3.
Production
Process Characterization
3.2.
Assumptions / Prerequisites
3.2.3.
Analysis of Variance Models (ANOVA)
3.2.3.3.
TwoWay Nested ANOVA
3.2.3.3.1.

TwoWay Nested ValueSplitting 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 twoway 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
twoway page. This produces the table below. From the Fvalues 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

Fvalue

Machine

.000303

4

.0000758

8.77 > 2.61

Operator(Machine)

.0000186

5

.00000372

.428 < 2.45

Residual

.000346

40

.0000087


Corrected Total

.000668

49



