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- Identify which variable is the response. - Residuals measure vertical differences on a scatter plot. - Use the sign to compare actual and predicted fuel use.

1. A residual is an observed \(y\)-value minus a predicted \(y\)-value. 2. Therefore, it has the same units as \(y\), which are gallons. 3. A residual of \(0.7\) means the trip used \(0.7\,\text{gallons}\) more fuel than the model predicted.

The residual is \(0.7\,\text{gallons}\). The observed fuel use was \(0.7\,\text{gallons}\) above the prediction.

- Compare each input with the smallest and largest observed \(x\)-values. - Inputs inside the range and outside the range have different labels. - Consider where the fitted relationship has direct data support.

1. \(x=25\) lies inside the observed range, so it is interpolation. 2. \(x=8\) lies below the observed range, so it is extrapolation. 3. \(x=55\) lies above the observed range, so it is extrapolation. 4. The extrapolations deserve more caution because the observed pattern may not continue outside the data range.

\(x=25\): interpolation \(x=8\): extrapolation \(x=55\): extrapolation The predictions at \(8\) and \(55\) require more caution.

- Identify the scope of the statement “residual equals zero.” - Separate one observation from the complete data set. - Think about what evidence is needed to judge overall fit.

1. A zero residual means the observed \(y\)-value for that one input equals the model’s prediction. 2. It does not describe the residuals of the other observations. 3. A model can pass exactly through one or several points while still fitting the full data set poorly.

The zero residual shows only that one observation lies on the fitted line. It does not prove that the model predicts every observation exactly or is appropriate overall.

- Generate one predicted value for each input. - Use the same subtraction order for every residual. - Check whether positive and negative errors balance.

1. The predicted values are \(14, 18, 22, 26, 30\). 2. The residuals \(y-\hat y\) are \(-1, 1, 0, 1, -1\). 3. Their sum is \(-1+1+0+1-1=0\). 4. Least-squares residuals from a model with an intercept balance to a sum of \(0\), apart from possible rounding.

Residuals: \(-1, 1, 0, 1, -1\) Sum: \(0\)

- Replace the predicted response with the target value. - Undo the intercept before undoing the rate. - Check the solution by substituting it back into the model.

1. Set the prediction equal to \(93\): \(93=7.5x+18\). 2. Subtracting \(18\) gives \(75=7.5x\). 3. Dividing by \(7.5\) gives \(x=10\).

\(10\) promotional posts

- Recall the balancing property of least-squares residuals. - Add the known signed errors before solving. - Keep the negative signs attached to their values.

1. The signed residuals from an unrounded least-squares line with an intercept sum to \(0\). 2. The known residuals sum to \(1.2-0.5+0.8-1.1=0.4\). 3. Therefore, \(0.4+r=0\), so \(r=-0.4\).

\(r=-0.4\)

- Use the model to fill every prediction first. - Keep residual defined as observed minus predicted. - Rearrange that relationship when the observed value is missing.

1. At \(x=2\), \(\hat y=8\), so \(y=8-1=7\). 2. At \(x=4\), \(\hat y=11\), so the residual is \(12-11=1\). 3. At \(x=6\), \(\hat y=14\), so \(y=14+2.5=16.5\).

<table><tr><th>\(x\)</th><th>Observed \(y\)</th><th>Predicted \(\hat y\)</th><th>Residual</th></tr><tr><td>\(2\)</td><td>\(7\)</td><td>\(8\)</td><td>\(-1\)</td></tr><tr><td>\(4\)</td><td>\(12\)</td><td>\(11\)</td><td>\(1\)</td></tr><tr><td>\(6\)</td><td>\(16.5\)</td><td>\(14\)</td><td>\(2.5\)</td></tr></table>

- Use the mean point as a point on the fitted line. - Substitute the known slope and both mean coordinates. - Solve the resulting equation for the constant term.

1. The least-squares line passes through \((12, 31)\). 2. Substitute into \(\hat y=1.8x+a\): \(31=1.8\cdot 12+a\). 3. Solving gives \(a=9.4\). 4. The equation is \(\hat y=1.8x+9.4\).

Y-intercept: \(9.4\) Regression equation: \(\hat y=1.8x+9.4\)

- Identify the special point through which the fitted line passes. - Compare the actual and predicted \(y\)-values at the mean \(x\)-value. - Apply the residual definition.

1. The fitted least-squares line passes through \((\bar x, \bar y)\). 2. At \(x=\bar x\), the predicted response is \(\hat y=\bar y\). 3. The residual is \(y-\hat y=\bar y-\bar y=0\).

