Fit a trend line to the data in Problem 4 above by the method of least squares. Also, obtain the trend value for the index of industrial production for the year 1987.

#### Solution

In the given problem, x = 10(even), two middle t – values are 1980 and 1981, h = 1

u = `"t - mean of two middle values"/("h"/(2)) = ("t" - 1980.5)/(1/2)` = 2(t – 1980.5)

We obtain the following table.

Year (t) |
Index of industrial production y_{t} |
u = 2 (t - 1980.5) |
u^{2} |
uy_{t} |
Trend value |

1976 | 0 | –9 | 81 | 0 | 0.1635 |

1977 | 2 | –7 | 49 | –14 | 1.0605 |

1978 | 3 | –5 | 25 | –15 | 1.9575 |

1979 | 3 | –3 | 9 | –9 | 2.8545 |

1980 | 2 | –1 | 1 | –2 | 3.7515 |

1981 | 4 | 1 | 1 | 4 | 4.6485 |

1982 | 5 | 3 | 9 | 15 | 5.5455 |

1983 | 6 | 5 | 25 | 30 | 6.4425 |

1984 | 7 | 7 | 49 | 49 | 7.3395 |

1985 | 10 | 9 | 81 | 90 | 8.2365 |

Total |
42 |
0 |
330 |
148 |

From the table, n = 10, `sumy_"t" = 42, sumu = 0, sumu^2 = 330, sumuy_"t" = 148`

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

∴ 42 = 10a' + b'(0) ...(i) and

148 = a'(0) + b'(330) ...(ii)

From (i), a' = `(42)/(10)` = 4.2

From (ii), b' = `(148)/(330)` = 0.4485

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

i.e., y_{t} = 4.2 + 0.4485 u, where u = 2(t – 1980.5)

∴ Now, For t = 1987, u = 2(1987 – 1980.5) = 2 x 6.5 = 13

∴ y_{t} = 4.2 + 0.4485 x 13 = 10.0305.