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Question
Fill in the blank :
The method of measuring trend of time series using only averages is _______
Solution
The method of measuring trend of time series using only averages is moving average.
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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.
Obtain the trend values for the above data using 3-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.
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 :
Fit a trend line to data in Problem 4 by the method of least squares.
Solve the following problem :
Obtain trend values for data in Problem 16 using 3-yearly moving averages.
Choose the correct alternative:
Moving averages are useful in identifying ______.
The complicated but efficient method of measuring trend of time series is ______
State whether the following statement is True or False:
Moving average method of finding trend is very complicated and involves several calculations
State whether the following statement is True or False:
Least squares method of finding trend is very simple and does not involve any calculations
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 |
- Obtain trend values for the above data using 5-yearly moving averages.
- Plot the original time series and trend values obtained above on the same graph.
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 |
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`