test of significance of sample correlation coefficient using probable error

Probable error(P.E)

The probable error of correlation coefficient is denoted by:

P.E.(r) = 0.6745 * \{\frac{1 – r²}{\{sqrt{n}\} }\} = 0.6745 * S.E.(r)

Where, r = the value of correlation coefficient

n = number of pairs of observations

P.E. is used in interpretation whether the calculated value of sample correlation coefficient ® is significant or not.

• If r< P.E.(r) , then it is not significant at all i.e. , there is no evidence of correlation.
• If r> 6 P.E.(r) , it is definitely significant.
• If P.E. (r) < r < 6 P.E. (r) , nothing can be concluded.

Also, probable error of correlation coefficient may be used to determine the lower and upper limits if population correlation coefficient within which correlation coefficient of population can be expected to lie.

Limits of correlation coefficient of population = r ± P.E.(r)

Remarks:

1. P.E. may lead to conclusions particularly, when pair of observations (n) , is small.
2. P.E. is meaningful when the following conditions exist.

• The sample must have been drawn from normal population.
• The selected sample observation are obtained by random sampling method.

Example:

Correlation coefficient between two variables with the pair of 10 observations is 0.81. Discuss if the value of r be significant or not. Also determine the limits of population correlation coefficient.

Solution:

We have given,

Pair of observation (n) = 10

The value of correlation coefficient (r) = 0.81

Than, probable error of r,

P.E. (r) = 0.6745 * \{\frac{1 – r²}{\{sqrt{n}\} }\} = 0.6745* \{\frac{1 – (0.81)²}{10}\} = 0.073

Now, 6 * P.E. (r) = 6 * 0.073 = 0.440

Since, r> 6 P.E. (r), we conclude that r is significant.

Again, limits of population correlation coefficient is ,

= r ± P.E. (r) = 0.81 ±0.0733 = (0.81 – 0.0733, 0.81 + 0.0733) = (0.7367, 0.8833)

Lower limit = 0.737

Upper limit = 0.833

Example :

Compute the Karl Pearson’s coefficient of correlation from the following data by the Karl Pearson’s method

Also,

• Calculate its probable error
• Interpret if the value of r is significant or not.
• Determine the limits within which the population correlation coefficient may be expected to lie.

Solution:

Let X be the price of tea and Y be the p[rice of coffee in rupees.

Computation of correlation coefficient

Karl Pearson’s correaltioon cofficient is,

r = \{\frac{n∑UV - ∑U * ∑V}{\{sqrt{n∑U² – (∑U)²}\} . \{sqrt{ n∑V² – (∑V)²}\} }\}

= \{\frac{8 * 241 – (-11) * (-13)}{ \{sqrt{8 * 207 – (-11)²}\} . \{sqrt{8 * 295 – (-13)²}\} }\}

= \{\frac{1928 - 143}{ \{sqrt{1535}\}. \{sqrt{2191}\} }\}

= 0.9733

• Probable error of correlation coefficient is given by:

P.E. (r) = 0.6745 * \{\frac{1 – r²}{\{sqrt{n}\} }\}

=0.6745* \{\frac{1 – (0.9733)²}{\{sqrt{8}\} }\}

=0.0125

• Significance of r:

6 * P.E. (r) = 6* 0.0125

= 0.0753

Since, r is much greater than 6 * P.E. (r) , the value of r is highly significant.

• Limit of population correlation coefficient

r ± 6 * P.E. (r) =0.9733 ± 0.0753

=(0.9733 – 0.0753, 0.9733 + 0.07533)

= (0.8990, 1.048)

=(0.8980,1.0)

(References)

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