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Correlation and linear regression · Predict in the right direction

Choose a regression model and justify its prediction

Identify the explanatory role before choosing a line. Then check the prediction range, model and population. A regression estimate is not a guaranteed individual outcome.

  1. Choose from the modelling setup

    Does the question prescribe an explanatory variable? Keep that response model when it does. With neither role prescribed, regress the requested response on the supplied predictor.

    y=a+bxy=a+bx

    x=c+dyx=c+dy

    x prescribed explanatoryKeep the y-on-x model

    For y from x, substitute into y=a+bxy=a+bx. For x from y, solve this same model backwards: x=(y−a)/bx=(y-a)/b when b≠0b\ne0. This is inverse calibration, with its own uncertainty.

    y prescribed explanatoryKeep the x-on-y model

    For x from y, substitute into x=c+dyx=c+dy. For y from x, solve this same model backwards: y=(x−c)/dy=(x-c)/d when d≠0d\ne0. Do not switch fits merely because the requested letter changes.

    Neither role prescribedFit the variable being predicted

    For y from x, use y on x. For x from y, use x on y. The two fits minimise residuals in different directions; reversing a fitted equation does not generally produce the other regression.

    A zero fitted gradient cannot give a unique inverse estimate. These role conventions select the model; they do not guarantee that inverse calibration is appropriate in every statistical setting.

  2. Check the range and context

    For direct prediction, compare the known explanatory input with its observed range. Use the inequality for the chosen explanatory variable. An inverse calibration needs a separate check of the estimated input and calibration context.

    xmin⁡≤x0≤xmax⁡x_{\min}\le x_0\le x_{\max}

    ymin⁡≤y0≤ymax⁡y_{\min}\le y_0\le y_{\max}

    Inside the explanatory rangeInterpolation still needs support

    Being inside the range does not guarantee accuracy. Check whether the scatter supports the chosen model and whether the sample and measurements are relevant. Two observations alone cannot establish an underlying linear relationship.

    Outside the explanatory rangeExplain the extrapolation risk

    The relationship may change beyond the observed inputs. A large correlation coefficient within the sample cannot validate its continuation outside that range.

    Every predictionKeep the context and units

    Check the population, the variable definitions and the conditions of measurement. Association does not establish causation. State which limitation matters for this estimate rather than giving a generic assurance of reliability.

    Keep the fitted variable and its units in the answer. Use the coefficients supplied by the question, or retain calculator precision when fitting a model yourself.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Same supplied value, different modelling setup

The unit example supplies y=1.188x+3.042y=1.188x+3.042 and x=0.736y−1.009x=0.736y-1.009. Estimate x when y=11y=11: first with x prescribed explanatory, then with neither role prescribed. Predict whether the two estimates must agree.

Show reasoning

  1. 1. Keep x as the prescribed explanatory variable

    When x is prescribed explanatory, keep the y-on-x model and solve it backwards. This is inverse calibration, not a new x-on-y fit.

    x=11−3.0421.188≈6.70.x=\frac{11-3.042}{1.188}\approx6.70.

  2. 2. With neither role prescribed, fit the requested response

    Now x is the response to be estimated from the supplied y value, so use the x-on-y equation.

    x=0.736(11)−1.009=7.087≈7.09.\begin{aligned}x&=0.736(11)-1.009\\&=7.087\approx7.09.\end{aligned}

  3. 3. Explain the difference

    The modelling assumptions changed. The two regressions minimise residuals in different response directions; rearranging one does not generally produce the other. Use the equations supplied by the question without refitting or replacing their coefficients.

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Check the range of the correct explanatory variable

In the same unit example, y is prescribed explanatory and its observed values range from 7 to 22. Estimate x at y=30y=30. Does the supplied correlation r=0.93r=0.93 make this prediction reliable?

Show reasoning

  1. 1. Use the x-on-y model

    Substitute the known explanatory y value into the supplied x-response fit.

    x=0.736(30)−1.009=21.071≈21.1.\begin{aligned}x&=0.736(30)-1.009\\&=21.071\approx21.1.\end{aligned}

  2. 2. Compare y with the observed y range

    The input 30 is outside 7 to 22, so this is extrapolation. A close association within the sample does not establish that the model continues outside it. An accurately supplied input does not remove this model uncertainty.

    30>22.30>22.
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