3.
Production
Process Characterization
3.2.
Assumptions / Prerequisites
3.2.3.
Analysis of Variance Models (ANOVA)
3.2.3.3.
Two-Way Nested ANOVA
3.2.3.3.1.
|
Two-Way Nested Value-Splitting Example
|
|
| Example: Operator is nested within
machine. |
The data table below contains data collected from five different
lathes, each run by two different operators. Note we are concerned here
with the effect of operators, so the layout is nested. If we were concerned
with shift instead of operator, the layout would be crossed. The measurement
is the diameter of a turned pin.
| Machine |
Operator |
Sample
|
|
1
|
2
|
3
|
4
|
5
|
| 1 |
Day |
.125
|
.127 |
.125 |
.126 |
.128 |
| Night |
.124 |
.128 |
.127 |
.126 |
.129 |
|
2
|
Day |
.118 |
.122 |
.120 |
.124 |
.119 |
| Night |
.116 |
.125 |
.119 |
.125 |
.120 |
| 3 |
Day |
.123 |
.125 |
.125 |
.124 |
.126 |
| Night |
.122 |
.121 |
.124 |
.126 |
.125 |
|
4
|
Day |
.126 |
.128 |
.126 |
.127 |
.129 |
| Night |
.126 |
.129 |
.125 |
.130 |
.124 |
| 5 |
Day |
.118 |
.129 |
.127 |
.120 |
.121 |
| Night |
.125 |
.123 |
.114 |
.124 |
.117 |
|
|
For the nested two-way case, just as in the crossed
case, the first thing we need to do is to sweep the cell means
from the data table to obtain the residual values. We then sweep the nested
factor (Operator) and the top level factor (Machine) to obtain the table
below.
|
|
| Machine |
Operator |
Common |
Machine |
Operator |
|
Sample
|
| |
1
|
2
|
3
|
4
|
5
|
| 1 |
Day |
.12404
|
.00246
|
-.0003
|
|
-.0012 |
.0008
|
-.0012 |
-.0002 |
.0018
|
| Night |
.0003
|
|
-.0028 |
.0012
|
.002
|
-.0008 |
.0022
|
|
2
|
Day |
-.00324
|
-.0002
|
|
-.0026 |
.0014
|
-.0006 |
.0034
|
-.0016
|
| Night |
.0002
|
|
-.005 |
.004
|
-.002
|
.004
|
-.001
|
| 3 |
Day |
.00006
|
.0005
|
|
-.0016 |
.0004
|
.0004
|
-.0006 |
.0014
|
| Night |
-.0005
|
|
-.0016 |
-.0026 |
.0004
|
.0024
|
.0014
|
|
4
|
Day |
.00296
|
.0002
|
|
-.0012 |
.0008
|
-.0012 |
-.002
|
.0018
|
| Night |
-.0002
|
|
-.0008 |
.0022
|
-.0018 |
.0032
|
-.0028 |
| 5 |
Day |
-.00224
|
.0012
|
|
-.005
|
.006
|
.004
|
-.003
|
-.002
|
| Night |
-.0012
|
|
.0044
|
.0024
|
-.0066
|
.0034
|
-.0036
|
|
|
| What does this table tell us? |
By looking at the residuals
we see that machines 2 and 5 have the greatest variability. There does not
appear to be much of an operator effect but there is clearly a strong machine
effect.
|
| Calculate sums of squares and mean
squares |
We can calculate the values for the ANOVA table according
to the formulae in the table on the nested
two-way page. This produces the table below. From the F-values we
see that the machine effect is significant but the operator effect is
not. (Here it is assumed that both factors are fixed).
|
|
|
Source
|
Sums of Squares
|
Degrees of Freedom
|
Mean Square
|
F-value
|
|
Machine
|
.000303
|
4
|
.0000758
|
8.77 > 2.61
|
|
Operator(Machine)
|
.0000186
|
5
|
.00000372
|
.428 < 2.45
|
|
Residual
|
.000346
|
40
|
.0000087
|
|
|
Corrected Total
|
.000668
|
49
|
|
|
|
