# 1. Suppose 20 respirator masks are examined every thirty minutes and the number of defective units…

1. Suppose 20 respirator masks are examined every thirty minutes and the number of defective units are recorded per 8-hour shift. The total number examined on a given shift is equal to n = 20(16) = 320. Table 18.5 gives the results for 30 such shifts. The center line for the p-chart is equal to the proportion of defectives for the 30 shifts, and is given by total number of defectives divided by the total number examined for the 30 shifts, or

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1. Suppose 20 respirator masks are examined every thirty minutes and the number of defective units…
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2. For the data in Table 18.5, the centerline is given by n-p = 320(.0075) =  2.4 and the control limits are LCL = 2.4 – 4.63 = -2.23, which we take as 0, and UCL = 2.4 + 4.63 = 7.03. If 8 or more defectives are found on a given shift, the process is out of control. The MINITAB solution is found by using the pull-down sequence

3. For the data in Table 18.5, the centerline is given by = 320(.0075) = 2.4 and the control limits are LCL = 2.4 – 4.63 = -2.23, which we take as 0, and UCL= 2.4 + 4.63 = 7.03. If 8 or more defectives are found on a given shift, the process is out of control. The MINITAB solution is found by using the pull-down sequence

 Shift # Number Proportion Number Proportion 1 Defective Defective Defective Defective 2 Xi Pi = X/n Shift Xi Pi = X/n 3 1 0.003125 16 2 0.006250 4 2 0.006250 17 0 0.000000 5 2 0.006250 18 4 0.012500 6 0 0.000000 19 1 0.003125 7 4 0.012500 20 7 0.021875 8 4 0.012500 21 4 0.012500 9 4 0.012500 22 1 0.003125 10 6 0.018750 23 0 0.000000 11 4 0.012500 24 4 0.012500 12 0 0.000000 25 4 0.012500 13 0 0.000000 26 3 0.009375 14 0 0.000000 27 2 0.006250 15 1 0.003125 28 0 0.000000 Shift # 0 0.000000 29 0 0.000000 1 7 0.021875 30 5 0.015625