# The data obtained on X, the length of time in weeks that a promotional project has been in progress at a small business, - Mathematics

Sum

The data obtained on X, the length of time in weeks that a promotional project has been in progress at a small business, and Y, the percentage increase in weekly sales over the period just prior to the beginning of the campaign.

 X 1 2 3 4 1 3 1 2 3 4 2 4 Y 10 10 18 20 11 15 12 15 17 19 13 16

Find the equation of the regression line to predict the percentage increase in sales if the campaign has been in progress for 1.5 weeks.

#### Solution

Here, X = Length of time in weeks,
Y = Percentage increase in weekly sales

 X = xi Y = yi x_i^2 xi yi 1 10 1 10 2 10 4 20 3 18 9 54 4 20 16 80 1 11 1 11 3 15 9 45 1 12 1 12 2 15 4 30 3 17 9 51 4 19 16 76 2 13 4 26 4 16 16 64 30 176 90 479

From the table, we have

n = 12, ∑ xi = 30, ∑ yi = 176, sum x_i^2 = 90, ∑ xi yi = 479

∴ bar x = (sum x_i)/"n" = 30/12 = 2.5

bar y = (sum y_i)/"n" = 176/12 = 14.67

Now, "b"_"YX" = (sum"x"_"i" "y"_"i" - "n" bar "x" bar "y")/(sum "x"_"i"^2 - "n" bar "x"^2)

= (479 - 12 xx 2.5 xx 14.67)/(90 - 12 xx (2.5)^2)

= (479 - 440.1)/(90 - 75) = 38.9/15 = 2.59

Also,

"a" = bar y - "b"_"YX"  bar x

= 14.67 - 2.59 × 2.5 = 14.67 - 6.475 = 8.195

∴ a ≈ 8.2

∴ The regression equation of percentage increase in weekly sales (Y) on length of weeks (X) is

Y = a + bYX + X

i.e., Y = 8.2 + 2.59 X

For X = 1.5, we get

Y = 8.2 + 2.59(1.5) = 8.2 + 3.885 = 12.085

∴ Increase in sales is 12.085% if the campaign has been in progress for 1.5 weeks.

Is there an error in this question or solution?
Chapter 3: Linear Regression - Miscellaneous Exercise 3 [Page 54]

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 3 Linear Regression
Miscellaneous Exercise 3 | Q 4.01 | Page 54

Share