= Solution
The <least-squares estimator> minimizes $(Y-Xb)^T(Y-Xb)$. Differentiation gives $X^TX\hat\beta=X^TY$, and the full column <rank> of $X$ makes $X^TX$ positive definite. Hence
$$
\boxed{\hat\beta=(X^TX)^{-1}X^TY.}
$$
The <fitted values> are $\hat Y=X\hat\beta=HY$, where the <hat matrix> is $H=X(X^TX)^{-1}X^T$. The <regression residuals> are $\hat\varepsilon=Y-\hat Y=(I-H)Y$. Since $H$ is the <orthogonal projection matrix> onto the column space of $X$, $(I-H)X=0$ and
$$
\boxed{\hat\varepsilon\sim N_n(0,\sigma^2(I-H)).}
$$
This 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 $X^T\hat\varepsilon=0$ 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 $R_i=\alpha+\beta a_i+\varepsilon_i$ 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 is
$$
R_i^{0.94}=\alpha+\beta a_i+\varepsilon_i,\qquad \varepsilon_i\overset{\mathrm{iid}}\sim N(0,\sigma^2).
$$
Its 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 mean
$$
\hat m=-3.62509+30.87295(3)=88.99376.
$$
Simply 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> gives
$$
\boxed{\widehat{\mathbb E[R\mid a=3]}\simeq\hat m^q+\frac12q(q-1)\hat m^{q-2}s^2=118.5227.}
$$
Here 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.
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