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प्रश्न
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
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उत्तर
In the given problem, n = 12 (even), two middle t − values are 1976 and 1977, h = 1
u = `("t" - "mean of two middle values")/("h"/2)`
= `("t" - 1976.5)/(1/2)`
= 2(t − 1976.5)
We obtain the following table:
|
Year t |
Poduction yt |
u = 2(t − 1976.5 | u2 | uyt | Trend Value |
| 1971 | 1 | – 11 | 121 | – 11 | 0.1535 |
| 1972 | 0 | – 9 | 81 | – 0 | 0.7165 |
| 1973 | 1 | – 7 | 49 | – 7 | 1.2795 |
| 1974 | 2 | – 5 | 25 | – 10 | 1.8425 |
| 1975 | 3 | – 3 | 9 | – 9 | 2.4055 |
| 1976 | 2 | – 1 | 1 | – 2 | 2.9685 |
| 1977 | 4 | 1 | 1 | 4 | 3.5315 |
| 1978 | 6 | 3 | 9 | 18 | 4.0945 |
| 1979 | 5 | 5 | 25 | 25 | 4.6575 |
| 1980 | 1 | 7 | 49 | 7 | 5.2205 |
| 1981 | 4 | 9 | 81 | 36 | 5.7835 |
| 1982 | 10 | 11 | 121 | 110 | 6.3465 |
| Total | 39 | 0 | 572 | 161 |
From the table, n = 12, ∑yt = 39, ∑u = 0, ∑u2 = 572, ∑uyt = 161
The two normal equations are:
∑yt = na' + b'∑u and ∑uyt = a'∑u + b'∑u2
∴ 39 = 12a' + b'(0) ......(i)
and 161 = a'(0) + b'(572) ......(ii)
From (i), a′ = `39/12` = 3.25
From (ii), b′ = `161/572` = 0.2815
∴ The equation of the trend line is yt = a′ + b′u
i.e., yt = 3.25+ 0.2815 u,
where u = 2(t − 1976.5)
संबंधित प्रश्न
Obtain the trend line for the above data using 5 yearly moving averages.
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 |
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.
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 :
Graphical method of finding trend is very complicated and involves several calculations.
State whether the following is True or False :
Moving average method of finding trend is very complicated and involves several calculations.
State whether the following is True or False :
Least squares method of finding trend is very simple and does not involve any calculations.
State whether the following is True or False :
All the three methods of measuring trend will always give the same results.
Solve the following problem :
Fit a trend line to data in Problem 4 by the method of least squares.
Solve the following problem :
The percentage of girls’ enrollment in total enrollment for years 1960-2005 is shown in the following table.
| 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.
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 :
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.
Solve the following problem :
Following tables shows the wheat yield (‘000 tonnes) in India for years 1959 to 1968.
| Year | 1959 | 1960 | 1961 | 1962 | 1963 | 1964 | 1965 | 1966 | 1967 | 1968 |
| Yield | 0 | 1 | 2 | 3 | 1 | 0 | 4 | 1 | 2 | 10 |
Fit a trend line to the above data by the method of least squares.
Choose the correct alternative:
Moving averages are useful in identifying ______.
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 | – | – |
Obtain the trend values for the following data using 5 yearly moving averages:
| Year | 2000 | 2001 | 2002 | 2003 | 2004 |
| Production xi |
10 | 15 | 20 | 25 | 30 |
| Year | 2005 | 2006 | 2007 | 2008 | 2009 |
| Production xi |
35 | 40 | 45 | 50 | 55 |
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.
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.
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`
