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प्रश्न
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 |
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उत्तर
In the given problem, n = 8 (even), two middle t – values are 1980 and 1981, h – 1
u = `"t - mean of two middle values"/("h"/2) = ("t" - 1980.5)/(1/2)` = 2(t – 1980.5)
We obtain the following table.
| Year t |
No. of boxes (in ten thousands) yt |
u = 2(t – 1980.5) | u2 | uyt | Trend Value |
| 1977 | 1 | –7 | 49 | –7 | 1.5836 |
| 1978 | 0 | –5 | 25 | 0 | 2.5240 |
| 1979 | 3 | –3 | 9 | –9 | 3.4644 |
| 1980 | 8 | –1 | 1 | –8 | 4.4048 |
| 1981 | 10 | 1 | 1 | 10 | 5.3452 |
| 1982 | 4 | 3 | 9 | 12 | 6.2856 |
| 1983 | 5 | 5 | 25 | 25 | 7.2260 |
| 1984 | 8 | 7 | 49 | 56 | 8.1664 |
| Total | 39 | 0 | 168 | 79 |
From the table, n = 8, `sumy_"t" = 39, sumu = 0, sumu^2 = 168,sumuy_"t" = 79`
The two normal equations are: `sumy_"t" = "na"' + "b"' sumu "and" sumuy_"t", = a'sumu + b'sumu^2`
∴ 39 = 8a' + b'(0) ...(i) and
79 = a'(0) + b'(168) ...(ii)
From (i), a' = `(39)/(8)` = 4.875
From (ii), b' = `(79)/(168)` = 0.4702
∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 4.875 + 0.4702 u, where u = 2(t – 1980.5).
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संबंधित प्रश्न
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.
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We can use regression line for past data to forecast future data. We then use the line which_______.
Choose the correct alternative :
Which of the following is a major problem for forecasting, especially when using the method of least squares?
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 _______
Fill in the blank :
The complicated but efficient method of measuring trend of time series is _______.
State whether the following is True or False :
Graphical method of finding trend is very complicated and involves several calculations.
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Solve the following problem :
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| Year | 1960 | 1965 | 1970 | 1975 | 1980 | 1985 | 1990 | 1995 | 2000 | 2005 |
| Percentage | 0 | 3 | 3 | 4 | 4 | 5 | 6 | 8 | 8 | 10 |
Fit a trend line to the above data by graphical method.
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 :
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 :
Fit a trend line to data in Problem 13 by the method of least squares.
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 16 using 3-yearly moving averages.
The complicated but efficient method of measuring trend of time series is ______
The method of measuring trend of time series using only averages is ______
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
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`
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 publisher of a magazine wants to determine the rate of increase in the number of subscribers. The following table shows the subscription information for eight consecutive years:
| Years | 1976 | 1977 | 1978 | 1979 |
| No. of subscribers (in millions) |
12 | 11 | 19 | 17 |
| Years | 1980 | 1981 | 1982 | 1983 |
| No. of subscribers (in millions) |
19 | 18 | 20 | 23 |
Fit a trend line by graphical method.
