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Sum

**Solve the following problem :**

Fit a trend line to data in Problem 4 by the method of least squares.

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#### Solution

In the given problem, n = 12 (odd), middle t – value is 1976, h = 1

u = `"t - middle value"/("h"/2) = ("t" - 1976.5)/(1/2)` = 2(t – 1976.5)

We obtain the following table.

Year t |
Production y_{t} |
u = 2(t – 1976.5) |
u^{2} |
uy_{t} |
Trend Value |

1971 | 1 | –11 | 121 | –11 | 0.0900 |

1972 | 0 | –9 | 81 | 0 | 0.6494 |

1973 | 1 | –7 | 49 | –7 | 1.2088 |

1974 | 2 | –5 | 25 | –10 | 1.7682 |

1975 | 3 | – | 9 | –9 | 2.3276 |

1976 | 2 | –1 | 1 | –2 | 2.8870 |

1977 | 3 | 1 | 1 | 3 | 3.4464 |

1978 | 6 | 3 | 9 | 18 | 4.0058 |

1979 | 5 | 5 | 25 | 25 | 4.5652 |

1980 | 1 | 7 | 49 | 7 | 5.1246 |

1981 | 4 | 9 | 81 | 36 | 5.6840 |

1982 | 10 | 11 | 121 | 110 | 6.243 |

Total |
38 |
0 |
572 |
160 |

From the table, n = 12, `sumy_"t" = 38, sumu = 0, sumu^2 = 572,sumuy_"t" = 160`

The two normal equations are: `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t", = a'sumu + b'sumu^2`

∴ 38 = 12a' + b'(0) ...(i) and

160 = a'(0) + b'(572) ...(ii)

From (i), a' = `(38)/(12)` = 3.1667

From (ii), b' = `(160)/(572)` = 0.2797

∴ The equation of the trend line is y_{t} = a' + b'u

i.e., y_{t }= 3.1667 + 0.2797 u, where u = 2(t – 1976.5).

Concept: Measurement of Secular Trend

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