Test of Significance of Sample Correlation Coefficient using Probable Error

Probable error(P.E)

The likelihood of the correlation coefficient mistake is shown 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.

The lower and upper bounds of the population correlation coefficient that can be anticipated to exist can also be found using the likely error of correlation coefficient.

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

Remarks:

• P.E. may result in conclusions, especially when the number of observations per pair (n) is minimal.
• P.E. is meaningful when the following conditions exist.
  • The sample had to be from a healthy population.
  • The random sampling approach was used to obtain the chosen sample of observations.

Example:

The pair of 10 observations and the correlation coefficient between two variables is 0.81. Discuss the significance of the value of r. Likewise, ascertain the population correlation coefficient's boundaries.

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 :

Using the Karl Pearson's approach, determine the Karl Pearson's coefficient of correlation from the following data.

Also,

• Determine its likely mistake.
• Determine whether or not the value of r is significant.
• Determine the range of values that the population correlation coefficient is likely to fall inside.

Solution:

Let X represent the cost of tea and Y represent the cost 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