- Start from the general exponential form \(y=ab^x\). - Apply a logarithm to turn multiplication and exponents into addition and multiplication. - Identify which original variable remains unchanged in the resulting linear equation.

1. An exponential model can be written as \(y=ab^x\), where \(y>0\). 2. Take the natural logarithm of the response: \(\ln(y)=\ln(a)+x\ln(b)\). 3. This equation is linear in \(x\). Use transformed explanatory variable \(x^*=x\) and transformed response variable \(y^*=\ln(y)\).

Use \(x^*=x\) and \(y^*=\ln(y)\); plot \(\ln(y)\) against \(x\). Then \(y=ab^x\) becomes \(\ln(y)=\ln(a)+x\ln(b)\).

- Begin with the power form \(y=ax^b\). - Log both sides so the exponent becomes a coefficient. - Identify the transformed variable represented on each axis of the linear plot.

1. A power model has the form \(y=ax^b\). 2. Take the natural logarithm of both variables: \(\ln(y)=\ln(a)+b\ln(x)\). 3. This equation is linear in \(\ln(x)\). Use transformed explanatory variable \(x^*=\ln(x)\) and transformed response variable \(y^*=\ln(y)\).

Use \(x^*=\ln(x)\) and \(y^*=\ln(y)\); plot \(\ln(y)\) against \(\ln(x)\). Then \(y=ax^b\) becomes \(\ln(y)=\ln(a)+b\ln(x)\).

- Write the general logarithmic model \(y=a+b\ln(x)\). - Identify which variable appears inside the logarithm. - The response does not need to be transformed for this model form.

1. A logarithmic model has the form \(y=a+b\ln(x)\). 2. The response \(y\) is already linear in \(\ln(x)\). 3. Use transformed explanatory variable \(x^*=\ln(x)\) and transformed response variable \(y^*=y\). Plot \(y\) against \(\ln(x)\).

Use \(x^*=\ln(x)\) and \(y^*=y\); plot \(y\) against \(\ln(x)\). The model \(y=a+b\ln(x)\) is linear in the transformed explanatory variable.

- Identify whether the response variable was transformed. - A regression of \(y\) on \(\ln(x)\) already predicts \(y\) on its original scale. - Match the form \(a+b\ln(x)\) to its model type.

1. Only the explanatory variable was transformed; the response \(y\) remained on its original scale. 2. Therefore, no back-transformation of \(\hat y\) is needed. 3. The original-variable model is \(\hat y=7.00-0.70\ln(x)\), which is a logarithmic model.

\(\hat y=7.00-0.70\ln(x)\), a logarithmic model.

- Look for the panel in which curvature has been replaced by an approximately straight pattern. - Identify which variable was transformed in that panel. - Connect a linear relationship between \(x\) and \(\ln(y)\) to its original-scale model family.

1. Panel a) shows clear upward curvature on the original scale. 2. Panel b), which plots \(\ln(y)\) against \(x\), places the points on a straight-line pattern. 3. Panel c), which plots \(y\) against \(\ln(x)\), remains curved. 4. A linear relationship between \(\ln(y)\) and \(x\) corresponds to an exponential model for \(y\) on the original scale.

Panel b) best linearizes the data. It suggests an exponential model on the original scale.

- Compare shape, not merely whether the plotted values rise. - Identify the panel in which both positive variables have been logarithmically transformed. - Recall which original-scale model becomes linear on a log-log plot.

1. The original-scale plot in panel a) is curved. 2. Logging only the response in panel b) does not remove the curvature. 3. In panel c), the points for \(\ln(y)\) versus \(\ln(x)\) lie on a straight-line pattern. 4. A linear log-log relationship corresponds to a power model \(y=ax^b\) on the original scale.

Panel c) provides the straight-line pattern. It indicates a power model on the original scale.

- First compare the shapes of the three point clouds and decide which is closest to one straight line. - Do not choose a transformation just because its axis label looks familiar; use the displayed pattern. - After choosing a panel, identify which original variable was transformed and how it appears in the linearized plot.

1. Panel 1 rises quickly at first and then levels off, so the original-scale relationship is curved rather than linear. 2. In panel 2, the plotted points form a straight-line pattern when \(y\) is graphed against \(\ln(x)\). 3. Panel 3 remains curved when only the response is logged. 4. A straight-line relationship between \(y\) and \(\ln(x)\) corresponds to a logarithmic model of the form \(y=a+b\ln(x)\) on the original scale.

a) Panel 2. b) A logarithmic model, \(y=a+b\ln(x)\).

- What numbers are allowed inside the real natural logarithm? - Does zero meet that requirement?

1. Natural logarithms of positive numbers are defined, so \(\ln(4)\) is allowed. 2. \(\ln(0)\) is undefined, so the zero response cannot enter the proposed log-response regression as written.

Only \(y=4\). The natural log requires a positive input, and \(\ln(0)\) is undefined.

- Read P's horizontal coordinate on the transformed axis. - What power of ten gives that logarithm?

1. The graph places P at \(\log_{10}(x)=2\). 2. Undo the base-10 logarithm: \(x=10^2=100\).

\(x=100\).

- Decide which axis represents the response in a scatterplot. - Compare the mathematical labels on that axis across the two panels.

1. The response is the quantity on the vertical axis. 2. Panel a) labels that axis \(y\), while panel b) labels it \(\log_{10}(y)\). 3. Therefore panel b) transforms the response before fitting.

