Obtain trend values for data in Problem 1 using 3-yearly moving averagesSolution: Year IMR 3 yearlymoving total 3-yearly movingaverage(trend value) 1980 10 – – 1985 7 □ 7.33 1990 5 16 □ 1995 4 12 4 - Mathematics and Statistics

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

Year IMR 3 yearly
moving total
3-yearly moving
average

(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 yearly
moving total
3-yearly moving
average

(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
Concept: Measurement of Secular Trend
  Is there an error in this question or solution?
Chapter 2.4: Time Series - Q.5

RELATED QUESTIONS

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


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_______.


Choose the correct alternative :

Which of the following is a major problem for forecasting, especially when using the method of least squares?


Choose the correct alternative :

What is a disadvantage of the graphical method of determining a trend line?


Fill in the blank :

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


State whether the following is True or False :

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


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

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 :

Obtain trend values for data in Problem 10 using 3-yearly moving averages.


Solve the following problem :

Fit a trend line to data in Problem 16 by the method of least squares.


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 simplest 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:

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


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  

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`


Complete the table using 4 yearly moving average method.

Year Production 4 yearly
moving
total
4 yearly
centered
total
4 yearly centered
moving 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
xi
10 15 20 25 30
Year 2005 2006 2007 2008 2009
Production
xi
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 xt u = t - 1979 u2 uxt
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 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 = 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 xt = `square`


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