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jacqie · 2022年01月10日

为什么不是按照公式算出来加入两个变量后的R大于加入前的R就能验证需要加入两个变量?

NO.PZ2020010801000037

问题如下:

You are interested in understanding the determinants of the yield spread of corporate bonds above a maturity matched sovereign bond. You include three explanatory variables: the leverage defined as the ratio of long-term debt to the book value of assets, a dummy variable for high yield, and a measure of the volatility of the profitability of the issuer. You are interested in testing whether there are nonlinear effects of some of these variables, and so use a RESET test including both the squared and cubic term. The R2R^2 of the original model is 68.6%, and the R2R^2 from the model that includes both additional terms is 68.9%. You have 456 observations. What do you conclude about the specification of the model?

选项:

解释:

The RESET test examines whether the two additional explanatory variables that squared and cubed fitted values have zero coefficients. It is implemented using an F-test:

(0.6890.6862)/(10.6894566) F2,450(\frac{0.689-0.686}{2})/(\frac{1-0.689}{456-6})~F_{2,450}

The F-test examines the difference between the R2R^2 in the two models. The critical value for an F2,450F_{2,450} is 3.01 (F.INV. RT(0.05,2,450) in Excel). The value of the test statistic is 2.17, which is less than the critical value, and so the null that the coefficient on the squared and cubic terms is 0 is not rejected.


我用这个公式算加入两个变量后的R是68.69%>加入前的68.53%,为什么不能这么做,而需要用假设检验?

1 个答案

李坏_品职助教 · 2022年01月10日

嗨,从没放弃的小努力你好:


判断是否可以加入两个解释变量,不能只看R^2大小,因为算出来的新的R^2比加入前的68.53%大的那一点,必须符合统计学意义才行(必须拒绝F检验的原假设才有统计意义)。


结果发现无法拒绝H0,也就是算出来的大的这一点R^2,还不够大,不具备统计意义。

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