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

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Question

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.

Sum
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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.

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Chapter 3: Linear Regression - Miscellaneous Exercise 3 [Page 54]

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

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