Panel b), because its vertical axis is \(\log_{10}(y)\). Panel a) instead transforms the input on the horizontal axis.

- Which coefficient measures how the fitted log response changes with the log input? - The model describes relative changes; the intercept does not control their percentage size. - What direction of change would you expect from a positive coefficient?

1. Identify the elasticity as the coefficient of \(\ln(x)\): \(0.8\). 2. Estimate the percent change in the fitted response: \(0.8\cdot2\%=1.6\%\). 3. The result is approximate because the rule applies to small changes.

About \(1.6\%\) more.

- First find the output of the equation on the log scale. - Which operation reverses the natural logarithm? - Finish the calculation on the original signal-strength scale before rounding.

1. At \(x=2\), the fitted log signal strength is \(0.500-0.180\cdot2=0.140\). 2. Exponentiate the result: \(\tilde{y}=e^{0.140}\approx1.150\).

The back-transformed fitted signal strength is \(\tilde{y}\approx1.150\).

- Substitute \(4\) for \(x\) directly in the logarithmic model. - The response is already on its original scale, so do not exponentiate. - Round after evaluating the full expression.

1. Substitute \(x=4\) into the model: \(\hat y=14.00-1.40\ln(4)\). 2. Evaluate: \(\hat y\approx14.00-1.40\cdot1.386294\approx12.059\). 3. The predicted resale price is approximately \(12.059\) thousand dollars.

\(\hat y\approx12.059\), or approximately \(\$12{,}059\).

- Compare the magnitudes of the three transformed correlations. - Map a linear relationship between \(x\) and \(\ln(y)\) back to its original-scale model family. - Ask what a single correlation coefficient cannot reveal about the shape or spread of a scatterplot.

1. Compare the absolute correlation values: \(|0.61|=0.61\), \(|0.97|=0.97\), and \(|0.72|=0.72\). 2. The \((x,\ln(y))\) transformation has the largest \(|r|\), so it is the strongest candidate from the correlation summaries. 3. A linear relationship between \(\ln(y)\) and \(x\) corresponds to an exponential model for \(y\) as a function of \(x\). 4. Correlation measures linear association but does not by itself show whether curvature, nonconstant spread, or unusual points remain. The transformed scatterplot and residual plot should be checked before accepting the model.

The \((x,\ln(y))\) relationship is the strongest candidate from the correlations because \(|r|=0.97\) is largest. It suggests an exponential model on the original scale. The transformed scatterplot and residual plot should still be checked for linearity and other model problems.

- Compare the magnitudes, not the signs, of the transformed correlations. - Identify the original-scale model family associated with a linear log-log relationship. - A correlation close to \(1\) does not by itself rule out residual structure.

1. Compare the absolute correlation values: \(|0.88|=0.88\), \(|0.75|=0.75\), and \(|0.99|=0.99\). 2. The \((\ln(x),\ln(y))\) transformation has the largest \(|r|\), so it is the strongest candidate from the correlation summaries. 3. A linear relationship between \(\ln(y)\) and \(\ln(x)\) corresponds to a power model for \(y\) as a function of \(x\). 4. Before selecting the model, inspect the transformed scatterplot and residual plot for remaining curvature, changing spread, or unusual observations.

The \((\ln(x),\ln(y))\) relationship is the strongest candidate from the correlations because \(|r|=0.99\) is largest. It suggests a power model. The transformed scatterplot and residual plot should be checked before concluding that the power model is adequate.

- Compare the magnitudes of the negative correlations. - Map a negative linear relationship between \(x\) and \(\ln(y)\) to its original-scale model family. - Ask what a correlation coefficient cannot reveal about residual structure.

1. Compare the absolute correlation values: \(|-0.83|=0.83\), \(|-0.98|=0.98\), and \(|-0.90|=0.90\). 2. The \((x,\ln(y))\) transformation has the largest \(|r|\), so it is the strongest candidate from the correlation summaries. 3. A negative linear relationship between \(\ln(y)\) and \(x\) corresponds to an exponential decay model for \(y\). 4. Correlation alone cannot rule out remaining curvature, changing spread, or unusual observations. Inspect the transformed scatterplot and residual plot before accepting the model.

The \((x,\ln(y))\) relationship is the strongest candidate because \(|r|=0.98\) is largest. It suggests an exponential decay model. The transformed scatterplot and residual plot should still be checked before accepting the model.

- Match each transformation to a specific model form. - Compare exponential linearization with power-model linearization. - A transformation is justified by the resulting plot and residuals, not merely by the fact that the original data are curved.

1. A response-log transformation linearizes model forms in which \(\ln(y)\) is linear in \(x\), such as exponential models. 2. It does not linearize every curved relationship. For example, a power relationship requires a log-log plot of \(\ln(y)\) against \(\ln(x)\). 3. The transformed scatterplot and residual plot must be checked to determine whether the chosen transformation actually produces a suitable linear model.

Taking \(\ln(y)\) linearizes only certain model forms, such as exponential relationships. Other curves may require a different transformation, and the transformed scatterplot and residuals must still be checked.

- Exponentiation must undo the logarithm applied to the entire expression. - Use \(e^{u+v}=e^u e^v\), not \(e^{u+v}=u+e^v\). - Check the model at \(x=0\): \(\ln(y)=2\) requires \(y=e^2\), not \(3\).

