Q3 (20pts). Jaycola company introduces an automatic monitoring machine can measu
ID: 3175866 • Letter: Q
Question
Q3 (20pts). Jaycola company introduces an automatic monitoring machine can measure which the amount of cola quickly. Therefore, the random sample with more bottles can be obtained. With the increased sample size, it is impractical to analyze the amount of cola each bottle in the sample. in Nonconforming is defined when the amount of cola in a bottle is less than 347 and greater than 352. To set up a control chart of the fraction nonconforming bottles, 20 samples of n 30 bottles were selected. The data are shown as follows. ample Number of nonconforming bottles Sample fraction nonconforming 0.433 0.300 12 0.400 0.200 16 18 0.200 14 13 0.433 12 0.300 13 0.233 14 15 0.500 0.767 0.200 17 0.567 17 18 16 0.533 21 0.700 0.633 19 1. Complete the table. 2. Calculate p 3. Construct a 5-sigma p chart for the above data. That is, at are the control limits?Explanation / Answer
1. Sample fraction non conforming for sample 2=14/30=0.467; sample 6=8/30=0.267; sample 7=16/30=0.533; sample 8=18/30=0.6; sample 10=14/30=0.467.
2. Tabulate the given information.
n np p
30 13 0.433333
30 14 0.466667
30 9 0.300000
30 12 0.400000
30 6 0.200000
30 8 0.266667
30 16 0.533333
30 18 0.600000
30 6 0.200000
30 14 0.466667
30 13 0.433333
30 9 0.300000
30 7 0.233333
30 15 0.500000
30 23 0.766667
30 6 0.200000
30 17 0.566667
30 16 0.533333
30 21 0.700000
Siigma np=262, sigma n=600, nbar=30, sigma p=8.7333
Center line, pbar=sigma np/sigma n=262/600=0.4367.
3. Upper control limit, UCLp=pbar+5sqrt[pbar(1-pbar)/nbar]=0.4367+5sqrt[0.4367(1-0.4367)/30]=0.8895
Center line,pbar=0.4367
Lower control limit, LCLp=pbar-5sqrt[pbar(1-pbar)/nbar]=0.4367-5sqrt[0.4367(1-0.4367)/30]=-0.0161
30 19 0.633333
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