Compare fitted models and plotted residuals

Try

A synthetic response measured at times 0,1,2,3,4,5 has values 2,4,7,10,15,21. The increase is not constant, but noisy data rarely announce a unique formula. A line, quadratic and exponential are candidate descriptions. Comparing them requires a common response scale and a view of what each leaves unexplained.

A residual is observed response minus fitted response at the same input: e=y−ŷ. A positive residual places the observation above the fitted curve. Squaring and adding residuals gives SSE, the sum of squared errors on the specified scale. Its units are squared response units.

Predict

Predict whether a fitted line will tend to miss the ends and middle in opposite directions. Which candidate has more adjustable coefficients? Does the smallest training SSE prove the best future predictor? Decide what would make a model plausible beyond a small numerical advantage on these six observations.

Experiment

Fit, inspect residuals and challenge the model

Each row is one paired observation. Predict a model family and its residual pattern before revealing calculations. The default data are synthetic. State the units and sampling process supplied by the surrounding investigation; the tool itself does not assign physical units.

Paired observations, one x,y pair per line0,2 1,4 2,7 3,10 4,15 5,21

Use 3–40 pairs of decimal numbers, each between −100 and 100. Rows may repeat; they are separate observations, not automatically independent samples. Pairing and observation order are retained.

Model to inspectLinearQuadraticExponential (log-linear fit)Reset paired dataTry noisy linear dataTry constant yShow paired-data calculations

Record the raw pairs, model criterion, equation, residual display and context-dependent conclusion in your practice response. Computations use floating-point arithmetic; values are rounded for display. Editing data or selecting a model hides old results. This temporary tool neither saves responses nor awards mastery.

Use the default nonlinear data, which are the fixed worked example here. Reveal all three fits and inspect the scatter curves and residual plot for each. Record equations and original-response-scale SSEs. Then choose the noisy linear preset and repeat the comparison, explaining why the scientific question and pattern matter rather than always selecting the most flexible family.

Fit to the six worked pairs	Approximate equation	Original-scale SSE
Linear	ŷ=0.476190+3.742857x	9.676190
Quadratic	ŷ=2.142857+1.242857x+0.5x²	0.342857
Log-linear exponential	ŷ=2.374843 exp(0.459395x)	8.477956

The coefficients are rounded for reading; calculate residuals with the tool’s retained precision. At x=3 the line predicts approximately 11.704762 while the observed value is 10, so the residual is about −1.704762. That numerical difference corresponds to one plotted point below the zero line in the residual display.

Understand

Linear least squares chooses intercept and slope to minimize the original y-scale SSE. Quadratic least squares chooses three coefficients for the same criterion. The exponential candidate is obtained by fitting a line to log y and then exponentiating. It minimizes squared log residuals, not original-scale SSE. Comparing its original-scale residuals with the others is useful, but it does not turn that fitting procedure into nonlinear least squares on y. The method is unavailable for nonpositive y values, and adding an arbitrary constant changes the modeled response.

For the worked line, residuals in increasing x order are approximately 1.524,−0.219,−0.962,−1.705,−0.448,1.810. Positive ends and predominantly negative middle form a curved pattern, suggesting the line misses systematic structure. The quadratic residuals are much smaller and alternate around zero. That supports a tentative comparison, not a proof that a quadratic generated the data.

The extra quadratic coefficient cannot increase the least attainable training SSE because the family includes lines when its squared-term coefficient is zero. Flexibility therefore earns part of its advantage automatically. New observations, held-out predictive checks, measurement error and a plausible mechanism help decide whether the added complexity is useful. Six synthetic points cannot establish a universal family preference.

Interpret coefficients in context. If x is elapsed days and y is millimeters, the line’s slope is fitted millimeters per day. Its intercept describes the fitted value at day zero; here zero was observed, but the coefficient is still a fitted estimate rather than an exact initial condition. For a study observed only at days 5–10, an intercept at day zero would be extrapolation and might have no meaningful interpretation. The coefficient of x² has units millimeters per day squared; it does not represent a constant slope.

A residual plot should label input and residual units and show zero explicitly. Random-looking residuals are encouraging but do not verify independence, equal variability or population representativeness. A pattern may arise from omitted variables, measurement changes or a few unusual points, not only the mathematical family chosen.

Apply

Enter the fresh practice pairs and retain all three equations, comparable SSEs and residual artifacts. Recommend a tentative model and explain its limitations. State whether zero input is meaningful and whether the requested prediction lies inside the observed range. Do not silently remove nonpositive responses to enable the exponential option or compare a log SSE directly with an original-scale SSE.

The workbench has bounded inputs and temporary state. Copy the requested evidence into your response; a displayed curve is a calculation aid, not a stored report or an independent validation result.

Continue

Next use repeated sampling to investigate estimation error. Model assumptions remain central: a precise calculation under the wrong sampling frame cannot fix biased data.