Question
A logistics analyst fits an ordinary least-squares model to road trips, predicting litres of fuel \(L\) from distance \(x\) measured in kilometres:
\[\widehat L=6+0.10x.\]
The fitted slope has standard error 0.02 litres per kilometre, and the residual root-mean-square size is 2 litres.
The same trips are exported to a second report. Its distance variable \(m=(x-16)/1.6\) measures miles past a checkpoint 16 km from the depot; its fuel variable \(g=(L-4)/4\) measures gallons above a fixed 4-litre priming allowance. Use the report's rounded conventions of \(1\,\mathrm{mile}=1.6\,\mathrm{km}\) and \(1\,\mathrm{gallon}=4\,\mathrm{litres}\).
Without refitting the observations, give the converted regression line, the slope's standard error and the residual size. What fuel reading does the new line predict at \(m=32\), and has the statistical evidence for a nonzero slope changed?
Get hint
Hint 1
Express the old distance and fuel variables in terms of the new ones before substituting into the original line.
Check your answer
Solution
Changing units cannot change the fitted values for the underlying trips; it changes only their coordinates. Here \(x=16+1.6m\) and \(g=(L-4)/4\). Substitute the original fitted line into the new response:
\[
\widehat g=\frac{\widehat L-4}{4}
=\frac{6+0.10(16+1.6m)-4}{4}
=\frac{3.6+0.16m}{4}
=\boxed{0.90+0.04m}.
\]
The slope transforms as \(\widehat\beta_g=(1.6/4)\widehat\beta_L\), so its standard error is scaled by the same positive factor: \(0.02(1.6/4)=\boxed{0.008}\) gallons/mile. The response residual is \(e_g=e_L/4\) for each trip, giving residual RMS \(2/4=\boxed{0.5}\) gallons. Because both slope and slope standard error are multiplied by 0.4, the nonzero-slope test statistic is unchanged:
\[
t_g=\frac{0.04}{0.008}=5=\frac{0.10}{0.02}=t_L.
\]
At \(m=32\), \(\widehat g=0.90+0.04(32)=\boxed{2.18}\) gallons above the allowance. Equivalently, \(x=16+1.6(32)=67.2\) km, so \(\widehat L=6+0.1(67.2)=12.72\) L and \((12.72-4)/4=2.18\).
More generally, if \(x'=a+bx\) and \(y'=c+dy\) with nonzero \(b,d\), least squares is equivariant: predictions, coefficient estimates and their standard errors transform by algebraic substitution, squared residuals scale by \(d^2\), and \(R^2\) and absolute slope \(t\)-statistics stay the same. A shifted predictor changes the intercept's uncertainty through intercept–slope covariance, so one cannot obtain a new intercept standard error from the supplied slope SE alone. The smaller-looking numerical slope is not evidence of a weaker relationship.
Assumptions
The two reports use identical observations and an ordinary intercept-plus-slope OLS fit.
The exercise uses the stated rounded conversion factors as exact within the model.
Root-mean-square residual size refers to the same residual divisor in both reports.
Final Answer
Converted line: \(\hat g=0.90+0.04m\); slope SE 0.008 gallons/mile; residual RMS 0.5 gallons. At \(m=32\), \(\hat g=2.18\) gallons above allowance. The slope t-ratio remains 5.