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Fit a straight line trend by the method of least squares and find the trend value for the year 2008 for the following data : | Year | Production (in lakh tonnes) | | --- | --- | | 2001 | 30 | | 2002 | 35 | | 2003 | 36 | | 2004 | 32 | | 2005 | 37 | | 2006 | 40 | | 2007 | 36 |

CBSECBSE Class XII Board 2022Subjective· 3mImportance★★★★★
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With origin at 2004, ∑Y=246, ∑XY=29, ∑X2=28\sum Y=246,\ \sum XY=29,\ \sum X^2=28, so a=35.14, b=1.036a=35.14,\ b=1.036; for 2008 (X=4X=4) the trend value is 275/7≈39.29275/7\approx 39.29 lakh tonnes.

Least-squares line Y=a+bXY=a+bX with origin shifted so ∑X=0\sum X=0:   a=∑YN,b=∑XY∑X2\;a=\dfrac{\sum Y}{N},\qquad b=\dfrac{\sum XY}{\sum X^2}, where X=X= (year −- middle year), Y=Y= production, and N=N= number of years.

  1. There are N=7N=7 years; the middle year is 2004, so let X=year−2004X=\text{year}-2004.
YearYYXXXYXYX2X^2
200130−3-3−90-909
200235−2-2−70-704
200336−1-1−36-361
20043200000
2005371137371
2006402280804
200736331081089
Total24602928
  1. Compute the intercept: a=∑YN=2467=35.142…≈35.14a=\dfrac{\sum Y}{N}=\dfrac{246}{7}=35.142\ldots\approx 35.14. …

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