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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.
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संबंधित प्रश्न
Obtain the trend line for the above data using 5 yearly moving averages.
Choose the correct alternative :
Which of the following is a major problem for forecasting, especially when using the method of least squares?
The simplest method of measuring trend of time series is ______.
Fill in the blank :
The method of measuring trend of time series using only averages is _______
State whether the following is True or False :
Graphical method of finding trend is very complicated and involves several calculations.
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 :
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 :
Obtain trend values for data in Problem 13 using 4-yearly moving averages.
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
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 |
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 | – | – |
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.
Following table gives the number of road accidents (in thousands) due to overspeeding in Maharashtra for 9 years. Complete the following activity to find the trend by the method of least squares.
| Year | 2008 | 2009 | 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 |
| Number of accidents | 39 | 18 | 21 | 28 | 27 | 27 | 23 | 25 | 22 |
Solution:
We take origin to 18, we get, the number of accidents as follows:
| Year | Number of accidents xt | t | u = t - 5 | u2 | u.xt |
| 2008 | 21 | 1 | -4 | 16 | -84 |
| 2009 | 0 | 2 | -3 | 9 | 0 |
| 2010 | 3 | 3 | -2 | 4 | -6 |
| 2011 | 10 | 4 | -1 | 1 | -10 |
| 2012 | 9 | 5 | 0 | 0 | 0 |
| 2013 | 9 | 6 | 1 | 1 | 9 |
| 2014 | 5 | 7 | 2 | 4 | 10 |
| 2015 | 7 | 8 | 3 | 9 | 21 |
| 2016 | 4 | 9 | 4 | 16 | 16 |
| `sumx_t=68` | - | `sumu=0` | `sumu^2=60` | `square` |
The equation of trend 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=68,sumu=0,sumu^2=60,sumux_t=-44`
Putting these values in normal equations, we get
68 = 9a' + b'(0) ...(3)
∴ a' = `square`
-44 = a'(0) + b'(60) ...(4)
∴ b' = `square`
The equation of trend line is given by
xt = `square`
