The formula for Karl's Pearson's cCoefficient is:
r = n ∑ i = 0 n x i y i − ( ∑ i = 0 n x i ) ( ∑ i = 0 n y i ) [ n ∑ i = 0 n x i 2 − ( ∑ i = 0 n x i ) 2 ] [ n ∑ i = 0 n y i 2 − ( ∑ i = 0 n y i ) 2 ] r=\frac{n\sum_{i=0}^{n}x_iy_i-(\sum_{i=0}^{n}x_i)(\sum_{i=0}^{n}y_i)}{\sqrt{[n\sum_{i=0}^{n}x^2_i-(\sum_{i=0}^{n}x_i)^2][n\sum_{i=0}^{n}y^2_i-(\sum_{i=0}^{n}y_i)^2]}} r = [ n ∑ i = 0 n x i 2 − ( ∑ i = 0 n x i ) 2 ] [ n ∑ i = 0 n y i 2 − ( ∑ i = 0 n y i ) 2 ] n ∑ i = 0 n x i y i − ( ∑ i = 0 n x i ) ( ∑ i = 0 n y i )
Since there are 4 values fo y and 5 for x, we assume that the last value is for y 0.
We set n=5 and receive:
n ∑ i = 0 n x i y i = 5 ⋅ ( 10 ⋅ 5 + 10 ⋅ 6 + 11 ⋅ 4 + 12 ⋅ 3 + 12 ⋅ 0 ) = 950 n\sum_{i=0}^nx_iy_i=5\cdot(10\cdot5+10\cdot6+11\cdot4+12\cdot3+12\cdot0)=950 n ∑ i = 0 n x i y i = 5 ⋅ ( 10 ⋅ 5 + 10 ⋅ 6 + 11 ⋅ 4 + 12 ⋅ 3 + 12 ⋅ 0 ) = 950
( ∑ i = 0 n x i ) ( ∑ i = 0 n y i ) = ( 10 + 10 + 11 + 12 + 12 ) ( 5 + 6 + 4 + 3 + 0 ) = 55 ⋅ 18 = 990 (\sum_{i=0}^{n}x_i)(\sum_{i=0}^{n}y_i)=(10+10+11+12+12)(5+6+4+3+0)=55\cdot18=990 ( ∑ i = 0 n x i ) ( ∑ i = 0 n y i ) = ( 10 + 10 + 11 + 12 + 12 ) ( 5 + 6 + 4 + 3 + 0 ) = 55 ⋅ 18 = 990
n ∑ i = 0 n x i 2 − ( ∑ i = 0 n x i ) 2 = 5 ( 1 0 2 + 1 0 2 + 1 1 2 + 1 2 2 + 1 2 2 ) − ( 10 + 10 + 11 + 12 + 12 ) 2 = n\sum_{i=0}^{n}x^2_i-(\sum_{i=0}^{n}x_i)^2=5(10^2+10^2+11^2+12^2+12^2)-(10+10+11+12+12)^2= n ∑ i = 0 n x i 2 − ( ∑ i = 0 n x i ) 2 = 5 ( 1 0 2 + 1 0 2 + 1 1 2 + 1 2 2 + 1 2 2 ) − ( 10 + 10 + 11 + 12 + 12 ) 2 =
= 20 =20 = 20
n ∑ i = 0 n y i 2 − ( ∑ i = 0 n y i ) 2 = 5 ( 5 2 + 6 2 + 4 2 + 3 2 + 0 2 ) − ( 5 + 6 + 4 + 3 + 0 ) 2 = 106 n\sum_{i=0}^{n}y^2_i-(\sum_{i=0}^{n}y_i)^2=5(5^2+6^2+4^2+3^2+0^2)-(5+6+4+3+0)^2=106 n ∑ i = 0 n y i 2 − ( ∑ i = 0 n y i ) 2 = 5 ( 5 2 + 6 2 + 4 2 + 3 2 + 0 2 ) − ( 5 + 6 + 4 + 3 + 0 ) 2 = 106
r = 950 − 990 20 ⋅ 106 = − 0.8687 r=\frac{950-990}{\sqrt{20\cdot106}}=-0.8687 r = 20 ⋅ 106 950 − 990 = − 0.8687
Answer: -0.8687
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