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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Use the method of least squares to fit a trend line to the data in Problem 1 above. Also, obtain the trend value for the year 1975. - Mathematics and Statistics

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प्रश्न

Use the method of least squares to fit a trend line to the data in Problem 1 above. Also, obtain the trend value for the year 1975.

बेरीज
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उत्तर

In the given problem, n = 11 (odd), middle t- values is 1967, h = 1

u = `"t - middle value"/"h" = ("t" - 1967)/1` = t – 1967
We obtain the following table.

Year (t) Production(yt)
(in '000 tonnes)
u = t 1967 u2 uyt Trend value
1962 0 –5 25 0 –1.0002
1963 0 –4 16 0 0.0362
1964 1 –3 9 –3 1.0726
1965 1 –2 4 –2 2.1090
1966 4 –1 1 –4 3.1454
1967 2 0 0 0 4.1818
1968 4 1 1 4 5.2182
1969 9 2 4 18 6.2546
1970 7 3 9 21 7.2910
1971 10 4 16 40 8.3274
1972 8 5 25 40 9.3638
Total 46 0 110 114  

From the table, n = 11, `sumy_"t" = 46, sumu = 0, sumu^2 = 110,sumuy_"t" = 114`

The two normal equations are: `sumy_"t" = "na"' + "b"' sumu  "and" sumuy_"t", = a'sumu + b'sumu^2`

∴ 46 = 11a' + b'(0)          ...(i)   and
114 = a' = `(46)/(11)` = 4.1818

From (ii), b' = `(114)/(110)` = 1.0364

∴ The equation of the trend line is yt = a' + b'u
i.e., yt = 4.1818 + 1.0364 u, where u = t – 1967
Now, for t = 1975, u = 1975 – 1967 = 8
∴ yt = 4.1818 + 1.0364 x 8 = 12.473.

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Components of a Time Series
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Time Series - Exercise 4.1 [पृष्ठ ६६]

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