Suppose you collected data fro
 Suppose you collected data from 1986 to 2005 to estimate thefollowing demand function:
Q_{X} = b_{0} +b_{1} P_{X} + b_{2} P_{R} +b_{3} M
Assume R is a substitute and X is anormal 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 
tStatistic 
PValue 
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 
Rsquared 
0.872386 
Fstatistic 
36.45966 

Adjusted Rsquared 
0.848459 
PValue for Fstatistic 
0.0000 
a Conduct t tests to see if thecoefficients are statistically significant at the 5% level ofsignificance. Disregard the intercept. Use the pvalues. Asimple yes or no is not enough. Explain
b. Conduct an F test to see if the equation, as awhole, is statistically significant at the 5%. Use the pvalue. Asimple yes or no is not enough. Explain.
c. Interpret the value of Rsquare.
Answer:
a.The t statistics are already provided in the output. Thettest tests the alternate hypothesis that the coefficient of theindependent variable is 0 (or the independent variable isstatistically significant).
For Px, pvalue for the ttest = 0. At significance level of 5%,0 < 0.05. Therefore pvalue < significance level. So Px isstatistically significant. This is expected as the price of anormal 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 werebuying before. It is negatively related to quantity demanded of thenormal good.
For P_{R}, pvalue for the ttest = 0. At significancelevel of 5%, 0 < 0.05. Therefore pvalue < significancelevel. So P_{R} is statistically significant. This isexpected as the price of a complement increases (decreases), thepeople would be buying less (more) of the complement and more(less) of the normal good. It is positively related to thequanitity demanded of the normal good.
For M, pvalue for the ttest = 0. At significance level of 5%,0.0054 < 0.05. Therefore pvalue < significance level. So Mis statistically significant.
b.The F statistic is goven equal to 36.45966.
The F test tests that the equation as a whole is statisticallysignificant. pvalue = 0. SInce pvalue of 0 < significancelevel of 0.05, we conclude that the model is significant.
c. R square is 0.872386. This is close to 1. It means that themodel is a very good fit. 87.24% of variation in the quantitiydemanded is explained by the independent variables.