Question

Results: Intercept TV radio TV× radio Coefficient Std. Error t-statistic p-value 27.23<0.0001 12.70 <0.0001 3.240.0014 20.73 <0.0001 6.7502 0.0191 0.0289 0.0011 0.248 0.002 0.009 0.000

1) According to the model for sales vs TV interacted with radio, what is the effect of an additional $1 of radio advertising if TV=$50? (with 4 decimal accuracy)

2) What if TV=$250? (with 4 decimal accuracy)

Answer 1

The regression equation is:

Y =b0+b1(TV)+b2(Radio)+b3(TV x Radio)

where, Y =Sales; b0 = y-intercept =6.7502; b1 =coefficient of TV variable =0.0191; b2 =coefficient of Radio variable =0.0289; b3 =coefficient of interaction between TV and Radio variables =0.0011

By using partial derivatives concept, we can solve it.

We want to find the effect due to change in radio variable. So, we have to differentiate with respect to radio variable and thus, other variables (here, TV) remains constant:

Here is the image with the computation:

So, we have: Change in Y/Change in Radio =b2+b3(TV) $\implies$Change in Y/Change in Radio =0.0289+0.0011(TV)

1)

Now, by substituting TV =$50, we have:

Change in Y due to change in Radio =Change in Y/Change in Radio =0.0289+0.0011(50) =$0.0839

Thus, the effect of an additional $1 of radio advertising if TV=$50 is increase in sales by ​​​$0.0839

2)

Substitute TV =$250

Change in Y/Change in Radio =0.0289+0.0011(250) =$0.3039

Thus, the effect of an additional $1 of radio advertising if TV=$250 is increase in sales by $0.3039