Question

Problem 10-25

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		1013	986	994	977
2		977	1027	982	1024
3		1027	990	998	1017
4		1023	1025	1016	1006
5		1005	1026	975	991
6		983	998	990	988
7		991	999	1001	985
8		1025	1023	1014	1023
9		1019	1004	982	979
10		999	995	991	1010
11		970	992	1006	1012
12		1010	985	983	1030
13		1030	1002	1016	982
14		979	986	1016	988
15		1028	1006	1019	1002

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

Answer a) We will calculate the Mean and Range for the given samples as mentioned below:

Step 1: Col 5: Find Column of Total = Sum of Col 1 to Col 4

Step 2: Col 6: Find the column of Mean = Col 5 / 4

Step 3: Col 7: Find the maximum value from the Col 1 to Col 4

Step 4: Col 8: Find the Minimum value from the Col 1 to Col 4

Step 5: Col 9: Find the column of Range = Col 3 - Col 4

Answer b)

i) X = X =

Sum of Col 6 / 15 = 15030 / 15 = 1002.000 (Rounded to 3 Decimal Places)

ii) R-R-

Sum of Col 9 / 15 = 15030 / 15 = 32.933 (Rounded to 3 Decimal Places)

Answer c) Upper Control Limit (UCL) and Lower Control Limit (LCL) for X=X= Chart:

i) UCL = (X=X=) + A2 (R-R-)

Where X=X= is 1002, A2 =  Control chart constant = 0.729 (Derived from the Table of Control Charts), and R-R- = 32.9333

Hence, UCL = 1002 + 0.729 (32.933)

= 1002 + 24.008

= 1026.008 (Rounded to 3 Decimal Places)

ii) LCL = (X=X=) - A2 (R-R-)

Where X=X= is 1002, A2 =  Control chart constant for the subgroup size of 4 = 0.729 (Derived from the Table of Control Charts), and R-R- = 32.933

Hence, UCL = 1002 + 0.729 (32.933)

= 1002 - 24.008

= 977.992 (Rounded to 3 Decimal Places)

Answer d) Upper Control Limit (UCL) and Lower Control Limit (LCL) for R- Chart:

i) UCL = D4 * R-

Where D4 = Control Chart Constant for the subgroup size of 4 = 2.282 (Derived from the Table of Control Charts), and R-R- = 32.933

Hence, UCL = 2.282 X 32.933 = 75.154 (Rounded to 3 Decimal Places)

ii) LCL = D3 * R-

Where D3 = Control Chart Constant for the subgroup size of 4 = 0 (Derived from the Table of Control Charts), and R-R- = 32.933

Hence, LCL = 0 X 32.933 = 0