Question

Webster Chemical Company produces mastics and caulking for the construction industry. The product is blended in large mixers and then pumped into tubes and capped. Management is concerned about whether the filing process for tubes of caulking is in statistical control. The process should be centered on 8 ounces per tube, Several samples of eight tubes were taken, each tube was weighted and the weights in the Table 3.3 were obtained. Several samples of eight tubes were taken, each tube was weighted, and the weights in the following table were obtained.

a. Assume that only six samples are sufficient and develop the control charts for the mean and the range.

b. Plot the observations on the control chart and comment on your findings.

				Tube Number
Sample	1	2	3	4	5	6	7	8
1	7.98	8.34	8.02	7.94	8.44	7.68	7.81	8.11
2	8.33	8.22	8.08	8.51	8.41	8.28	8.09	8.16
3	7.89	7.77	7.91	8.04	8	7.89	7.93	8.09
4	8.24	8.18	7.83	8.05	7.9	8.16	7.97	8.07
5	7.87	8.13	7.92	7.99	8.1	7.81	8.14	7.88
6	8.13	8.14	8.11	8.13	8.14	8.12	8.13	8.14

Answer 1

a.

A B C D E F G H I I K L 3 Sample 2 1 7.98 8.33 7.89 8.24 7.87 8.13 2 3 8.34 8.02 8.22 8.08 7.77 7.91 8.18 7.83 8.137.92 8.14 8.11 Tube Number 4 5 7.94 8.44 8.51 8.41 8.04 8 8.05 7.9 7.99 8. 1 8.13 8.14 7.68 8.28 7.89 8.16 7.81 8.12 8 Mean Range 7.81 8.11 8.04 0.76 8.09 8.16 8.26 0.43 7.93 8.09 7.94 0.32 7.97 8.07 8.05 0.41 8.14 7.887.98 0.33 8.13 8.14 8.13 0.03 Average 8.066667 0.38 6

Formula

к Tube Number 2 8.44 8.41 2 3 Sample 1 4 1 7.98 5 2 8.33 6 3 7.89 74 8.24 8 5 7.87 8.13 8.34 8.22 7.77 8.18 8.13 8.14 8.02 8.08 7.91 7.83 7.94 8.51 8.04 8.05 7.99 8.13 7.68 8.28 7.89 8.16 7.81 8.12 7.81 8.09 7.93 7.97 8.14 8.13 Mean =AVERAGE(B4:14) =AVERAGE(B5:15) EAVERAGE(B6:16) =AVERAGE(B7:17) EAVERAGE(B8:18) =AVERAGE(B9:19) =AVERAGE(14:19) 8.11 8.16 8.09 8.07 7.88 8.14 Average Range =MAX(B4:14)-MIN(B4:14) =MAX(B5:15)-MIN(B5:15) =MAX(16:16) -MIN(B6:16) =MAX(B7:17)-MIN(B7:17) =MAX(B8:18)-MIN(B8:18) =MAX(B9:19)-MIN(B9:19) -AVERAGE(K4:K9) 7.9 8.1 8.14 7.92 8.11

For sample size 8, A2 = 0.373

Mean chart,

UCL = x-bar-bar + A2*R-bar = 8.066666667+0.373*0.38 = 8.208406667

LCL = x-bar-bar - A2*R-bar = 8.066666667-0.373*0.38 = 7.924926667

Range chart,

For sample size = 8, D4 = 1.864 and D3 = 0.136

UCL = D4*R-bar = 1.864*0.38 = 0.70832

LCL = D3*R-bar = 0.136*0.38 = 0.05168

b.

Mean chart

Mean chart 6 3 - Mean 4 — 5 LCL - Center line UCL -

According to mean chart, process is out of control as sample 2 has breached UCL

Range chart 1 2 Center line - — Range - UCL — LCL

According to R-chart, the process is out of control as sample 1 has breached UCL and sample 6 has breached LCL