2022 H2 Maths Paper 2, Question 10

There is a fairly strong negative linear correlation between distance driven, $d$, and price, $p$.

Explanation: the product moment correlation coefficient $r = - 0.78$ is negative and reasonably close to $- 1$, so cars in this sample that have been driven further tend to have a lower price, and this decreasing relationship is reasonably well described by a straight line.

The PMCC would be unchanged; it would remain $r = - 0.78$.

Explanation: converting units is a linear scaling of each variable — $d_{\text{km}} = 1.6\, d_{\text{miles}}$ and $p_{\text{SGD}} = 1.7\, p_{\text{GBP}}$, i.e. each variable is multiplied by a positive constant. The PMCC is unaffected by this kind of linear (origin-preserving, positive-scale) transformation of either variable.

This car should not be used in any analysis of the relationship between $d$ and $p$ for the standard model.

Explanation: being a special version with a bigger engine and extra features, it is not part of the same population as the rest of the data. Its price is likely to be higher for reasons unrelated to distance driven, so including it would bias any analysis of the relationship for the standard model.

We exclude the special car and compute $x = \frac{100\, 000}{d}$ for the remaining $6$ cars.

We plot $p$ against $x$ and draw a line of best fit by eye.

The distances used in the least squares regression line of $p$ on $\frac{100\, 000}{d}$ are the vertical distances from each plotted point to the line — that is, the difference between each observed price $p_{i}$ and the price predicted by the line at the corresponding $x_{i} = \frac{100\, 000}{d_{i}}$.

These distances are squared so that points above and below the line (positive and negative deviations) do not cancel out when summed, and so that points further from the line are penalised more heavily than points close to it. The line is chosen to make the sum of these squared vertical distances as small as possible, which is why the procedure is called the method of least squares.

By GC, performing linear regression of $p$ on $x = \frac{100\, 000}{d}$ for the $6$ remaining cars,

For $d = 5055$:

Substituting into the regression equation from (f):

This gives an estimated price of about £$19400$.

Explanation: this estimate is not likely to be reliable. The $6$ points used to fit the line have $x$ ranging only from about $2.20$ to $11.17$, and $x = 19.78$ lies well outside this range, so using the equation here is an extrapolation: the linear relationship may not hold beyond the data. Moreover, $d = 5055$ is exactly the special car excluded in (c), which is not of the standard specification the line was fitted to, so the estimate is doubly unreliable.