Question

. The production manager at Sunny Soda, Inc., is interested in tracking the quality of the company's 12-ounce bottle filling line. The bottles must be filled within the tolerances set for this product because the dietary information on th label shows 12 ounces as the serving size. The design standard for the product calls for a fill level of 12 0.10 ounces. The manager collected the accompanying sample data n fluid ounces per bottle) on the production process. 1 Click the icon to view the sample data. 2 Click the icon to view the table of factors for calculating three-sigma limits for the X-chart and R-chart Design an R- and an x-chart for this process. The UCL equals ounces and the LCLR equals ounces. (Enter your responses rounded to three decimal places.) The UCL equals ounces and the LCL equals ounces. (Enter your responses rounded to three decimal places.) a. Are the process average and range in statistical control? The process average is (1) statistical control and the process range is (2) statistical control. b. Is the process capable meeting the design standard at four-sigma quality? Explain. The C is equal to so the process (3)- capable. (Enter your response rounded to three decimal places.) Data Table Bottle 2 3 Sample 4 12.02 11.93 11.9511.90 1 11.93 12.04 12 1011 88 3 11.98 12.01 11.98 12.07 4 12.1211.94 11.95 12.12 12.12 11.97 12.06 11.93 11.91 12.0511.8911.94 11,89 12.04 12.01 12.09 9 12.06 11.95 11.96 11.88 10 12.09 12.12 11.98 12.08 11 12 12.07 12.02 11.91 12.05 15 11.97 12.11 11.98 11.99 2: More Info

Answer 1

	Bottle
Sample	1	2	3	4	X-bar	Range
1	12.02	11.93	11.95	11.9	11.95	0.12
2	11.91	11.94	11.92	12.08	11.96	0.17
3	11.93	12.04	12.1	11.88	11.99	0.22
4	11.98	12.01	11.98	12.07	12.01	0.09
5	12.12	11.94	11.95	12.12	12.03	0.18
6	12.12	11.97	12.06	11.93	12.02	0.19
7	11.91	12.05	11.89	11.94	11.95	0.16
8	12.06	11.9	11.9	12.09	11.99	0.19
9	11.89	12.04	12.01	12.09	12.01	0.20
10	12.06	11.95	11.96	11.88	11.96	0.18
11	12.09	12.12	11.98	12.08	12.07	0.14
12	12.07	12.02	11.91	12.05	12.01	0.16
13	11.95	11.98	12.12	11.98	12.01	0.17
14	11.91	12.04	11.94	12.11	12.00	0.20
15	11.97	12.11	11.98	11.99	12.01	0.14
				Averages	11.998	0.167
					X_double-bar	R-bar

For sample size n = 4,A₂ = 0.729, D₃ = 0, and D₄ = 2.282

UCL_R = D₄ * R_bar = 2.282 * 0.167 = 0.382
LCL_R = D₃ * R_bar = 0 * 0.167 = 0

UCL_X = X_double-bar + A₂ * R_bar = 11.998 + 0.729*0.167 = 12.120
LCL_X = X_double-bar - A₂ * R_bar = 11.998 - 0.729*0.167 = 11.876

a.

Process averages are in control. Process ranges are also in control.

b.

Estimate for process stdev = A₂ * R_bar * √n / 3 = 0.729*0.167*√4 / 3 = 0.081

Cpu = (USL - mean) / 3.stdev = (12.10 - 11.998) / (3*0.081) = 0.420
Cpl = (mean - LSL) / 3.stdev = (11.998 - 11.90) / (3*0.081) = 0.402

So, Cpk = Min(Cpu, Cpl) = 0.402 and the process is not capable.