1. Exponentiate the entire right side: \(y=e^{2+0.4x}\). 2. Use the exponent rule \(e^{u+v}=e^u e^v\): \(y=e^2e^{0.4x}=e^2(e^{0.4})^x\). 3. Therefore, the original model is exponential, \(y\approx7.389(1.492)^x\), not a sum of \(2\) and an exponential term.

The correct model is \(y=e^{2+0.4x}=e^2(e^{0.4})^x\approx7.389(1.492)^x\), not \(y=2+e^{0.4x}\).

- Match \(\ln(y)=A+B\ln(x)\) with \(\ln(a)+b\ln(x)\). - The transformed intercept is \(\ln(a)\). - Exponentiate the intercept while keeping the transformed slope as the power exponent.

1. The transformed intercept \(1.1\) equals \(\ln(a)\), not \(a\). 2. Exponentiate the intercept: \(a=e^{1.1}\approx3.004\). 3. The transformed slope is the power exponent, \(b=1.6\). Therefore, the original-scale model is \(y=e^{1.1}x^{1.6}\approx3.004x^{1.6}\).

The coefficient is \(a=e^{1.1}\approx3.004\), not \(1.1\). The original-scale model is \(y\approx3.004x^{1.6}\).

- Separate the direction of a correlation from the strength of its linear association. - Correlation summarizes linear association on the transformed scale but does not check residual patterns. - Use the known circle-area relationship to identify the scientifically appropriate model form.

1. The strength of a linear relationship depends on \(|r|\), not on whether \(r\) is positive. A correlation near \(-1\) is just as strongly linear as one near \(1\). 2. Even the largest \(|r|\) does not by itself guarantee that a transformed model is appropriate. The residual plot, domain restrictions, predictive performance, and mathematical or scientific context must also support the model. 3. For radius and circle area, the known relationship \(y=\pi x^2\) is a power model, so a log-log transformation is justified by both the model form and the data diagnostics.

The statement is incorrect. Compare correlations by \(|r|\), because strong negative and positive correlations can both indicate linearity. Then use residual plots and context to determine whether the transformation is appropriate; for circle area and radius, the power relationship \(y=\pi x^2\) supports a log-log model.

- Connect the slope in the log-scale equation to the one-step multiplier after exponentiation. - Use an inverse operation to find the slope from the multiplier. - Think about why a decay factor must give a negative log-scale slope.

1. A fitted log-response line \(\widehat{\ln(y)}=A+Bx\) gives a back-transformed curve \(\tilde{y}=e^A(e^B)^x\). 2. Thus \(e^B=0.88\), so \(B=\ln(0.88)\approx-0.128\). 3. Since the multiplier is below \(1\), the fitted response decreases as \(x\) increases, giving a negative log-scale slope.

\(B\approx-0.128\). The slope is negative because each additional unit of \(x\) multiplies the back-transformed fitted response by less than \(1\).

- Use the regression to find the fitted value of the transformed response. - What operation reverses the reciprocal transformation? - Check that the fitted reciprocal is positive before converting back.

1. At \(x=20\), the fitted reciprocal is \(\hat v=0.020+0.005\cdot20=0.120\). 2. Undo the reciprocal transformation: \(\tilde{y}=\frac{1}{0.120}\approx8.33\). 3. This is the reciprocal of the fitted transformed response, not necessarily the mean of the original response \(y\).

The back-transformed fitted value is \(\tilde{y}\approx8.33\).

- Replace the transformed variable by its definition in the fitted equation. - Compare what happens to the transformed value when the sign of \(x\) changes. - Use the rewritten model for both requested predictions.

1. Substitute \(u=x^2\) into the transformed linear model to get \(\hat y=7+0.4x^2\). 2. At \(x=-3\), \(\hat y=7+0.4\cdot9=10.6\). 3. At \(x=3\), the squared explanatory value is also \(9\), so \(\hat y=10.6\). 4. The transformation produces an original-scale model symmetric about \(x=0\).

The original-scale model is \(\hat y=7+0.4x^2\). It predicts \(10.6\) at both \(x=-3\) and \(x=3\).

- Replace the transformed predictor by its definition. - Then consider what happens to the transformed term when the original explanatory value becomes very large. - The constant term determines the long-run baseline once the transformed contribution becomes negligible.

1. Substitute \(u=\frac{1}{x^2}\) to obtain \(\hat y=5+\frac{120}{x^2}\). 2. As \(x\) becomes very large, \(\frac{120}{x^2}\) approaches \(0\). 3. Therefore, the predicted response approaches \(5\).

The original-scale model is \(\hat y=5+\frac{120}{x^2}\), and its predicted response approaches \(5\) as \(x\) increases.

- A useful transformation should remove systematic structure that the original linear fit leaves behind. - Focus on residual pattern rather than the overall height of the plotted points. - Random scatter around the zero line supports linear form more than a repeated curved pattern does.

1. Panel a) has a clear U-shaped residual pattern, so the original-scale straight line misses systematic curvature. 2. Panel b) has residuals scattered on both sides of \(0\) without an evident curved pattern. 3. Therefore, the transformed fit is better supported with respect to linear form.

The transformed fit in panel b) is better supported. Its residuals are patternless around \(0\), whereas panel a) shows clear curvature.

