11.9 Regression (Fuel Efficiency)

Step 1 — Ask which variable does the predicting. A car's weight is a physical property fixed by its design; how far it travels on a gallon of fuel responds to that weight. So weight predicts fuel efficiency, not the other way round.

Step 2 — Assign the roles. Weight is the independent variable \(x\), plotted horizontally. Fuel efficiency is the dependent variable \(y\), plotted vertically.

Step 3 — Predict the sign of the slope. A heavier car requires more energy to move, so it covers fewer miles per gallon. As \(x\) increases, \(y\) decreases, which is a negative slope.

Step 4 — Note why this matters before collecting. Deciding the expected direction in advance gives you a check on your own arithmetic. If your calculator returns a positive slope for these variables, suspect a data-entry error before you believe the result.

Answer: Weight is independent, fuel efficiency is dependent, and the slope should be negative.

Step 1 — Consider the range of the sample. Sports cars, pickups and hybrids are all at extremes of the weight-and-efficiency space, with few ordinary mid-size sedans between them. A sample clustered at the ends can make a relationship look far tighter than it is, because the middle — where cars actually disagree with the trend — is missing.

Step 2 — Consider the mix. Hybrids get high mileage for reasons that have little to do with weight: an electric motor and regenerative braking. Including several of them mixes two different mechanisms into one line, and the fitted slope then describes neither cleanly.

Step 3 — Name the general problem. The sample was chosen by the person who has a hypothesis about it. That is not a random sample of new model cars, so no conclusion drawn from it generalises to new model cars.

Answer: The extremes-only spread inflates the apparent strength of the relationship, and mixing in hybrids blends a second mechanism into the same line. Both stem from the selection being made by judgment rather than at random.

Step 1 — Place the first point. Weight 3,200 is toward the left of the horizontal axis; 28 mpg is high on the vertical axis. The point sits in the upper-left region.

Step 2 — Place the second point. Weight 5,400 is toward the right; 15 mpg is low. The point sits in the lower-right region.

Step 3 — Read the direction. Moving from left to right, the points drop. High values of weight occur with low values of fuel efficiency, which is a negative relationship.

Step 4 — State the limit of two points. Any two distinct points determine a line, so two points can always be joined by one. Only the full set of 20 shows whether the relationship is actually linear and how tightly the data follow it.

Answer: Upper-left and lower-right respectively, suggesting a negative relationship — though two points alone cannot establish one.

Step 1 — Write the equation. Substituting \(a\) and \(b\) into \(\hat{y} = a + bx\):

$$ \hat{y} = 47.1976 - 0.0064x $$

Step 2 — Test significance. The rule compares \(|r|\) against the critical value. Here \(|-0.8912| = 0.8912\) and the critical value is 0.444. Since \(0.8912 > 0.444\), the correlation is significant.

Step 3 — State what significance licenses. Because \(r\) is significant and the scatterplot shows a linear trend, the line may be used to predict fuel efficiency from weight — within the range of weights actually sampled.

Step 4 — Compute the coefficient of determination.

$$ r^{2} = (-0.8912)^{2} \approx 0.7942 $$

Step 5 — Interpret it. About 79% of the variation in fuel efficiency is explained by variation in weight, using the regression line. The remaining 21% is not — engine design, transmission, aerodynamics and tyre choice all matter and none of them is weight.

Answer: \(\hat{y} = 47.1976 - 0.0064x\); the correlation is significant since \(0.8912 > 0.444\); \(r^{2} \approx 0.7942\), so about 79% of the variation in mpg is explained by weight.

Step 1 — Substitute 3,000.

$$ \hat{y} = 47.1976 - 0.0064(3000) = 47.1976 - 19.20 = 27.9976 $$

About 28.0 miles per gallon.

Step 2 — Substitute 4,000.

$$ \hat{y} = 47.1976 - 0.0064(4000) = 47.1976 - 25.60 = 21.5976 $$

About 21.6 miles per gallon.

Step 3 — Check the direction. The heavier car has the lower predicted efficiency, which matches the negative slope and ordinary experience. That agreement is a free check on the arithmetic.

Step 4 — Note the size of the gap. A thousand extra pounds costs about 6.4 mpg on this line — which is exactly \(1000 \times 0.0064\), the slope scaled up.

Answer: About 28.0 mpg at 3,000 pounds and about 21.6 mpg at 4,000 pounds.

Step 1 — Round each to four decimal places. \(a = 47.1976\), \(b = -0.0064\), \(r = -0.8912\).

Step 2 — Predict with the rounded values.

$$ \hat{y} = 47.1976 - 0.0064(3600) = 47.1976 - 23.04 = 24.1576 $$

about 24.2 mpg.

Step 3 — Predict with the full-precision values.

$$ \hat{y} = 47.1976382 - 0.00641197(3600) \approx 47.19764 - 23.08309 = 24.11455 $$

about 24.1 mpg.

Step 4 — Compare. Rounded to one decimal place the answers differ, 24.2 against 24.1. The slope was rounded in the fourth decimal, but it is multiplied by 3,600, so that small change is magnified roughly 3,600-fold.

Step 5 — State the working rule. Keep full precision inside the calculator and round only the final answer. Round the coefficients first and the error rides along with every prediction you make.