The residual is \(0\).

- Keep signs for part a). - Remove signs without squaring for part b). - Square each residual separately for part c), then compare what each operation does to a large error.

1. The signed sum is \(-2+1+0+3-2=0\). 2. The sum of absolute residuals is \(2+1+0+3+2=8\). 3. The sum of squared residuals is \(4+1+0+9+4=18\). 4. The signed sum can cancel positive and negative errors. Absolute residuals measure total error size without cancellation, while squared residuals give extra weight to larger errors.

a) \(0\) b) \(8\) c) \(18\) d) Signed errors can cancel; absolute errors measure total distance; squared errors emphasize large residuals.

- Use the residual to recover the prediction before using the model. - Pay attention to the subtraction order in the residual definition. - Solve the resulting linear equation for the input.

1. Since residual equals observed minus predicted, the predicted value is \(40-3=37\). 2. Set the model equal to \(37\): \(2.5x+7=37\). 3. Solving gives \(2.5x=30\), so \(x=12\).

The observation has input \(x=12\).

- Identify the quantity the plot is named for. - Ask what information the original scatter plot already shows. - Think about how model errors should be centered for inspection.

1. A residual plot must display the errors left after fitting the model. 2. The vertical coordinate should be \(y-\hat y\), the residual for each observation. 3. Plotting observed \(y\)-values simply recreates a version of the original scatter plot and does not reveal model error around zero.

The vertical axis should show residuals \(y-\hat y\). This makes patterns in the model’s prediction errors visible around the zero line.

- Check whether successive rates of change match. - Use a point with \(x\)-coordinate zero to identify the intercept. - Determine each residual once the exact pattern is found.

1. Every increase of \(2\) in \(x\) produces an increase of \(3\) in \(y\), so the slope is \(\frac{3}{2}=1.5\). 2. The point \((0, -2)\) gives the intercept \(-2\), so \(\hat y=1.5x-2\). 3. Every observation lies exactly on this line, so every residual is \(0\). 4. The sum of squared residuals is \(0\).

Regression line: \(\hat y=1.5x-2\) Sum of squared residuals: \(0\)

- Find the model’s prediction at the given input. - Translate “zero residual” into a relationship between actual and predicted values. - Do not use the current observed value after finding the required target.

1. At \(x=6\), the model predicts \(\hat y=5.2\cdot 6-1=30.2\). 2. A zero residual requires observed \(y\) to equal predicted \(\hat y\). 3. The corrected \(y\)-value must be \(30.2\).

\(30.2\)

- Distinguish an observed count from a model’s expected value. - Think about whether averages of whole-number counts must be whole numbers. - Match the final reporting precision to the context.

1. A regression prediction represents an expected or typical response, not a guarantee of one exact observed count. 2. An average or fitted value can be noninteger even when every actual count is a whole number. 3. In context, the result can be reported as about \(48\) customers, while retaining \(47.6\) for calculations.

The decimal is a valid fitted average. Report it as a prediction of about \(48\) customers.

- First compute each prediction from \(\hat y=2x+1\). - Use residual \(=\) observed \(-\) predicted, preserving the signs. - A useful linear model should leave residuals without a clear systematic pattern.

1. The model predictions are \(3,5,7,9,11,13\). 2. Subtract prediction from observation: \(4-3=1\), \(4-5=-1\), \(8-7=1\), \(8-9=-1\), \(12-11=1\), and \(12-13=-1\). 3. Plot a) places these residuals at the corresponding \(x\)-values, so it is the correct residual plot. 4. The residuals alternate systematically between positive and negative rather than appearing randomly around zero. This pattern suggests that the linear model misses structure in the data.

a) \(1,-1,1,-1,1,-1\) b) Plot a). c) No. The alternating residual pattern is systematic, so the linear model does not capture all of the structure.

- Add the claimed signed residuals. - Recall the balancing property of least-squares residuals. - Compare the two totals.

1. Six residuals of \(2\) would have signed sum \(6\cdot 2=12\). 2. Residuals from an unrounded least-squares line with an intercept must sum to \(0\). 3. Therefore, the claimed residual list cannot come from that least-squares fit.

It is impossible because the residuals would sum to \(12\), not \(0\).

- Use the sign of \(r\) for direction and its magnitude for strength. - Square the decimal coefficient before converting to a percentage. - Distinguish explained variation from counting points on a line.