- Compare the overall shape of the same observations before and after the response transformation. - Treat straightness and residual spread as separate model features. - In the residual plot, compare the vertical spread near small \(x\) with the spread near large \(x\).

1. Panel 1 bends upward strongly as \(x\) increases, so a straight-line model is not a good description on the original response scale. 2. In panel 2, the log-transformed responses follow an overall straight increasing trend. Thus the transformation improves the linear form. 3. Panel 3 shows that residuals lie progressively farther from \(0\) as \(x\) increases, producing a fan-shaped pattern. 4. The changing residual spread indicates nonconstant variance. Therefore, the transformation improves linearity but does not make the fitted linear model fully adequate.

a) Yes. Panel 1 is strongly curved, while panel 2 is much more nearly linear after replacing \(y\) with \(\ln(y)\). b) No. Panel 3 shows increasing residual spread as \(x\) increases, so nonconstant variance remains.

- Treat the two reported correlation magnitudes as close rather than automatically choosing the larger one. - Look for systematic structure in each residual plot. - A good transformation should support the assumptions of the fitted linear model, not merely maximize one summary statistic.

1. Panel a) has a systematic U-shaped residual pattern, so Transformation A leaves curvature that a linear model does not explain. 2. Panel b) has residuals scattered around \(0\) without a comparable systematic pattern. 3. Therefore, Transformation B is better supported even though its \(|r|\) is slightly smaller. 4. Correlation summarizes linear association strength but does not reveal residual curvature or other model-condition problems.

Transformation B is better supported. Panel b) has patternless residuals, while panel a) retains curvature. The small difference in \(|r|\) cannot override the residual diagnostic.

- How can you return from a fitted logarithm to the original scale? - Calculate the average directly from the two original responses. - Must the back-transformed value match that average?

1. Undo the logarithm: \(\tilde y=e^{\ln(3)}=3\). 2. The original responses have mean \((1+9)\div2=5\). 3. The values differ. Back-transforming a fitted log response does not necessarily give the original-scale mean.

Back-transformed fit: \(3\). Original-response mean: \(5\). They are not equal.

- Read the two response heights rather than assuming a percent change from the slope alone. - What percent is the input increase from 1 to 1.21? - Multiply that input percent by the log-log slope only for the approximation.

1. The graph gives fitted responses \(1\) and \(1.1\). 2. Their percent change is \(\frac{1.1-1}{1}\cdot100\%=10\%\). 3. The input rises by \(21\%\), so elasticity gives the approximation \(0.5\cdot21\%=10.5\%\).

The fitted responses are \(1\) and \(1.1\). Their increase is exactly \(10\%\); the elasticity approximation is \(10.5\%\).

- Look for a curved pattern that stays in the residuals after fitting a line. - Compare patterns, not the numerical sizes of residuals on two different scales. - Think about the difference between the mean of logarithms and the logarithm of a mean.

1. The original-area residuals are positive at the first and last days and negative in the middle. This curved pattern suggests a line is a poor fit for area. 2. The log-area residuals are much more irregular around zero, so a line fits the logged areas more plausibly. 3. Exponentiating a fitted log value does not generally give the mean area on the original scale.

Panel b) is more suitable for a line fitted on its scale. No: back-transforming a fitted log value does not automatically give the average original area.

- Read the two input values from the number line and find their difference. - What is the change in a natural logarithm when the original positive value doubles? - The slope compares the change on the log scale with the change in input.

1. Read the marked inputs as \(x=0\) and \(x=2\), an increase of 2 units. 2. Doubling the back-transformed fitted response raises its fitted logarithm by \(\ln(2)\), so \(b=\frac{\ln(2)}{2}\approx0.346574\). 3. To three decimals, \(b\approx0.347\).

\(b\approx0.347\).

- Which axis shows the logarithm of the cell count? - Read the two heights on that axis before converting to counts. - Compare the original counts using a ratio, not a difference.

1. The plotted response is \(\log_{10}(y)\). Its value is \(1\) at A and \(3\) at B. 2. Undo the logarithms: the original counts are \(10^1=10\) and \(10^3=1000\). 3. The factor is \(1000\div10=100\).

The cell count at B is \(100\) times the count at A.

- The regression predicts the logarithm of the bacteria count. How can you reverse that transformation? - Separate the exponent's constant part from the part involving time. - Match the resulting expression with a coefficient times a fixed base raised to \(x\).

1. Exponentiate the fitted log value: \(\tilde{y}=e^{0.800+0.120x}\). 2. Separate the constant and \(x\) terms: \(\tilde{y}=e^{0.800}(e^{0.120})^x\). 3. Evaluate \(a=e^{0.800}\approx2.226\) and \(b=e^{0.120}\approx1.127\). 4. The back-transformed fitted curve is \(\tilde{y}\approx2.226(1.127)^x\). This does not automatically estimate the mean bacteria count without bias.

\(\tilde{y}\approx2.226(1.127)^x\), the back-transformed fitted curve.

- Which two variables were logged before fitting the line? - Undo the logarithm of the fitted response, then simplify the factor involving \(\ln(x)\). - Look for an exponent rule that turns an exponential expression involving \(\ln(x)\) into a power of \(x\).

