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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 leas - Mathematics and Statistics

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प्रश्न

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`

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उत्तर

Let the equation of trend line be y = a + bx   .....(i)

Here n = 7(odd), middle year is 1995 and h = 5

x = `("t" - "middle year")/"h"`

= `("t" - 1995)/5`

We obtain the following table:

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

From the table, n = 7, Σyt = 30, Σx = 0, Σx2 = 28, Σxyt = – 44

The normal equations are

Σy = na + bΣx

∴ 30 = 7a + bΣx

As, Σx = 0, a = `30/7` = 4.2857

Also, Σxy = aΣx + bΣx

∴ – 44 = aΣx + b′(28)

As, Σx = 0, b =`(-44)/28` =  – 1.5714

∴ The equation of trend line is y = a + bx

∴ The equation of trend line is y = 4.2857 – 1.5714x

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Measurement of Secular Trend
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2.4: Time Series - Q.5

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.


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Index 0 2 3 3 2 4 5 6 7 10

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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_______.


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Least squares method of finding trend is very simple and does not involve any calculations.


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Year 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982
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Production 1 0 1 2 3 2
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Year 1975 1976 1977 1978 1979 1980 1981 1982 1983
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The secular trend component of time series represents irregular variations


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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

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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

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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.


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