There are various ways of reporting the results of regression analysis, but in this text we shall use the following format, employing the consumption-income example of Chapter 3 as an illustration:

Y = 24.4545 + 0.5091X se = (6.4138) (0.0357) t = (3.8128) (14.2605) p = (0.002571) (0.000000289)

In Eq. (5.11.1) the figures in the first set of parentheses are the estimated standard errors of the regression coefficients, the figures in the second set are estimated t values computed from (5.3.2) under the null hypothesis that r2 = 0.9621 df = 8 F18 = 202.87

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

Testing

146 PART ONE: SINGLE-EQUATION REGRESSION MODELS

the true population value of each regression coefficient individually is zero (e.g., 3.8128 = 24.4545 ^ 6.4138), and the figures in the third set are the estimated p values. Thus, for 8 df the probability of obtaining a t value of 3.8128 or greater is 0.0026 and the probability of obtaining a t value of 14.2605 or larger is about 0.0000003.

By presenting the p values of the estimated t coefficients, we can see at once the exact level of significance of each estimated t value. Thus, under the null hypothesis that the true population intercept value is zero, the exact probability (i.e., thep value) of obtaining a t value of 3.8128 or greater is only about 0.0026. Therefore, if we reject this null hypothesis, the probability of our committing a Type I error is about 26 in 10,000, a very small probability indeed. For all practical purposes we can say that the true population intercept is different from zero. Likewise, the p value of the estimated slope coefficient is zero for all practical purposes. If the true MPC were in fact zero, our chances of obtaining an MPC of 0.5091 would be practically zero. Hence we can reject the null hypothesis that the true MPC is zero.

Earlier we showed the intimate connection between the F and t statistics, namely, F1rk = t^. Under the null hypothesis that the true ft = 0, (5.11.1) shows that the F value is 202.87 (for 1 numerator and 8 denominator df) and the t value is about 14.24 (8 df); as expected, the former value is the square of the latter value, except for the roundoff errors. The ANOVA table for this problem has already been discussed.

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