Advertisements
Advertisements
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 |
Advertisements
Solution
In the given problem, n = 15 (odd), middle t – values is 1969, h = 1
u = `("t" - "middle value")/"h"`
= `("t" - 1969)/1`
= t – 1969
We obtain the following table:
Year |
Production y_{t} |
u = t − 1969 | u^{2} | uy_{t} | Trend Value |
1962 | 0 | − | 49 | 0 | − 0.6 |
1963 | 0 | − 6 | 36 | 0 | 0.2 |
1964 | 1 | − 5 | 25 | − 5 | 1 |
1965 | 1 | − 4 | 16 | − 4 | 1.8 |
1966 | 2 | − 3 | 9 | − 6 | 2.6 |
1967 | 3 | − 2 | 4 | − 6 | 3.4 |
1968 | 4 | − 1 | 1 | − 4 | 4.2 |
1969 | 5 | 0 | 0 | 0 | 5 |
1970 | 6 | 1 | 1 | 6 | 5.8 |
1971 | 8 | 2 | 4 | 16 | 6.6 |
1972 | 9 | 3 | 9 | 27 | 7.4 |
1973 | 9 | 4 | 16 | 36 | 8. |
1974 | 8 | 5 | 25 | 40 | 9 |
1975 | 9 | 6 | 36 | 54 | 9.8 |
1976 | 10 | 7 | 49 | 70 | 10.6 |
Total | 75 | 0 | 280 | 224 |
From the table, n = 15, ∑y_{t} = 75, ∑u = 0, ∑u^{2} = 280, ∑uy_{t} = 224
The two normal equations are:
∑y_{t} = na' + b'∑u and ∑uy_{t} = a' ∑u + b'∑u^{2}
∴ 75 = 15a' + b'(0) ......(i)
and
224 = a′(0) + b′(280) .....(ii)
From (i), a′ = `75/15` = 5
From (ii), b′= `224/280` = 0.8
∴ The equation of the trend line is y_{t} = a′ + b′u
i.e., y_{t} = 5 + 0.8 u, where u = t – 1969
Now, for t = 1975, u = 1975 – 1969 = 6
∴ y_{t} = 5 + 0.8 × 6 = 9.8
RELATED QUESTIONS
Obtain the trend values for the above data using 3-yearly moving averages.
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 _______
Solve the following problem :
Following data shows the number of boxes of cereal sold in years 1977 to 1984.
Year | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 |
No. of boxes in ten thousand | 1 | 0 | 3 | 8 | 10 | 4 | 5 | 8 |
Fit a trend line to the above data by graphical method.
Solve the following problem :
Fit a trend line to data by the method of least squares.
Year | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 |
Number of boxes (in ten thousands) | 1 | 0 | 3 | 8 | 10 | 4 | 5 | 8 |
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 19 using 3-yearly moving averages.
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:
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
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 |
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.
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 | x^{2} | 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 + bΣx^{2}
As, Σx = 0, b =`square`
∴ The equation of trend line is y = `square`
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 | – | – |
Fit equation of trend line for the data given below.
Year | Production (y) | x | x^{2} | 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 + bΣx^{2}
As Σx = 0, b = `square`
Substitute values of a and b in equation (i) the equation of trend line is `square`
To find trend value for the year 2016, put x = `square` in the above equation.
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 | – | – |
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 complicated but efficient method of measuring trend of time series is ______.
Fit a trend line to the following data by the method of least square :
Year | 1980 | 1985 | 1990 | 1995 | 2000 | 2005 | 2010 |
IMR | 10 | 7 | 5 | 4 | 3 | 1 | 0 |