1. The positive sign and magnitude \(0.84\) indicate a strong positive linear association. 2. \(r^2=(0.84)^2=0.7056\), or about \(70.6\%\). 3. In this regression setting, about \(70.6\%\) of the variation in the response is accounted for by its linear relationship with the explanatory variable. 4. Correlation and its square summarize linear association and explained variation; neither counts the percentage of observations lying exactly on a line.

a) A strong positive linear association. b) \(r^2=0.7056\approx70.6\%\). About \(70.6\%\) of the variation in the response is accounted for by the linear model. c) Neither \(r\) nor \(r^2\) is a percentage of points that lie exactly on the regression line.

- Store the full regression coefficients even if only four decimal places are displayed. - Substitute \(32.4\) into the stored regression model. - Round only the final predicted value, then compare with the result from prematurely rounded coefficients.

1. The exact least-squares coefficients are \(\frac{59}{600}\) for the slope and \(\frac{27}{100}\) for the intercept. Displayed to four decimal places, the line is \(\hat y=0.0983x+0.2700\). 2. Use the unrounded coefficients for the prediction: \(\hat y=\frac{59}{600}(32.4)+\frac{27}{100}=3.456\). 3. Rounded to the nearest hundredth, the predicted speed is \(3.46\) miles per hour. 4. Substituting into the displayed rounded equation gives \(0.0983(32.4)+0.2700=3.45492\), which would round to \(3.45\). This difference shows why the unrounded coefficients must be used.

Regression line: \(\hat y=0.0983x+0.2700\) Predicted speed using unrounded coefficients: approximately \(3.46\) miles per hour

- Trace the residuals from left to right in each panel. - Look for a curve, trend, or changing spread rather than only positive and negative balance. - Prefer the model whose residuals appear randomly scattered around \(0\).

1. Model A’s residuals are scattered above and below \(0\) without a clear order-based pattern. 2. Model B’s residuals increase steadily from negative to positive as \(x\) increases. 3. The systematic trend for Model B indicates that its line misses structure in the data, so Model A is more appropriate.

Model A is more appropriate because its residuals are scattered around \(0\) without a clear pattern. Model B has a systematic increasing residual trend.

- Follow the residuals from left to right rather than reading them independently. - Translate positive and negative residuals into points above or below the fitted line. - Decide whether the pattern is random or structured.

1. The residuals are negative at small and large \(x\)-values and positive near the middle. 2. Negative residuals mean the line predicts too high; positive residuals mean it predicts too low. 3. The residuals form an upside-down U-pattern, indicating systematic curvature in the original relationship. 4. A linear model is not appropriate because the errors are not randomly scattered around zero.

The original data likely follow a curved, arch-shaped relationship. The linear model is not appropriate because the residuals form a systematic upside-down U-pattern.

- Compare each labeled point’s horizontal position with the main cluster. - Compare each point’s vertical distance from the fitted line. - Recall that leverage concerns unusual \(x\)-position, while residual concerns vertical error.

1. Point P has an \(x\)-value near the center of the main cluster but lies far above the fitted line, so it is primarily a vertical outlier. 2. Point Q has an \(x\)-value far beyond the rest of the data but lies close to the line, so it is a high-leverage point. 3. Point P has the larger residual because residuals measure vertical observed-minus-predicted distance.

Point P is the vertical outlier and has the larger residual. Point Q is the high-leverage point.

- Evaluate both equations at the same input. - Keep all displayed digits through the first calculation. - Compare the final predictions by subtraction.

1. Using the full coefficients gives \(\hat y=1.2746\cdot 85+3.918=112.259\). 2. Using the rounded equation gives \(\hat y=1.3\cdot 85+3.9=114.4\). 3. The early rounding changes the prediction by \(114.4-112.259=2.141\).

Full-coefficient prediction: \(112.259\) Prematurely rounded prediction: \(114.4\) Rounding error: \(2.141\)

- Use the displayed equation consistently for all four predictions. - Keep the residual subtraction order unchanged. - Compare the small total with what exact least-squares coefficients would produce.

1. The rounded predictions are \(1.59, 3.97, 7.54, 9.92\). 2. The residuals are \(-0.59, 1.03, -0.54, 0.08\). 3. Their sum is \(-0.02\). 4. The exact least-squares coefficients make the signed residuals sum to zero; rounding the coefficients changes the predictions slightly.

Residuals: \(-0.59, 1.03, -0.54, 0.08\) Sum: \(-0.02\) The small nonzero sum is caused by rounded regression coefficients.

- Apply each error rule separately. - Notice how squaring changes the influence of one large residual. - Use the criterion named in part c).

