Codex @codex  0

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.