ANSWER
Minitab used
Stat------ANOVA -- General Linear Model----Fit General Linear Model
To test the effect of strip, calculated F= 1.16, P=0.377 which is > 0.05 level of significance. Strip has no significant effect on time.
To test the effect of position, calculated F= 2.10, P=0.143 which is > 0.05 level of significance. position has no significant effect on time.
To test the effect of electrode, calculated F= 15.83, P=0.000 which is < 0.05 level of significance. electrode has a significant effect on time.
Minitab output:
General Linear Model: time versus strip, position, electrode
Method
Factor coding |
(-1, 0, +1) |
Factor Information
Factor |
Type |
Levels |
Values |
strip |
Fixed |
5 |
1, 2, 3, 4, 5 |
position |
Fixed |
5 |
1, 2, 3, 4, 5 |
electrode |
Fixed |
5 |
A, B, C, D, E |
Analysis of Variance
Source |
DF |
Adj SS |
Adj MS |
F-Value |
P-Value |
strip |
4 |
0.3136 |
0.07840 |
1.16 |
0.377 |
position |
4 |
0.5696 |
0.14240 |
2.10 |
0.143 |
electrode |
4 |
4.2896 |
1.07240 |
15.83 |
0.000 |
Error |
12 |
0.8128 |
0.06773 |
||
Total |
24 |
5.9856 |
Model Summary
S |
R-sq |
R-sq(adj) |
R-sq(pred) |
0.260256 |
86.42% |
72.84% |
41.06% |
Coefficients
Term |
Coef |
SE Coef |
T-Value |
P-Value |
VIF |
Constant |
2.8240 |
0.0521 |
54.25 |
0.000 |
|
strip |
|||||
1 |
-0.024 |
0.104 |
-0.23 |
0.822 |
1.60 |
2 |
-0.184 |
0.104 |
-1.77 |
0.103 |
1.60 |
3 |
0.096 |
0.104 |
0.92 |
0.375 |
1.60 |
4 |
-0.024 |
0.104 |
-0.23 |
0.822 |
1.60 |
position |
|||||
1 |
-0.104 |
0.104 |
-1.00 |
0.338 |
1.60 |
2 |
-0.184 |
0.104 |
-1.77 |
0.103 |
1.60 |
3 |
0.136 |
0.104 |
1.31 |
0.216 |
1.60 |
4 |
0.216 |
0.104 |
2.07 |
0.060 |
1.60 |
electrode |
|||||
A |
0.596 |
0.104 |
5.73 |
0.000 |
1.60 |
B |
-0.564 |
0.104 |
-5.42 |
0.000 |
1.60 |
C |
-0.284 |
0.104 |
-2.73 |
0.018 |
1.60 |
D |
0.316 |
0.104 |
3.04 |
0.010 |
1.60 |
Regression Equation
time |
= |
2.8240 - 0.024 strip_1 - 0.184 strip_2 + 0.096 strip_3 - 0.024
strip_4 + 0.136 strip_5 |
-------------------------------------
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