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प्रश्न
Solve the following problem :
Fit a trend line to the data in Problem 7 by the method of least squares.
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उत्तर
In the given problem, n = 10 even), middle t – value is 1980 and 1985, h = 5
u = `"t - mean of two middle values"/("h"/2) = ("t" - 1985.5)/(5/2) = (2("t" – 1985.5))/(5)`
We obtain the following table.
| Year t |
Percentage of enrolment yt |
u = `(2("t" – 1982.5))/(5)` | u2 | uyt | Trend Value |
| 1960 | 0 | –9 | 81 | 0 | 0.8187 |
| 1965 | 3 | –7 | 49 | –21 | 1.7701 |
| 1970 | 3 | –5 | 25 | –15 | 2.7215 |
| 1975 | 4 | –3 | 9 | –12 | 3.6729 |
| 1980 | 4 | –1 | 1 | –4 | 4.6243 |
| 1985 | 5 | 1 | 1 | 5 | 5.5757 |
| 1990 | 6 | 3 | 9 | 18 | 6.5271 |
| 1995 | 8 | 5 | 25 | 40 | 7.4785 |
| 2000 | 8 | 7 | 49 | 56 | 8.4299 |
| 2005 | 10 | 9 | 81 | 90 | 9.3813 |
| Total | 51 | 0 | 330 | 157 |
From the table, n = 10, `sumy_"t" = 51, sumu = 0, sumu^2 = 330,sumuy_"t" = 157`
The two normal equations are: `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t", = a'sumu + b'sumu^2`
∴ 51 = 10a' + b'(0) ...(i) and
157 = a'(0) + b'(330) ...(ii)
From (i), a' = `(51)/(10)` = 5.1
From (ii), b' = `(157)/(330)` = 0.4757
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 5.1 + 0.4757 u, where u = `(2("t" – 1985.5))/(5)`.
APPEARS IN
संबंधित प्रश्न
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.
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.
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.
Choose the correct alternative :
We can use regression line for past data to forecast future data. We then use the line which_______.
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 _______
State whether the following is True or False :
Least squares method of finding trend is very simple and does not involve any calculations.
Solve the following problem :
The following table shows the production of pig-iron and ferro- alloys (‘000 metric tonnes)
| Year | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 |
| Production | 0 | 4 | 9 | 9 | 8 | 5 | 4 | 8 | 10 |
Fit a trend line to the above data by graphical method.
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 |
Solve the following problem :
Fit a trend line to data in Problem 4 by the method of least squares.
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 :
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 :
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 :
Fit a trend line to data in Problem 16 by the method of least squares.
Choose the correct alternative:
Moving averages are useful in identifying ______.
The method of measuring trend of time series using only averages is ______
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
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 | 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 | – | – |
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 | – | – |
The complicated but efficient method of measuring trend of time series is ______.
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.
