Question

(a) Find the coefficient of correlation and obtain the lines of regression from the data given below : x : 22 26 29 30 31 31 34 35 y : 20 20 21 29 27 24 27 31​

Answer 1

I hope it's help you I can't understand

Answer 2

x : 22 26 29 30 31 31 34 35

y : 20 20 21 29 27 24 27 31​

calculating coefficient of correlation

x : 22 26 29 30 31 31 34 35 y : 20 20 21 29 27 24 27 31​    

mean of x = 29.75

mean of y = 24.875

(xi−¯x) or values of (x - mean of x) = -7.75, -3.75, -0.75, 0.25, 1.25, 1.25, 4.25, 5.25

(yi−¯y) or values of (y - mean of y) =-4.875, -4.875, -3.875, 4.125, 2.125, 0.125, 2.125,  6.125  

(xi−¯x)² = 60.0625, 14.0625, 0.5625, 0.0625, 1.5625, 1.5625, 18.0625, 27.5625

(yi−¯y)² = 23.765625, 23.765625, 15.015625, 17.015625, 4.515625, 0.765625, 4.515625, 37.515625

(xi−¯x)(yi−¯y) = 37.78125, 18.28125, 2.90625, 1.03125, 2.65625, -1.09375, 9.03125, 32.15625

Σ(xi−¯x)(yi−¯y) = 102.75

Σ(xi−¯x)² = 123.5

∑(yi−¯y)² =126.875

coefficient of correlation or r= Σ(xi−¯x)(yi−¯y)/√Σ(xi−¯x)²∑(yi−¯y)²

 = 102.75/√123.5*126.875

=0.8208

to calculate regression line

byx = n∑xy−∑x∑y/n∑x ²−(∑x) ²

bxy = n∑xy−∑x∑y/n∑y² −(∑y)²

n∑xy= 48184

∑x∑y= 47362

n∑x ²=57632

n∑y²=40616

(∑x) ²=56644

(∑y)²=39601

byx= 822/988=0.83198

bxy=822/1015 =0.8098

regression lines :

line of y on x or y−yˉ =byx (x− xˉ )

⇒ y-24.875= 0.83198(x-29.75)

⇒y-24.87=0.832x- 24.752

⇒x-(y/0.832)= -0.1478

line of x on y or (x− xˉ )=bxy​ (y− yˉ​ )

⇒(x-29.75)=0.8098( y-24.875)

⇒x-29.75= 0.8098y-24.0931

⇒x-5.65690=0.8098y