1. Exponentiate the predicted log volume: \(\tilde{y}=e^{1.250+1.320\ln(x)}\). 2. Use the laws of exponents: \(\tilde{y}=e^{1.250}e^{1.320\ln(x)}=e^{1.250}x^{1.320}\). 3. Since \(e^{1.250}\approx3.490\), the back-transformed power curve is \(\tilde{y}\approx3.490x^{1.320}\).

\(\tilde{y}\approx3.490x^{1.320}\).

- Only positive mass values can be used in this log-log equation. - Substitute the mass before undoing the response transformation. - Keep extra digits in the log prediction until after exponentiating.

1. The fitted log area at \(x=3\) is \(0.820+0.900\ln(3)\approx1.808751\). 2. Taking the exponential gives \(\tilde{y}=e^{1.808751\ldots}\approx6.103\).

\(\tilde{y}\approx6.103\), the back-transformed fitted surface area.

- Put the number of days into the fitted log-response equation. - Convert the result back to the scale of thousands of visits. - Keep track of the response unit when writing the final count.

1. At day \(6\), the fitted log number of visits is \(1.180+1.200\ln(6)\approx3.330111\). 2. The back-transformed fit is \(\tilde{y}=e^{1.180+1.200\ln(6)}\approx27.941\) thousand visits. 3. This corresponds to about \(27{,}941\) visits; exponentiation alone does not guarantee an unbiased prediction of the mean count.

\(\tilde{y}\approx27.941\) thousand visits, or about \(27{,}941\) visits.

- In the exponentiated curve, the log intercept sets the value at time zero. - A fixed increase of one time unit adds the same amount to the fitted log count each time. - Convert the associated multiplier to a percentage by comparing it with \(1\).

1. Exponentiating the fitted log equation gives \(\tilde{y}=e^{1.200}(e^{0.095})^x\). 2. At \(x=0\), \(a=e^{1.200}\approx3.320\). 3. Increasing time by one unit multiplies \(\tilde{y}\) by \(b=e^{0.095}\approx1.100\). 4. The exact multiplier gives a relative increase of \((e^{0.095}-1)\cdot100\%\approx10.0\%\) in the back-transformed fitted count.

\(a\approx3.320\), \(b\approx1.100\). The back-transformed fitted count increases by about \(10.0\%\) per time unit.

- Find the exponentiated log intercept at the starting time. - A negative slope on the log scale gives a multiplier below \(1\) on the original scale. - Use the unrounded factor to find the percent decrease.

1. At time zero, the back-transformed fitted concentration is \(a=e^{0.400}\approx1.492\). 2. Advancing one time unit changes the exponent by \(-0.220\), multiplying the fitted value by \(b=e^{-0.220}\approx0.803\). 3. Since \(b<1\), this is a decrease. Using the unrounded factor gives \((1-e^{-0.220})\cdot100\%\approx19.7\%\).

Starting value \(a\approx1.492\); decay factor \(b\approx0.803\). The back-transformed fitted concentration decreases by about \(19.7\%\) each time unit.

- A straight line on log-log axes corresponds to which power of the original explanatory variable? - The fitted intercept must be exponentiated to return to the original scale. - Relate a small relative change in the input to the shape of the fitted power curve, rather than treating the slope as an additive change.

1. Exponentiate the fitted log response: \(\tilde{y}=e^{0.80}x^{1.25}\). 2. The coefficient is \(a=e^{0.80}\approx2.226\) and the exponent is \(b=1.25\). 3. A small \(1\%\) increase in spending corresponds to approximately a \(1.25\%\) increase in the back-transformed fitted sales curve. The exact increase for a \(1\%\) change would be \(((1.01)^{1.25}-1)\cdot100\%\approx1.252\%\).

\(\tilde{y}\approx2.226x^{1.25}\). The elasticity \(1.25\) means that a \(1\%\) increase in advertising spending corresponds to approximately a \(1.25\%\) increase in the back-transformed fitted sales.

- Check for an exact tie before naming one transformation. - Map \((\ln(x),y)\) to a logarithmic model and \((\ln(x),\ln(y))\) to a power model. - When correlations are tied, compare residual behavior and scientific plausibility.

1. Compare the absolute correlations. The \((\ln(x),y)\) and \((\ln(x),\ln(y))\) relationships tie at \(|r|=0.95\), while \((x,\ln(y))\) has \(|r|=0.94\). 2. The tie means correlation alone does not identify a unique best transformation. 3. The tied candidates suggest a logarithmic model and a power model, respectively. Residual plots, model assumptions, predictive performance, and scientific context are needed to choose between them.

There is no unique choice from the correlations alone. The \((\ln(x),y)\) and \((\ln(x),\ln(y))\) relationships tie at \(|r|=0.95\), suggesting logarithmic and power models, respectively. Use residual plots and context to decide.

- This model uses base-\(10\) logarithms, so what operation reverses the transformation? - Rewrite the power of \(10\) as a product of a constant and a factor raised to \(x\). - Compare the growth factor with \(1\) to interpret the percent change.

1. Reverse the common logarithm: \(\tilde{y}=10^{1.200+0.080x}=10^{1.200}(10^{0.080})^x\). 2. The starting coefficient is \(a=10^{1.200}\approx15.849\), and the multiplier is \(b=10^{0.080}\approx1.202\). 3. The back-transformed fit rises by about \((10^{0.080}-1)\cdot100\%\approx20.2\%\) for each one-unit increase in \(x\). It need not be an unbiased prediction of the mean response.

