Calculate Coefficient of correlation using step deviation method: X 30 40 60 70 100 Y 90 110 140 150 160

X dx = X - A dx' dx'^2 Y dy = Y - A dy' dy'^2 dx'.dy' 30 -30 -3 9 90 -50 -5 25 15 40 -20 -2 4 110 -30 -3 9 6 60 0 0 0 140 0 0 0 0 70 10 1 1 150 10 1 1 1 100 40 4 16 160 20 2 4 8 =300 N=5 '=0 '^2=30 '=-5 '^2=39 'dy'=30 r= N. 'dy'-( ')( ') N. '^2-( ')^2( N. '^2-( ')^2 =5 30-(0)(-5) 5 30-(0)^25 39-(-5)^2 =150 150170 =150 12.24 13.03 =150 159.48 =0.94

Calculate Coefficient of correlation using step deviation method:…

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Question

Calculate Coefficient of correlation using step deviation method:

X	$30$	$40$	$60$	$70$	$100$
Y	$90$	$110$	$140$	$150$	$160$

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Answer

$X$	$dx = X - A$	$dx'$	$dx'^2$	$Y$	$dy = Y - A$	$dy'$	$dy'^2$	$dx'.dy'$
$30$	-30	-3	9	90	-50	-5	25	15
40	$-20$	-2	4	110	-30	-3	9	6
60	0	0	0	140	0	0	0	0
70	10	1	1	150	10	1	1	1
100	40	4	16	160	20	2	4	8
$\Sigma\text{X}=300\ \text{N}=5$		$\Sigma\text{dx}'=0$	$\Sigma\text{dx}'^2=30$			$\Sigma\text{dy}'=-5$	$\Sigma\text{dy}'^2=39$	$\Sigma\text{dx}'\text{dy}'=30$

$\text{r}=\frac{\text{N.}\Sigma\text{dx}'\text{dy'}-(\Sigma\text{dx}')(\Sigma\text{dy'})}{\sqrt{\text{N.}\Sigma\text{dx}'^2-(\Sigma\text{dx}')^2(\sqrt{\text{N.}\Sigma\text{dy}'^2-(\Sigma\text{dy'})^2}}}$
$=\frac{5\times30-(0)(-5)}{\sqrt{5\times30-(0)^2}\sqrt{5\times39-(-5)^2}}$
$=\frac{150}{\sqrt{150}\sqrt{170}}=\frac{150}{12.24\times13.03}=\frac{150}{159.48}=0.94$

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