(ultimately determining their strength). Below is the results of this day\'s tes
ID: 3204454 • Letter: #
Question
(ultimately determining their strength). Below is the results of this day's tests.
7
b. Is this process in statistical control? Why or why not?
Any assistance would be greatly appreciated.
The Langley tire company measures their tire strength on a scale from 1 to 10 every 4 hours. They do this by randomly selecting three tires for each test period and destroying them(ultimately determining their strength). Below is the results of this day's tests.
Hour A B C 4 8 6 7 8 6 8 8 12 7 8 7 16 6 87
a. Find the 3 sigma UCL and LCL for the mean and range charts.b. Is this process in statistical control? Why or why not?
Any assistance would be greatly appreciated.
Explanation / Answer
Part a
Here, we have to find the 3 sigma UCL and LCL for the mean and range charts.
The UCL and LCL for mean chart is given as below:
UCL = X-double-bar + A2R-bar
LCL = X-double-bar – A2R-bar
The UCL and LCL for range chart is given as below:
UCL = D4R-bar
LCL = D3R-bar
Calculation table is given as below:
Hour
A
B
C
Xbar
Range
4
8
6
7
7
2
8
6
8
8
7.333333
2
12
7
8
7
7.333333
1
16
6
8
7
7
2
X-double-Bar =
7.222222
R-bar = 1.75
We are given n = 4
The coefficients A2, D3 and D4 are given as below:
A2 = 0.73
D3 = 0
D4 = 2.28
Now, plug all values and find out the limits given as below:
The UCL and LCL for mean chart is given as below:
UCL = X-double-bar + A2R-bar = 7.222222 + 0.73*1.75 = 8.499722
LCL = X-double-bar – A2R-bar = 7.222222 – 0.73*1.75 = 5.944722
The UCL and LCL for range chart is given as below:
UCL = D4R-bar = 2.28*1.75 = 3.99
LCL = D3R-bar = 0*1.75 = 0.00
Part b
All range points are lies between 0.00 and 3.99.
All Xbar points are lies between 5.944722 and 8.499722.
So, this process is in statistical control.
Hour
A
B
C
Xbar
Range
4
8
6
7
7
2
8
6
8
8
7.333333
2
12
7
8
7
7.333333
1
16
6
8
7
7
2
X-double-Bar =
7.222222
R-bar = 1.75
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