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Assignment â€“ Statistical Quality Control
In this assignment, student will learn how to use Minitab in solving process variation by creating R chart,
S chart, and X-bar chart. To understand concepts of process control chart, go to Unit Two/Supplement,
folder â€œMinitabâ€, and read the handouts.
Some of this variation is due to a wide variety of causes that are inherent to the process like natural
variation in the process. These causes are called common causes or chance causes. When common
causes are the only source of variation, the process is said to be in a state of statistical control, or in
control. These could be machines that are malfunctioning, operator error, fluctuations in ambient
conditions, and variations in the properties of raw materials. These factors are called special causes or
assignable causes. Special causes generally produce a higher level of variability than do common causes,
this variability is considered unacceptable. When a process is operating in the presence of one or more
special factors, it is said to be out of statistical control (Navidi, 2008).
Problem: Control chart for variables
The quality engineer in charge of a salt packaging process is concerned about the moisture content in
packages of salts. Assume that five packages of salt are sampled every 15 minutes for eight hours (a
total of 32 samples), and that the moisture content in each package is measured as a percentage of total
weight. The data are presented below:
Sample
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Sample Values
2.53
2.66
2.69
2.38
2.67
2.23
2.1
2.26
2.64
2.42
2.64
1.63
2.58
2.69
2.31
2.39
3.03
2.68
2.86
3.22
2.71
2.8
2.95
3.54
3.14
2.84
2.85
3.29
2.82
3.71
3.17
3.07
2.81
3.21
2.99
2.65
3.11
2.74
2.83
2.74
2.76
2.85
2.54
2.63
1.88
2.34
2.1
2.51
2.56
2.95
3.01
2.6
2.27
2.72
3.09
2.59
3.77
3.25
3.36
3.14
2.95
2.79
2.59
3.03
2.59
2.32
2.21
2.47
2.43
2.58
2.51
2.12
3.01
2.4
2.54
3.09
2.6
3.31
2.8
3.35
2.95
3.63
3.04
2.8
3.01
2.68
2.23
2.48
2.26
2.61
2.54
2.28
2.36
2.67
2.23
2.46
2.63
2.48
3.39
2.87
3.22
3.59
3.37
3.7
2.85
2.95
3.03
2.49
2.87
2.93
Mean (x)
2.308
2.498
2.394
2.346
2.498
2.402
2.704
2.432
2.63
2.874
2.918
3.052
3.154
3.266
3.242
3.342
2.972
2.836
2.896
2.754
2.66
2.58
Range
0.78
0.35
0.57
0.48
0.28
1.32
0.78
0.29
0.76
0.74
0.79
0.95
0.97
0.74
0.89
0.63
0.4
0.34
0.52
0.54
0.64
0.61
SD (s)
0.303
0.149
0.230
0.196
0.111
0.525
0.327
0.108
0.274
0.294
0.320
0.375
0.390
0.267
0.358
0.298
0.160
0.137
0.221
0.198
0.265
0.226
23
24
25
26
27
28
29
30
31
32
2.27
2.4
2.41
2.4
2.56
2.21
2.56
2.42
2.62
2.21
2.54
2.62
2.72
2.33
2.47
2.61
2.26
2.37
2.11
2.15
2.82
2.84
2.29
2.4
2.11
2.59
1.95
2.13
2.47
2.18
2.11
2.5
2.35
2.02
2.43
2.24
2.26
2.09
2.27
2.59
2.69
2.51
2.63
2.43
2.85
2.34
2.4
2.41
2.49
2.61
2.486
2.574
2.48
2.316
2.484
2.398
2.286
2.284
2.392
2.348
0.71
0.44
0.43
0.41
0.74
0.4
0.61
0.33
0.51
0.46
0.293
0.168
0.186
0.169
0.266
0.191
0.225
0.161
0.201
0.231
Using Minitab to perform R chart, S chart, and X-bar chart, and answer the following questions:
1. Creating R chart with UCL, R-bar, and LCL
2. Creating s chart with UCL, R-bar, and LCL
3. From the above R chart and s chart, which sample is a signal of outside of the control limit?
What should be a next step? Explain
4. After deleting the out-of-control sample, re-compute the control limits and create a new R
chart, s chart, and X-bar chart
5. From the X-bar chart, explain the chart, if the process is not in control, what should be a next
step?
