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Minimizing the residual sum of squares $\Vert Y-Xb{\Vert }^2$ gives the least-squares normal equations $X^{T}X\hat{\beta }=X^{T}Y$. Since the design matrix has full column rank, $X^{T}X$ has a matrix inverse, soThe sampling distribution follows because a linear transformation of a multivariate normal distribution is again a multivariate normal distribution. Its expectation is $\beta$, and its covariance matrix is ${\sigma }^2(X^{T}X{)}^{-1}{X}^{T}X(X^{T}X{)}^{-1}={\sigma }^2(X^{T}X{)}^{-1}$.

The hat matrix is $H=X(X^{T}X{)}^{-1}{X}^T$. It is the orthogonal projection matrix onto the column space of $X$: $H^T=H$, $H^2=H$ and $HX=X$. Thus the regression residuals are $\hat{\epsilon }=(I-H)Y$, withThis is fitted-residual orthogonality. The vectors are jointly normal, so independence of uncorrelated jointly normal variables strengthens both zero-covariance statements to independence.

For the quadratic linear regression, the columns of the design matrix are $1,x,x^2$. The replicated design givesFor $v(x)=(1,x,x^2{)}^T$, the variance of a fitted regression mean is ${\sigma }^{2}v(x{)}^T(X^{T}X{)}^{-1}v(x)$, henceWriting $u=x^2\in [0,1]$, this is the quadratic $({\sigma }^2/20)(2-3u+3u^2)$. Its derivative vanishes at $u=1/2$; it decreases before that point and increases afterwards. Therefore the maximum is ${\sigma }^2/10$ at $x=-1,0,1$, and the minimum is ${\sigma }^2/16$ at $x=\pm 1/\sqrt{2}$.

At $100^{\circ }\mathrm{C}$, $x=0$ and the expected yield is $\alpha$. The unbiased residual variance estimate is $s^2=2.43/(30-3)=0.09$. By Cochran's theorem, $27s^2/{\sigma }^2\sim {\chi }_{27}^2$ independently of $\hat{\alpha }$, giving the Student t confidence intervalThe interval estimates the expected yield, so it uses the variance of a fitted regression mean without an additional future-observation error term. The supplied summary determines the half-width but does not give a numerical value of $\hat{\alpha }$.

The least-squares estimator minimizes $(Y-Xb{)}^T(Y-Xb)$. Differentiation gives $X^{T}X\hat{\beta }=X^{T}Y$, and the full column rank of $X$ makes $X^{T}X$ positive definite. HenceThe fitted values are $\hat{Y}=X\hat{\beta }=HY$, where the hat matrix is $H=X(X^{T}X{)}^{-1}{X}^T$. The regression residuals are $\hat{\epsilon }=Y-\hat{Y}=(I-H)Y$. Since $H$ is the orthogonal projection matrix onto the column space of $X$, $(I-H)X=0$ andThis is a singular multivariate normal distribution supported on the $(n-p)$-dimensional orthogonal complement of the column space: its components are generally correlated. The equality $\hat{Y}=X\hat{\beta }$ puts the fitted vector in that column space, while $X^T\hat{\epsilon }=0$ expresses fitted-residual orthogonality. No residual component lies in an explanatory direction that could improve the least-squares fit.

For the calibration model, write $R_i$ for a reading and $a_i$ for the known amount. The first fit assumes $R_i=\alpha +\beta a_i+{\epsilon }_i$ with independent $N(0,{\sigma }^2)$ errors. Its residual-versus-fitted plot has systematic changes in group means and an increasing within-group spread. These suggest both a nonlinear mean relationship and heteroscedasticity, rather than the random scatter expected from a satisfactory normal linear model.

The Box–Cox transformation plot is a profile log-likelihood for the power parameter $\lambda$. Its maximum is near $0.94$; the horizontal line gives an approximate 95% likelihood-ratio confidence interval, which excludes the untransformed value $1$. Since affine rescaling of the response does not change the fitted mean space, fitting the raw power $R_i^{0.94}$ is equivalent to using $(R_i^{0.94}-1)/0.94$ with an intercept. The second model isIts residual spread is more even, but observation 17 has a conspicuously negative residual near $-9.95$. This motivates checking for an outlier or recording error. The third fit omits that observation; omission should be justified by that investigation, since choosing exclusions from residuals can affect inference. Some group pattern still remains, so the transformation alone does not certify a perfect model.

At an amount of $3$ nanograms, the third fit predicts the transformed meanSimply applying the inverse power gives ${\hat{m}}^{1/0.94}=118.5187$. The requested expected reading needs bias correction after an inverse transformation, since $\mathbb{E}[g(Z)]\ne g(\mathbb{E}[Z])$ for nonlinear $g$. With $q=1/0.94$ and $s=2.832$, a second-order Taylor expansion givesHere the correction uses the residual variance, rather than the variance of the estimated fitted mean. The power-transformed Gaussian model is an approximation for positive readings; the fitted mean is over 31 residual standard deviations above zero, so the negative tail has negligible effect on this local approximation. The ordinary inverse-transformed prediction is therefore essentially $118.52$ at the displayed precision.

The regression residual vector is $e=Y-X\hat{\beta }=(I-H)Y$, where $H=X(X^{T}X{)}^{-1}{X}^T$ is the hat matrix. Both $H$ and $I-H$ are symmetric idempotent matrices, and $(I-H)X=0$. Therefore $e=(I-H)\epsilon$ and the linear image of a multivariate normal vector givesThis multivariate normal distribution is singular, with rank $n-p$: fitted-residual orthogonality forces $X^{T}e=0$. Specifically, $\mathrm{Var}(e_i)={\sigma }^2(1-h_{ii})$ and $\mathrm{Cov}(e_i,e_j)=-{\sigma }^2{h}_{ij}$ for distinct observations. Thus regression residuals are generally neither independent nor identically distributed, even though the model errors are. Also $\mathrm{RSS}/{\sigma }^2\sim {\chi }_{n-p}^2$ and is independent of $\hat{\beta }$, which justifies the standard errors and exact Student t-tests below.

Because both vectors are linear transformations of the same multivariate normal response, $(\hat{Y},e)$ is jointly multivariate normal. Its cross-covariance matrix isZero cross-covariance implies independence for jointly multivariate normal vectors, including singular ones. Therefore the fitted values and the entire vector of regression residuals are independent. The fitted-residual orthogonality identity gives the zero covariance; the normal distribution assumption is what upgrades it to independence.

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.