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

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

Year 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985
Index 0 2 3 3 2 4 5 6 7 10

The following table shows the index of industrial production for the period from 1976 to 1985. using the year 1976 as the base year. Obtain the trend values for the following data using 4 yearly centered moving averages:

Year 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985
Index 0 2 3 3 2 4 5 6 7 10
Chart
Sum
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Solution

Construct the following table for finding 4 yearly centered moving average

Year t Index yt 4–yearly moving total 4–yearly moving averages 2 unit
moving total
4 yearly centred moving average (trend values)
1976 0        
           
1977 2 8 2    
           
1978 3 10 2.5 4.5 2.25
           
1979 3 12 3 5.5 2.75
           
1980 2 14 3.5 6.5 3.25
           
1981 4 17 4.25 7.75 3.875
           
1982 5 22 5.5 9.75 4.875
           
1983 6 28 7 12.5 6.25
           
1984 7        
           
1985 10        
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Measurement of Secular Trend
  Is there an error in this question or solution?
Chapter 4: Time Series - Exercise 4.1 [Page 66]

RELATED QUESTIONS

Fit a trend line to the data in Problem 7 by the method of least squares. Also, obtain the trend value for the year 1990.


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 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.


Choose the correct alternative :

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


Fill in the blank :

The complicated but efficient 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.


Solve the following problem :

The following table shows the production of pig-iron and ferro- alloys (‘000 metric tonnes)

Year 1974 1975 1976 1977 1978 1979 1980 1981 1982
Production 0 4 9 9 8 5 4 8 10

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


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

Year 1974 1975 1976 1977 1978 1979 1980 1981 1982
Production 0 4 9 9 8 5 4 8 10

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Year 1974 1975 1976 1977 1978 1979 1980 1981 1982
Production 0 4 9 9 8 5 4 8 10

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The percentage of girls’ enrollment in total enrollment for years 1960-2005 is shown in the following table.

Year 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005
Percentage 0 3 3 4 4 5 6 8 8 10

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


Solve the following problem :

Fit a trend line to the data in Problem 7 by the method of least squares.


Solve the following problem :

Obtain trend values for the data in Problem 7 using 4-yearly moving averages.


Solve the following problem :

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Solve the following problem :

Following table shows the number of traffic fatalities (in a state) resulting from drunken driving for years 1975 to 1983.

Year 1975 1976 1977 1978 1979 1980 1981 1982 1983
No. of deaths 0 6 3 8 2 9 4 5 10

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


Solve the following problem :

Obtain trend values for data in Problem 16 using 3-yearly moving averages.


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Choose the correct alternative:

Moving averages are useful in identifying ______.


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


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


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)
0 0 1 1 2 3 4 5
Year 1970 1971 1972 1973 1974 1975 1976  
Production
(million barrels)
6 8 9 9 8 7 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  
  1. Obtain trend values for the above data using 5-yearly moving averages.
  2. Plot the original time series and trend values obtained above on the same graph.

Following table shows the all India infant mortality rates (per ‘000) for years 1980 to 2010

Year 1980 1985 1990 1995
IMR 10 7 5 4
Year 2000 2005 2010  
IMR 3 1 0  

Fit a trend line by the method of least squares

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`


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

Years 1966 1967 1968 1969 1970
Gross Capital information 20 25 25 30 35
Years 1971 1972 1973 1974 1975
Gross Capital information 30 45 40 55 65

Obtain trend values using 5-yearly moving values.


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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