Question

Day	Sample (n)	Number (np)	Proportion (p)
1	500	12	0.024
2	500	15	0.030
3	500	19	0.038
4	500	13	0.026
5	500	9	0.018
6	500	26	0.052
7	500	18	0.036
8	500	14	0.028
9	500	17	0.034
10	500	18	0.036
11	500	16	0.032
12	500	24	0.048
13	500	11	0.022
14	500	31	0.062
15	500	16	0.032
16	500	10	0.020
17	500	16	0.032
18	500	17	0.034
19	500	20	0.040
20	500	15	0.030
21	500	8	0.016
22	500	13	0.026
23	500	12	0.024
24	500	17	0.034
25	500	18	0.036

Use the collected data above, please construct a p chart and an np chart and explain how do they both look?

Answer 1

For p chart, define p.bar as the mean of the proportions given. Then the control limits for p chart are

$UCL=\bar{p}+3\sqrt{\bar{p} (1-\bar{p})/n}\\ CL=\bar{p}\\ LCL=\bar{p}-3\sqrt{\bar{p} (1-\bar{p})/n}\\$

We compute p.bar=.0324 and hence

UCL=.05616, CL=.0324 and LCL=.00864

P Chart 1 0.06 UCL=0.05616 0.05 0.04 Proportion AWAY P=0.0324 0.03 0.02 0.01 LCL =0.00864 1 3 T 7 5 9 11 13 15 17 19 21 23 25 Sample

From the chart, we find that the 14 th observation falls outside the control limits.

For np chart, define p.bar as the mean of the proportions given. Then the control limits for np chart are

$UCL=n\bar{p}+3\sqrt{n\bar{p} (1-\bar{p})}\\ CL=n\bar{p}\\ LCL=n\bar{p}-3\sqrt{n\bar{p} (1-\bar{p})}\\$

We compute p.bar=.0324 and hence

UCL=28.08, CL=16.2 and LCL=4.32

NP Chart 1 30 UCL-28.08 25 20 Sample Count WY nP=16.2 15 10 л LCL =4.32 1 3 T 9 5 7 11 13 15 17 19 21 23 25 Sample

From the chart, we find that the 14 th observation also falls outside the control limits.

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