Suppose you collected data fro
- Suppose you collected data from 1986 to 2005 to estimate thefollowing demand function:
QX = b0 +b1 PX + b2 PR +b3 M
Assume R is a substitute and X is anormal good. The estimation results are reported below:
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Dependent Variable: QX |
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Method: Least Squares |
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Date: 07/23/06 Time: 11:10 |
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Sample: 1 20 |
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Included observations: 20 |
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Variable |
Coefficient |
Std. Error |
t-Statistic |
P-Value |
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C |
19.777074 |
1.784179 |
11.0846 |
0.0000 |
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PX |
-13.79169 |
2.303381 |
-5.987587 |
0.0000 |
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PR |
5.069939 |
0.844016 |
6.006920 |
0.0000 |
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M |
0.257197 |
0.080063 |
3.212410 |
0.0054 |
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R-squared |
0.872386 |
F-statistic |
36.45966 |
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Adjusted R-squared |
0.848459 |
P-Value for F-statistic |
0.0000 |
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a Conduct t tests to see if thecoefficients are statistically significant at the 5% level ofsignificance. Disregard the intercept. Use the p-values. 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 p-value. Asimple yes or no is not enough. Explain.
c. Interpret the value of R-square.
Answer:
a.The t statistics are already provided in the output. Thet-test tests the alternate hypothesis that the coefficient of theindependent variable is 0 (or the independent variable isstatistically 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 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 PR, p-value for the t-test = 0. At significancelevel of 5%, 0 < 0.05. Therefore p-value < significancelevel. So PR 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, p-value for the t-test = 0. At significance level of 5%,0.0054 < 0.05. Therefore p-value < 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. p-value = 0. SInce p-value 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.
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