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Write $q=n-p$. In this normal linear model, the ordinary least squares criterion is $\Vert Y-Xb{\Vert }^2$. Differentiating with respect to $b$ gives the normal equations for linear least squares, $X^{T}Xb=X^{T}Y$. Full column rank of the design matrix makes $X^{T}X$ invertible, and completing the square proves that the unique minimum isSince $\hat{\beta }=\beta +(X^{T}X{)}^{-1}{X}^Tϵ$, its sampling distribution is the multivariate normal distributionLet $H=X(X^{T}X{)}^{-1}{X}^T$ be the hat matrix. It is symmetric and idempotent with rank $p$, and $HX=X$. The regression residual vector is $(I-H)Y=(I-H)ϵ$, soAn orthogonal projection of a Gaussian vector onto the residual space gives $\mathrm{RSS}/{\sigma }^2\sim {\chi }_q^2$. The residual projection and fitted projection have zero cross-covariance, hence are independent because they are jointly normal. In particular, $s^2$ is an unbiased estimator of ${\sigma }^2$ and is independent of $\hat{\beta }$. The maximum-likelihood estimator of ${\sigma }^2$ instead divides by $n$ and is not unbiased.

For the null hypothesis ${\beta }_1=0$, put $v_{11}=((X^{T}X{)}^{-1}{)}_{11}$. Under that null hypothesis, ${\hat{\beta }}_1/(\sigma \sqrt{v_{11}})$ is standard normal and independent of $q{s}^2/{\sigma }^2$. Therefore the Student t-test usesA two-sided level-$\alpha$ test rejects when $|T|>t_{q,1-\alpha /2}$. The denominator is the estimated standard error of the regression coefficient.

The fitted linear regression includes an intercept and three covariates:Thus $n=25$, $p=4$ and the residual degrees of freedom are $21$. The fitted conditional mean isEach slope measures a conditional association with the response while the other two covariates are held fixed. Increasing wind speed by one mile per hour corresponds to a fitted decrease of $0.527410$ in the concentration units used. Its standard error is $0.224904$, giving $t=-2.345$ and a two-sided p-value of $0.0289$: evidence of a negative conditional association at the 5% level. The temperature slope is $0.124991$ per degree Fahrenheit, with standard error $0.075173$, $t=1.663$ and p-value $0.1112$. The insolation slope is $-0.005259$ per langley per day, with standard error $0.006637$, $t=-0.792$ and p-value $0.4370$. Neither of these last two conditional effects is significant at 5%; that does not establish that either true slope is zero. These are associations, not a demonstration of causal effects.

The intercept estimates the mean at zero values of all three covariates. Its value $3.784916$ has a large standard error $7.979647$, giving $t=0.474$ and p-value $0.6402$. The zero-covariate combination may be outside the observed range, so this test may have little practical relevance. Every coefficient test uses a Student t-distribution with $21$ degrees of freedom.

The reported residual standard deviation $s=1.844$ measures unexplained variation in concentration units; the subsequent table gives $\mathrm{RSS}=71.428$, consistent with $s=\sqrt{71.428/21}$ after rounding. The five-number residual summary describes the spread, from $-2.3052$ to $3.4033$, with median $-0.4990$. An intercept forces the residual sum to zero, not the residual median. These five numbers alone cannot establish normal errors, constant variance or independence; residual-versus-fitted and quantile-quantile plots would be more informative, and successive days also warrant checking serial dependence. The coefficient of determination $R^2=0.6533$ means that 65.33% of the sample response sum of squares about its mean is explained by this fit. Its adjusted version iswhich accounts for the number of estimated regression coefficients.

