Show all your work. 1. A quality department inspects the final products (rulers)
ID: 330736 • Letter: S
Question
Show all your work.
1. A quality department inspects the final products (rulers) by taking 3 samples with 4 observations as presented below. Assuming that the standard deviation of data is 0.2 cms, develop control limits of 3 standard deviations.
Samples of Shampoo Bottle Volume in Ounces
Observation
1
2
3
1
19.7
19.7
19.7
2
20.6
20.2
18.7
3
18.9
18.9
21.6
4
20.8
20.7
20.0
Mean
20.0
19.875
20.0
2. A quality manager monitors the quality of the company’s production process. The product is expected to be filled between 16.00 ± 0.30. The quality department collected the data as follows.
Observations
Sample
1
2
3
4
1
16.40
16.11
15.90
15.78
2
15.97
16.10
16.20
15.81
3
15.91
16.00
16.04
15.92
4
16.20
16.21
15.93
15.95
5
15.87
16.21
16.34
16.43
6
15.43
15.49
15.55
15.92
7
16.43
16.21
15.99
16.00
8
15.50
15.92
16.12
16.02
9
16.13
16.21
16.05
16.01
10
15.68
16.43
16.20
15.97
a. Show if the process is in statistical control in terms of mean and range.
b. Show if the process is capable of meeting the design standard.
3. The table below shows the # of defective products discovered in 5 random samples of 20 observations each. Develop a 3-sigma control chart.
Sample
Number of Defects
Number of Observations in Sample
1
1
20
2
2
20
3
2
20
4
1
20
5
0
20
Total
6
100
4. The management of a hotel is worried about the number of customer complaints. A detail analysis shows the number of complaints in 100 feedback per week over the past 12 weeks.
a. Develop control charts with z = 3.
b. Is the process in control?
Week
Number of complaints
1
4
2
5
3
6
4
6
5
3
6
2
7
6
8
7
9
3
10
4
11
3
12
4
5. Compute the Cpand Cpk measure of process capability for the following machine and interpret the findings.
Machine data:
USL = 100
LSL = 70
Process ? = 5
Process ? = 80
Samples of Shampoo Bottle Volume in Ounces
Observation
1
2
3
1
19.7
19.7
19.7
2
20.6
20.2
18.7
3
18.9
18.9
21.6
4
20.8
20.7
20.0
Mean
20.0
19.875
20.0
Explanation / Answer
X-bar-bar = Average of Mean Sample observations = (20+19.875+20)/3 = 19.96
standard deviation = 0.02
Number of observation = 4
s = standard deviation/sqrt(Number of observation) = 0.02/sqrt(4) = 0.01
z = 3 for 3 standard deviation
UCL = X-bar-bar + z*s = 19.96 + 3*0.01 = 19.99
LCL = = X-bar-bar - z*s = 19.96 - 3*0.01 = 19.93