# 16-04-01 - UCL:= p 3 ⋅ se p =.115383 LCL:= max p 3 ⋅ se...

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16.4.1 Construct a p -chart from the data in DS 16.4.1, which are the number of defective items found in random samples of n := 100 items taken at k := 30 time points. Begin by calculating p for each of those sampling records. k n x p 1 100 5 0.05 2 100 6 0.06 3 100 2 0.02 4 100 4 0.04 5 100 4 0.04 6 100 8 0.08 7 100 5 0.05 8 100 5 0.05 9 100 4 0.04 10 100 5 0.05 11 100 3 0.03 12 100 7 0.07 13 100 6 0.06 14 100 4 0.04 15 100 3 0.03 16 100 7 0.07 17 100 6 0.06 18 100 3 0.03 19 100 6 0.06 20 100 4 0.04 21 100 2 0.02 22 100 4 0.04 23 100 4 0.04 24 100 6 0.06 25 100 5 0.05 26 100 10 0.1 27 100 9 0.09 28 100 2 0.02 29 100 4 0.04 30 100 7 0.07 AVG 100 5 0.05 MAX 100 10 0.1 MIN 100 2 0.02

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(a) Is there any reason to believe that the process was out of control while this data set was collected? p := \$d32 = .05 se p := p ⋅ 1 - p n se p = .021794
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Unformatted text preview: UCL := p + 3 ⋅ se p = .115383 LCL := max p- 3 ⋅ se p , 0 = 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30-0.02 0.02 0.04 0.06 0.08 0.1 p-Chart k p All of these data points fall within their acceptable ranges and so the manufacturer ought to retain them. (b) How many defective items would need to be discovered in a future sample for the process to be considered to be out of control? ceiling UCL ⋅ n = 12. The rounding upward takes place because the process requires a whole number of measured defects....
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16-04-01 - UCL:= p 3 ⋅ se p =.115383 LCL:= max p 3 ⋅ se...

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