Consider the following data

Year (x)	2010	2011	2012	2013	2014
Profit (Rs. in thousands) (y)	10	12	14	16	13

The equation of straight line trend by method of least square for the above data is given by

The line of best fit using the Least Squares Method has the form $y = a + bx$.

Year (x)	2010	2011	2012	2013	2014
Profit (y)	10	12	14	16	13

Number of data points: $n = 5$

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Calculate the required sums:

$\sum x = 2010 + 2011 + 2012 + 2013 + 2014 = 10,060$

$\sum y = 10 + 12 + 14 + 16 + 13 = 65$

$\sum xy = (2010 \times 10) + (2011 \times 12) + (2012 \times 14) + (2013 \times 16) + (2014 \times 13)$

$\sum xy = 20,100 + 24,132 + 28,168 + 32,208 + 26,182 = 130,790$

$\sum x^2 = 2010^2 + 2011^2 + 2012^2 + 2013^2 + 2014^2$

$\sum x^2 = 4,040,100 + 4,044,121 + 4,048,144 + 4,052,169 + 4,056,196 = 20,240,730$

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The slope is given by:

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

$b = \dfrac{5(130,790) - (10,060)(65)}{5(20,240,730) - (10,060)^2}$

$b = \dfrac{653,950 - 653,900}{101,203,650 - 101,203,600}$

$b = \dfrac{50}{50}$

$b = 1$

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The intercept is given by:

$a = \dfrac{\sum y - b(\sum x)}{n}$

$a = \dfrac{65 - 1(10,060)}{5}$

$a = \dfrac{65 - 10,060}{5}$

$a = \dfrac{-9,995}{5}$

$a = -1999$

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The equation of the straight line trend is:

$y = a + bx$

$y = -1999 + x$

$y = x - 1999$