Answer to Question #350586 in Statistics and Probability for sergu

Question

9. The data for the car rental company is given:

Company                    Cars                Revenue

   A                            63                   7

   B                            29                   3.9

   C                            20.8                2.1

   D                           19.1                2.8

   E                            13.4                1.4

   F                             8.5                  1.5

Compute the linear correlation coefficient for the data and find the equation of regression line.

Answer 1

$X=[ 63, 29, 20.8, 19.1, 13.4, 8.5]$

$Y=[7, 3.9, 2.1, 2.8, 1.4, 1.5]$

$n=6$

1. Linear correlation coefficient $r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2-(\sum x)^2]\cdot[n\sum y^2-(\sum y)^2]}}$

$\sum x=63+29+20.8+19.1+13.4+8.5=153.8$

$\sum y = 7+3.9+2.1+2.8+1.4+1.5=18.7$

$\sum x^2=63^2+29^2+20.8^2+19.1^2+13.4^2+8.5^2=5859.26$

$\sum y^2= 7^2+3.9^2+2.1^2+2.8^2+1.4^2+1.5^2=80.67$

$\sum xy=63\cdot7+29\cdot3.9+20.8\cdot2.1+19.1\cdot2.8+13.4\cdot1.4+8.5\cdot1.5=682.77$

$(\sum x)^2=153.8^2=23654.44$

$(\sum y)^2=18.7^2=349.69$

$(\sum x)(\sum y)=153.8\cdot18.7=2876.06$

$r=\frac{6\cdot682.77-2876.06}{\sqrt{[6\cdot5859.26-23654.44]\cdot[6\cdot80.67-349.69]}}=0.9819798$

2. Equation of regression line $y=a+bx$,

where $a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}$, $b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}$.

$a=\frac{18.7\cdot5859.26-153.8\cdot682.77}{6\cdot5859.26-23654.44}=0.39632105$

$b=\frac{6\cdot682.77-2876.06}{6\cdot5859.26-23654.44}=0.10612531$

$y=0.39632105+0.10612531x$