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Obtain the trend values for the following data using 5 yearly moving averages: Year 2000 2001 2002 2003 2004 Productionxi 10 15 20 25 30 Year 2005 2006 2007 2008 2009 Productionxi 35 40 45 50 55 - Mathematics and Statistics

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प्रश्न

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
सारिणी
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उत्तर

Year Production 5-Yearly
Moving
total
5-yearly
Moving
average
(Trend value)
2000 10
2001 15
2002 20 100 20
2003 25 125 25
2004 30 150 30
2005 35 175 35
2006 40 200 40
2007 45 235 45
2008 50
2009 55
shaalaa.com
Measurement of Secular Trend
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2021-2022 (March) Set 1

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संबंधित प्रश्न

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.


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

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.


Fill in the blank :

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


State whether the following is True or False :

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


Solve the following problem :

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

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

Solve the following problem :

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


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 13 using 4-yearly moving averages.


Solve the following problem :

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

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

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.


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.


Choose the correct alternative:

Moving averages are useful in identifying ______.


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: 

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


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.

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

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  

Following table shows the amount of sugar production (in lakh tonnes) for the years 1931 to 1941:

Year Production Year Production
1931 1 1937 8
1932 0 1938 6
1933 1 1939 5
1934 2 1940 1
1935 3 1941 4
1936 2    

Complete the following activity to fit a trend line by method of least squares:


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.


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`


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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