Source of variation SS* df MSS+

V uf

*SS means sum of squares.

"•"Mean sum of squares, which is obtained by dividing SS by their df.

Gujarati: Basic I I. Single-Equation I 5. Two-Variable I I © The McGraw-Hill

Econometrics, Fourth Regression Models Regression: Interval Companies, 2004 Edition Estimation and Hypothesis

Testing

CHAPTER FIVE: TWO VARIABLE REGRESSION: INTERVAL ESTIMATION AND HYPOTHESIS TESTING 141

1 df in the numerator and (n — 2) df in the denominator. (See Appendix 5A, Section 5A.3, for the proof. The general properties of the F distribution are discussed in Appendix A.)

What use can be made of the preceding F ratio? It can be shown18 that

(Note that j2 and a2 appearing on the right sides of these equations are the true parameters.) Therefore, if j2 is in fact zero, Eqs. (5.9.2) and (5.9.3) both provide us with identical estimates of true a2. In this situation, the explanatory variable X has no linear influence on Y whatsoever and the entire variation in Y is explained by the random disturbances 4. If, on the other hand, jj2 is not zero, (5.9.2) and (5.9.3) will be different and part of the variation in Y will be ascribable to X. Therefore, the F ratio of (5.9.1) provides a test of the null hypothesis H0: jj2 = 0. Since all the quantities entering into this equation can be obtained from the available sample, this F ratio provides a test statistic to test the null hypothesis that true 2 is zero. All that needs to be done is to compute the F ratio and compare it with the critical F value obtained from the F tables at the chosen level of significance, or obtain the p value of the computed F statistic.

To illustrate, let us continue with our consumption-income example. The ANOVA table for this example is as shown in Table 5.4. The computed F value is seen to be 202.87. The p value of this F statistic corresponding to 1 and 8 df cannot be obtained from the F table given in Appendix D, but by using electronic statistical tables it can be shown that the p value is 0.0000001, an extremely small probability indeed. If you decide to choose the level-of-significance approach to hypothesis testing and fix a at 0.01, or a 1 percent level, you can see that the computed F of 202.87 is obviously significant at this level. Therefore, if we reject the null hypothesis that 2 = 0, the probability of committing a Type I error is very small. For all practical

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