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Fit equation of trend line for the data given below.

Year |
Production (y) |
x |
x^{2} |
xy |

2006 | 19 | – 9 | 81 | – 171 |

2007 | 20 | – 7 | 49 | – 140 |

2008 | 14 | – 5 | 25 | – 70 |

2009 | 16 | – 3 | 9 | – 48 |

2010 | 17 | – 1 | 1 | – 17 |

2011 | 16 | 1 | 1 | 16 |

2012 | 18 | 3 | 9 | 54 |

2013 | 17 | 5 | 25 | 85 |

2014 | 21 | 7 | 49 | 147 |

2015 | 19 | 9 | 81 | 171 |

Total |
177 |
0 |
330 |
27 |

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

Here n = `square` (even), two middle years are `square` and 2011, and h = `square`

The normal equations are Σy = na + bΣx

As Σx = 0, a = `square`

Also, Σxy = aΣx + bΣx^{2}

As Σx = 0, b = `square`

Substitute values of a and b in equation (i) the equation of trend line is `square`

To find trend value for the year 2016, put x = `square` in the above equation.

y = `square`

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#### Solution

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

Here n = **10** (even), two middle years are **2010** and 2011, and h = **2**

The normal equations are Σy = na + bΣx

As Σx = 0,

∴ 177 = 10a + b(0)

∴ a = `177/10`

a = **17.7**** **

Also, Σxy = aΣx + bΣx^{2}

As Σx = 0,

27 = a(0) + b(330)

∴ b = `27/330`

= 0.08

b = **0.1**

Substitute values of a and b in equation (i) the equation of trend line is **y = 17.7 + 0.1x**

To find trend value for the year 2016, put x = **11** in the above equation.

y = 17.7 + 1.1

y = **18.8**

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#### RELATED QUESTIONS

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**Choose the correct alternative :**

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**Fill in the blank :**

The method of measuring trend of time series using only averages is _______

**State whether the following is True or False :**

Moving average method of finding trend is very complicated and involves several calculations.

**State whether the following is True or False :**

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**Solve the following problem :**

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Year |
1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |

Production |
0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 10 |

Obtain trend values for the following data using 4-yearly centered moving averages.

Year |
1971 | 1972 | 1973 | 1974 | 1975 | 1976 |

Production |
1 | 0 | 1 | 2 | 3 | 2 |

Year |
1977 | 1978 | 1979 | 1980 | 1981 | 1982 |

Production |
3 | 6 | 5 | 1 | 4 | 10 |

**Solve the following problem :**

Following data shows the number of boxes of cereal sold in years 1977 to 1984.

Year |
1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 |

No. of boxes in ten thousand |
1 | 0 | 3 | 8 | 10 | 4 | 5 | 8 |

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**Solve the following problem :**

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**Solve the following problem :**

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**Solve the following problem :**

Following tables shows the wheat yield (‘000 tonnes) in India for years 1959 to 1968.

Year |
1959 | 1960 | 1961 | 1962 | 1963 | 1964 | 1965 | 1966 | 1967 | 1968 |

Yield |
0 | 1 | 2 | 3 | 1 | 0 | 4 | 1 | 2 | 10 |

Fit a trend line to the above data by the method of least squares.

The complicated but efficient method of measuring trend of time series is ______

The method of measuring trend of time series using only averages is ______

**State whether the following statement is True or False:**

The secular trend component of time series represents irregular variations

**State whether the following statement is True or False:**

Moving average method of finding trend is very complicated and involves several calculations

Following table shows the amount of sugar production (in lac tons) for the years 1971 to 1982

Year |
1971 | 1972 | 1973 | 1974 | 1975 | 1976 |

Production |
1 | 0 | 1 | 2 | 3 | 2 |

Year |
1977 | 1978 | 1979 | 1980 | 1981 | 1982 |

Production |
4 | 6 | 5 | 1 | 4 | 10 |

Fit a trend line by the method of least squares

Obtain trend values for data, using 4-yearly centred moving averages

Year |
1971 | 1972 | 1973 | 1974 | 1975 | 1976 |

Production |
1 | 0 | 1 | 2 | 3 | 2 |

Year |
1977 | 1978 | 1979 | 1980 | 1981 | 1982 |

Production |
4 | 6 | 5 | 1 | 4 | 10 |

The following table gives the production of steel (in millions of tons) for years 1976 to 1986.

Year |
1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 | 1985 | 1986 |

Production |
0 | 4 | 4 | 2 | 6 | 8 | 5 | 9 | 4 | 10 | 10 |

Obtain the trend value for the year 1990

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

Year |
1976 | 1977 | 1978 | 1979 | 1980 | 1981 |

Production |
0 | 4 | 4 | 2 | 6 | 8 |

Year |
1982 | 1983 | 1984 | 1985 | 1986 | |

Production |
5 | 9 | 4 | 10 | 10 |

The following table shows the production of gasoline in U.S.A. for the years 1962 to 1976.

Year |
1962 | 1963 | 1964 | 1965 | 1966 | 1967 | 1968 | 1969 |

Production (million barrels) |
0 | 0 | 1 | 1 | 2 | 3 | 4 | 5 |

Year |
1970 | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 | |

Production (million barrels) |
6 | 7 | 8 | 9 | 8 | 9 | 10 |

- Obtain trend values for the above data using 5-yearly moving averages.
- Plot the original time series and trend values obtained above on the same graph.

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 | – | – |

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 |

The complicated but efficient method of measuring trend of time series is ______.

Following table gives the number of road accidents (in thousands) due to overspeeding in Maharashtra for 9 years. Complete the following activity to find the trend by the method of least squares.

Year |
2008 | 2009 | 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 |

Number of accidents |
39 | 18 | 21 | 28 | 27 | 27 | 23 | 25 | 22 |

**Solution:**

We take origin to 18, we get, the number of accidents as follows:

Year |
Number of accidents x_{t} |
t |
u = t - 5 |
u^{2} |
u.x_{t} |

2008 | 21 | 1 | -4 | 16 | -84 |

2009 | 0 | 2 | -3 | 9 | 0 |

2010 | 3 | 3 | -2 | 4 | -6 |

2011 | 10 | 4 | -1 | 1 | -10 |

2012 | 9 | 5 | 0 | 0 | 0 |

2013 | 9 | 6 | 1 | 1 | 9 |

2014 | 5 | 7 | 2 | 4 | 10 |

2015 | 7 | 8 | 3 | 9 | 21 |

2016 | 4 | 9 | 4 | 16 | 16 |

`sumx_t=68` | - | `sumu=0` | `sumu^2=60` | `square` |

The equation of trend 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=68,sumu=0,sumu^2=60,sumux_t=-44`

Putting these values in normal equations, we get

68 = 9a' + b'(0) ...(3)

∴ a' = `square`

-44 = a'(0) + b'(60) ...(4)

∴ b' = `square`

The equation of trend line is given by

x_{t} = `square`