Question

The manager of a company wants to find a measure which he can use to fix the monthly wages of persons applying for a job in the production department. As an experimental project, he collected data of 7 persons from that department referring to years of service and their monthly income : 

Years of service	11	7	9	5	8	6	10
Income (` in thousands)	10	8	6	5	9	7	11

Find regression equation of income on the years of service. 

Answer 1

For the given data we have the following table 

Year of service	income	xy	`X^2`	`Y^2`
11	10	110	121	100
7	8	56	49	64
9	6	54	81	36
5	5	25	25	25
8	9	72	64	81
6	7	42	36	49
10	11	110	100	121
56	56	469	476	496

Here `n =7,∑x=56,∑y=56,∑xy=469,∑x^2=476,∑y^2=496` 

Now  `barx=∑x/n=56/7=8` 

`bary=∑y/n=56/7=8`  

The regression of y on x is given by 

`b_yx=(n∑_xy-∑x∑y)/(n∑x^2-(∑x)^2)` 

=`(7xx469-56xx56)/(7xx476-(56)^2)` 

=`(3283-3136)/(3332-3136)` 

=`147/196=0.75` 

Hence the regression line of y on x is given by 

`y-bary=b_(yx)(x-barx)` 

`y-8=0.75(x-8)`

`y-8=0.75x-6.00`

`y=0.75x+2`

 or `y=3/4x+2`

⇒ `4y=3x+8`