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		1000	992	1014	1026
2		1006	977	1027	1016
3		985	1029	1018	992
4		1001	1006	971	1000
5		1028	970	979	1025
6		997	1026	1007	1003
7		1016	1003	987	1006
8		995	1029	978	986
9		1027	995	1003	1008
10		991	987	981	996
11		994	1000	978	971
12		999	983	973	1005
13		1003	993	974	1002
14		1005	972	994	1000
15		1012	1018	982	1014

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

X=X=
R−R−

c. Determine the UCL and LCL for a X−X−chart. (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.

Answer 1

Question – a:

Sample Number	Mean	Range
1	1008	34
2	1006.5	50
3	1006	44
4	994.5	35
5	1000.5	58
6	1008.25	29
7	1003	29
8	997	51
9	1008.25	32
10	988.75	15
11	985.75	29
12	990	32
13	993	29
14	992.75	33
15	1006.5	36

Question – b:

Finding x-double-bar and R-bar

x-double bar = sum of all mean/total samples = 14988.75/15 = 999.250

R-bar = sum of all range/total samples = 536/15 = 35.733

Question – c:

UCL and LCL for X-bar chart:

We know that,

UCL(X-bar) = x-double-bar + A2*R-bar

LCL(X-bar) = x-double-bar – A2*R-bar

Here sub group size = 4. So, A2 = 0.729

x-double bar = 999.25

R-bar = 35.733

So,

UCL = 999.25 + 35.733*0.729 = 999.25 + 26.049 = 1025.299

LCL = 999.25 - 35.733*0.729 = 999.25 - 26.049 = 973.201

UCL(X-bar) = 1025.299

LCL(X-bar) = 973.201

Question – d:

UCL and LCL for R-chart

For R-chart, we know that,

UCL = R-bar*D4

LCL = R-bar*D3

For subgroup size n = 4,

D3 = 0

D4 = 2.282

R-bar = 35.733

So,

UCL = 35.733*2.282 = 81.543

LCL = 35.733*0 = 0

UCL(R) = 81.543

LCL(R) = 0

Question – e:

As, all data and range are within the specified limit for UCL and LCL of X-bar and R-chart, so

The process is in statistical control.

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