An engineer suspects that the surface finish of a metal part is influenced by th
ID: 3055496 • Letter: A
Question
An engineer suspects that the surface finish of a metal part is influenced by the feed rate and the depth of cut.He selects three feed rates and four depths of cut.He then conducts a factorial experiment and obtains the following data:
Depth of Cut
Feed Rate
Surface Finish
1
0.15
0.2
74
2
0.15
0.2
64
3
0.15
0.2
58
4
0.18
0.2
79
5
0.18
0.2
68
6
0.18
0.2
73
7
0.2
0.2
82
8
0.2
0.2
86
9
0.2
0.2
92
10
0.25
0.2
99
11
0.25
0.2
104
12
0.25
0.2
96
13
0.15
0.25
92
14
0.15
0.25
86
15
0.15
0.25
88
16
0.18
0.25
98
17
0.18
0.25
104
18
0.18
0.25
86
19
0.2
0.25
99
20
0.2
0.25
104
21
0.2
0.25
95
22
0.25
0.25
104
23
0.25
0.25
112
24
0.25
0.25
99
25
0.15
0.3
98
26
0.15
0.3
98
27
0.15
0.3
102
28
0.18
0.3
104
29
0.18
0.3
99
30
0.18
0.3
95
31
0.2
0.3
108
32
0.2
0.3
112
33
0.2
0.3
99
34
0.25
0.3
114
35
0.25
0.3
107
36
0.25
0.3
111
a- Perform an Analysis of Variance (ANOVA) on the response variable of Surface Finish, and draw conclusions.Use alpha = .05
Depth of cut is not significant, and feed rate is significant
Depth of cut is significant, and feed rate is significant
Depth of cut is not significant, and feed rate is not significant
Depth of cut is significant, and feed rate is not significant
b- What is the p-value for feed rate?
c- What is the p-value for depth of cut?
d- Plot the residuals on a normal probability scale. Does the residual analysis appear satisfactory (Normal Distribution)?
Is the residual Normal: Yes or No
1- Yes, the residuals are a Normal Distribution
2- No, the residuals are not a Normal Distribution
Depth of Cut
Feed Rate
Surface Finish
1
0.15
0.2
74
2
0.15
0.2
64
3
0.15
0.2
58
4
0.18
0.2
79
5
0.18
0.2
68
6
0.18
0.2
73
7
0.2
0.2
82
8
0.2
0.2
86
9
0.2
0.2
92
10
0.25
0.2
99
11
0.25
0.2
104
12
0.25
0.2
96
13
0.15
0.25
92
14
0.15
0.25
86
15
0.15
0.25
88
16
0.18
0.25
98
17
0.18
0.25
104
18
0.18
0.25
86
19
0.2
0.25
99
20
0.2
0.25
104
21
0.2
0.25
95
22
0.25
0.25
104
23
0.25
0.25
112
24
0.25
0.25
99
25
0.15
0.3
98
26
0.15
0.3
98
27
0.15
0.3
102
28
0.18
0.3
104
29
0.18
0.3
99
30
0.18
0.3
95
31
0.2
0.3
108
32
0.2
0.3
112
33
0.2
0.3
99
34
0.25
0.3
114
35
0.25
0.3
107
36
0.25
0.3
111
Explanation / Answer
We use minitab to solve this question.
Minitab Output-
Regression Analysis: Surface Finish versus Depth of Cut, Feed Rate
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 2 5231.8 2615.92 56.40 0.000
Depth of Cut 1 2149.2 2149.17 46.34 0.000
Feed Rate 1 3082.7 3082.67 66.47 0.000
Error 33 1530.5 46.38
Lack-of-Fit 9 777.8 86.42 2.76 0.023
Pure Error 24 752.7 31.36
Total 35 6762.3
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant -3.92 9.30 -0.42 0.676
Depth of Cut 212.3 31.2 6.81 0.000 1.00
Feed Rate 226.7 27.8 8.15 0.000 1.00
Regression Equation
Surface Finish = -3.92 + 212.3 Depth of Cut + 226.7 Feed Rate
________________________________________________________________________________________________
a)
From the analysis of variance on the response variable of surface finish,
b)
P-value for Depth of cut is 0.000 < 0.05 indicate significance.
c)
P-value for feed rate is 0.000 < 0.05 indicate significance.
Therefore,
a)
Depth of cut is significant, and feed rate is significant.
d)
From the plot of the residuals on a normal probability scale shown below, Observe that all points lies within the stright line therefore it shows normality.
1) Yes, the residulas are a Normal Distribution.