Question

1. Suppose you collected data from 1986 to 2005 to estimate the following demand function:

Q_X = b₀ + b₁ P_X + b₂ P_R + b₃ M

Assume R is a substitute and X is a normal good. The estimation results are reported below:

Dependent Variable: QX
Method: Least Squares
Date: 07/23/06   Time: 11:10
Sample: 1 20
Included observations: 20
Variable	Coefficient	Std. Error	t-Statistic	P-Value
C	19.777074	1.784179	11.0846	0.0000
PX	-13.79169	2.303381	-5.987587	0.0000
PR	5.069939	0.844016	6.006920	0.0000
M	0.257197	0.080063	3.212410	0.0054
R-squared	0.872386	F-statistic		36.45966
Adjusted R-squared	0.848459	P-Value for F-statistic		0.0000

           

            a   Conduct t tests to see if the coefficients are statistically significant at the 5% level of significance. Disregard the intercept. Use the p-values. A simple yes or no is not enough. Explain

           

            b.   Conduct an F- test to see if the equation, as a whole, is statistically significant at the 5%. Use the p-value. A simple yes or no is not enough. Explain.

                       

c. Interpret the value of R-square.

Answer 1

a.The t statistics are already provided in the output. The t-test tests the alternate hypothesis that the coefficient of the independent variable is 0 (or the independent variable is statistically significant).

For Px, p-value for the t-test = 0. At significance level of 5%, 0 < 0.05. Therefore p-value < significance level. So Px is statistically significant. This is expected as the price of a normal good will affect the demand of the good. If price increases, people would not be able to buy as much of the good s they were buying before. It is negatively related to quantity demanded of the normal good.

For P_R, p-value for the t-test = 0. At significance level of 5%, 0 < 0.05. Therefore p-value < significance level. So P_R is statistically significant. This is expected as the price of a complement increases (decreases), the people would be buying less (more) of the complement and more (less) of the normal good. It is positively related to the quanitity demanded of the normal good.

For M, p-value for the t-test = 0. At significance level of 5%, 0.0054 < 0.05. Therefore p-value < significance level. So M is statistically significant.

b.The F statistic is goven equal to 36.45966.

The F test tests that the equation as a whole is statistically significant. p-value = 0. SInce p-value of 0 < significance level of 0.05, we conclude that the model is significant.

c. R square is 0.872386. This is close to 1. It means that the model is a very good fit. 87.24% of variation in the quantitiy demanded is explained by the independent variables.