#### 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.

#### Solution

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`