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Solve the following problem : Following data shows the number of boxes of cereal sold in years 1977 to 1984.Fit a trend line to the above data by graphical method. - Mathematics and Statistics

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Question

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

Fit a trend line to the above data by graphical method.

Graph

Solution

Taking year on X-axis and number of boxes on Y-axis, we plot the points for number of boxes corresponding to years. Joining these points, we get the graph of the time series. We fit the trend line as shown in the following graph.

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Measurement of Secular Trend
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Chapter 4: Time Series - Miscellaneous Exercise 4 [Page 69]

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RELATED QUESTIONS

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


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Year 1962 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976
Production
(Million Barrels)
0 0 1 1 2 3 4 5 6 7 8 9 8 9 10

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


The simplest method of measuring trend of time series is ______.


State whether the following is True or False :

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


State whether the following is True or False :

Least squares method of finding trend is very simple and does not involve any calculations.


State whether the following is True or False :

All the three methods of measuring trend will always give the same results.


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Production 0 4 9 9 8 5 4 8 10

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Year 1971 1972 1973 1974 1975 1976
Production 1 0 1 2 3 2
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Production 3 6 5 1 4 10

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Least squares method of finding trend is very simple and does not involve any calculations


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IMR 10 7 5 4
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IMR 3 1 0  

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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 x2 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 + bΣx2

As, Σx = 0, b =`square`

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


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

Year 2000 2001 2002 2003 2004
Production
xi
10 15 20 25 30
Year 2005 2006 2007 2008 2009
Production
xi
35 40 45 50 55

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1931 1 1937 8
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1933 1 1939 5
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1935 3 1941 4
1936 2    

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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 xt t u = t - 5 u2 u.xt
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 xt =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

xt = `square`


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