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The ordinary least squares objective has gradient $2X^{T}X\beta -2X^{T}Y$. The normal equation therefore givesThe Gram matrix $X^{T}X$ is a positive-definite matrix because $X$ has full column rank. More explicitly, for any $b$,since $X^T(Y-X\hat{\beta })=0$. This proves that the displayed solution is the unique global minimum.

The fitted values are $\hat{Y}=X\hat{\beta }=HY$, where the hat matrix isThe inverse of the symmetric matrix $X^{T}X$ is symmetric, so $H^T=H$. Multiplication gives $H^2=X(X^{T}X{)}^{-1}(X^{T}X)(X^{T}X{)}^{-1}{X}^T=H$. Thus $H$ is the orthogonal projection onto the column space of the design matrix.

The regression residual vector is $e=Y-\hat{Y}=GY$ with $G=I_n-H$. Consequently $G^T=G$, $G^2=I_n-2H+H^2=G$, and the residual sum of squares isBy fitted-residual orthogonality, $GH=0$. Using $\mathrm{Cov}(Y)={\sigma }^2{I}_n$, the cross-covariance matrix isBoth vectors are jointly normal, so they are also independent by independence of uncorrelated jointly normal variables. The zero covariance calculation itself only needs the common error variance and lack of error correlations.

For the first wind fit, let $v_i$ be wind velocity and $Y_i$ electrical output. The simple linear regression is $Y_i=\alpha +\beta v_i+{\epsilon }_i$, with independent ${\epsilon }_i\sim N(0,{\sigma }^2)$. It has two fitted mean parameters and $25-2=23$ residual degrees of freedom. The missing analysis of variance entries are therefore velocity degrees of freedom $1$, velocity mean square $8.9296$, and residual degrees of freedom $23$. The residual mean square is $1.2816/23\simeq 0.05572$, and $8.9296/(1.2816/23)\simeq 160.25$, consistent with the printed value after rounding.

Because the model includes a regression intercept, the coefficient of determination is the explained sum of squares divided by the total centered sum of squares:A large coefficient of determination does not rule out a wrong mean function. In the PDF's first residual-versus-fitted plot, residuals are negative at both ends and positive in the middle. This curved pattern agrees with the visibly flattening output-versus-velocity relationship and motivates a polynomial regression with a quadratic term.

The second wind model is $Y_i=\alpha +{\beta }_1{v}_i+{\beta }_2{v}_i^2+{\epsilon }_i$, again with independent common-variance normal errors. Its fitted mean is $-1.155898+0.722936v_i-0.038121v_i^2$. The line marked (A) is a two-sided Student t-test of $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$, conditional on retaining the intercept and linear term. Under the null hypothesis,Its two-sided p-value is $6.59\times 10^{-8}$. This is far below $0.05$, so reject a purely linear mean in favour of the quadratic fit. Equivalently, the extra-term nested-model F-test has $F=t^2\simeq 63.15$ and null law $F_{1,22}$.

The third fit is a reciprocal-predictor regression, $Y_i=a+b/v_i+{\epsilon }_i$, with estimated mean $2.9789-6.9345/v_i$. It has two mean parameters rather than the quadratic model's three. Its residual standard error is smaller, $0.09417$ versus $0.1227$, and its coefficient of determination is larger, $0.9800$ versus $0.9676$. The corresponding residual sums of squares are approximately $23(0.09417{)}^2=0.20396$ and $22(0.1227{)}^2=0.33122$. The reciprocal fit is preferable on both these fit measures and parsimony. The two models are not nested, so an ordinary extra-term F-test between them is inappropriate. On the common normal-error likelihood, the difference ${\mathrm{AIC}}_3-{\mathrm{AIC}}_2=25\log (0.20396/0.33122)-2\simeq -14.12$ also favours the third model.

Neither lower residual sum of squares nor a higher coefficient of determination establishes adequate assumptions. The quadratic residual-versus-fitted plot removes the original pronounced curvature; the reciprocal plot also has no comparably obvious mean trend. The low-output residuals appear somewhat more spread out, so check scale-location plots and residuals against velocity for heteroscedasticity. Q-Q plots assess normality; residuals against observation order assess serial dependence; regression leverage and Cook's distance identify influential observations. Further cross-validation, replicate observations at comparable velocities, and prediction errors would help choose between their extrapolation behaviours. Both fitted shapes should be judged principally over the observed positive-velocity range.

The command fits the simple linear regressionusing ordinary least squares. The fitted conditional mean is a straight line in time. The regression residuals are mainly positive at the beginning and end and negative in the middle, so the mean function needs curvature.

A direct improvement is polynomial regression of degree two:The nested-model F-test tests $H_0:{\beta }_2=0$ against a nonzero quadratic coefficient. If $N$ measurements were used, its statistic isA small reported $p$-value supports the quadratic term. The new regression residuals should then be checked for remaining systematic structure, changing variance and temporal dependence, because the F-test assumes independent errors of common variance. Radioactive decay also motivates an exponential mean, but the displayed curvature already justifies this simple nested comparison.