Question: The selling price of an item and sales volume in of thousands of items is given in the table below.
a) Find the equation to line of best fit, using least square regression.
Selling Price in dollars 
60 
80 
100

120

140 
160 
180

200

220

240

Sales in thousands of Numbers

400 
350 
300 
275 
250 
210 
190 
150 
100 
50 
b) Also estimate the sales volume when the selling price is 175 dollars.
Solution:
We need to find the line of best fit of the model y' = a + bx. That
is the task is to find the values of the intercept 'a' and slope 'b'.
Let
us rewrite the table in order to find the summed up values that can be
plugged in the formulas for the slope 'b' and Y' intercept 'a'.
Price x dollars 
Sales y 
xy

x^{2 }

y^{2 }

60

400 
24,000 
3,600 
160,000 
80

350 
28,000 
6,400 
122,500 
100

300 
30,000 
10,000

90,000

120

275 
33,000 
14,400 
75,625 
140

250

35,000 
19,600 
62,500

160

210 
33,600 
25,600 
44,100

180

190 
34,200 
32,400 
36,100

200

150 
30,000 
40,000 
22,500 
220

100 
22,000 
48,400 
10,000 
240

50 
12,000 
57,600 
2,500

∑x = 1,500 
∑y = 2,275 
∑xy = 281,800 
∑x^{2} = 258,000 
∑y^{2} = 625,825 
We have the required sums to be substituted in the formula.
a =
$\frac{(\sum y)(\sum x^{2})(\sum x)(\sum xy)}{n(\sum x^{2})(\sum x)^{2}}$ =
$\frac{(2275)(258000)(1500)(281800)}{10(258000)(1500)^{2}}$ =
497.73b =
$\frac{n(\sum xy)(\sum x)(\sum y)}{n(\sum x^{2})(\sum x)^{2}}$ =
$\frac{10(281800)(1500)(2275)}{10(25800)(1500)^{2}}$ =
1.80Hence the required regression line is
y' = 497.73  1.80x.
b) To estimate the sales when the price x = 175 dollars, substitute x =175 in the regression equation and calculate y'.
y' = 497.73  1.80(175) =
182.73 This means 182,730 items are expected to be sold at a price of 175 dollars.