29) A large beverage company would like to use statistical process control to mo

Question

29) A large beverage company would like to use statistical process control to monitor a new bottling machine designed to load liquid into 350-ml bottles. This company knows that the exact amount the machine places in each bottle can naturally vary by a small amount, but does not have any more specific information about the process. The company operated this new machine under careful supervision, confident that the machine was under complete control, for 7 hours. Each hour, a sample of six bottles was taken off the line and the amount of liquid in each bottle was carefully measured. This is the resulting data Sample Sample Mean 350.85 ml 350.80 ml 351.20 ml 351.00 ml 350.62 ml 351.12 ml 351.50 ml Sample Range 0.6 ml 0.8 ml 0.5 ml 1.0 ml 0.8 ml 0.7 ml 0.8 ml 4 a. Where should the control limits be placed on a mean chart intended to monitor this machine in the future, using the same sampling procedure that produced this data? b. where should the control limits be placed on a range chart intended to monitor this machine in the future, using the same sampling procedure that produced this data?

Explanation / Answer

Solution:
The Control limits for Xbar chart can be calculated as
UCL = Xdoublebar + A2*Rbar
LCL = Xdoublebar - A2*Rbar
Xdoublebar = (350.85+350.80+351.2+351+351+350.62+351.12+351.5)/7 = 351.0128
Rbar = (0.6+0.8+0.5+1+0.8+0.7+0.8)/7 = 0.742857
Given sample =6 so A2 = 0.483
So UCL = 351.0128 + 0.483*0.742857 = 351.0128 + 0.358799 = 351.371599

LCL = 351.0128 - 0.483*0.742857 = 351.0128 - 0.358799 = 350.65400
So LCL and UCL for Mean chart is 350.65 to 351.371599

Solution(b)
Rbar = 0.7428
UCL and LCL for Rbar is
UCL = D4*Rbar = 2.004*0.7428 = 1.48857
LCL = D3*Rbar = 0*0.7428 = 0
So For range chart LCL and UCL are 0 and 1.48857 Respectively

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