Predict from a transformed regression model

The model is for a non-negative response and is written as the square of y. After substituting x = 4, what is the final step?

Reuse the logarithmic-input model. Move the transformed coordinate from 1 to 2 and compare the original input with the response. The response remains on its original scale

When the original input grows from 10 to 100, what changes in this model?

Predicting from a transformed regression

A transformed regression predicts on the scale used by its fitted equation. Start with the original explanatory value, apply any required input transformation, then undo a response transformation before reporting the prediction

1. 1Start with the original input Use the stated explanatory-variable value before deciding whether the explanatory side needs a transformation \(\displaystyle x=4\Rightarrow\text{start with }4\)
2. 2Transform the input when required If the explanatory side is transformed, apply that operation before substituting into the fitted line. This input-square calculation is a separate example \(\displaystyle x'=x^{2}=4^{2}=16\)
3. 3Substitute into the line For the response-square example, the input remains x=4. Evaluate the fitted equation to find the predicted response on its stated transformed scale \(\displaystyle \widehat{y^{2}}=5+4\times4=21\)
4. 4Undo the response transformation Return to the original response scale and use its stated domain to choose an inverse branch \(\displaystyle y\ge0,\quad\widehat{y}=\sqrt{21}\approx4.6\)

Data or model

\(\displaystyle \widehat{y^2}=5+4x,\quad x=4\)

Calculate the predicted squared response for the stated input

Data or model

\(\displaystyle \widehat{y^2}=21,\quad y\ge0\)

Return to the original non-negative response, to one decimal place

Reporting the transformed response as the original response. One of these lines is right. Which one?

Identify the changed side before calculating

An input transformation changes what you substitute; a response transformation changes how you interpret and invert the result. Use numerical values in the declared units; logarithms require positive numerical arguments

1. 1Three input routes Transform the explanatory input as shown for each model. Each result is already on the original response scale \(\displaystyle \begin{aligned}\hat y&=a+bx^2 &&\text{substitute }x^2\\\hat y&=a+b\log_{10}(x),\ x>0 &&\text{substitute }\log_{10}(x)\\\hat y&=a+\frac{b}{x},\ x\ne0 &&\text{substitute }\frac1x\end{aligned}\)
2. 2Three response routes First evaluate the fitted linear value. Apply the inverse for the stated response transformation and check its domain. A negative fitted value is allowed for a log-response \(\displaystyle \begin{aligned}z&=a+bx\\y^2=z,\ y\ge0,\ z\ge0&\Rightarrow\hat y=\sqrt z\\\log_{10}(y)=z&\Rightarrow\hat y=10^z\\\frac1y=z,\ z\ne0&\Rightarrow\hat y=\frac1z\end{aligned}\)
3. 3Check range and purpose separately Compare the original input with the inclusive original observed interval. Interpolation can still be inaccurate if the model is poor. A finite, positive output does not validate extrapolation or establish causation

Predict and explain three input transformations

Use these invented models exactly as stated. All inputs are dimensionless settings and all responses are readings in units. Construct a table with the original query, transformed coordinate, prediction and range judgement

1. 1Produce and verify Choose a calculator or spreadsheet and record its formulas. Make a labelled plot of B using the original input axis and another using log₁₀(x). Identify which curve or line and which interval each plot represents. Predict how a tenfold increase in x changes B, then verify two examples
2. 2Save your work before opening the model Record your formulas, substituted values, full stored precision, labelled plot and contextual range/domain judgement. Complete external-tool work, plots and written explanations are not automatically marked; review them yourself or with a teacher

Model response: Predict and explain three input transformations

Equivalent correct formulas and suitable plot windows are acceptable

1. 1Predictions A: x²=9 and ŷ=29, an interpolation. B: log₁₀(10)=1 gives ŷ=10, an interpolation; log₁₀(1000)=3 gives ŷ=16, an extrapolation because 1000>100. C: 1/x=1/4 gives ŷ=5, an interpolation. No response inverse is needed
2. 2Plots and verification For B use a raw window x=0..1100,y=0..18, drawing the curve only for x>0; an observed-range view x=0..110 may reveal the shape more clearly. For the transformed plot use u=−0.5..3.5,y=0..18 and line y=7+3u. Mark the observed interval u=0..2, and distinguish the extension to u=3. Do not label the equal spacing in u as equal spacing in original x.
3. 3Transfer Each tenfold increase adds 1 to log₁₀(x), hence adds 3 response units: x=1→10 predicts 7→10, and x=10→100 predicts 10→13. This is an additive change in the predicted response, not multiplication by 10.

Data or model

\(\displaystyle \hat y=2+3x^2,\quad x=3\)

Predict from input-square model A at x=3.

Data or model

\(\displaystyle \hat y=7+3\log_{10}(x),\quad x=1000\)

Predict from model B at x=1000.

Model B was fitted over original x=1..100. What does its prediction at x=1000 represent?

Data or model

\(\displaystyle \hat y=2+12/x,\quad x=4\)

Predict from reciprocal-input model C at x=4.

