Compare pollution models and audit a damaged regression fragment

Plot average traffic flow horizontally and the corresponding hourly pollution reading vertically. The original response shows a curved increasing pattern; the independently drawn second panel shows the squared-response transformation considered below. The sampling unit is an hour, not eight different stations or eight vehicles.

Using all eight pairs gives $0.9588684475\ldots$, matching the printed $0.959$. Despite the large positive value, inspect the curved scatter before selecting a straight response model.

Square each pollution reading while retaining its matching traffic flow. The coefficient is $0.9951588471\ldots$, matching the printed $0.995$. This transformation is of the response, not of flow.

The original scatter curves and flattens, while the squared-response scatter is close to a straight rising line and has a correlation closer to 1. Together these support the transformed candidate. This compares fit to this sample; it does not establish a causal mechanism or a universally valid law.

The full independently calculated regression is $y^2=c+dx$, with

$\begin{aligned}c&=-47.98514143459646,\\d&=0.027897204434110024.\end{aligned}$

matching the printed rounded form $y^2=0.0279x-48.0$. With the unrounded coefficients,

$\begin{aligned}&y\\ & =\sqrt{c+2000d}\\ & =2.794506653\ldots\approx2.8.\end{aligned}$

Take the positive root, consistent with the supplied positive pollution scale and observations. 2000 vehicles per hour is within 1796–2850; this is interpolation. The source prints 2.8 without a precision instruction, so that displayed result is preserved; 2 s.f. states its precision rather than inventing a printed requirement.

The sampled hourly observations do not control possible factors such as weather or other pollution sources. Those are possible unmeasured influences, not data supplied by the question. A transformed coefficient close to 1 does not establish that changing traffic alone causes the observed pollution change.

Both genuine regression lines pass through the mean point. From the first equation,

$\begin{aligned}&\bar y\\ & =2.5(4.4)+3.8\\ & =14.8.\end{aligned}$

Write the fragment as $x=1.5y-k$; then

$4.4=1.5(14.8)-k\quad\Rightarrow\quad k=17.8,$

which reproduces the printed missing-number answer. However, for a nondegenerate paired dataset, the product of the two regression slopes equals $r^2\le1$. Here $2.5(1.5)=3.75>1$, so these cannot both be the least-squares regressions of the same data. The formal mean-point completion is retained, but the source fragment is mathematically inconsistent as stated. No corrected slope or replacement dataset is invented.