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Given below is the data of workers welfare expenses (in lakh ₹) in steel industries during 2016 – 2020 : | Year | 2016 | 2017 | 2018 | 2019 | 2020 | |---|---|---|---|---|---| | Workers welfare expenses (in lakh ₹) | 160 | 185 | 220 | 300 | 510 | Find the best fitted trend line by the method of least squares and tabulate the trend values.

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★
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Taking origin at 2018 so ∑X=0\sum X=0, least squares gives a=Yˉ=275a=\bar{Y}=275 and b=∑XY∑X2=81.5b=\dfrac{\sum XY}{\sum X^2}=81.5; the trend line y=275+81.5Xy=275+81.5X yields trend values 112,193.5,275,356.5,438112,193.5,275,356.5,438.

Least-squares line y=a+bxy=a+bx. With deviations X=x−xˉX=x-\bar{x} (so ∑X=0\sum X=0): a=∑Yna=\dfrac{\sum Y}{n} and b=∑XY∑X2b=\dfrac{\sum XY}{\sum X^2}.

  1. Build the computation table with X=year−2018X=\text{year}-2018:
YearYY (expenses)XXX2X^2XYXY
2016160−2-24−320-320
2017185−1-11−185-185
2018220000
201930011300
2020510241020
Total∑Y=1375\sum Y=1375∑X=0\sum X=0∑X2=10\sum X^2=10∑XY=815\sum XY=815
  1. Intercept: a=∑Yn=13755=275.a=\dfrac{\sum Y}{n}=\dfrac{1375}{5}=275. …

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