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.