Subject: Business Statistics
It is possible to calculate the correlation between two variables simultaneously, or bi-variate correlation.
A correlation table or bivariate frequency table, which displays the frequency distributions of two linked variables, is frequently used to classify data when the number of observations in a bivariate distribution is infinitely large. To examine correlation between two grouped series, a correlation table is required. Two variables, x and y, each have a class interval, one of which is stated in the captions and the other in the stubs to the left of the table. The correlation coefficient of the bivariate distribution is then calculated using the formula below.
Where,
N = total frequency
V = Y-B
B= assume mean of variable Y
U = X-A
A = assume mean of variable X
where ,
N = total frequency
V = \{\frac{Y - B}{K}\}
B = assume mean of variable Y
K = class size of variable Y
U = \{\frac{X - A}{h}\}
A = assume mean of variable X
h = class size of variable X
Steps:
Example:
Utilizing the following bivariate frequency distribution, calculate the coefficient of correlation. also figure out the likely error.
Sales revenue (Rs. In lakh)
|
Advertising expenditure in Rs.
|
|||
5000-10000 |
10000-15000 |
15000-20000 |
20000-25000 |
|
75-125 |
4 |
1 |
- |
- |
125-175 |
7 |
6 |
2 |
1 |
175-225 |
1 |
3 |
4 |
2 |
225-275 |
1 |
1 |
3 |
4 |
So;ution:
U’= \{\frac{X - 150}{50}\}
V’ = \{\frac{Y – 12.5}{5}\}
Adv. Exp(Rs. ‘000) |
5-10 |
10-15 |
15-20 |
20-25 |
f |
fU’ |
fU’2 |
fU’V’ |
||||||
Mid-value Y |
7.5 |
12.5 |
17.5 |
22.5 |
||||||||||
Sales revenue |
Mid-value |
V’ U’ |
-1 |
0 |
1 |
2 |
||||||||
75-125 |
100 |
-2 |
8 4 |
0 1 |
- |
- |
5 |
-10 |
20 |
8 |
||||
125-175 |
150 |
-1 |
7 7 |
0 6 |
-2 2 |
-2 1 |
16 |
-16 |
16 |
3 |
||||
175-225 |
200 |
0 |
0 1 |
0 3 |
0 4 |
0 2 |
10 |
0 |
0 |
0 |
||||
225-275 |
250 |
1 |
-1 1 |
0 1 |
3 3 |
8 4 |
9 |
9 |
9 |
10 |
||||
f |
13 |
11 |
9 |
7 |
N=40 |
∑fU’= -17 |
∑fU’2=45 |
∑fU’V’=21 |
||||||
fV’ |
-13 |
0 |
9 |
14 |
∑fV’=10 |
|||||||||
fV’2 |
13 |
0 |
9 |
28 |
∑fV’2=50 |
|||||||||
fU’V’ |
14 |
0 |
1 |
6 |
∑ |
Fv’U’=21 |
||||||||
Karl Pearson’s coefficient of correlation is given by:
r= \{\frac{N ∑f U’.V’ – (∑fU’) (∑fV’) }{\{sqrt{N. ∑fU’2—(∑fU’)2 }\} \{sqrt{N. ∑fV’2—(∑fV)'2 }\} }\}
= \{\frac{40(21) – (-17)(10)}{ \{sqrt{40(45) – (-17)2}\} \{sqrt{40(50) – (10)2}\} }\}
Therefore, r= 0.596
Probable error = 0.6745 * \{\frac{1 – r2}{ \{sqrt{N}\}}\}
= 0.6745 * \{\frac{1 – (0.596)2}{ \{sqrt{40}\} }\}
=0.69
Refrence
Chaudary, A.K. (2061).Business statistics. kathmandu:Bhundipuran Prakshan
Dhakal Bashanta (2014).Business Statistics,Buddha academic publisher
Sthapit, Azaya Bikram(2006),Business Statistics,Asmita publication
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