\(\tilde{y}\approx15.849(1.202)^x\). Each additional time unit multiplies the back-transformed fitted response by about \(1.202\), an increase of about \(20.2\%\).

- Recall the factor that changes a base-\(10\) logarithm into a natural logarithm. - Apply the same conversion to the intercept and the slope, not just one coefficient. - You are changing the scale of the fitted log response, not the underlying observations.

1. The same logged response can be rescaled using \(\ln(y)=\ln(10)\log_{10}(y)\). 2. Multiply the fitted common-log equation by \(\ln(10)\): \(\widehat{\ln(y)}=1.300\ln(10)+0.200\ln(10)x\). 3. Thus \(A\approx2.993\) and \(B\approx0.461\).

\(\widehat{\ln(y)}\approx2.993+0.461x\).

- Convert the target back-transformed value to the same scale as the fitted response. - Isolate the logarithm of the explanatory value. - Check that your final input is positive, as the model requires.

1. If the back-transformed fitted value is \(20\), its fitted log value equals \(\ln(20)\). 2. Solve \(\ln(20)=1.100+1.400\ln(x)\) for \(\ln(x)\): \(\ln(x)=\frac{\ln(20)-1.100}{1.400}\). 3. Exponentiate to obtain \(x=e^{\frac{\ln(20)-1.100}{1.400}}\approx3.873\).

\(x\approx3.873\).

- Write the observed log response minus its fitted log response. - How does subtracting two logarithms translate to a comparison on the original scale? - Convert the factor to a percent after removing the log transformation.

1. The residual gives \(\ln(y)-\widehat{\ln(y)}=0.200\). 2. Exponentiating yields \(\frac{y}{e^{\widehat{\ln(y)}}}=e^{0.200}\approx1.221\). 3. Therefore, the observed \(y\) is about \((e^{0.200}-1)\cdot100\%\approx22.1\%\) above the back-transformed fitted value.

The observed response is approximately \(1.221\) times the back-transformed fitted value, or \(22.1\%\) larger.

- Check the domain of every transformation before applying it to the data. - Look specifically at the smallest explanatory value in the data set. - Ask whether changing the variable before taking a logarithm leaves the original model family unchanged.

1. The proposed log-log transformation requires positive explanatory values because \(\ln(x)\) is defined only for \(x>0\) in this context. 2. The observation with \(x=0\) cannot be transformed using \(\ln(x)\), so the proposed transformation cannot be applied to all five observations as stated. 3. Replacing \(x\) with \(x+1\) creates a different transformed variable and therefore a different original-scale model. It is not a harmless way to fit the same power model. 4. Any shifted transformation would need a substantive modeling justification and a new interpretation.

No. The observation with \(x=0\) makes \(\ln(x)\) undefined, so the log-log transformation cannot be applied to all five observations as stated. Replacing \(x\) by \(x+1\) changes the model rather than preserving the original power-model interpretation.

- Recall how logarithms with different bases are related. - Ask whether multiplying all transformed response values by one positive constant changes the shape of their association with \(x\). - Separate a change in vertical scale from a change in model form.

1. The two transformed responses are related by \(\ln(y)=\ln(10)\log_{10}(y)\). 2. Thus, changing from common logarithms to natural logarithms multiplies every transformed response by the same positive constant. 3. A positive linear rescaling changes the vertical scale and fitted coefficients but does not change whether the transformed association is linear or the pattern of residuals apart from scale. 4. Therefore, the logarithm base should not be chosen by looking for a straighter pattern; either base represents the same linearizing transformation.

The statement is incorrect. Natural logs and common logs differ only by a positive constant factor, so they produce the same linearity pattern. The axis scale and regression coefficients change, but one base cannot make the relationship intrinsically straighter than the other.

- Express the new response as a constant multiple of the old response. - Logarithms turn this multiplication into an addition. Which part of the fitted line would that change? - The per-unit multiplier comes from the coefficient of \(x\), not from the intercept.

1. Multiplying each original response by \(100\) adds \(\ln(100)\) to each log response: \(\ln(y_c)=\ln(100)+\ln(y)\). 2. Thus the new fitted equation is \(\widehat{\ln(y_c)}=0.800+\ln(100)+0.120x\approx5.405+0.120x\). 3. The log-scale slope stays \(0.120\), since the unit conversion changes only the log intercept. 4. The back-transformed multiplier per unit of \(x\) remains \(e^{0.120}\); the percent growth does not change.

\(\widehat{\ln(y_c)}\approx5.405+0.120x\). The slope stays \(0.120\), and the back-transformed multiplier per unit of \(x\) stays \(e^{0.120}\).

- Do not choose a transformation from one summary statistic alone. - Compare the residual patterns after each transformation. - Identify the original-scale model family associated with transforming both positive variables logarithmically.

1. A large transformed correlation alone is not enough to establish that a linear model is appropriate. 2. Model A has residual curvature, so its transformed relationship still has systematic nonlinearity. 3. Model B has a slightly smaller transformed correlation but a residual plot consistent with linear form and roughly constant spread. 4. A linear relationship between \(\ln(y)\) and \(\ln(x)\) corresponds to a power model on the original scale.

Model B is better supported. Its residual plot is consistent with a linear model, and the log-log transformation suggests an original-scale power model.

- Compare the model at \(x\) and at a constant multiple of \(x\). - Use a logarithm property to separate the multiplier from the original explanatory value. - Subtract the two predictions and look for terms that cancel.

