Question

You are given the following table of output pertaining to T-Shirt sales data from different stores in Pennsylvania. The variable shirts is the number of T-Shirts sold per month in 100s, and price is the price in dollars of the T-shirt.

Table 1: OLS regression Price and T-Shirts sold
Dependent variable:
shirts
price -0.40∗∗∗
(0.10)
Constant 7.00∗∗∗
(0.06)
Observations 1120
R2 0.20
Adjusted R2 0.20
Residual Std. Error 7.26 (df = 998)
F Statistic 12.15∗∗∗ (df = 1; 998)
Note: ∗p<0.1; ∗∗p<0.05; ∗∗∗p<0.01

(a) Use the information in the table to write the regression equation. (3 points)
(b) How much of the variation in shirts is explained by this model? Interpret this finding. (3 points)
(c) Suppose one of the stores is located in Erie. The price in Erie was $15.00, and shirts was equal to 2. Use the output in the table to calculate the predicted number of T-shirts sold. (3 points)
(d) What is the residual for Erie? (2 points)

Answer 1

Ans (a) As shirts are dependent variable, thus it is denoted by y

Then, price is the independent variable and is denoted by x

So, the regression equation is written as

Where c is the constant , e is the error and is the coefficient of x (price).

Therefore, the regression equation is :

(b) Since the R² = 0.20 , therefore 20% of the variation in shirts is explained by this model. Since the R² is quite low, therefore the regression model is a weak predictor for calculating the number of T-shirts sold.


(c)
   Here, x = 15 , therefore y = 7 - (0.4*15)

= 7- 6

= 1
Therefore, the predicted number of T-Shirts sold is 100.


(d) Predicted number of T-shirt sold = 1 and actual number of shirt sold = 2

So, residual is calculated as Actual value - Predicted Value

Residual value = actual value - predicted value

= 2 - 1

= 1