Let y = sales at a fast-food outlet (1000s of $), x1 = number of competing outle

Question

Let y = sales at a fast-food outlet (1000s of $), x1 = number of competing outlets within a 1-mile radius, x2 = population within a 1-mile radius (1000s of people), and x3 be an indicator variable that equals 1 if the outlet has a drive-up window and 0 otherwise. Suppose that the true regression model is Y = 8.00 1.1x1 + 6.8x2 + 15.2x3 +

(a) What is the mean value of sales when the number of competing outlets is 3, there are 7000 people within a 1-mile radius, and the outlet has a drive-up window? (Round your answer to one decimal place.) X thousand dollars

(b) What is the mean value of sales for an outlet without a drive-up window that has two competing outlets and 6000 people within a 1-mile radius? (Round your answer to one decimal place.) X thousand dollars

(c) Interpret 3.

It takes $15,200 to build a drive-up.

A drive-up will add $15,200 to total sales.

There are on average 15.2 fast-food outlets with drive-ups in a 1 mile radius.

Approximately 1 in 15.2 fast-food outlets have drive-ups.

Explanation / Answer

The true regression model is give by Y = 8.00 1.1*x1 + 6.8*x2 + 15.2*x3 +

Where x1 = number of competing outlets within a 1-mile radius,

x2 = population within a 1-mile radius (1000s of people), and

x3 is the indicator variable that equals 1 if the outlet has a drive-up window and 0 otherwise.

is the error term

a) The mean value of sales when the number of competing outlets is 3, there are 7000 people within a 1- mile radius, and the outlet has a drive-up window is given by

Y = 8.00 1.1*(3) + 6.8*(7 )+ 15.2*(1) = 67.5 i.e., 67.5 thousand dollars

b) The mean value of sales for an outlet without a drive-up window that has two competing outlets and 6000 people within a 1-mile radius is

Y = 8.00 1.1*(2) + 6.8*(6 )+ 15.2*(0) = 46.6 i.e., 46.6 thousand dollars

c) The interpretation of 3 = 15.2 is A drive- up will add $15,200 to total sales Y

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