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 n_ij observation value j service type of i bank, and k=1,...n_ij such that;

, is the error term.

n_{i. =} and n_..=

Using the data table easily, for example,

n_1.= n_{i. =} == n₁₁ + n₁₂ + n₁₃ =6

n_2.= = = 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+ + ₁(0)+₂(1) + (0) + (0) +(0) +(0)+(1) +(0)+ ₂₂₁

y₂₃₁ = 3+ + ₁(0)+₂(1) + (0) + (0) +(0) +(0)+(0) +(1)+ ₂₃₁

y₂₃₂ = 8+ + ₁(0)+₂(1) + (0) + (1) +(0) +(0)+(0) +(1)+ ₂₃₂

y₂₃₃ = 10+ + ₁(0)+₂(1) + (0) + (1) +(0) +(0)+(0) +(1)+ ₂₃₃

The result is like shown above. Let’s write the indicator matrix(design matrix) changed into matrix form.

= = +

We can show these equation system by matrix and vectors like below:

Y = X +

Y = Vector of observations

X = The matrix of coefficients consisting of 0 and 1.

= The unknown parameter vector consisting of three sub vectors(=())

= Vector of errors

In the system above, we see that each sub vector of each equation has one member at least. It causes the X indicator matrix to go into a certain order:

1) It only consists of 0 and 1 values.

2) The X matrix won’t be a full rank matrix either it is square or rectangle because of its special structure.(The first column is equal to the sum of not only the 2. ,3., but also the 4., 5.,6., 7., 8., 9. columns.)

This means that we will come across to models not having full rank. We will solve this models with a new method called “g-inverse”.

()= (2)

For the solution of normal equations

=

must be like this. But (2) has just one solution for the full rank regression models. ,is an estimable parameter. In another way, it is

E()=.

In our model, it can be seen easily that matrix(design matrix~indicator matrix )is not full rank. 0 and 1 called “dummy variable” have a special meaning. The dispersal of 0 and 1 inshows us how the terms of models line up among the observations, in other ways what kind of classification these observation have. An endless solutain is about to be made for whenis not full rank.And to tell about the estimated vector for just one , some writers like Kempthorne, Federer, Steel-Torrie have chosen the way of making the solution single(Y=X+). They have made abbreviations like . And for the second, they have made the solution by “g-inverse(generalized inverse). They have showed that the solution of the system will be

=

if the equation system is consistent. Obtaining the G generalized inverse matrix

and is a general solution of the normal equations, gives us the

equality.

Z is a vector made of the unknown. There are different methods related to calculating the g-inverses.Now let’s go back to our model and try to find a solution for .This solution’s

is this. Let’s express the matrix here numerically.

= =

The degree of indicator matrix is 9 and rank is 6. Because;

The sum of the 2. and 3. lines gives the first line.

The sum of the 4., 5., and 6. lines gives the second line

The sum of the 7., 8. and 9. lines gives the third line. Due to this linear relation, rank.

And we can say that, for hierarchic classification with two factors model, the rank of X’X matrix is equal to the number of its lower classes. And its total level which we will express with q is 6 . Considering that p is the number showing the equations of factors , in , q is number showing the equations and the number 1 showing the equation, it’s found that when the degree of X’X matrix is calculated by m=1+p+q, the rank is X’X=1+p+q-(1+p)=q .

Also normal equations are made by taking the solution vector of (1+a) member of zero(=0) . In other words, a limit is put. And this is

.

=[ 0 0 0 4 6 8 8 9 7]

is the line vector that is formed by the cell averages.The g-inverse(generalized inverse matrix) that responses to X’X matrix is,

assuming that D(1/n_ij) is a matrix whose diagonal members are ½, ½, ½,1/3, 1 and 1/3 it is a diagonal matrix provided from the sub(lower) square matrix of the sixth degree of the X’X matrix;

matrix is G-generalized inverse matrix. For = is that, let’s calculate the multiply.

= and = .

Variance Analysis Of The Model

Let’s calculate the sum of the squares of the model considering the observation data in Tablo-1.

R()=SSM=n_..=13.(7²)=637

R(=SSR==

₌ 8²/2+12²/2+16²/2+24²/3+9²/1+21²/3

=652

R(=R(-R()=652-637=15

SST==5²+3²+9²+3²+8²+8²+9²+9²+6²+9²+3²+8²+10²

=704

and SSE=SST-SSR=704-652=52

Let’s try to explain the meanings of the R() functions(Reduction) used here:

R(): y_i= reduction according to the last model.The average of sum of the squares.

R(:joining of the factor to the model after the factor. Average of the squares

SST: Sum of the squares

SSE: sum of the squares of errors

Variance Analysis (V.A) Table

Source of variation	s.d	Sum of Square	Average of squares	Statistics of F
Average	1	R()=637	637/1=637	F()=637/(SSE/N-r)=85,73
After the average model	b-1=6-1=5	R(=15	15/5=3	F(Rm)=3/(SSE/N-r)=0,4
Error	N-b=13-6=7	SSE=52	52/7=7,43
Sum	13	SST=704

Looking at the table values of the F statistics in %5 significance level,

F_1,7,0.05=5,59 and F_6,7,0.05=3,87

is seen. Commenting on the result of this test like below is possible.

Comment: In %5 significance level F_1,7,0.05=5,59 value is smaller than the calculated F(R_m)=85,73 value. And H₀=E()=0 hypothesis is rejected

E(y_ijk)= model can explain the changes in “y” much more than E(y_ijk)= model explain. But the hypothesis can be accepted because F_6,7,0.05=3,87 is bigger than the calculated F(R_m)=0,4 value. And E(y_ijk)= model can explain the changes in “y” much more than E(y_ijk)= model explain. Explanition of the model with only the average is not enough. (Banks) factors and all of these levels(commercial activities) must also join to the model.

Conclusion

=[ 0 0 0 4 6 8 8 9 7]

parameter of the estimable vector shows us that customers of the public banks( ) don’t benefit from the service provided them in the result of “Credit card activities” while the special banks provide this service more well.(=4 ; =8)

It’s observed that the level of “Consumer credit” provided to their customers by special banks is very useful and it has created a more advantageous condition according to the public banks. Also we can say that level of “Commercial credits” provided by public banks are more advantageous than that provided by commercial banks.(). It can certainly be shown in the model that doing and without joining the model the advantages and disadvantages cannot be tested.

References

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