1. Model A’s absolute-error sum is \(2+0+0+2=4\), and Model B’s is \(1+1+1+1=4\). 2. Model A’s squared-error sum is \(4+0+0+4=8\), while Model B’s is \(1+1+1+1=4\). 3. Least squares prefers Model B because it has the smaller sum of squared residuals. 4. Both residual lists sum to \(0\), so the comparison does not conflict with the balancing property of least-squares residuals.

a) Both absolute-error sums are \(4\). b) Model A: \(8\); Model B: \(4\) c) Model B is preferred by least squares.

- Compare the vertical width of the residual cloud at small and large fitted values. - Check the center and the spread as separate features. - Consider whether one typical error size describes the whole range.

1. The residuals remain centered near \(0\), but their vertical spread increases from left to right. 2. This creates a fan-shaped pattern. 3. The model’s typical prediction error is small for low fitted values and much larger for high fitted values, so prediction accuracy is not consistent across the range.

The residuals form a fan-shaped pattern with increasing variability. The model’s prediction accuracy is not consistent across the response range.

- Read the residuals from low to high \(x\). - Translate negative and positive residuals into overprediction and underprediction. - Decide whether the fitted line should rise faster or more slowly.

1. Negative residuals at low \(x\)-values mean the line predicts too high there. 2. Positive residuals at high \(x\)-values mean the line predicts too low there. 3. The fitted line rises too slowly across the data, so its slope is too small.

The line’s slope is too small. It overpredicts at low \(x\)-values and underpredicts at high \(x\)-values.

- Follow the residuals in time order. - Compare the first cycle with the later cycle. - Separate balancing around \(0\) from random scatter around \(0\).

1. The residuals repeat a positive-to-negative cycle rather than appearing randomly. 2. A near-zero average residual only shows that positive and negative errors balance overall. 3. The repeated pattern shows that time of day produces systematic structure not captured by the line, so a simple linear trend is inadequate.

The linear model is inadequate because the residuals follow a repeating daily cycle. Balanced positive and negative errors do not remove that systematic pattern.

- Interpret the residual sign separately for each marker type. - Check whether each group is centered around \(0\). - Consider which omitted category could explain the systematic split.

1. Nearly all North residuals are positive, so the line usually underpredicts delivery times for Warehouse North. 2. Nearly all South residuals are negative, so the line usually overpredicts delivery times for Warehouse South. 3. Warehouse membership is associated with a systematic shift not represented by the single line. 4. Separate models or a model that includes warehouse as an additional explanatory factor should be considered.

The line systematically underpredicts one warehouse and overpredicts the other. Warehouse membership should be incorporated, possibly through separate fitted lines.

- Separate “unusual” from “incorrect.” - Trace the observation back to its source. - Consider whether the data point belongs to the population the model is intended to describe.

1. Check the original record for a data-entry, unit, or measurement error. 2. Check whether the observation came from a legitimate but unusual condition that differs from the rest of the data. 3. A large residual identifies an unusual response relative to the model, but it does not show that the observation is incorrect. 4. The point should be removed only for a defensible data-quality or scope reason, not merely to improve the fit.

Verify the recorded value and units, and investigate the conditions under which the observation was collected. A large residual marks an unusual point but does not prove the point is erroneous.

- Treat the two predictions as points on the fitted line. - Find the rate of change before the intercept. - Use the recovered model to compare observed and predicted values.

1. The slope is \(\frac{42-18}{8-2}=4\). 2. Using \(18=4\cdot 2+b\) gives \(b=10\), so \(\hat y=4x+10\). 3. At \(x=5\), the prediction is \(30\). 4. The residual is \(33-30=3\).

a) \(\hat y=4x+10\) b) Residual: \(3\)

- Find where the predicted response reaches its contextual lower limit. - Compare that input with the range used to fit the model. - Separate an algebraic solution from a justified modeling domain.

1. Set the prediction equal to zero: \(45-0.8x=0\). 2. Solving gives \(x=\frac{45}{0.8}=56.25\). 3. Beyond \(56.25\) hours, the equation predicts negative thickness, which is impossible. 4. More importantly, the model is supported only from \(5\) to \(30\) hours, so predictions far beyond that range are unverified even before they become negative.

The equation reaches \(0\) at \(x=56.25\) hours. The model should be used mainly within the observed range of \(5\) to \(30\) hours because distant extrapolations are unsupported and eventually impossible.

- Substitute the mean \(x\)-value into each candidate equation. - Compare each prediction with the mean \(y\)-value. - Decide whether passing through one point uniquely determines a line.

