)2}
n = sample size.
Calculations:
Summary of Excel calculations is given below:
Mean Xbar
14.37
SD = s = cap
1.50
3s
4.50
LCL = Xbar - 3s
9.87
UCL = Xbar + 3s
18.87
Part (a)
Control Chart
LCL = 9.87
UCL = 18.87 ANSWER 1
All the 26 xi values are within the two limits. It is reasonable to assume that the process is stable. ANSWER 2
Part (b)
Mean = 14.37
Standard deviation = 1.50 ANSWER 3
Mean Xbar
14.37
SD = s = cap
1.50
3s
4.50
LCL = Xbar - 3s
9.87
UCL = Xbar + 3s
18.87