0
$\begingroup$

Note: $b_1$ is the estimate for $β_1$

What I tried:

$ \operatorname{var}(b_1) = \operatorname{var}( Σ(x_iy_i)/Σx_i^2)$

$ \operatorname{var}(b_1)= (1/(Σx_i^2)^2) * Σx_i * \operatorname{var}(y_i)$

Since $ \operatorname{var}(y_i) = σ^2$

$ \operatorname{var}(b_1) = σ^2/Σx_i^2$

Is this correct?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.