Advertisements
Advertisements
प्रश्न
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.
उत्तर
In the given problem, n = 11 (odd), middle t- values is 1981, h = 1
u = `"t - middle value"/"h" = ("t" - 1981)/(1)` = t – 1981
We obtain the following table.
| Year t |
Production yt |
u = t–1981 | u2 | uyt | Trend Value |
| 1976 | 0 | –5 | 25 | 0 | 1.6819 |
| 1977 | 4 | –4 | 16 | –16 | 2.4728 |
| 1978 | 4 | –3 | 9 | –12 | 3.2637 |
| 1979 | 2 | –2 | 4 | –4 | 4.0546 |
| 1980 | 6 | –1 | 1 | –6 | 4.8455 |
| 1981 | 8 | 0 | 0 | 0 | 5.6364 |
| 1982 | 5 | 1 | 1 | 5 | 6.4273 |
| 1983 | 9 | 2 | 4 | 18 | 7.2182 |
| 1984 | 4 | 3 | 9 | 12 | 8.0091 |
| 1985 | 10 | 4 | 16 | 40 | 8.8 |
| 1986 | 10 | 5 | 25 | 50 | 9.5909 |
| Total | 62 | 0 | 110 | 87 |
From the table, n = 11, `sumy_"t" = 62, sumu = 0, sumu^2 = 110, sumuy_"t" = 87`
The two normal equations are : `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t" = "a"' sumu + "b"'sumu^2`
∴ 62 = 11a' + b'(0) ...(i) and
87 = a'(0) + b'(110) ...(ii)
From (i), a' = `(62)/(11)` = 5.6364
From (ii), b' = `(87)/(110)` = 0.7909
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 5.6364 + 0.7909 u, where u = t – 1981
∴ Now, For t = 1990, u = 1990 – 1981= 9
∴ yt = 5.6364 + 0.7909 x 9 = 12.7545.
APPEARS IN
संबंधित प्रश्न
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 |
Fill in the blank :
The method of measuring trend of time series using only averages is _______
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 |
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 |
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 the data in Problem 7 using 4-yearly moving averages.
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.
Obtain trend values for data in Problem 19 using 3-yearly moving averages.
Choose the correct alternative:
Moving averages are useful in identifying ______.
The simplest 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
Following table shows the amount of sugar production (in lac tons) for the years 1971 to 1982
| 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 |
Fit a trend line by the method of least squares
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 | 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`
Fit equation of trend line for the data given below.
| Year | Production (y) | x | x2 | 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 + bΣx2
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`
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 ______.
