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The below text is from statistical learning Page 76-77. The data used in the following example can be found here . I just want to know

how to calculate $RSS_0$ (It is different from TSS)

As listed in the point 1 below the square of each t-statistics is the corresponding F-statistics please explain this point. (What I understand from this is that in a simple linear model with one variable the square of t-statistics of a variable is equal to it's F-statistics. Am I correct ? My other question is, Is this also valid for multiple regression models as I found if I do the sum of squares of all values of t-statistics in the figure below (except intercept ) is not equal to F-statistics? Please clarify what the authors want to convey here)

Sometimes we want to test that a particular subset of q of the coefficients are zero. This corresponds to a null hypothesis $H_0 : > β_{p−q+1} = β_{p−q+2} = ...= β_p =0$

where for convenience we have put the variables chosen for omission at the end of the list. In this case we fit a second model that uses all the variables except those last q. Suppose that the residual sum of squares for that model is $RSS_0$. Then the appropriate F-statistic is $$F=\frac{(RSS_0-RSS)/q}{RSS/(n-p-1)}$$

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Notice that in the table above, for each individual predictor a t-statistic and a p-value were reported. These provide information about whether each individual predictor is related to the response, after adjusting for the other predictors. It turns out that each of these are exactly $equivalent^1$ to the F-test that omits that single variable from the model, leaving all the others in—i.e. q=1 in $$F=\frac{(RSS_0-RSS)/q}{RSS/(n-p-1)}$$

So it reports the partial effect of adding that variable to the model.

1.The square of each t-statistic is the corresponding F-statistic

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Yes. This is correct. You can interchange the t and F statistics as from a theoretical point of view the F distribution arises for the square of a random variable with a t distribution. Similarly the square of a normal random variable has a $\chi^2$ distribution, so you can expect the square of a z statistic to be $\chi^2$ distributed. What the author wants to convey in that table is that if you took the t statistic value and squared it, it would be exactly equal to the F statistic as calculated from the given formula.

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The square of t-statistic equal to partial F-test.

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