1. At explanatory value \(x\), the prediction is \(12.0+4.0\ln(x)\). 2. At \(3x\), the prediction is \(12.0+4.0\ln(3x)=12.0+4.0\ln(3)+4.0\ln(x)\). 3. The predicted change is therefore \(4.0\ln(3)\approx4.394\), independent of the starting value of \(x\).

Multiplying \(x\) by \(3\) increases the predicted response by approximately \(4.394\) units.

- Compare the two back-transformed fitted values as a ratio. - Count how many unit steps in \(x\) separate those values. - The slope on the log scale is connected to the natural logarithm of the multiplier.

1. The input increases by \(3\), so the ratio of back-transformed fitted values is \(b^3=\frac{54.0}{20.0}=2.7\). 2. The multiplier is \(b=2.7^{1/3}\approx1.392\). 3. In the fitted log-response equation, the slope is \(B=\ln(b)=\frac{\ln(2.7)}{3}\approx0.331\).

The one-unit multiplier is \(b\approx1.392\), and the fitted log-response slope is \(B\approx0.331\).

- Express the original explanatory variable in terms of the new unit. - Substitute before simplifying the power. - Distinguish the coefficient of a power model from its exponent.

1. Since \(u=100x\), \(x=\frac{u}{100}\). 2. Substituting gives \(\hat y=3.2\left(\frac{u}{100}\right)^{1.5}=\frac{3.2}{100^{1.5}}u^{1.5}=0.0032u^{1.5}\). 3. The exponent remains \(1.5\), so the slope of a regression of \(\ln(y)\) on \(\ln(u)\) remains \(1.5\). 4. Changing the explanatory-variable unit changes the coefficient but not the power exponent.

\(\hat y=0.0032u^{1.5}\). The log-log slope remains \(1.5\).

- Which quantity is on the horizontal axis of the transformed line? - Find the original input that has a logarithm of zero. - Undo the transformation of the fitted response to give a value on the original scale.

1. The explanatory variable on the fitted straight-line plot is \(\ln(x)\), so its intercept occurs where \(\ln(x)=0\). 2. This means \(x=1\), not \(x=0\). The logarithm is not even defined at \(x=0\). 3. At \(x=1\), the fitted log response is \(1.500\), giving \(\tilde{y}=e^{1.500}\approx4.482\).

The intercept occurs at \(x=1\), since \(\ln(1)=0\). The back-transformed fitted response there is \(\tilde{y}\approx4.482\).

- Compare the model at \(x\) and at \(2x\) using a ratio. - Express the stated multiplicative change as a power of \(2\). - Recall what the exponent becomes after taking logarithms of a power model.

1. In a power model, doubling \(x\) multiplies the prediction by \(2^b\). 2. The given condition therefore gives \(2^b=5\). 3. Taking logarithms gives \(b=\frac{\ln(5)}{\ln(2)}\approx2.322\). 4. In the log-log linearized model, the slope is the power exponent, so the transformed slope is also approximately \(2.322\).

\(b\approx2.322\). The slope of the log-log regression is also approximately \(2.322\).

- Express the old time variable in terms of the shifted one. - Separate the constant power of the exponential base from the part involving the new variable. - A shift in the time origin changes the starting level but not the growth factor per time step.

1. Since \(u=x-3\), \(x=u+3\). 2. Substituting gives \(\hat y=5.00(1.20)^{u+3}=5.00\cdot(1.20)^3(1.20)^u=8.64(1.20)^u\). 3. The one-year multiplicative factor remains \(1.20\) because shifting the time origin does not change the growth rate. 4. The log-linear slope remains \(\ln(1.20)\approx0.182\); only the intercept changes.

\(\hat y=8.64(1.20)^u\). The multiplicative factor remains \(1.20\), and the log-linear slope remains approximately \(0.182\).

- Replace each explanatory value by its reciprocal and compare how the response changes across the transformed pairs. - Look for a constant rate of change after the transformation. - For the prediction, transform the new explanatory value before using the linear relationship.

1. The transformed explanatory values are \(u=\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}\). 2. For each transformed point, \(y-3=24u\), so the points lie on the line \(\hat y=3+24u\). 3. When \(x=8\), \(u=\frac{1}{8}\). 4. The prediction is \(\hat y=3+24\cdot\frac{1}{8}=6\).

The reciprocal transformation gives the linear model \(\hat y=3+24u\), where \(u=\frac{1}{x}\). For \(x=8\), the predicted response is \(6\).

- Evaluate the transformed explanatory value at the three requested \(x\)-values. - Compare changes on the transformed scale before comparing predictions. - Interpret the slope with respect to the variable that appears linearly in the fitted equation.

1. The predictions are \(11\) at \(x=1\), \(14\) at \(x=4\), and \(17\) at \(x=9\). 2. The predicted response increases by \(3\) from \(1\) to \(4\) and by \(3\) from \(4\) to \(9\). 3. In both cases, \(\sqrt{x}\) increases by \(1\), even though the raw increases in \(x\) are different. 4. Thus, the coefficient \(3\) is the change in predicted response for a one-unit increase in \(\sqrt{x}\), not for a one-unit increase in \(x\).

Each interval increases the prediction by \(3\). The coefficient \(3\) applies to a one-unit increase in \(\sqrt{x}\), so the model does not imply a constant change in \(y\) for equal raw changes in \(x\).

