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Question
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.
Solution
a. We construct the following table to obtain 5 yearly moving average:
Year 't' |
Production (millions of barrels) yt |
5-yearly moving total |
5-yearly moving averages trend value |
1962 | 0 | – | – |
1963 | 0 | – | – |
1964 | 1 | 4 | 0.8 |
1965 | 1 | 7 | 1.4 |
1966 | 2 | 11 | 2.2 |
1967 | 3 | 15 | 3.0 |
1968 | 4 | 20 | 4.0 |
1969 | 5 | 25 | 5.0 |
1970 | 6 | 30 | 6.0 |
1971 | 7 | 35 | 7.0 |
1972 | 8 | 38 | 7.6 |
1973 | 9 | 41 | 8.2 |
1974 | 8 | 44 | 8.8 |
1975 | 9 | – | – |
1976 | 10 | – | – |
b. Taking year on X-axis and production trend on Y-axis, we plot the points for production corresponding to years to get the graph of time series and plot the point for trend values corresponding to years to get the graph of trend as shown below:
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RELATED QUESTIONS
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Choose the correct alternative:
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State whether the following statement is True or False:
Moving average method of finding trend is very complicated and involves several 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
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 |
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`
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 | – | – |
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:
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 |
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`