**The following table gives the aptitude test scores and productivity indices of 10 workers selected at random.**

Aptitude score (X) |
60 | 62 | 65 | 70 | 72 | 48 | 53 | 73 | 65 | 82 |

Productivity Index (Y) |
68 | 60 | 62 | 80 | 85 | 40 | 52 | 62 | 60 | 81 |

Obtain the two regression equations and estimate the productivity index of a worker whose test score is 95.

#### Solution

Here, X = Aptitude score, Y = Productivity index

X = xi | Y =yi | `"x"_"i" - bar"x"` | `bar"y"_"i" - bar"y"` | `("x"_"i" - bar"x")^2` | `("y"_"i" - bar"y")^2` | `("x"_"i" - bar"x")("y"_"i" - bar"y")` |

60 | 68 | -5 | 3 | 25 | 9 | -15 |

62 | 60 | -3 | -5 | 9 | 25 | 15 |

65 | 62 | 0 | -3 | 0 | 9 | 0 |

70 | 80 | 5 | 15 | 25 | 225 | 75 |

72 | 85 | 7 | 20 | 49 | 400 | 140 |

48 | 40 | -17 | -25 | 289 | 625 | 425 |

53 | 52 | -12 | -13 | 144 | 169 | 156 |

73 | 62 | 8 | -3 | 64 | 9 | -24 |

65 | 60 | 0 | -5 | 0 | 25 | 0 |

82 | 81 | 17 | 16 | 289 | 256 | 272 |

650 |
650 |
- |
- |
894 |
1752 |
1044 |

From the table, we have

n = 10, ∑ x_{i} = 650, ∑ y_{i} = 650

∴ `bar"x" = (sum "x"_"i")/"n" = 650/10 = 65`

`bar"y" = (sum "y"_"i")/"n" = 650/10 = 65`

Since the mean of X and Y are whole numbers, we will use the formula

`"b"_"YX" = (sum ("x"_"i" - bar"x")("y"_"i" - bar"y"))/(sum("x"_"i" - bar"x")^2) and "b"_"XY" = (sum ("x"_"i" - bar"x")("y"_"i" - bar"y"))/(sum("y"_"i" - bar"y")^2)`

From the table, we have

`sum ("x"_"i" - bar"x")("y"_"i" - bar"y") = 1044, sum ("x"_"i" - bar"x")^2 = 894, sum ("y"_"i" - bar"y") = 1752`

`"b"_"YX" = (sum ("x"_"i" - bar"x")("y"_"i" - bar"y"))/(sum("x"_"i" - bar"x")^2) = 1044/894 = 1.16`

Now, `"a" = bar"y" - "b"_"YX" bar"x"`

= 65 - 1.16 × 65 = 65 - 75.4 = - 10.4

∴ The regression equation of productivity index (Y) on Aptitude score (X) is

Y = a + b_{YX} X

∴ Y = - 10.4 + 1.16 X

For X = 95,

Y = - 10.4 + 1.16(95) = - 10.4 + 110.2 = 99.8

∴ The productivity index of worker with a test score of 95 is 99.8.