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Worked Examples · Example 5
Q.

Fit a straight-line trend by the method of least squares to the following data and write down the trend value for each year.

Year20132014201520162017
Sales (Rs lakh)1218202327
Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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There are n=5n = 5 years (odd), so we take the middle year 2015 as origin and let XX be the deviation in years: X=−2,−1,0,1,2X = -2, -1, 0, 1, 2. Then ∑X=0\sum X = 0 and the normal equations reduce to a=∑Yna = \dfrac{\sum Y}{n} and b=∑XY∑X2b = \dfrac{\sum XY}{\sum X^2}.

YearYYXXXYXYX2X^2
201312−2-2−24-244
201418−1-1−18-181
201520000
2016231231
2017272544
Total10003510

Hence

a=∑Yn=1005=20,b=∑XY∑X2=3510=3.5.a = \frac{\sum Y}{n} = \frac{100}{5} = 20, \qquad b = \frac{\sum XY}{\sum X^2} = \frac{35}{10} = 3.5.

The trend line is Yt=20+3.5XY_t = 20 + 3.5X (origin 2015, unit of XX = 1 year).

Trend values (substitute each XX):

YearXXYt=20+3.5XY_t = 20 + 3.5X
2013−2-220−7=13.020 - 7 = 13.0
2014−1-120−3.5=16.520 - 3.5 = 16.5
2015020.020.0
2016120+3.5=23.520 + 3.5 = 23.5
2017220+7=27.020 + 7 = 27.0

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