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

The correlation coefficient is given by [tex] \frac{S_{xy}}{\sqrt{S_{xx} S_{yy} } } [/tex]

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

[tex]S_{xx}=(2^{2}+{3^{2}+5^{2})- \frac{(2+3+5)^{2} }{3} = 34.7 [/tex]

[tex]S_{yy}=( 47^{2}+2^{2}+26^{2} )[/tex][tex]- \frac{(47+2+26)^{2} }{3} [/tex]

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

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virtuematane virtuematane · Answer 2 · 2019-04-23T09:34:32+02:00

Answer:

The correlation coefficient is:

-0.290742

Step-by-step explanation:

The formula for the correlation coefficient is given by :

[tex]r=\dfrac{\sum{XY}}{\sqrt{\sum{X^2}\sum{Y^2}}}---------(1)[/tex]

where,

[tex]X=x-x'\\and\\Y=y-y'[/tex]

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,

[tex]x'=\dfrac{2+3+5}{3}\\\\x'=\dfrac{10}{3}[/tex]

and,

[tex]y'=\dfrac{47+2+26}{3}\\\\y'=\dfrac{75}{3}\\\\y'=25[/tex] )

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

r= -0.290742