Statistical Process Control Assignment 5: Control Charts

This assignment solution provides answers to Statistical Process Control (SPC) Assignment 5, addressing questions on control charts, process capability, and sampling plans. The solution includes calculations for control limits (UCL and LCL) for both X-bar and R charts, using provided data and standard values. It covers process capability calculations including Cp and Cpk indices, and percentage of tolerance. Additionally, the solution presents a random sampling plan for an eight-hour shift, detailing the procedure for selecting sample times using random numbers. The document also includes control charts for the questions, providing a comprehensive analysis of the statistical process control concepts.

Process Control
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ANS: For n = 4, d2=2. 059 , D4=2 . 282, D3=0 , A2=0. 729
a. So, standard deviation
s= R
¿
d2
= 0 . 0096
2 . 059 =0 . 0047
b. UCLR =D4 R
¿
=2 .282∗0 .0096=0 . 022 and LCRR=D 3∗R
¿
=0
c. UCLx
¿ =x
=
+ A2 ¿ R
¿
=4 . 725+0 .729∗0 . 0096=4 . 732 and
LCLx
¿ =x
=
− A2 ¿ R
¿
=4 .725−0 .729∗0 . 0096=4 .718
ANS: For n = 5, d2=2. 326 , D4=2 . 114 , D3=0 , A2=0 . 577
a. Process capability PCR= 6 s
tolerance where
s= R
¿
d2
= 0 . 042
2 .326 =0 .181
Now, tolerance=USL−LSL=6 . 450−6 .350=0 .1 and PCR=6∗0 . 181
0 .1 =10. 86
Again, c p= tolerance
6 s = 1
10 . 86 =0 .092
Percentage of tolerance = 0.181 * 100 = 18.1%.
2

b. Capability index c pk= x
=
−LSL
3 s = 6. 391−6. 350
3∗0. 181 =0 . 0755 and percentage of tolerance =
1
c pk
X 100= 1
0 .0755 X 100=1324 . 5 %
c. We know that z=3∗c pk=0 . 2265
P (|z|≥0 .23 ) =2∗0. 409=0 . 818
Hence, 0.818 X 100 = 81.8% products were out of specification.
ANS: R
¿
= ∑ R
k = 0 . 028
25 =0 .0011 , x
=
=∑ x
−
k =18 . 6472
25 =0 . 7459
Table 1: Standard Values
n A2 D3 D4 d2
3 1.023 0 2.574 1.693
UCLR =D4 R
¿
=2 .574∗0 . 0011=0 . 00029 LCRR=D3∗R
¿
=0
UCLx
¿ =x
=
+ A2 ¿ R
¿
=0. 7459+1 .023∗0 . 0011=0 . 747
LCLx
¿ =x
=
− A2 ¿ R
¿
=0 .7459−1. 023∗0 . 0011=0 .7447
Control charts included below (Also scanned image file attached)
3

ANS: Random sampling utilizing random numbers for an eight hour shift starting from
3:00 P.M. has been constructed as follows. First digit from the left hand side was considered for
hour and last two digits were considered for minutes. It was noted that the process started at 3:00
P.M. and the shift can run to 11:00 P.M. only. It was also noted that each sample consists of five
parts which took 15 minutes to complete. Hence, from ordered time only the appropriate time
slots were identified with at least a gap of 15 minutes to build the sampling plan.
Table 2: Sampling Plan of 10 Samples
Random Number (s) Sample Time Ordered Sample Time Sampling Plan
838, 918, 470, 422 03:00:00 PM 04:22:00 07:22:00 PM 03:07:00 PM 03:07:00 PM
007, 03:00:00 PM 00:07:00 03:07:00 PM 03:22:00 PM 03:22:00 PM
287, 956, 143 03:00:00 PM 01:43:00 04:43:00 PM 03:58:00 PM 03:58:00 PM
676, 203 03:00:00 PM 02:03:00 05:03:00 PM 04:43:00 PM 04:43:00 PM
533, 03:00:00 PM 05:33:00 08:33:00 PM 05:03:00 PM 05:03:00 PM
410, 03:00:00 PM 04:10:00 07:10:00 PM 06:57:00 PM 06:57:00 PM
641, 03:00:00 PM 06:41:00 09:41:00 PM 06:58:00 PM
733, 03:00:00 PM 07:33:00 10:33:00 PM 07:04:00 PM
357, 03:00:00 PM 03:57:00 06:57:00 PM 07:10:00 PM
404, 03:00:00 PM 04:04:00 07:04:00 PM 07:22:00 PM 07:22:00 PM
816, 358 03:00:00 PM 03:58:00 06:58:00 PM 08:33:00 PM 08:33:00 PM
694, 928, 645 03:00:00 PM 06:45:00 09:45:00 PM 09:41:00 PM 09:41:00 PM
753, 699, 079, 853, 745 03:00:00 PM 07:45:00 10:45:00 PM 09:45:00 PM
961, 657 03:00:00 PM 06:57:00 09:57:00 PM 09:57:00 PM 09:57:00 PM
Procedure
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Control charts of question 14
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