Given a table of historical demand for a product as follows:
[tex]\begin{tabular}
{|c|c|}
&Demand\\[1ex]
April&60\\
May&55\\
June&75\\July&60\\August&80\\September&75\\
\end{tabular}[/tex]
The linear regression equation is given by
[tex]\hat{Y}=bx+a[/tex]
where:
[tex]b= \frac{n\Sigma xy-\Sigma x\Sigma y}{n\Sigma x^2-(\Sigma x)^2} [/tex]
and
[tex]a= \frac{1}{n} (\Sigma y-b\Sigma x)[/tex]
We calculate the required values using the following table, where
April = 1, May = 2, and so on.
[tex]\begin{tabular}
{|c|c|c|c|}
X &Y&X^2&XY\\[1ex]
1&60&1&60\\
2&55&4&110\\
3&75&9&225\\
4&60&16&240\\
5&80&25&400\\
6&75&36&450\\[1ex]
\Sigma X=21&\Sigma Y=405&\Sigma X^2=91&\Sigma XY=1,485
\end{tabular}[/tex]
Thus,
[tex]b= \frac{6(1,485)-21(405)}{6(91)-(21)^2} \\ \\ = \frac{8,910-8,505}{546-441} = \frac{405}{105} \\ \\ \approx3.86[/tex]
and
[tex]a= \frac{1}{6} (405-(3.86)(21)) \\ \\ = \frac{1}{6} (405-81)= \frac{1}{6} (324) \\ \\ =54[/tex]
Therefore, the trend line for the historical data is given by [tex]\hat{Y}=3.86x+54[/tex]
