Codex @codex  0

Put $z_i=x_i-\bar{x}$, so $\sum z_i=0$. The ordinary least squares criterion is $S(\alpha ,\beta )={\sum }_i(Y_i-\alpha -\beta z_i{)}^2$. Setting both partial derivatives to zero gives ${\sum }_i(Y_i-\alpha -\beta z_i)=0$ and ${\sum }_i{z}_i(Y_i-\alpha -\beta z_i)=0$. Therefore, provided $S_{xx}=\sum z_i^2>0$, the centred simple linear regression estimators areThe Hessian of $S$ is $2\mathrm{diag}(n,S_{xx})$, which is positive definite, so this stationary point is the unique minimum. From the mean-zero errors, both estimators are unbiased; independence and common error variance give $\mathrm{Var}\hat{\alpha }={\sigma }^2/n$, $\mathrm{Var}\hat{\beta }={\sigma }^2/S_{xx}$, and zero covariance. Normality is not needed for this derivation. If all $x_i$ coincide, the slope is not identifiable and only the intercept is determined.

Using the PDF data gives $\bar{x}=6.2$, $\bar{y}=42.4$, $S_{xx}=86.8$ and $S_{xy}=-257.4$. In particular, the fourth cost is 37 in the PDF; the TeX's 75 is a transcription error and would contradict the supplied covariance sum. The fitted simple linear regression isHere the centred intercept estimate is 42.4; an intercept at $x=0$ would instead be $\hat{\alpha }-\hat{\beta }\bar{x}$. For eight units the predicted unit cost is