correlation coefficient for bi-variate data

Karl Pearson’s correlation in bivariate frequency table

When the number of observations in a bivariate distribution is finitely large the data are often classified into two way frequency distribution called a correlation table or bivariate frequency table which shows the frequency distributions of two related variables. The correlation table is necessary to study correlation between two grouped series. The class interval for two variables x an y , one is listed in the captions and another is listed in the stubs at the left of the table. Then the correlation coefficient of bivariate distribution is computed by using following formula.

• r = \{\frac{N ∑f UV – (∑fU) (∑fV) }{\{sqrt{N. ∑fU^2—(∑fU)² }\} \{sqrt{N. ∑fV^2—(∑fV)² }\} }\}

Where,

N = total frequency

V = Y-B

B= assume mean of variable Y

U = X-A

A = assume mean of variable X

• r= \{\frac{N ∑f U’.V’ – (∑fU’) (∑fV’) }{\{sqrt{N. ∑fU’^2—(∑fU’)² }\} \{sqrt{N. ∑fV’^2—(∑fV)'² }\} }\}

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:

• List the class interval of two variables X and Y, one is in column heading and another is in row heading.
• Calculate mid-points of class intervals of variables X and Y and then take deviations (or steps-deviations) from their assumed means which are denoted by U and V ( or U’ and V’) respectively.
• For each class of X, add the frequencies of total cells. Similar for Y.
• Multiply the frequency of X variable, with the corresponding value of U and the products are summed up to obtain ∑fV.
• Again multiply fU with U and fV with V.
• Multiply, f, U and V of each cell and write the figure so obtained in the right-hand corner of each cell.
• All the values in the top corner sequences are added to get the last column (oir row) fU.V to obtain ∑fU.V
• Substitute all sum of values in formula to calculate r.

Example:

Calculate the coefficient of correlation from the following bivariate frequency distribution .also , calculate the probable error.

So;ution:

U’= \{\frac{X - 150}{50}\}

V’ = \{\frac{Y – 12.5}{5}\}

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

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