Answer: \(a = 47.1976\), \(b = -0.0064\), \(r = -0.8912\); the rounded line gives about 24.2 mpg and the full-precision line about 24.1 mpg, because a slope rounded in the fourth decimal is multiplied by a four-digit weight.

Step 1 — Apply the definition literally. For each additional pound of weight, predicted fuel efficiency falls by 0.0064 miles per gallon, on average.

Step 2 — Scale to something imaginable. Multiply both sides by 1,000: for each additional 1,000 pounds of weight, predicted fuel efficiency falls by about 6.4 miles per gallon, on average.

Step 3 — Explain why the raw version is unhelpful. Nobody compares cars one pound at a time. A difference of 0.0064 mpg is far below what any driver could notice or any test could measure reliably, so the sentence is true and conveys nothing about the size of the effect.

Step 4 — Note what the scaling does not change. Scaling the interpretation does not change the line, the slope, or any prediction. It changes only the unit the sentence is phrased in.

Answer: For each additional 1,000 pounds, fuel efficiency drops about 6.4 mpg on average. The per-pound form is correct but too small a unit to be meaningful to a reader.

Step 1 — Say what the slope actually measures. The slope is the change in \(y\) per one-unit increase in \(x\). It is expressed in the units of the problem — dollars per page in one case, miles per gallon per pound in the other.

Step 2 — Note that the units are not comparable. A page and a pound are different things. Rescaling weight from pounds to tons would multiply this slope by 2,000 without changing a single data point or the relationship at all.

Step 3 — Identify what DOES measure strength. The correlation coefficient \(r\), and its square. Textbooks gave \(r = 0.8412\) and this lab \(r = -0.8912\); squaring, about 71% versus about 79% of the variation explained. By that measure the fuel-efficiency relationship is slightly stronger, the opposite of the classmate's conclusion.

Step 4 — Address the negative sign. The sign records direction only. A negative correlation of \(-0.8912\) is exactly as strong as a positive one of \(0.8912\).

Answer: Slope magnitude depends on the units and says nothing about strength; \(r\) and \(r^{2}\) measure strength, and by those the fuel-efficiency relationship is marginally the stronger of the two.

Step 1 — Check the range first. The sampled weights run from about 2,400 to 5,800 pounds. A weight of 10,000 pounds lies far outside that range, so any prediction there is extrapolation.

Step 2 — Compute it anyway, to see what happens.

$$ \hat{y} = 47.1976 - 0.0064(10000) = 47.1976 - 64.00 = -16.8024 $$

Step 3 — Judge the answer against reality. The line predicts about \(-16.8\) miles per gallon. A negative fuel efficiency is impossible; a vehicle cannot travel a negative distance on a gallon of fuel.

Step 4 — Explain why the line did that. A straight line with a negative slope, extended far enough, must eventually cross zero and keep falling. Nothing in the data says real vehicles behave that way — a 10,000-pound truck gets poor mileage, not negative mileage.

Step 5 — Answer the question. No, we cannot predict the fuel efficiency of a 10,000-pound car from this line. The weight is outside the range of the sample, and the equation returns a value that is not merely uncertain but impossible.

Answer: No. The prediction is \(-16.8\) mpg, which cannot occur, because 10,000 pounds lies far outside the sampled range of 2,400 to 5,800 pounds.

Step 1 — Identify the outlier. The point at (4,100 lb, 42 mpg) is the outlier. Nineteen points follow a downward trend; on that trend a 4,100-pound car would be predicted near \(47.1976 - 0.0064(4100) \approx 20.9\) mpg, so 42 mpg sits about 21 mpg above the line.

Step 2 — Consider the likeliest explanation. A car this heavy achieving 42 mpg is almost certainly a hybrid or plug-in, whose efficiency comes from an electric drivetrain rather than from being light.

Step 3 — Decide what to do. Do not delete it merely because it spoils the line. It is real data about a real car, and it is telling you something true: weight is not the only thing that determines fuel efficiency.

Step 4 — Name the defensible options. Either keep it and report it, noting that the relationship holds for conventional drivetrains and that hybrids sit above the line; or restrict the study to non-hybrid vehicles and say so explicitly. What is not defensible is quietly dropping it to raise \(r\).

Answer: (4,100 lb, 42 mpg) is the outlier, most likely a hybrid. Keep it and report it, or narrow the study's stated scope — but do not delete it silently.

Step 1 — Judge whether the line fits. Partly. The line describes the light cars well, and the heavy ones poorly, so a single yes-or-no answer would hide what the plot is showing.

Step 2 — Name the pattern. The spread of the residuals grows as \(x\) increases. That violates the constant-error-variance condition from Section 11.4, which requires that residuals not consistently widen as the explanatory variable increases.

Step 3 — Say what it means for predictions. Predictions for light cars are more trustworthy than predictions for heavy ones, even though a single \(r\) value summarises the whole plot with one number and hides that difference.

Step 4 — Say what the correlation implies. A strong negative \(r\) still tells you heavier cars get worse mileage on average, and that is what most people would expect. It does not tell you the line predicts equally well everywhere.

Answer: The line fits well at low weights and poorly at high ones; the widening residuals violate constant error variance, so predictions for heavy cars carry more uncertainty than the single \(r\) suggests.