Compare used-car prices with reciprocal mileage

The supplied $r=-0.78$ indicates fairly strong negative linear association, but the seven plotted observations show a decreasing curve that levels off. Thus a decreasing relationship is supported; a coefficient of this size alone does not make a straight-line price model appropriate.

The new variables are $1.6d$ and $1.7p$. Multiplying each variable by a positive constant changes the units without changing correlation. These are the conversion rates supplied for this exercise.

Its larger engine and extra features affect price in addition to mileage. The remaining cars are intended to represent the ordinary version, so treating this distinct specification as comparable would mix populations. The exclusion is supported by the supplied specification information, rather than merely by an inconvenient position on the scatter. Use only the first six pairs in (d)–(g).

Retain the pairings and use the unrounded transformed values for fitting:

A sensible by-eye line runs through the middle of the transformed points. The diagram illustrates $p=9000+550w$ as one possible hand-drawn choice, not a unique answer or the calculator regression requested in (f).

For a candidate line $\hat p=a+bw$, mark the vertical segment at each observed $w_i$, from the fitted price $\hat p_i$ to the observed price $p_i$. The signed residual is $e_i=p_i-\hat p_i$; the geometric segment has length $|e_i|$. Squaring prevents positive and negative residuals from cancelling and weights larger discrepancies more heavily. Least squares chooses the coefficients minimizing $\sum_{i=1}^{6}e_i^2$. The second diagram marks these vertical segments for the computed regression; horizontal or perpendicular distances solve different fitting problems.

Fitting $p$ on $w=100000/d$ with the six comparable cars gives $p=a+bw$, with

$\begin{aligned}a&=9398.492473025944,\\b&=505.73025297673377.\end{aligned}$

The coefficient is $r=0.9869962799\ldots$, which rounds to the printed 0.987. Rounding the fitted intercept and slope gives $p=9400+506w$, equivalently the printed $p=9400+50600000/d$. The original negative coefficient concerned $d$; the transformed variable $1/d$ decreases as $d$ increases, so the positive transformed coefficient is consistent with decreasing price as mileage increases.

Using the full coefficients from (f),

$p=a+b\frac{100000}{5055}=19403.0474281\ldots,$

which gives the printed £19400 to 3 significant figures. The target mileage 5055 is below the comparable sample range 8954–45452, so this is extrapolation; the transformed input is also beyond the observed reciprocal range. The actual £22500 car at 5055 miles was the special version and remains excluded. It does not provide a comparable observation that validates this prediction for an ordinary car.