4.M Focus on Modeling (Chapter Supplement)
Ideas in this lesson
- Fit and interpret exponential regressions
- Fit power models and respect physical prediction domains
- Use raw and transformed plots to compare model families
- Fit bounded logistic growth and invert target counts
- Fit attenuation and solve ecological threshold inequalities
Fit and interpret exponential regressions · Worked example
An invented emissions-control study records percentage reduction as input and dollars per vehicle as output. These are not current quotations.
Use every observation. Draw a raw scatter plot, use least squares on log output versus input to fit \(y=C b^{x}=C e^{kx}\), derive the relation between the two bases, and overlay the fitted curve on the raw data. Predict cost at 90% reduction and interpolate at 57%. Compare the extra cost from 60% to 66% with that from 72% to 78%, and explain diminishing returns. Do not treat the input percentages as fractions in this fitted equation. State the domain, units and limitations; give coefficients to six decimals and predictions to three decimals. Show written work; selecting a summary alone is not a complete response. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.
| Reduction (percent) | Cost (dollars per vehicle) |
|---|---|
| \(30\) | \(9.379\) |
| \(36\) | \(11.831\) |
| \(42\) | \(16.7\) |
| \(48\) | \(21.261\) |
| \(54\) | \(29.148\) |
| \(60\) | \(36.824\) |
| \(66\) | \(50.811\) |
| \(72\) | \(68.243\) |
| \(78\) | \(86.765\) |
| \(84\) | \(120.654\) |
Answer: Fitted \(y\approx 2.227431e^{0.047296x}\); predictions 157.198 and 33.007. Interpretation and graph are required.
Positive outputs are required. \(y=C e^{k x}\Rightarrow\ln y=\ln C+kx\). Logs use numerical values in the specified units.
| Reduction (percent) | Input | Log output |
|---|---|---|
| \(30\) | \(30\) | \(2.238473\) |
| \(36\) | \(36\) | \(2.470723\) |
| \(42\) | \(42\) | \(2.815409\) |
| \(48\) | \(48\) | \(3.056874\) |
| \(54\) | \(54\) | \(3.372386\) |
| \(60\) | \(60\) | \(3.60615\) |
| \(66\) | \(66\) | \(3.928113\) |
| \(72\) | \(72\) | \(4.223075\) |
| \(78\) | \(78\) | \(4.463203\) |
| \(84\) | \(84\) | \(4.792927\) |
Fit an ordinary least-squares line \(Y=A+B Z\) to the transformed data, where \(Y=\ln y\) and \(Z=x\). This minimizes \(\sum_i(Y_i-A-BZ_i)^{2}\), not squared errors in the original output scale.
\(B=\dfrac{\sum_i(Z_i-\bar Z)(Y_i-\bar Y)}{\sum_i(Z_i-\bar Z)^{2}},\qquad A=\bar Y-B\bar Z\). Numerically \(A\approx 0.800849,\ B\approx 0.047296\).
\(C=e^{A}\approx 2.227431\), and \(k=B\approx 0.047296,\ b=e^{k}\approx 1.048433\). The recovered exponential curve is \(y\approx 2.227431e^{0.047296x}\). Coefficients are rounded for display, not for later calculations.
The transformed squared-error sum is 0.00518277. The original-output squared-error sum of this same fitted curve is 14.413763. These are different objectives and have different scales; an original-scale nonlinear fit can have different coefficients.
\(b^{x}=e^{x\ln b}\Rightarrow k=\ln b\). The fitted input unit is fixed, so \(b\) is the factor per one input unit.
\(\widehat y(90)\approx 157.198,\qquad\widehat y(57)\approx 33.007\), in cost (dollars per vehicle). The first prediction is outside the observed input interval and is extrapolation; the second lies inside and is interpolation.
The model costs an additional 12.482 dollars per vehicle for 60% to 66%, but 22.018 for 72% to 78%. Equal percentage-point improvements become more expensive, so reduction obtained per additional dollar diminishes. Percent reduction is physically between 0 and 100; a finite exponential fit does not establish the cost or feasibility of exactly 100% reduction.
The fitted curve need not pass through every observation. These rounded noisy measurements describe an approximate relationship; use the model only where its context and domain remain plausible.
Written-work review:
- Use all 10 supplied rows. Draw the raw scatter plot and give the transformed table with every input and natural-log output, using the stated units rather than silently rescaling the data.
- Derive the semilog linearization and state that the fitted line minimizes squared log-output residuals. Compute its intercept and slope using least squares, exponentiate the intercept, and derive the relation between the base-e rate and the alternative exponential base.
- Report coefficients to six decimals and the predictions at inputs 90 and 57 to three decimals with the stated output units. Label the former extrapolation and the latter interpolation relative to the observed interval; use unrounded coefficients for calculations.
- Give both the log-response and original-response squared-error sums of this fitted curve. Distinguish evaluation in the original scale from minimizing that scale's objective; the two fitting methods need not agree.
- Overlay the fitted curve on the actual raw observations with labeled, consistently scaled axes. Do not force the curve through every noisy point.
- Compute the two additional costs, each as a difference of fitted dollar-per-vehicle costs, for the stated equal percentage-point improvements. Explain diminishing returns as less extra reduction per additional dollar, not as decreasing total cost.
- Use percentage numbers as fitted inputs rather than fractions; restrict physical reduction to zero through 100 percent and explain that the fit does not establish cost or feasibility at complete reduction. Preserve dollars per vehicle in forecasts and differences.
- Numerical grading: accept mathematically equivalent fitted forms. For linearized coefficients, transformed tables and final results, allow two units in the last decimal displayed in the worked solution; for a scientific-notation value apply that tolerance to its mantissa. Nonlinear original-response coefficients may differ by 0.00001 if the recomputed original-response SSE agrees within 0.000001 and forecasts within 0.002 output units. Keep extra intermediate digits and evaluate every supplied row; a selected summary without the fitting work, requested displays and interpretation is not a complete response.