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

Answer:

r=0.4 (Same value)

Step-by-step 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:

$0.4=r=\frac{b(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2-(\sum y)^2]}}$

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:

$r_f=\frac{2b(\sum xy)-2(\sum x)(\sum y)}{\sqrt{4[n\sum x^2 -(\sum x)^2][n\sum y^2-(\sum y)^2]}}$

And symplyfing we see this:

$r_f=2\frac{b(\sum xy)-(\sum x)(\sum y)}{2\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2-(\sum y)^2]}}$

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

$r_f=\frac{b(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2-(\sum y)^2]}}=0.4$

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