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Obtain trend values for data, using 3-yearly moving averages

Solution:

Year |
IMR |
3 yearlymoving total |
3-yearly movingaverage (trend value) |

1980 | 10 | – | – |

1985 | 7 | `square` | 7.33 |

1990 | 5 | 16 | `square` |

1995 | 4 | 12 | 4 |

2000 | 3 | 8 | `square` |

2005 | 1 | `square` | 1.33 |

2010 | 0 | – | – |

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#### Solution

Year |
IMR |
3 yearlymoving total |
3-yearly movingaverage (trend value) |

1980 | 10 | – | – |

1985 | 7 | 22 |
7.33 |

1990 | 5 | 16 | 5.33 |

1995 | 4 | 12 | 4 |

2000 | 3 | 8 | 2.67 |

2005 | 1 | 4 |
1.33 |

2010 | 0 | – | – |

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Production(Million Barrels) |
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1962 | 1963 | 1964 | 1965 | 1966 | 1967 | 1968 | 1969 |

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**Solution: **Let us fit equation of trend line for above data.

Let the equation of trend line be y = a + bx .....(i)

Here n = 7(odd), middle year is `square` and h = 5

Year |
IMR (y) |
x |
x^{2} |
x.y |

1980 | 10 | – 3 | 9 | – 30 |

1985 | 7 | – 2 | 4 | – 14 |

1990 | 5 | – 1 | 1 | – 5 |

1995 | 4 | 0 | 0 | 0 |

2000 | 3 | 1 | 1 | 3 |

2005 | 1 | 2 | 4 | 2 |

2010 | 0 | 3 | 9 | 0 |

Total |
30 |
0 |
28 |
– 44 |

The normal equations are

Σy = na + bΣx

As, Σx = 0, a = `square`

Also, Σxy = aΣx + bΣx^{2}

As, Σx = 0, b =`square`

∴ The equation of trend line is y = `square`

Complete the table using 4 yearly moving average method.

Year |
Production |
4 yearly moving total |
4 yearly centered total |
4 yearly centeredmoving average(trend values) |

2006 | 19 | – | – | |

`square` | ||||

2007 | 20 | – | `square` | |

72 | ||||

2008 | 17 | 142 | 17.75 | |

70 | ||||

2009 | 16 | `square` | 17 | |

`square` | ||||

2010 | 17 | 133 | `square` | |

67 | ||||

2011 | 16 | `square` | `square` | |

`square` | ||||

2012 | 18 | 140 | 17.5 | |

72 | ||||

2013 | 17 | 147 | 18.375 | |

75 | ||||

2014 | 21 | – | – | |

– | ||||

2015 | 19 | – | – |

**Obtain the trend values for the following data using 5 yearly moving averages:**

Year |
2000 |
2001 |
2002 |
2003 |
2004 |

Production x _{i} |
10 | 15 | 20 | 25 | 30 |

Year |
2005 |
2006 |
2007 |
2008 |
2009 |

Production x _{i} |
35 | 40 | 45 | 50 | 55 |

**Complete the following activity to fit a trend line to the following data by the method of least squares.**

Year |
1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 |

Number of deaths |
0 | 6 | 3 | 8 | 2 | 9 | 4 | 5 | 10 |

**Solution:**

Here n = 9. We transform year t to u by taking u = t - 1979. We construct the following table for calculation :

Year t |
Number of deaths x_{t} |
u = t - 1979 |
u^{2} |
ux_{t} |

1975 | 0 | - 4 | 16 | 0 |

1976 | 6 | - 3 | 9 | - 18 |

1977 | 3 | - 2 | 4 | - 6 |

1978 | 8 | - 1 | 1 | - 8 |

1979 | 2 | 0 | 0 | 0 |

1980 | 9 | 1 | 1 | 9 |

1981 | 4 | 2 | 4 | 8 |

1982 | 5 | 3 | 9 | 15 |

1983 | 10 | 4 | 16 | 40 |

`sumx_t` =47 | `sumu`=0 | `sumu^2=60` | `square` |

The equation of trend line is x_{t}= a' + b'u.

The normal equations are,

`sumx_t = na^' + b^' sumu` ...(1)

`sumux_t = a^'sumu + b^'sumu^2` ...(2)

Here, n = 9, `sumx_t = 47, sumu= 0, sumu^2 = 60`

By putting these values in normal equations, we get

47 = 9a' + b' (0) ...(3)

40 = a'(0) + b'(60) ...(4)

From equation (3), we get a' = `square`

From equation (4), we get b' = `square`

∴ the equation of trend line is x_{t} = `square`