Answer to Question #200451 in Statistics and Probability for Raghav

Question #200451

The sales of a small factory since 2008 are as follows:

Year Sales (in ₹ lakhs)

  • 2008 8
  • 2009 10
  • 2010 9
  • 2011 11
  • 2012 11
  • 2013 12

Using 2008 as the zero year, find the least-square trend-line equation.


1
Expert's answer
2021-06-06T16:17:52-0400
xˉ=ixin=156=2.5\bar{x}=\dfrac{\sum_ix_i}{n}=\dfrac{15}{6}=2.5

yˉ=iyin=61610.166667\bar{y}=\dfrac{\sum_iy_i}{n}=\dfrac{61}{6}\approx10.166667

SSxx=i(xixˉ)2=ixi2nxˉ2SS_{xx}=\sum_i(x_i-\bar{x})^2=\sum_ix_i^2-n\cdot\bar{x}^2

=556(2.5)2=17.5=55-6\cdot(2.5)^2=17.5


SSyy=i(yiyˉ)2=iyi2nyˉ2SS_{yy}=\sum_i(y_i-\bar{y})^2=\sum_iy_i^2-n\cdot\bar{y}^2

=6316(616)2=65610.833333=631-6\cdot(\dfrac{61}{6})^2=\dfrac{65}{6}\approx10.833333


SSxy=i(xixˉ)(yiyˉ)=ixiyinxˉyˉSS_{xy}=\sum_i(x_i-\bar{x})(y_i-\bar{y})=\sum_ix_iy_i-n\cdot\bar{x}\bar{y}

=1656(2.5)(616)=12.5=165-6\cdot(2.5)(\dfrac{61}{6})=12.5

b=SSxySSxx=12.517.5=570.714286b=\dfrac{SS_{xy}}{SS_{xx}}=\dfrac{12.5}{17.5}=\dfrac{5}{7}\approx0.714286


a=yˉbxˉ=61657(2.5)=176218.380952a=\bar{y}-b\bar{x}=\dfrac{61}{6}-\dfrac{5}{7}(2.5)=\dfrac{176}{21}\approx8.380952


y=8.380952+0.714286xy=8.380952+0.714286x




r=SSxySSxxSSyy0.9078413r=\dfrac{SS_{xy}}{\sqrt{SS_{xx}}\sqrt{SS_{yy}}}\approx0.9078413

r20.8241758r^2\approx0.8241758



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