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 00 vau AWNA 998 READINGS (IN OHMS) 1027 994 977 994 975 1013 999 1017 988 1016 974 997 998 1024 1006 1010 990 1012 990 1000 1016 1001 1030 1000 983 979 971 973 982 975 1030 992 1028 991 998 997 1026 972 1021 990 1021 1028 992 1021 998 996 970 1027 993 996 1022 981 1014 983 977 993 986 983 WNHO 996

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 Xbar and Rbar (Round your answers to 3 decimal places.)

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

d. Determine the UCL and LCL for Rbar chart. (Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 3 decimal places.)

e. What comments can you make about the process?
process is in statistical control

process is out of statistical control

Answer 1

a)

X1 x3 Range 1027 975 988 998 990 1016 1000 973 x2 994 1013 1016 1024 1012 998 983 982 1028 1026 1021 998 993 981 993 977 999 974 1006 990 1001 979 975 991 x4 994 1017 997 1010 1000 1030 971 1030 998 1021 992 970 996 X-bar 998.00 1001.00 993.75 1009.50 998.00 1011.25 983.25 990.00 1002.25 1004.00 1007.75 996.25 1003.00 1000.00 984.75 992 972 997 990 1021 1027 1022 977 12 13 14 15 1028 996 996 1014 34 983 41 16 986 983 X-double bar R-bar 998.85 38.07

b)

X-bar = 998.85

R-bar = 38.07

c)

From table of control chart constant, A2 = 0.729, D3=0, D4=2.282

X-bar Chart

$UCL = \bar {\bar X} + A_2*\bar R = 998.85 + 0.729*38.07 = 1026.60$

$LCL = \bar {\bar X} - A_2*\bar R = 998.85 - 0.729*38.07 = 971.10$

d) R-bar chart

$UCL = D_4 * \bar R = 2.282*38.07 = 86.86$

$LCL = D_3 * \bar R = 0*38.07 = 0$

e)

Mean and Range for all samples are between UCL and LCL of X-bar and R-bar chart respectively. So process is in statistical control.