Question

17. (10points) One hundred product labels are inspected every day for surface nonconformities. Results for the past 25 days are 22, 29, 25, 17, 20, 16, 34, 11, 31, 29, 15, 10, 33, 23, 27, 15, 17, 17, 19, 22, 23, 27, 29, 33, and 21. Plot the points on graph paper (run chart) and determine if the process is stable. Determine the trial central line and control limits. (Hint: count of defects/unit)

Answer 1

SOLUTION:-

Number of product labels inspected , n= 100.

For surface non-conformities the result for 25 days are tabulated below and fraction non-conforming.

Fraction of non-conforming products (p)

np р3

Substitute number non-conforming value 22 in. the above relation:

$p=\frac{22}{100}=0.22.$

Fraction of non-conforming products (p)

np р3

Substitute number non-conforming value 29 in, the above relation:

$p=\frac{29}{100}=0.29.$

Fraction of non-conforming products (p)

np р3

substitute number non-conforming value 25 in. the above relation:

25 = 0.25. 100

The values are tabulated by using given data and proportion of nonconforming is also tabulated:

sub group	Number of products inspected (n)	Number of non conforming products np	Proportion of non conforming products
1	100	22	0.22
2	100	29	0.29
3	100	25	0.25
4	100	17	0.17
5	100	20	0.2
6	100	16	0.16
7	100	34	0.34
8	100	11	0.11
9	100	31	0.31
10	100	29	0.29
11	100	15	0.15
12	100	10	0.1
13	100	33	0.33
14	100	23	0.23
15	100	27	0.27
16	100	15	0.15
17	100	17	0.17
18	100	17	0.17
19	100	19	0.19
20	100	22	0.22
21	100	23	0.23
22	100	27	0.27
23	100	29	0.29
24	100	33	0.33
25	100	21	0.21
	$\sum n=2500$	$\sum np=565$

The center line can be calculated using the below formula:

$\bar{p}=\frac{\sum np}{\sum n}$

Substitute _{$\sum n=2500$} and  _{$\sum np=565$}

$\bar{p}=\frac{565}{2500}=0.226$

The center line is 0.226.

Calculate the upper control limit,

$UCL=\bar{p}+\sqrt[3]{\frac{\bar{p}(1-\bar{p})}{n}}$

Substitute the _{$\bar{p}=0.226,n=100$}   in the above relation.

$UCL=0.226+\sqrt[3]{\frac{0.226(1-0.226)}{100}}=0.351.$

Upper control limit is 0.351.

Calculate the lower control limit,

$LCL=\bar{P}-\sqrt[3]{\frac{\bar{p}(1-\bar{p})}{n}}$

Substitute the _{$\bar{p}=0.226,n=100$} in the above relation.

$LCL=0.226-\sqrt[3]{\frac{0.226(1-0.226)}{100}}=0.101$

Lower control limit is 0.351.

Tabulate the values of upper control limit, lower control limit and center line:

Sub group	Number of products inspected n	Number of non-conforming products (np)	Proportion of non conforming products	UCL	Centre line	LCL
1	100	22	0.22	0.351	0.226	0.101
2	100	29	0.29	0.351	0.226	0.101
3	100	25	0.25	0.351	0.226	0.101
4	100	17	0.17	0.351	0.226	0.101
5	100	20	0.2	0.351	0.226	0.101
6	100	16	0.16	0.351	0.226	0.101
7	100	34	0.34	0.351	0.226	0.101
8	100	11	0.11	0.351	0.226	0.101
9	100	31	0.31	0.351	0.226	0.101
10	100	29	0.29	0.351	0.226	0.101
11	100	15	0.15	0.351	0.226	0.101
12	100	10	0.1	0.351	0.226	0.101
13	100	33	0.33	0.351	0.226	0.101
14	100	23	0.23	0.351	0.226	0.101
15	100	27	0.27	0.351	0.226	0.101
16	100	15	0.15	0.351	0.226	0.101
17	100	17	0.17	0.351	0.226	0.101
18	100	17	0.17	0.351	0.226	0.101
19	100	19	0.19	0.351	0.226	0.101
20	100	22	0.22	0.351	0.226	0.101
21	100	23	0.23	0.351	0.226	0.101
22	100	27	0.27	0.351	0.226	0.101
23	100	29	0.29	0.351	0.226	0.101
24	100	33	0.33	0.351	0.226	0.101
25	100	21	0.21	0.351	0.226	0.101

Create the spread sheet as follows:

RUN CHART 0.4 0.35 proportion of non conforming products 0.3 0.25 0.2 UCL 0.15 Centre line 0.1 0.05 -LCL 20 10 30 SUB GROUP PROPORTON OF NONCONFORMING

Therefore, the center line, UCL, and , LCL are determined.