Sum

**Solve the following problem :**

Fit a trend line to data in Problem 16 by the method of least squares.

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#### Solution

In the given problem, n = 7 (odd), middle t – value is 1995, h = 5

u = `"t - middle value"/"h" = ("t" - 1995)/(1)`

We obtain the following table.

Year t |
Infant mortality rate y_{t} |
u = `("t" - 1995)/(5)` |
u^{2} |
uy_{t} |
Trend Value |

1980 | 10 | –3 | 9 | –30 | 8.9999 |

1985 | 7 | –2 | 4 | –14 | 7.4285 |

1990 | 5 | –1 | 1 | –5 | 5.8571 |

1995 | 4 | 0 | 0 | 0 | 4.2857 |

2000 | 3 | 1 | 1 | 3 | 2.7143 |

2005 | 1 | 2 | 4 | 2 | 1.1429 |

2010 | 0 | 3 | 9 | 0 | –0.4285 |

Total |
30 |
0 |
28 |
–44 |

From the table, n = 7, `sumy_"t" = 10, sumu = 0, sumu^2 = 28,sumuy_"t" = – 44`

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

∴ 30 = 7a' + b'(0) ...(i) and

– 44 = a'(0) + b'(28) ...(ii)

From (i), a' = `(30)/(7)` = 4.2857

From (ii), b' = `(-44)/(28)` = 1.5714

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

i.e., y_{t }= 4.2857 – 1.5714 u, where u = `("t" - 1995)/(5)`.

Concept: Measurement of Secular Trend

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