1. Suppose 20 respirator masks are examined every thirty minutes and the number of defective units…
 September 24, 2021 /
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1. Suppose 20 respirator masks are examined every thirty minutes and the number of defective units are recorded per 8hour 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 pchart 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
2. For the data in Table 18.5, the centerline is given by np = 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 pulldown 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 pulldown sequence
Shift # 
Number 
Proportion 

Number 
Proportion 
1 
Defective 
Defective 

Defective 
Defective 
2 
X_{i} 
P_{i} = X/n 
Shift 
X_{i} 
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 