= Solution
Because `model2` uses a <logarithmic transformation>, the logarithm of the payment ratio is the difference of two predicted log payments. The cars have equal `Brand` and `Bonus` values and differ by one `Kilometres` level, so its fitted mean is the displayed mileage <regression coefficient>, $\widehat\beta_K=0.057132$.
For two independent future payments, the <prediction interval> must include both the <standard error> $0.032654$ of $\widehat\beta_K$ and two independent residual errors. The estimated variance of the predicted log ratio is therefore
$$
0.032654^2+2(0.7817)^2.
$$
Using the residual degrees of freedom shown in the output, a 95% <prediction interval for a ratio of log-normal responses> is
$$
\boxed{
\left[
\exp\!\left\{0.057132-t_{284,0.975}\sqrt{0.032654^2+2(0.7817)^2}\right\},
\exp\!\left\{0.057132+t_{284,0.975}\sqrt{0.032654^2+2(0.7817)^2}\right\}
\right].
}
$$
With $t_{284,0.975}\simeq1.97$, this is approximately $[0.120,9.33]$. The interval is wide because it predicts the ratio of two individual future payments; an interval for the ratio of their fitted mean responses would omit the $2\widehat\sigma^2$ term.
Back to article page