Submit your assignment in Word Document as a file attachment in Discussion Board, under forum
â€œAssignment Four- SQCâ€. The final product will include answers of these 5 questions and diagrams from
Minitab.
Note:
To create R chart, S chart, X-bar chart in Minitab, follow these steps
1.
2.
3.
4.
5.
6.
7.
1.
Open the data file (attached with this assignment), Assn-SQC
Choose Stat > Control Chart > Variable chart for subgroup
To work on R chart, move cursor down and select â€œRâ€
In a box, click (then you will see the cursor), then click on the data set (C2), click on â€œselectâ€
In Subgroup sizes, enter 5
Then click â€œOKâ€
For s chart and X-bar chart, follow the above step
Definition and more details of SQC with examples are posted in Blackboard, Unit Two/Supplement,
Minitab/SQC.
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Assignment – Design of Experiment (DOE)
In this assignment, student will learn how to use Minitab in solving DOE problems, focusing on twofactor experiments. To understand concepts of DOE and how to download Minitab free version, go to
Unit Two/Supplement, folder â€œMinitabâ€, and read the two handouts.
Design of Experiment (DOE) is a powerful tool that can be used in solving problems in experimental
studies. The purpose is to determine whether varying the level of two factors affect the response. DOE
can also be used to confirm suspected input/output relationships and to develop a predictive equation
suitable for performing what-if analysis. For example, it may be desirable to understand the effect of
temperature and pressure on the strength of a glue bond (ASQ-DOE by Dean Christolear).
Problem: a chemical engineer is studying the effects of various reagents and catalysts on the yield of a
certain process. Yield is expressed as a percentage of a theoretical maximum. Four runs of the process
were made for each combination of three reagents and four catalysts. The results are presented in
below table.
Catalyst
A
B
C
D
R1
86.8
82.4
86.7
83.5
71.9
72.1
80
77.4
65.5
72.4
76.6
66.7
63.9
70.4
77.2
81.2
Reagent
R2
93.4
85.2
94.8
83.1
74.5
87.1
71.9
84.1
66.7
77.1
76.7
86.1
73.7
81.6
84.2
84.9
R3
77.9
89.6
89.9
83.7
87.5
82.7
78.3
90.1
72.7
77.8
83.5
78.8
79.8
75.7
80.5
72.9
Using Minitab to perform Two-way ANOVA and answer the following questions:
1. How many degrees of freedom are there for the effect of catalyst type?
2. How many degrees of freedom are there for the effect of reagent type?
3. How many degrees of freedom are there for interactions?
4.
5.
6.
7.
How many degrees of freedom are there for error?
Construct an ANOVA table, using alpha 0.05 or 95% level of confidence
Is the additive model plausible? Provide the value of the test statistic and the P-value.
Is it plausible that the main effects of catalyst are all equal to 0? Provide the value of the test
statistic and the P-value.
8. Is it plausible that the main effects of reagent are all equal to 0? Provide the value of the test
statistic and the P-value.
9. Construct 2 diagrams: 1) main effects plot for yield and 2) Interaction plot for yield.
Submit your assignment in Word Document as a file attachment in Discussion Board, under forum
â€œAssignment Three- DOEâ€. The final product will include answers of these 9 questions, together with
tables/diagrams from Minitab.
Note:
To perform a two-way ANOVA in Minitab, use Stat > ANOVA > General Linear Model > Fit General Linear
Model. Suppose your response is called A and your factors are B and C.
1.
2.
3.
4.
5.
Choose Stat > ANOVA > General Linear Model > Fit General Linear Model.
In Responses, enter Yield.
In Factors, enter Catalyst and Reagent.
Click Model.
In Factors and covariates, select both B and C. To the right of Interactions through order, choose
6. Click OK in each dialog box.
To create a diagram on main effects plot
1.
2.
3.
4.
Choose Stat > ANOVA > Main effect plot
In Responses, enter Yield.
In Factors, enter Catalyst and Reagent.
Click OK in each dialog box.
To create a diagram on interaction plot
1.
2.
3.
4.
5.
Choose Stat > ANOVA > Interaction plot
In Responses, enter Yield.
In Factors, enter Catalyst and Reagent.
Check on display full interaction plot matrix
Click OK in each dialog box.
Definition and more details of DOE with examples are posted in Blackboard, Unit Two/Supplement,
Minitab/DOE.
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