We're doing simple linear regression. Anova decomposition. $$SS=RegSS+RSS$$ or better $$(n-1)s_Y^2=\hat{b}_1^2s_{XX}+(n-2)s^2$$

We know that when the fit is good, then RegSS will be large. Therefore we use the test hypothesis $H_0:\beta_1=0$.

Now my book says "Under the null hypothesis, $\beta_1=0$ and since we know $$\hat{\beta}_1 \sim N\left(\beta_1, \frac{\sigma^2}{s_{XX}}\right)$$ we have that $\hat{\beta}_1 \sim \left(0, \frac{\sigma^2}{s_{XX}}\right)$."

Then goes on finding a statistics which is $N(0,1)$ and then dividing by a certain factor, gets an $F$ distribution.

My question is: if we are under the null hypothesis, that says that the simpler model (the constant one $Y_i \sim N(\beta_0, \sigma^2)$) is correct, i.e. $\beta_1=0$, then why can we still use the estimator for $\beta_1$? Surely if we are under $H_0$, then $\beta_1$ has no meaning and also the meaning of $\beta_0$ changes, indeed now $\beta_0$ is estimated by $\bar{Y}$ rather than $\hat{\beta}_0 \sim N\left(\beta_0, \frac{1}{n}+\frac{\bar{x}^2}{s_{XX}}\right)$. So if the meaning AND the distribution of the estimator of $\beta_0$ change completely, surely so should for the estimator of $\beta_1$. Not only that, but if we want to be more precise, we should have

  • $\beta_0^{(1)}$ and $\beta_1^{(1)}$ which are the parameters of the linear regression model, estimated by $\hat{\beta}_0^{(1)}$ and $\hat{\beta}_1^{(1)}$.
  • $\beta_0^{(0)}$ and $\beta_1^{(0)}$ which are the parameters under the null hypothesis (i.e. simpler model) and are estimated by $\hat{\beta}_0^{(0)}$ and $\hat{\beta}_1^{(0)}$.

So not only, under the null hypothesis, the previous $\hat{\beta}_1$ makes no sense, but also, it would be a totally different object than the one in the regression model.


Notice that whether or not your null hypothesis is accepted or rejected, is not affecting your model at all - your current model and all statistics does not change, and it readily incorporates the scenarios that you null hypothesis could not be rejected, but it's just "how likely" it's gonna happen. Even if we could not reject null hypothesis, it does not mean your coefficients are now definitely $0$, but it just mean it has a really low probability to be $0$ and thus is statistically significant.


To answer the question in the comment: I think your understanding of the book is a bit off. The book is correct, but your understanding below is off the right track. What the book arrives is that $\hat{\beta}_1$ is of a normal distribution with mean equals to a constant $\beta_1$. And based on this, now it tries to have a test with hypothesis $\beta_1 = 0$ - this test is only about the r.v. $\hat{\beta}_1$, but is not about the regression model. And for the test you could plug in the null hypothesis early (as the book), or you could do it later - it does not matter.

$$\hat{\beta}_1 \sim N\left(\beta_1, \frac{\sigma^2}{s_{XX}}\right)$$ $$\frac{\hat{\beta}_1-\beta_1}{\sqrt{\frac{\sigma^2}{s_{XX}}}} \sim N\left(0,1\right)$$

All the transformation I did above is just to re-organize the random variable to make it more convenient to do the test, but if you do not change it, you could still do it.

Then you could construct either t-test or F-test based upon this normal distribution of $N(0,1)$, and you'll notice the test statistics you construct will contain a parameter of $\beta_1$. Now you could do the null test for $\beta_1=0$ by just plugging in the value of $\beta_1$, which is $0$, and compare value you get v.s. the critical value given certain significance level.

But notice all the above is just for a hypothesis test to a random variable, we plugged in $\beta_1=0$ because this is what we want to test. But all the test is not gonna change our linear regression model, and it is a separate hypothesis test for a random variable we derived from the regression model.

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  • $\begingroup$ So then why does the estimator of $\beta_0$ change? $\endgroup$ – Euler_Salter May 7 '17 at 12:54
  • $\begingroup$ @Euler_Salter Sorry for the late reply, was out today. I've updated my post, in the hope to answer your question. In sum, I think you are a bit confused about hypothesis test. $\endgroup$ – Yujie Zha May 7 '17 at 22:18

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