Do we include the intercept term when predicting in simple linear regression?

Question

In "Introduction to statistical learning wit R", there's a simple linear regression fit on a 'TV' feature where the target class is the 'Sales'.

Beta 0 is 7.03 and Beta 1 is 0.0475

Now we want to predict how much sales will increase when 1000 dollars is spent on TV advertising. I naturally thought that this would

1000 * 0.0475 + 7.03 

which equals to 54.6

The book however says that it's 47.5 which i presume is because

1000 * 0.0475

I then ran my own linear regression model on the data, and asked it to predict on the 1000 dollars and it agreed with me that it's 54.6

which is it?

Answer 1

predict how much sales will increase

They are asking how much sales will increase. $\beta_1$ shows how much sales will change for every dollar spent on TV advertising. So the increase is $1000 \times \beta_1 = 1000 \times 0.0475 = 47.5$.

What you have calculated is the predicted value with 1000 dollars spent total on TV advertising: $\hat{y} = \beta_0 + \beta_1X$, or $sales = 7.03 + 0.0465 \times X$.

When total dollars spent on TV advertising is 1000, then the predicted value is what you have: $\hat{y} = 7.03 + .0465 \times 1000 = 54.60$

Your answer was the predicted value when X = 1000. What you were asked for was the increase in the predicted value when X increases by 1000.

You could do it between any two predicted values:

• When X = 5000, $\hat{y}$ = 244.53.

• When X = 6000 (i.e., increases by 1000), $\hat{y}$ = 292.03

• $\hat{y}_{X=6000} - \hat{y}_{X=5000}$ = 292.03 - 244.53 = 47.5.