Question

What is the correlation coefficient with the following data points: (2,47), (3,2), (5,26)?

Answer 1

Answer:

The correlation coefficient is:

    -0.290742

Step-by-step explanation:

The formula for the correlation coefficient is given by :

$r=\dfrac{\sum{XY}}{\sqrt{\sum{X^2}\sum{Y^2}}}---------(1)$

where,

$X=x-x'\\and\\Y=y-y'$

where x' and y' are the mean of x and y entries respectively.

Now,

     x         y         X        Y          XY          X^2        Y^2

    2       47       -4/3     22       -88/3        16/9       484

    3        2        -1/3     -23        23/3        1/9         529

    5       26      5/3        1           5/3         25/9         1

-----------------------------------------------------------------------------------

  ∑XY= -20

 ∑X^2=42/9    

 ∑Y^2=1014

( Since,

$x'=\dfrac{2+3+5}{3}\\\\x'=\dfrac{10}{3}$

and,

$y'=\dfrac{47+2+26}{3}\\\\y'=\dfrac{75}{3}\\\\y'=25$  )

Hence,on putting all the values in equation (1) we get:

           r= -0.290742

Answer 2

The correlation coefficient is given by $\frac{S_{xy}}{\sqrt{S_{xx} S_{yy} } }$

$S_{xy}=$∑$xy- \frac{∑x}{n}}$ =$S_{xy}= (2*47)(3*7)(5*26)- \frac{(10*75)}{3}$=

$S_{xx}=(2^{2}+{3^{2}+5^{2})- \frac{(2+3+5)^{2} }{3} = 34.7$

$S_{yy}=( 47^{2}+2^{2}+26^{2} )$$- \frac{(47+2+26)^{2} }{3}$

$S_{xy}$\frac{250}{ \sqrt{34.7+2884} }= 0.79 [/tex]