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Solve the following problem : Obtain trend values for data in Problem 13 using 4-yearly moving averages.

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Question

Solve the following problem :

Obtain trend values for data in Problem 13 using 4-yearly moving averages.

Sum
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Solution

Construct the following table for finding 4-yearly centred moving averages.

Year
t
No. of deaths 
yt
4–yearly moving average 4–yearly moving Averages 2 unit moving total 4–yearly centred moving averages trend value
1975 0        
           
1976 6        
    17 4.25    
1977 3     9 4.5
    19 4.75    
1978 8     10.25 5.125
    22 5.5    
1979 2     11.25 5.625
    23 5.75    
19880 9     1075 5.375
    20 5    
1981 4     12 6
    28 7    
1982 5        
           
1983 10        
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Measurement of Secular Trend
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Chapter 4: Time Series - Miscellaneous Exercise 4 [Page 70]

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Balbharati Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 4 Time Series
Miscellaneous Exercise 4 | Q 4.15 | Page 70

RELATED QUESTIONS

Obtain the trend line for the above data using 5 yearly moving averages.


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.


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Index 0 2 3 3 2 4 5 6 7 10

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We can use regression line for past data to forecast future data. We then use the line which_______.


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


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


State whether the following is True or False :

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


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Year 1974 1975 1976 1977 1978 1979 1980 1981 1982
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The following table gives the production of steel (in millions of tons) for years 1976 to 1986.

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Use the method of least squares to fit a trend line to the data given below. Also, obtain the trend value for the year 1975.

Year 1962 1963 1964 1965 1966 1967 1968 1969
Production
(million barrels)
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6 8 9 9 8 7 10  

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Year 1980 1985 1990 1995
IMR 10 7 5 4
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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`


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Year Production (y) x x2 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 + bΣx2

As Σx = 0, b = `square`

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

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y = `square`


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1933 1 1939 5
1934 2 1940 1
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1936 2    

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The following table shows gross capital information (in Crore ₹) for years 1966 to 1975:

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The publisher of a magazine wants to determine the rate of increase in the number of subscribers. The following table shows the subscription information for eight consecutive years:

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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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