Question

# 19,Suppose that you monitor the fraction defectives in ounces of a process that fills beer cans....

19,Suppose that you monitor the fraction defectives in ounces of a process that fills beer cans. The data given below were collected in the production process. Each day, 20 cans are inspected. What control chart would be the best for process monitoring? Calculate UCL and LCL of the relevant chart.

Days     # of defectives

1   2

2   3

3       5

4       1

5   2

 x-bar chart, UCL = 0.042 and LCL = 0 c chart, UCL = 0.1462 and LCL = 0 p chart, UCL = 0.1462 and LCL = 0 p chart, UCL = 0.3556 and LCL = 0 p chart, UCL = 0.3556 and LCL = -0.0956

21.

In a process, cycle time is the critical-to-quality characteristic being monitored. The data given below were collected from 15 cycle times in the process. Using histogram, identify what time interval has the highest frequency in cycle time? (# of columns = K = 6)

13 minutes 13 minutes    12 minutes

14 minutes    11 minutes    10 minutes

2 minutes.    11 minutes    11 minutes

7 minutes    7 minutes       9 minutes

9 minutes       6 minutes       7 minutes

 10-12 minutes 2-4 minutes 12-14 minutes 4-6 minutes none of the above

19) For the process monitoring the p-chart would be best as we are monitoring the fraction defectives

Sample size(n) = 20

Number of samples = 5

Total number of observation (n ) = Sample size x number of samples = 20 x 5 = 100

Sum of number of defectives(np)= 2+3+5+1+2 = 13

P-bar = np/ n = 13/100 = 0.13

Sp = √{[P-bar(1-P-bar)] / n}

= √ {[0.13(1-0.13)] / 20}

= √ [(0.13 x 0.87) /20]

= √(0.1131/20)

= √0.005655

= 0.075199734

UCL = P-bar + 3(Sp) = 0.13 + (3x0.075199734) = 0.13 + 0.2256 = 0.3556

LCL = P-bar - 3(Sp) = 0.13 - (3x0.075199734) = 0.13 - 0.2256 = -0.0956 = 0 (when the LCL value is negative it is taken as 0)

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