Question

Using simple linear regression, calculate the trend line for the historical data. say the x axis is april = 1, may = 2, and so on, while the y axis is demand. (round your intercept value to the nearest whole number and slope value to 2 decimal places.

Answer 1

Given a table of historical demand for a product as follows:

$\begin{tabular} {|c|c|} &Demand\\[1ex] April&60\\ May&55\\ June&75\\July&60\\August&80\\September&75\\ \end{tabular}$

The linear regression equation is given by

$\hat{Y}=bx+a$

where:

$b= \frac{n\Sigma xy-\Sigma x\Sigma y}{n\Sigma x^2-(\Sigma x)^2}$
and
$a= \frac{1}{n} (\Sigma y-b\Sigma x)$

We calculate the required values using the following table, where
April = 1, May = 2, and so on.

$\begin{tabular} {|c|c|c|c|} X &Y&X^2&XY\\[1ex] 1&60&1&60\\ 2&55&4&110\\ 3&75&9&225\\ 4&60&16&240\\ 5&80&25&400\\ 6&75&36&450\\[1ex] \Sigma X=21&\Sigma Y=405&\Sigma X^2=91&\Sigma XY=1,485 \end{tabular}$

Thus,

$b= \frac{6(1,485)-21(405)}{6(91)-(21)^2} \\ \\ = \frac{8,910-8,505}{546-441} = \frac{405}{105} \\ \\ \approx3.86$

and

$a= \frac{1}{6} (405-(3.86)(21)) \\ \\ = \frac{1}{6} (405-81)= \frac{1}{6} (324) \\ \\ =54$

Therefore, the trend line for the historical data is given by $\hat{Y}=3.86x+54$