I'm tasked with this question but unable to proceed on.
Q: Calculate the linear product moment correlation coefficient between x and m for these samples:
$$ \Sigma x=205,\\ \Sigma m=1240, \\ \Sigma m^2=78,320,\\ \Sigma xm=13,020, \\ \Sigma(x−\bar x)^2=126 $$
However when I was given the formula in the MF15 (http://www.seab.gov.sg/content/syllabus/alevel/2016Syllabus/ListMF15.pdf), I couldn't use $\Sigma(x−\bar x)^2$ for the second formula as in the above link.
Hence I was thinking of manipulating it but unsure of how I can do so. Would really appreciate your help. Thanks.
It is $\sum (x_i-\overline x)^2$
First you have multiply out the brackets: $\sum (x_i^2- 2\left( x_i\right)\overline x+\overline x ^2)$
Each summand in the brackets gets a sigma sign. The constants $2, \overline x$ and $\overline x^2$ can factored out.
$= \sum x_i^2-2\overline x\sum x_i+\overline x^2\sum 1$
$= \sum x_i^2-2\overline x\cdot n\cdot \overline x+n\overline x^2= \sum x_i^2-2\cdot n \overline x ^2+n\overline x^2=\sum x_i^2+n\overline x^2=\sum x_i^2-\frac1n\cdot \left( \sum x_i \right)^2$
$=126$
Now you can use the second formula straight forward.