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It is one of the exercise in "Introduction to Multiple Linear Regression". Consider the usual linear regression model \begin{equation} y = \beta_0+ \beta_1 x +\epsilon \end{equation} where $\beta_0$ is known.

It can be calculated that the length of the interval of $\beta_1$ with known intercept is $$L_1 = 2t_{\alpha/2,n-1}\sqrt{\frac{MS_{Res1}}{\sum_{i=1}^nx_i^2}}$$ and the length of the interval of $\beta_1$ with unknown intercept is $$L_2 = 2t_{\alpha/2,n-2}\sqrt{\frac{MS_{Res2}}{\sum_{i=1}^n(x_i-\bar{x})^2}}$$ where $MS_{Res1}$ and $MS_{Res2}$ are th mean residual sum of squares of the model with known and unknown $\beta_0$ respectively.

Is the $(1-\alpha)\%$ interval of $\beta_1$ narrower than the estimator for the case where both slope and intercept are unknown?

SP SINGH
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  • Assuming the sums have $n$ terms (not just two terms), note that $\sum_i (x_i-\bar x)^2=\sum_i x_i^2-n(\bar x)^2$. Also think how the critical t stat changes with df. Now compare them. – Golden_Ratio Mar 09 '22 at 05:09
  • @Golden Ratio: $t_{\alpha/2,n-1}$(known intercept)< $t_{\alpha/2,n-2}$ (unknown intercept). What can be said about their MSRes? – SP SINGH Mar 09 '22 at 05:42

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