Variance analysis of advantages and disadvantages of the service that active financial associations provide their customers solution by the model of hierarchy classification with two factors and an application
Doç. Dr. Adnan MAZMANOĞLU
Department of Mathematics
Marmara University.
Göztepe Campus/İstanbul
E-Mail:amazman@marun.edu.tr
Key Word : Linear Models, Two-Way Nested Classification
V.A. : Variance
Analysis
The
“data table” below is formed by the results of analysis made face to face with
some customers chosen random and among different geographical regions and different status, to test the advantage of
the level of consumer credit, credit
card and commercial credits(level of 
to factor in 
) which are not only included in the economical activities of
the public and special banks(factor ) but also assumed by our model, provide their customers.
Table :1: Observation Table
|
Observation Values |
|||||
|
Banks |
Bank type |
Points |
Sum(Observation) |
Number |
Average |
|
Public |
Credit
card (1) |
5, 3 |
8 |
2 |
4 |
|
Consumer
credit(2) |
9, 3 |
12 |
2 |
6 |
|
|
Commercial (3) credit |
8,8 |
16 |
2 |
8 |
|
|
|
Sum |
36 |
6 |
6 |
|
|
Special Banks |
Credit
card (1) |
9, 9, 6 |
24 |
3 |
8 |
|
Consumer
credit(2) |
9 |
9 |
1 |
9 |
|
|
Commercial (3) credit |
3, 8, 10 |
21 |
3 |
7 |
|
|
Sum |
54 |
7 |
8 |
||
|
General Sum : 90 13 7 |
|||||
In this table, having
shown the credit card, consumer credit and commercial credits service types
among the busiest activities of the public and special banks, and finding out how
much the customers are satisfied with this services, have directed us to the
two-factored hierarchic(nested) variance analyses model in which the different
levels of hierarchicly set two factors will be tested
Model
Looking at the data
table we can understand that the most suitable model is
(1)
. Let’s try to explain this parameters:
: is the k observation
value of the i type of bank in the j
service level
: is the type of i bank
: shows the effect of the i type bank and j
type service type
The model consist of two factors 
and and these are one within the other. The factor has two levels
named public and special banks, and the factor has six levels,
three of them in the first factor of and three of them in
the second factor of .
p=2; i=1,...p
q1=3; j=1,...q1
q2=3
Assuming that there are nij observation
value j service type of i bank, and
k=1,...nij such that;
, is the error term.
ni. = 
and n..=
Using the data table easily,
for example,
n1.= ni.
= =
= n11 + n12 + n13 =6
n2.= 
= 
= n21 + n22 + n23 = 7
n.. = 
=6+7=13
Let’s write the normal equations of these thirteen
observations using the (1) model.
y111 =
5=
+ 1 + 
+ 
111
y112 =
3= + 1 + + 112
y121 =
9= + 1 + 
+ 121
y122 =
3= + 1 + + 122
y131 =
8= + 1 + 
+ 131
y132 =
8= + 1 + + 132
y211 =
9= + 2 + 
+ 211
y212 =
9= + 2 + + 212
y213 =
6= + 2 + + 213
y221 =
9= + 2 + 
+ 221
y231 =
3= + 2 + 
+ 231
y232 =
8= + 2 + + 232
y233 = 10= + 2 + + 233
If we write the models consisting of the (1, 0)
indicator values of these equations again,
y111 = 5+ + 1(1)+2(0) + (1) + (0) +(0) +(0)+(0) +(0)+ 111
y112 = 3+ + 1(1)+2(0) + (1) + (0) +(0) +(0)+(0) +(0)+ 112
y121 = 9+ + 1(1)+2(0) + (0) + (1) +(0) +(0)+(0) +(0)+ 121
y122 = 3+ + 1(1)+2(0) + (0) + (1) +(0) +(0)+(0) +(0)+ 122
y131 = 8+ + 1(1)+2(0) + (0) + (0) +(1) +(0)+(0) +(0)+ 131
y132 = 8+ + 1(1)+2(0) + (0) + (1) +(1) +(0)+(0) +(0)+ 132
y211 = 9+ + 1(0)+2(1) + (0) + (0) +(0) +(1)+(0) +(0)+ 211
y212 = 9+ + 1(0)+2(1) + (0) + (0) +(0) +(1)+(0) +(0)+ 212
y213 = 6+ + 1(0)+2(1) + (0) + (0) +(0) +(1)+(0) +(0)+ 213
y221 = 9+ +