Correlation and Regression Analysis

$X\\ 1\\ 2\\ 3\\ 4\\ 5\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \overline { x } \cfrac { 15 }{ 5 } =3$                $Y\\ 7\\ 6\\ 5\\ 4\\ 3\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \overline { y } =\cfrac { 25 }{ 5 } =5$                  $Y=y-\overline { y } \\ \quad 02\\ \quad 01\\ \quad 00\\ -1\\ -1\\ \ _ \ _ \ _ \ _ \ _ \ _ \\ \quad 0$                    $XY\\ -4\\ -1\\ \quad 0\\ -1\\ -4\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { XY=-10 } $                 ${ X }^{ 2 }\\ 4\\ 1\\ 0\\ 1\\ 4\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { { X }^{ 2 }=10 } $        

${ Y }^{ 2 }\\ 4\\ 1\\ 0\\ 1\\ 4\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { { Y }^{ 2 }=10 } $

Therefore, $r=\cfrac { \sum { XY }  }{ \sqrt { \sum { { X }^{ 2 }\sum { { Y }^{ 2 } }  }  }  } \\ =\cfrac { -10 }{ \sqrt { 10*10 }  } \\ =\cfrac { -10 }{ 10 } \\ =-1.$

$\sum x=125,\ \sum{{x}^{2}}=650,\ \sum{xy}=334$  $\sum y=100,\ \sum{{y}^{2}}=508$

Corrected $\sum y=100-14-6+12+8=100$

Corrected $\sum x^2=650-(6)^2-(8)^2+(6)^2+(8)^2=650$

Corrected $\sum y^2=460-(14)^2-(6)^2+(12)^2+(8)^2=436$

Corrected $\sum xy=508-6\times 18-8\times 6+8\times 12+6\times 8=520$

The formula for Pearson product moment correlation is

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

    $=\dfrac{25\times 520-125\times 100}{\sqrt{[25\times 650-(125)^2][25\times 436-(100)^2]}}$

   $=\dfrac{13000-12500}{\sqrt{[16250-15625][10900-10000]}}$

   $=\dfrac{500}{\sqrt{[625][900]}}$

   $=\dfrac{500}{(25)(30)}$

   $=\dfrac{2}{3}$

   $=0.667$

$3x+2y=25$                  ----- ( 2 )

Multiplying  equation ( 1 ) by $2$ we get,

$\Rightarrow$  $12x+2y=60$         ---- ( 3 )

Subtracting equation ( 2 ) from equation ( 3 ) we get,

$\Rightarrow$  $9x=35$

$\therefore$  $x=\dfrac{35}{9}$

Substituting value of $x$ in equation ( 1 ) we get

$\Rightarrow$  $6\times\dfrac{35}{9}+y=30$

$\Rightarrow$  $\dfrac{70}{3}+y=30$

$\Rightarrow$  $y=30-\dfrac{70}{3}$

$\Rightarrow$  $y=\dfrac{90-70}{3}$

$\therefore$  $y=\dfrac{20}{3}$

$\therefore$  $\overline{x}=\dfrac{35}{9}$ and $\overline{y}=\dfrac{20}{3}$

Now, we are going to find $r:$

$6x+y=30$                            [ Given ]

$6x=30-y$

$x=\dfrac{30}{6}-\dfrac{y}{6}$

$\Rightarrow$  $b _{xy}=\dfrac{-1}{6}$

$3x+2y=25$              [ Given ]

$2y=25-3x$

$y=\dfrac{25}{2}-\dfrac{3}{2}x$

$\Rightarrow$  $b _{yx}=\dfrac{-3}{2}$

$\Rightarrow$  Coefficient of correlation $(r)=\sqrt{b _{xy}\times b _{yx}}$

                                                        $=\sqrt{\dfrac{-1}{6}\times \dfrac{-3}{2}}$

                                                        $=\sqrt{\dfrac{1}{4}}$

                                                        $=\dfrac{1}{2}$

                                                        $=0.5$

$\therefore$  $r=0.5$