Question

In: Operations Management

Problem 13-9 Resistors for electronic circuits are manufactured on a high-speed automated machine. The machine is...

Problem 13-9

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 1010 991 985 986
2 995 996 1009 994
3 990 1003 1015 1008
4 1015 1020 1009 998
5 1013 1019 1005 993
6 994 1001 994 1005
7 989 992 982 1020
8 1001 986 996 996
9 1006 989 1005 1007
10 992 1007 1006 979
11 996 1006 997 989
12 1019 996 991 1011
13 981 991 989 1003
14 999 993 988 984
15 1013 1002 1005 992

a. Calculate the mean and range for the above samples. (Round "Mean" to 2 decimal places and "Range" to the nearest whole number.)

b. Determine X=X= and R−R− . (Round your answers to 3 decimal places.)

c. Determine the UCL and LCL for a X−X− chart. (Round your answers to 3 decimal places.)

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.)

e. What comment can you make about the process?

o The process is in statistical control.

o The process is out of statistical control.

Solutions

Expert Solution

Sample Ob. 1 ob 2 ob 3 ob 4 Mean Range
1 1010 991 985 986 993 25
2 995 996 1009 994 998.5 15
3 990 1003 1015 1008 1004 25
4 1015 1020 1009 998 1010.5 22
5 1013 1019 1005 993 1007.5 26
6 994 1001 994 1005 998.5 11
7 989 992 982 1020 995.75 38
8 1001 986 996 996 994.75 15
9 1006 989 1005 1007 1001.75 18
10 992 1007 1006 979 996 28
11 996 1006 997 989 997 17
12 1019 996 991 1011 1004.25 28
13 981 991 989 1003 991 22
14 999 993 988 984 991 15
15 1013 1002 1005 992 1003 21

Mean of means xbarbar =999.1

Mean of range values Rbar =21.73

UCLxbar = xbarbar + A2*Rbar = 999.1+0.729*21.73 =1014.94

LCLxbar = xbarbar - A2*Rbar =999.1-0.729*21.73 =983.25

UCLR = D4*Rbar =2.282*21.73 =49.58

LCLR = D3*Rbar =0

Process is in control as all values are within limits

keosha answered 7 months ago
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