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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(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
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online