Question

please answer number 24 part A and for part B answer it in equation form, not graph. thank you.

Process Capability Sample 24. The production manager at Sunny Soda, Inc. is inter- ested 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 the label shows 12 ounces as the serving size. The design standard for the product calls for a fill level of 12.00 0.10 ounces. The manager collected the following sample data (in fluid Ounces per bottle) on the production process: 12.03 11.95 12.09 12.01 12.00 11.92 11.91 OBSERVATION 2 3 12.00 12.00 12.04 11.99 11.96 11.97 11.94 12.09 11.99 12.05 12.00 12.06 11.99 12.06 12.00 12.05 12.05 12.01 12.01 12.03 12.00 12.10 11.97 12.03 11.95 11.97 Sample 3 12.00 11.91 OBSERVATION 2 11.97 12.10 11.94 12.10 12.02 11.97 12.09 12.05 11.92 12.12 11.98 1206 12.08 11.96 11.99 11.95 11.98 12.02 12.00 11.89 12.10 12.08 12.05 12.08 a. Are the process average and range in statistical control? b. Is the process capable of meeting the design standard at four-sigma quality? Explain. 11.94

Answer 1

Part A=

                                         Range      X bar

12.00 11.97 12.10 12.08         0.13     12.04

11.91 11.94 12.10 11.96          0.19     11.98

11.89 12.02 11.97 11.99         0.13    11.97

12.10 12.09 12.05 11.95         0.15     12.05

12.08 11.92 12.12 12.05          0.20     12.04

11.94 11.98 12.06 12.08          0.14     12.02

12.09 12.00 12.00 12.03         0.09     12.03

12.01 12.04 11.99 11.95        0.09    12.00

12.00 11.96 11.97 12.03        0.07     11.99

11.92 11.94 12.09 12.00         0.17     11.99

11.91 11.99 12.05 12.10         0.19     12.01

12.01 12.00 12.06 11.97        0.09     12.01

11.98 11.99 12.06 12.03         0.08    12.02

12.02 12.00 12.05 11.95          0.10    12.01

12.00 12.05 12.01 11.97        0.08    12.01

R bar Sum of range/N =0.13

X bar= X double bar = X bar/N =12.01

For N=4, D4=2.282, D3=0,

UCLR= D4(R bar)=2.282*0.13 = 0.2891

LCLR= D3(R bar)=0*0.13=0

As values of Ranges fall within the category of 0 and 0.2891 thus Range is within control.

For X bar=

For N=4,A2 = 0.729)

UCL Xbar=X double bar+A2(R bar) = 12.01 + 0.729 x 0.13 = 12.1018

LCL Xbar=X double bar-A2(R bar)= 12.01 – 0.729 x 0.13 = 11.9172

As all X bar values are with in the range of 11.9172 and 12.1018, so process is in control

Part B=

Standard deviation = R-bar / d2 = 0.13 / 2.059 = 0.06

Cpk = min{(12.01 – 11.90)/(3 x 0.06) , (12.10 – 12.01)/(3 x 0.06) } = 0.50

in order to have 5-sigma capability , it is important to have Cpk value of at least 1.67.

Therefore the process is NOT CAPABLE of meeting the specifications

Reason= We hav to find out Cp

Cp = (12.10 – 11.90)/(6 x 0.06) = 0.56

Thus the issue is that the process has a large standard deviation