## Related questions with answers

Following are forecasts of worldwide annual cell phone handset sales:

$\begin{matrix} \text{Year x} & \text{3} & \text{5} & \text{7}\\ \text{Sales y (millions)} & \text{500} & \text{600} & \text{800}\\ \end{matrix}$

(x=3 represents 2003). Complete the following table and obtain the associated regression line. (Round coefficients to 2 decimal places.)

$\begin{matrix} & \text{} & \text{x} & \text{y} & \text{xy} & \text{}{x^{2}}\\ & \text{} & \text{3} & \text{500} & \text{} & \text{}\\ & \text{} & \text{5} & \text{600} & \text{} & \text{}\\ & \text{} & \text{7} & \text{800} & \text{} & \text{}\\ \text{Totals} & \text{} & \text{} & \text{} & \text{}\\ \end{matrix}$

Use your regression equation to project the 2008 sales.

Solution

VerifiedGiven the set of points

$(\mathrm{x}_{1},y_{1}), (x_{2},y_{2}), \ldots, (x_{n},y_{n})$

The regression line (least squares line, best-fit line):

$y=mx+b,$

where $m$ and $b$ are computed as follows.

$m=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^{2})-(\sum x)^{2}}\qquad b=\frac{\sum y-m(\sum x)}{n}$

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