The stepwise selection by the Akaike information criterion searches between the intercept-only model and the specified model with all three main effects. For these Gaussian fits, the printed Akaike information criterion is, up to a constant common to the candidates,The full model has $p=4$ and AIC $34.246$. Removing insolation increases the residual sum of squares by $2.136$ but decreases AIC to $32.982$, so that removal is chosen. Removing temperature or wind instead would give AIC $35.337$ or $38.060$ and is worse. In the next step the candidate models include deletions and restoration of insolation. Keeping wind and temperature has the smallest AIC: removing temperature gives $33.389$, restoring insolation gives $34.246$, and removing wind gives $37.960$. Hence the search stops, givingThe accompanying F-tests compare a larger model with a one-parameter reduction: $F=\Delta \mathrm{RSS}/s^2$, where $s^2$ is estimated from the larger model. For example, removing insolation gives $2.136/(71.428/21)=0.628$, equal to the corresponding squared test statistic. Deletions from the second-stage model use $73.564/22$ in the denominator, whereas restoring insolation uses the full-model denominator. These tests describe the comparisons; they do not determine the AIC choice. In particular, temperature is retained despite its deletion-test p-value $0.15016$. AIC balances fit and complexity, and a stepwise optimum need not be a global optimum over a richer model class. Ordinary coefficient inference after choosing a model does not account for the selection itself.

For positive responses, the Box–Cox transformation is $T_{\lambda }(y)=(y^{\lambda }-1)/\lambda$ for $\lambda \ne 0$, and $T_0(y)=\log y$. Its profile likelihood refits the linear regression and variance at each $\lambda$ and includes the transformation Jacobian. The graph peaks at roughly $\lambda =0.3$–$0.4$. The approximate 95% likelihood-ratio confidence interval contains values a little below zero through approximately one, using the cutoff ${\ell }_{\max }-{\chi }_{1,0.95}^2/2$. Thus a logarithmic transformation ($\lambda =0$) and a square-root transformation ($\lambda =1/2$) are plausible, while leaving the response on its original scale ($\lambda =1$) is at or very near the upper confidence boundary. The broad profile supports considering a mild power transformation; it does not determine a sharply estimated power. The raster does not justify more precise endpoints or a decisive assertion about which side of the boundary $\lambda =1$ lies.

If a department has population $P$, $C_{mpers}$ crimes against persons and $C_{mprop}$ crimes against property, cancellation of the common numerator givesThus $y$ is a property-to-person crime count ratio, not its reciprocal. Both counts must be positive for the printed ratio to be defined.

For the reconstructed 85-row data, fit a normal linear model for the Box–Cox transformation $T_{\lambda }(y)$, including four region contrasts plus Literacy, Donations and Wealth. With $z_i(\lambda )=(y_i^{\lambda }-1)/\lambda$ and $z_i(0)=\ln y_i$, the profile likelihood, up to constants, isThe second term is the transformation's Jacobian determinant; omitting it would incorrectly compare different response scales. Maximization gives $\hat{\lambda }=0.143$, with approximate 95% likelihood-ratio confidence interval $[-0.120,0.406]$. The logarithmic transformation is inside this interval, whereas the raw-scale choice $\lambda =1$ is far outside: the twice-log-likelihood drops at zero and one are $1.14$ and $39.54$, respectively. A log-response model is therefore a simple supported choice.

In the full log model, partial F-tests give $F_{4,77}=13.15$ for Region ($p=3.3\times 10^{-8}$), $F_{1,77}=27.99$ for Wealth ($p=1.1\times 10^{-6}$), and $p=0.786,0.682$ for Literacy and Donations. Dropping Literacy and Donations together gives $F_{2,77}=0.109$ and $p=0.897$. A parsimonious final normal linear model, with C as reference and $W=\mathrm{Wealth}/10$, isThe standard errors in that order are $0.1056,0.1105,0.1158,0.1102,0.1104,0.01539$; residual standard deviation is $0.3212$ on 79 degrees of freedom and $R^2=0.626$. Use residual-versus-fitted and quantile-quantile plots, influence diagnostics and checks for remaining curvature to assess the working Gaussian model; the small p-values alone do not validate its assumptions.

Holding Region fixed, worsening wealth rank by ten positions multiplies the fitted conditional median ratio by $e^{-0.08539}=0.918$. Relative to C at the same wealth rank, the fitted median multipliers are $1.081$ for E, $1.375$ for N, $0.563$ for S and $1.212$ for W. Region and wealth rank explain much of the variation; Literacy and Donations add little once those terms are included. These are associations in the reconstructed dataset. Back-transforming the fitted log mean gives a conditional median under lognormal errors; estimating an original-scale mean requires bias correction after an inverse transformation. Also, the public archive's metadata describe some historical measures differently from the exam wording, so the fitted Donations effect is interpreted only on its recorded scale.

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.