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Fit a straight line trend by method of least squares to the following data and find the trend values : | Year : | 2010 | 2012 | 2013 | 2014 | 2015 | 2016 | 2019 | | --- | --- | --- | --- | --- | --- | --- | --- | | Sales (in lakh ₹) : | 65 | 68 | 70 | 72 | 75 | 67 | 73 |

CBSECBSE Class XII Board 2023Subjective· 5mImportance★★★★★
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Least-squares trend Y^=69.89+0.747 x\hat Y=69.89+0.747\,x (with x=Year−2014x=\text{Year}-2014); trend values ≈66.90,68.40,69.15,69.89,70.64,71.39,73.63\approx 66.90,68.40,69.15,69.89,70.64,71.39,73.63 lakh ₹.

Straight-line trend Y^=a+bx\hat Y=a+bx. Normal equations: ∑Y=na+b∑x\sum Y=na+b\sum x and ∑xY=a∑x+b∑x2\sum xY=a\sum x+b\sum x^{2}, where xx is the coded year.

  1. Code the year as x=Year−2014x=\text{Year}-2014, giving x=−4,−2,−1,0,1,2,5x=-4,-2,-1,0,1,2,5 for 2010, 2012, 2013, 2014, 2015, 2016, 2019.
YearxxYYxYxYx2x^2
2010−4-465−260-26016
2012−2-268−136-1364
2013−1-170−70-701
20140072000
2015117575751
201622671341344
2019557336536525
Total∑x=1\sum x=1∑Y=490\sum Y=490∑xY=108\sum xY=108∑x2=51\sum x^2=51
  1. Normal equations (n=7n=7): 7a+b=4907a+b=490 and a+51b=108.a+51b=108.
  2. From the first, b=490−7ab=490-7a; substitute: a+51(490−7a)=108⇒−356a=−24882⇒a=69.89.a+51(490-7a)=108\Rightarrow -356a=-24882\Rightarrow a=69.89.
  3. Then b=490−7(69.89)=0.747.b=490-7(69.89)=0.747. Trend line: Y^=69.89+0.747 x.\hat Y=69.89+0.747\,x.
  4. Trend values Y^=a+bx\hat Y=a+bx:
YearxxY^\hat Y (lakh ₹)
2010−4-466.9066.90

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