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Consider the simple linear regression model $ y_i = \beta_0 + \beta_1{x_i} + \epsilon_i$

Suppose in an OLS, the $t$-statistic for the null hypothesis $\beta_1 = 0$ is $1.92$, what is the $F$-statistic for the overall significance of the model?

I am aware that $F$ statistic for testing exclusion of a single variable is equal to the square of the corresponding $t$ statistic: however, I can't seem to find how to link that information to this question. Am I missing information?

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    $\begingroup$ Consider what the null hypothesis would be for the “significance of the entire model”. $\endgroup$
    – Dave
    Commented Nov 28, 2021 at 22:45
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    $\begingroup$ Oh it would just be $\beta_1$ = 0, wouldn't it? $\endgroup$
    – Giacomo Oliva
    Commented Nov 28, 2021 at 23:11
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    $\begingroup$ That’s what I would say, and that’s what R gives you in a “summary” of a linear model. And then the alternative hypothesis would be $\ne$. // This is a subtle point, but good for you for not writing $\hat\beta_1=0$. $\endgroup$
    – Dave
    Commented Nov 28, 2021 at 23:16
  • $\begingroup$ Thank you! I spent way too much time looking for an answer when it was sitting right in front me $\endgroup$
    – Giacomo Oliva
    Commented Nov 28, 2021 at 23:19
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    $\begingroup$ Since you pretty much solved the problem yourself, you might consider posting a self-answer. (Cross Validated does not consider it arrogant or poor etiquette to do so. In fact, I have posted several self-answers.) $\endgroup$
    – Dave
    Commented Nov 28, 2021 at 23:37

1 Answer 1

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In the simple linear regression model, testing the significance of the model requires testing the hypothesis: $$H_0 \colon \beta_1 = 0 \\ H_0 \colon \beta_1 \neq 0 $$

Given that we are whether a single parameter is equal to 0, the $F$ statistic will simply equal the square of the corresponding $t$ statistic. This means that if the $t$-statistic for the null is $1.92$, the $F$-statistic will simply be $1.92^2$.

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    $\begingroup$ Not sure how this answers your question. $\endgroup$
    – Adrian Keister
    Commented Nov 29, 2021 at 15:36
  • $\begingroup$ The edit might should have made it clearer $\endgroup$
    – Giacomo Oliva
    Commented Nov 29, 2021 at 17:25

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