Resistors for electronic circuits are manufactured on a high-speed automated mac
ID: 376003 • Letter: R
Question
Resistors for electronic circuits are manufactured on a high-speed automated machine. The machine is set up to produce a large run of resistors of 1,000 ohms each. Use Exhibit 10.13. To set up the machine and to create a control chart to be used throughout the run, 15 samples were taken with four resistors in each sample. The complete list of samples and their measured values are as follows: Use three-sigma control limits. SAMPLE NUMBER READINGS (IN OHMS) 1 986 987 1014 1001 2 994 993 973 972 3 983 1022 981 983 4 1022 989 1007 994 5 1015 988 1020 1027 6 1013 998 990 1002 7 986 998 977 1021 8 995 1026 1016 1002 9 1023 991 993 990 10 1002 1012 1008 998 11 996 987 1017 1023 12 972 1018 1005 1018 13 1007 1007 992 998 14 978 982 986 984 15 1012 1023 1001 1003 a. Calculate the mean and range for the above samples. (Round "Mean" to 2 decimal places and "Range" to the nearest whole number.) Sample Number Mean Range 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 b. Determine formula37.mml and 1formula216.mml. (Round your answers to 3 decimal places.) formula37.mml 1formula216.mml c. Determine the UCL and LCL for a 11formula252.mmlchart. (Round your answers to 3 decimal places.) UCL LCL d. Determine the UCL and LCL for R-chart. (Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 3 decimal places.) UCL LCL e. What comments can you make about the process? The process is in statistical control. The process is out of statistical control.
Explanation / Answer
Please find below table which calculates Mean and Range for each sample :
SERIAL NUMBER
SAMPLE
1
2
3
4
MEAN
RANGE
1
986
987
1014
1001
997
28
2
994
993
973
972
983
22
3
983
1022
981
983
992.25
41
4
1022
989
1007
994
1003
33
5
1015
988
1020
1027
1012.5
39
6
1013
998
990
1002
1000.75
23
7
986
998
977
1021
995.5
44
8
995
1026
1016
1002
1009.75
31
9
1023
991
993
990
999.25
33
10
1002
1012
1008
998
1005
14
11
996
987
1017
1023
1005.75
36
12
972
1018
1005
1018
1003.25
46
13
1007
1007
992
998
1001
15
14
978
982
986
984
982.5
8
15
1012
1023
1001
1003
1009.75
22
SUM =
15000.25
435
Following formula may be noted :
Mean for any sample = Sum of sample values / 4
Range for each sample = Maximum value in that sample – Minimum value in that sample
Therefore,
Xbar-bar = Mean of sample means = Sum of all sample means / 15 ( i.e number of samples ) = 15000.25/15 = 1001.01
Rbar = Mean of Range values = Sum of all range values / 15 (i.e number of samples ) = 435/15 = 29
Following are the value so constants derived from standard table for Xbar chart and Range chart for sample size, n = 4 :
A2 = 0.729
D4 = 2.282
D3 = 0
Accordingly,
Control Limits for Xbar chart :
Upper Control Limit = UCL = Xbar-bar + A2.Rbar = 1001.01 + 0.729x 29 = 1001.01 + 21.14 = 1022.15
Lower Control Limit = LCL = Xbar-bar – A2.Rbar = 1001.01 – 0.729x 29 = 1001.01 – 21.14 = 979.87
Control Limits for Range chart :
Upper Control Limit -= D4.Rbar = 2.282 x 29 = 66.178
Lower Control Limit = D3.Rbar = 0
As per control limits for Xbar chart, for the process to be in control all data must be within control range of 979.87 - 1022.15
However, there are sample data e.g 972, 973,977, 1023, 1026, 1027 which are outside the above control limits
It therefore can be concluded that the process is out of statistical control
SERIAL NUMBER
SAMPLE
1
2
3
4
MEAN
RANGE
1
986
987
1014
1001
997
28
2
994
993
973
972
983
22
3
983
1022
981
983
992.25
41
4
1022
989
1007
994
1003
33
5
1015
988
1020
1027
1012.5
39
6
1013
998
990
1002
1000.75
23
7
986
998
977
1021
995.5
44
8
995
1026
1016
1002
1009.75
31
9
1023
991
993
990
999.25
33
10
1002
1012
1008
998
1005
14
11
996
987
1017
1023
1005.75
36
12
972
1018
1005
1018
1003.25
46
13
1007
1007
992
998
1001
15
14
978
982
986
984
982.5
8
15
1012
1023
1001
1003
1009.75
22
SUM =
15000.25
435
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.