- Identify the response variable used to compute each \(R^2\). - Ask whether the unexplained variation is being measured on the same scale. - Consider what must happen before predictions from the transformed model can be evaluated on the original scale.

1. The two \(R^2\) values describe variation on different response scales: \(y\) for the first model and \(\ln(y)\) for the second. 2. A larger \(R^2\) on the transformed scale does not directly measure smaller prediction error on the original \(y\)-scale. 3. Model comparison should also consider residual patterns on the fitted scale, the appropriateness of the transformation, and prediction performance after back-transformation.

No. Because the models use different response scales, their \(R^2\) values do not directly compare prediction accuracy for the original response \(y\).

- Turn a percentage increase in \(x\) into a multiplication factor. - The slope on the log-log scale becomes the exponent in the back-transformed power curve. - Use that exponent to find a response multiplier before converting to percent.

1. Raising \(x\) by \(25\%\) multiplies it by \(1.25\). 2. The back-transformed fitted curve is proportional to \(x^{0.8}\), so the fitted response is multiplied by \(1.25^{0.8}\approx1.19544\). 3. Its percentage increase is \((1.25^{0.8}-1)\cdot100\%\approx19.54\%\).

The back-transformed fitted response increases by approximately \(19.54\%\).

- Use logarithm rules to bring the exponent down as a coefficient. - Do not confuse the logarithm of a given fitted curve with a new regression fitted directly to logged observations. - Think about what sum of squared differences is minimized before and after logging the response.

1. Apply the base-\(10\) logarithm to the given curve: \(\log_{10}(\hat y)=\log_{10}(5)+x\log_{10}(1.2)\). 2. Since \(\log_{10}(5)\approx0.69897\) and \(\log_{10}(1.2)\approx0.07918\), the logged curve is \(\log_{10}(\hat y)\approx0.69897+0.07918x\). 3. This is an algebraic rewriting of the supplied curve. A regression separately fitted to \(\log_{10}(y)\) may differ, because transforming the response changes the criterion minimized by least squares.

\(\log_{10}(\hat y)\approx0.69897+0.07918x\). No: logging an original-scale fitted curve does not mean a separate least-squares fit to \(\log_{10}(y)\) would yield the same line.

- Compare the overall shape in panels 1 and 2 before deciding what the transformation accomplished. - Identify which two variables are logarithmically transformed in panel 2. - Use panel 3 to distinguish a broad residual pattern from one unusually large residual. - Keep an outlying residual distinct from leverage or influence.

1. Panel 1 has clear curvature, while panel 2 is much closer to a straight-line pattern apart from one unusual observation. Thus the log-log transformation improves the linear form. 2. A straight-line relationship between \(\ln(y)\) and \(\ln(x)\) corresponds to a power model \(y=ax^b\) on the original scale. 3. Panel 3 shows the same observation as a much larger positive residual than the others. 4. That observation is an outlier in the transformed response direction. Its source should be investigated, and the model can be refit with and without it to assess its effect. A large residual by itself does not establish high leverage or influence.

a) The log-log transformation makes the overall relationship much more nearly linear, supporting a power model. b) One transformed observation has an unusually large positive residual. Investigate its source and assess how the fitted model changes with and without it; residual size alone does not establish influence.

- Treat \(\ln(y)\) as the vertical coordinate when finding the transformed-line slope. - Use one transformed point to solve for \(A\). - Convert \(A\) and \(B\) with \(a=e^A\) and \(b=e^B\).

1. Find the slope of the transformed line: \(B=\frac{\ln(10.985)-\ln(5.000)}{4-1}\approx0.262364\). 2. Use the point \((1,\ln(5.000))\) to find the intercept: \(A=\ln(5.000)-0.262364\cdot1\approx1.347074\). 3. Back-transform the parameters: \(a=e^A\approx3.846\) and \(b=e^B=1.300\). 4. Therefore, the exponential model is \(y\approx3.846(1.300)^x\).

\(y\approx3.846(1.300)^x\).

- On a log-log line, the slope equals the power exponent \(b\). - Use one transformed point to solve for \(\ln(a)\). - Exponentiate \(\ln(a)\) to recover the original-scale coefficient.

1. The slope of the log-log line is the power exponent: \(b=\frac{\ln(22.229)-\ln(6.762)}{\ln(7)-\ln(2)}\approx0.949963\). 2. Use the first point to find the transformed intercept: \(\ln(a)=\ln(6.762)-0.949963\ln(2)\approx1.252854\). 3. Back-transform the intercept: \(a=e^{1.252854}\approx3.500\). 4. Therefore, the power model is \(y\approx3.500x^{0.950}\).

\(y\approx3.500x^{0.950}\).

- Think about a ratio of fitted values after two different input values. - Changes in \(x\) add on the log scale but multiply after exponentiation. - Use a logarithm to solve for the required input change.

1. Increasing \(x\) by \(\Delta x\) multiplies the exponentiated fitted value by \(e^{-0.180\Delta x}\), regardless of the starting \(x\). 2. Halving it means \(e^{-0.180\Delta x}=0.5\). 3. Take logarithms: \(\Delta x=\frac{\ln(0.5)}{-0.180}\approx3.851\).

\(x\) must increase by approximately \(3.851\) units to halve the back-transformed fitted value.