Question

A set of n = 15 pairs of scores (X and Y values) produces a correlation of r = 0.40. If each of the X values is multiplied by 2 and the correlation is computed for the new scores, what value will be obtained for the new correlation?a. r = 0.20b. r = 0.40c. r = 0.80 (my incorrect answer)d. cannot be determined without knowing all the X and Y scores

Answer 1

r=0.4 (Same value)

Explanation:

The correlation coefficient is unaffected by the scale of relation.

Correlation is a "statistical measure that indicates the extent to which two or more variables fluctuate together". And is always between -1 and 1. 1 indicates perfect linear relationship and -1 perfect inverse linear relationship. The formula for the correlation is given by:

`0.4=r=(b(\sum xy)-(\sum x)(\sum y))/(√([n\sum x^2 -(\sum x)^2][n\sum y^2-(\sum y)^2]))`

Applying this formula we got that the correlation coeffcient it's 0.4. Now if we multiply all the x values by 2 we have this:

`r_f=(2b(\sum xy)-2(\sum x)(\sum y))/(√(4[n\sum x^2 -(\sum x)^2][n\sum y^2-(\sum y)^2]))`

And symplyfing we see this:

`r_f=2(b(\sum xy)-(\sum x)(\sum y))/(2√([n\sum x^2 -(\sum x)^2][n\sum y^2-(\sum y)^2]))`

We can cancel the 2 on the numerator and denominator and we got the same formula equal to 0.4.

`r_f=(b(\sum xy)-(\sum x)(\sum y))/(√([n\sum x^2 -(\sum x)^2][n\sum y^2-(\sum y)^2]))=0.4`

So for this reason the correlation coefficient it's not affected by scale changes on the independent or dependent variables.