Predict and validate three response transformations

Use these invented models exactly. The variables are dimensionless indices. For the first query in each row, calculate the transformed response and original response to one decimal place. Investigate the remaining queries and state both their range position and mathematical and contextual validity

1. 1Check the inverse Record the substituted linear value before applying the inverse. Substitute each unrounded original prediction back through its response transformation. Do not treat a negative log-response like a negative squared response
2. 2Save your work before opening the model Record your formulas, substituted values, full stored precision, labelled plot and contextual range/domain judgement. Complete external-tool work, plots and written explanations are not automatically marked; review them yourself or with a teacher

Model response: Predict and validate three response transformations

Equivalent correct formulas and suitable plot windows are acceptable

1. 1Three interpolations At x=2, D gives z=9 and y-hat=3.0; E gives z=1.5 and y-hat=10^1.5≈31.6; F gives z=0.3 and y-hat=10/3≈3.3. All are within the original input intervals. Squaring 3, logging 10^1.5 and reciprocating 10/3 recover the respective intermediate values
2. 2Outside the data For D at x=−1, z=−3 has no real square-root prediction. For E at x=−5, z=−2 gives y-hat=0.01, mathematically positive but an extrapolation requiring evidence. For F at x=5, z=0 has no reciprocal; at x=6, z=−0.1 gives −10, violating the stated positive response domain. A calculator output alone cannot make these predictions appropriate
3. 3Branch and precision The non-negative domain selects +√z for D. If a separate model explicitly required y≤0, the negative root would be selected instead. Keep unrounded intermediate values and round only the requested final prediction; inverse checks using rounded outputs will generally be approximate

Data or model

\(\displaystyle \widehat{\log_{10}(y)}=1.5\)

Model E gives a log-response of 1.5. Report the original response to one decimal place

Data or model

\(\displaystyle \widehat{1/y}=0.3\)

Model F gives a reciprocal response of 0.3. Report the original response to one decimal place

Which outside-range calculation is mathematically valid for the stated positive response?

Continue the two fitted models from 4B

Reuse the original 4B fits and their observed input interval. Predict the squared-input model to three decimal places and the logged-response model to one decimal place. Construct the original-scale log-response curve using the observed pairs (1, 3), (2, 30), (3, 30) and (4, 300), label observations separately from predictions, and justify each range judgement

1. 1Compare two explicit precision contracts Repeat both calculations using these separately defined approximate models. Explain why neither result should silently replace a full-fit calculation \(\displaystyle \hat y=2.081+0.922x^2,\qquad\hat v=-0.0229+0.6000x\quad(x=5)\)
2. 2Record technology and evidence Store full coefficients or the exact expressions, enter the query in original units, and use the correct inverse. Save the command/formula and a back-substitution check. Explain what the inverse-fitted curve does and does not minimise
3. 3Save your work before opening the model Record your formulas, substituted values, full stored precision, labelled plot and contextual range/domain judgement. Complete external-tool work, plots and written explanations are not automatically marked; review them yourself or with a teacher

Model response: Continue the two fitted models from 4B

Equivalent correct formulas and suitable plot windows are acceptable

1. 1Full-fit predictions At x=4, the squared-input model gives 4345/258≈16.841, an interpolation at the inclusive endpoint. The log-response model gives y-hat=3×10^(−1/2+3x/5): at x=2 it is 3×10^0.7≈15.0; at x=5 it is 3×10^2.5≈948.7. The first is interpolation and the second extrapolation. The logged prediction is an intermediate, not the original response
2. 2Displayed equations define different models Using 2.081+0.922×16 gives 16.833. Using 10^(−0.0229+0.6000×5) gives approximately 948.6369, hence 948.6. These correctly answer the separate displayed-equation tasks. They do not replace 16.841 and 948.7 when the task requires the original full fit
3. 3Plot and interpretation Plot original observations (1,3),(2,30),(3,30),(4,300) in x=0..5.5,y=0..1000. Draw the inverse-fitted curve solid over x=1..4 and dashed to x=5; Q is a calculated prediction, not an observation. The original-scale curve minimised squared log-response residuals before inversion, not original-scale squared residuals. Without further assumptions it is not automatically an arithmetic conditional mean

Data or model

\(\displaystyle \hat y=179/86+(119/129)x^2,\quad x=4\)

Use the full input-square fit at x=4. Report three decimal places

Data or model

\(\displaystyle \hat y=2.081+0.922x^2,\quad x=4\)

Here the displayed equation itself defines the model. Predict at x=4 to three decimal places

Data or model

\(\displaystyle \widehat{\log_{10}(y)}=\log_{10}(3)-1/2+(3/5)x,\quad x=5\)

Use the full log-response fit at x=5. Report the original response to one decimal place

What does the dashed extension and point Q represent?

For a separate square-input model fitted over original x=−3..2, classify a query at x=3.

What can be claimed about the inverse of a fitted log-response line without additional assumptions?