I'm attempting to prove that the $1-\alpha$ confidence interval for $\beta_i$ in a linear model is given by

$$\hat{\beta} \pm t_{n-p} se(\hat{\beta_i})$$.

I am starting from the fact that for a null hypothesis $H_0:\beta_j=b$, we have the test statistic

$$\frac{\hat{\beta_j} - b}{se(\hat{\beta{i}})}\sim t_{n-p}$$

From this we know

$$P(-t_{n-p}>\frac{\hat{\beta_j} - b}{se(\hat{\beta{i}})}>t_{n-p}) = 1-\alpha$$

And then we get

$$P(b-t_{n-p}se(\hat{\beta{i}})>\hat{\beta_j}>b+t_{n-p}se(\hat{\beta{i}})) = 1-\alpha$$

But this isn't the same as the confidence interval given. I don't know where I'm going wrong. Do I need to define the hypothesis differently? Even if I replace $b$ with $\beta_i$ I don't see how this gives the right interval.

  • $\begingroup$ Remember t* is the upper (1-C)/2 critical value for the t distribution with n-p degrees of freedom, t(n-p). $\endgroup$
    – AJKOER
    Commented Oct 21, 2020 at 22:27

1 Answer 1


I think you may be confusing $\hat \beta_i$ with $\beta_i$ and $b_i$:

  • $\beta_i$ - unknown parameter
  • $\hat \beta_i$ and $b_i$ is the same thing - an estimator for an unknown parameter.

Hint 1: Try to use only $\beta_i$ and $\hat \beta_i$

Hint 2: Then take look at the test statistic.

  • $\begingroup$ Do I need to replace $\hat{\beta_j}$ with $\beta_j$ and $b$ with $\hat{\beta_j}$ in the test statistic? $\endgroup$
    – Sparsity
    Commented Oct 21, 2020 at 22:06
  • $\begingroup$ Will it lead to correct result? $\endgroup$
    – cure
    Commented Oct 21, 2020 at 22:07
  • $\begingroup$ Yes, but I don't understand the reasons why this set up is correct. $\endgroup$
    – Sparsity
    Commented Oct 22, 2020 at 7:57
  • $\begingroup$ Maybe this helps, that we generally want to calculate an interval in which in 95% cases the real value is. The best value to start (to create interval around it) is seemingly an estimator. If this is not hintful enough, you can trying providing some more information. You can try rewriting a question or creating another one, which would not be 'self-study' since you no more ask about possible homework, but search for deeper understanding. $\endgroup$
    – cure
    Commented Oct 22, 2020 at 15:22

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