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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

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प्रश्न

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:

सारिणी
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उत्तर

Let yt be the trend line represented by the equation

yt = a + bt

Let u = `(t - "Midvalue")/h`

Midvalue = 1936, h = 1

∴ u = `(t - 1936)/1` = t – 1936

Year (t) yt u u2 u.yt
1931 1 – 5 25 – 5
1932 0 – 4 16 0
1933 1 – 3 09 – 3
1934 2 – 2 04 – 4
1935 3 – 1 01 – 3
1936 2 0 00 0
1937 8 1 01 8
1938 6 2 04 12
1939 5 3 09 15
1940 1 4 16 04
1941 4 5 25 20
  `sumy_t` = 33 `sumu` = 0 `sumu^2` = 110 44

The equation of trend line becomes,

yt = a' + b'u     .......(1)

The normal equations are

`sumy_t = na^' + b^'sumu`  .......(2)

`sumu.y_t = u^'sumu + b^'sumu^2`  ......(3)

From equation (2), we get

∴ Normal equations are

33 = 11a' + 0.b'

⇒ 11a' = 33

⇒ a' = 3

From equation (3), we get

44 = a'.0 + 110.b'

⇒ b' = `44/110` = 0.4

∴ b' = 0.4

∴ The equation of the trend line is given by yt = 3 + (0.4)u.

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Measurement of Secular Trend
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2021-2022 (March) Set 1

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2013 17 5 25 85
2014 21 7 49 147
2015 19 9 81 171
Total 177 0 330 27

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

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xi
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xi
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Complete the following activity to fit a trend line to the following data by the method of least squares.

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Here n = 9. We transform year t to u by taking u = t - 1979. We construct the following table for calculation :

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1975 0 - 4 16 0
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  `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`


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
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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`


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