Illustration 28
Year |
2005 |
2006 |
2007 |
2008 |
2009 |
2010 |
Advertising Outlay |
120 |
130 |
110 |
100 |
170 |
140 |
Sales Proceeds |
200 |
500 |
900 |
700 |
1200 |
800 |
Analyse the least square method of deterioration with the data given in the tablet.
Solution
Least Square Method
Let us now scrutinise in some detail the deterioration method of least squares. To commence with we have schemed the data of sales proceeds and relative advertising outlay of several levels in the diagram. A deterioration line that apparently fits the data has been framed.
It may be noted that from this diagram that a point on the deterioration line relating to the advertising outlay of a year fluctuates or deviates from the observed or actual values of sales proceeds. It is this vertical divergence that depicts the blunder in the approximated value of the sales proceeds.
It is the sum of the square of these blunders refers that is reduced by the choice of
parameters ‘a and b’ in this deterioration technique among advertising
outlay and sales proceeds can be presented as below:
^ ^
Yt = a
+ bXt + et
Illustration 29
Year |
Advertising Outlay |
Sales Proceeds |
2005 |
120 |
200 |
2006 |
130 |
500 |
2007 |
110 |
900 |
2008 |
100 |
700 |
2009 |
170 |
1200 |
2010 |
140 |
800 |
Calculate the mean values of actual advertising outlay and sales proceeds and also show the estimated value of the intercept and β Co-efficient to show the association among advertising outlay and sales proceeds. What would be the estimated equation of the demand function.
Solution
Year |
Advertising Outlay |
Sales Proceeds |
_ |
_ |
_ |
_ _ |
2005 |
120 |
200 |
-8.33 |
-516.67 |
69.39 |
4303.86 |
2006 |
130 |
500 |
1.67 |
-216.67 |
2.79 |
-361.84 |
2007 |
110 |
900 |
-18.33 |
183.33 |
335.99 |
-3360.44 |
2008 |
100 |
700 |
-28.33 |
-16.67 |
802.59 |
472.26 |
2009 |
170 |
1200 |
41.67 |
483.33 |
1736.39 |
1805.56 |
2010 |
140 |
800 |
11.67 |
83.33 |
136.19 |
972.46 |
Σ |
770 |
4300 |
0 |
0 |
3083.34 |
3831.86 |
_ _ _ _
Mean is ascertained by computing X and Y where X and Y are calculated as follows:
Number of years (n) = 6;
_ _
X = ΣXt and Y = ΣYt
n n
^ ^
Estimated Equation of the demand function Y = a + bXt
The association of the outlay and proceeds are calculated by the following formula:
^ ^
a is the estimated value of the intercept and b is
the estimated β Co-efficient.
To ascertain these estimates, the formulae used are as follows:
^ _ ^_
a = Y
- bX
^ t
= last year_ _
b = Σ (Xt – X)
(Yt – Y)
t
= first year
t
= last year_
Σ (Xt – X)^2
t
= first year
_
X = ΣXt = 770 = 128.33
n 6
_
Y = ΣYt = 4300 = 716.67
n 6
_
Σ(Xt – X)= 0
_
Σ(Yt – Y)= 0
^ t
= 2010_ _
b = Σ (Xt – X)
(Yt – Y) = 3831.86 = 1.24
t
= 2005 3083.34
t
= 2010_
Σ (Xt – X)^2
t
= 2005
^ _ ^_
a = Y – bX = 716.67 – 1.24*128.33
= 557.54
_
Y = a + bXt = 557.54
+ 1.24X
Illustration 30
Estimate the Demand Price Association with the given details in the below tablet.
Restaurant |
1 |
2 |
3 |
4 |
5 |
Price |
72 |
76 |
52 |
64 |
56 |
Items Served per day |
360 |
340 |
480 |
420 |
400 |
Solution
Let us assume the price of items to be P and Items served per day to be I.
The estimated demand function will be I = a – bP
Numbers 1 to 6 represents number of hotels
Pi is the price meal and Ii is the number of meals per day in a restaurant.
Restaurant |
Price P in $ |
Meals per day (I) |
_ |
_ |
_ |
_ _ |
1 |
72 |
360 |
8 |
-40 |
16 |
320 |
2 |
76 |
340 |
12 |
-60 |
144 |
-720 |
3 |
52 |
480 |
-12 |
80 |
144 |
-960 |
4 |
64 |
420 |
0 |
20 |
0 |
0 |
5 |
56 |
400 |
-8 |
0 |
16 |
0 |
Σ |
320 |
2000 |
0 |
0 |
320 |
-1360 |
n = number
of restaurants
_
P = ΣP = 320 = 64
n 5
_
I = ΣI = 2000 = 400
n 5
_ _
Σ(Pi – P) = 0
and Σ(Ii – I) = 0
_
Σ(Pi – P)^2 = 320
_ _
(Pi – P) (Ii – I) = -1360
^ _ _
b = Σ (Pi – P)
(Ii – I)
_
Σ (Pi – P)^2
= -1360 = -
4.25
320
^ _ ^ _
a = I – b
P
= 400 – 4.25 * 64
= 400 – 272 = 128
_
Therefore the estimated demand function will be as follows: I
= 128 – 4.25P
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