1. Line A gives \(2\cdot 5+2=12\), so it passes through the mean point. 2. Line B gives \(5+7=12\), so it also passes through the mean point. 3. Line C gives \(2\cdot 5+3=13\), so it cannot be the least-squares line. 4. The mean-point property eliminates C but cannot choose between A and B without more information.

Line C is impossible. The mean-point property alone cannot determine whether A or B is the actual regression line.

- Apply the temperature conversion to the entire predicted response. - Distinguish transforming a value from transforming a difference. - Check which part of the conversion affects residual size.

1. Convert the prediction: \(\hat F=1.8(0.4x+10)+32=0.72x+50\). 2. A temperature difference scales by \(1.8\), while the added \(32\) cancels in a residual. 3. The Fahrenheit residual is \(1.8\cdot 2=3.6^\circ\text{F}\).

a) \(\hat F=0.72x+50\) b) \(3.6^\circ\text{F}\)

- Decide whether a vertical shift changes rate of change. - Apply the shift to both observed and predicted responses. - Compare the subtraction defining a residual before and after the shift.

1. Increasing every \(y\)-value by \(100\) shifts the response data and every prediction upward by \(100\). 2. The slope stays \(2.4\), and the intercept becomes \(-7+100=93\). 3. The new line is \(\hat y=2.4x+93\). 4. Each residual is unchanged because both observed and predicted values increase by the same amount.

New regression line: \(\hat y=2.4x+93\) Every residual remains unchanged.

- Express hours in terms of minutes. - Replace the original input variable rather than changing the response. - Check that sixty minutes produces the same predicted change as one hour.

1. Because \(m=60x\), the hour value is \(x=\frac{m}{60}\). 2. Substitute into the original model: \(\hat y=6\left(\frac{m}{60}\right)+4\). 3. Simplifying gives \(\hat y=0.1m+4\).

\(\hat y=0.1m+4\)

- Enter each ordered pair as a separate observation. - Distinguish repeated inputs from every input being identical. - Use all six responses in the regression calculation.

1. Technology gives \(\hat y=2.75x+0.1667\), with the intercept rounded to four decimal places. 2. The \(x\)-variable still varies across \(1, 2, 3\), so a linear relationship can be estimated. 3. Repeated \(x\)-values represent multiple responses at the same input and contribute to the least-squares fit normally.

\(\hat y=2.75x+0.1667\) Repeated \(x\)-values are allowed because \(x\) is not constant across the entire data set.

- Look for whether the response rises overall, falls overall, or balances symmetrically. - Find the average \(y\)-value. - A zero slope does not require all \(y\)-values to be equal.

1. The correlation coefficient is \(r=0\). 2. The mean response is \(\bar y=4\), and the regression slope is \(0\). 3. The least-squares line is \(\hat y=4\). 4. The horizontal line is valid because the response varies but has no net linear increase or decrease with \(x\).

\(r=0\) \(\hat y=4\)

- Identify which variable the model predicts. - Keep the explanatory variable fixed when comparing actual and predicted values. - Distinguish vertical distance from shortest geometric distance.

1. In regression of \(y\) on \(x\), a residual compares the observed \(y\)-value with the predicted \(y\)-value at the same \(x\)-value. 2. This difference is measured vertically, parallel to the y-axis. 3. The shortest perpendicular distance generally changes both \(x\) and \(y\) and is not the least-squares residual used by this model.

The vertical distance represents the residual. A residual is \(y-\hat y\) at the same \(x\)-value, not the shortest perpendicular distance to the line.

- Keep the required vertical coordinate fixed. - Ask what horizontal quantity is being compared with the errors. - Consider which plot would make changing error spread easiest to see.

1. Both horizontal-axis choices are valid because the vertical axis contains the same residuals. 2. Plotting residuals against \(x\) helps reveal patterns tied directly to the explanatory variable, such as curvature over the input range. 3. Plotting residuals against \(\hat y\) helps reveal whether error spread changes as the predicted response grows.

Yes. Residuals versus \(x\) emphasize patterns across the input range, while residuals versus \(\hat y\) emphasize patterns across predicted response levels.

- Squared errors require squared unit conversions. - Convert both totals before comparing their numbers. - Check whether the conversion factor itself must be squared.

1. Since \(1\,\text{m}=100\,\text{cm}\), \(1\,\text{m}^2=10{,}000\,\text{cm}^2\). 2. Model B’s total is \(2500\div 10{,}000=0.25\,\text{m}^2\). 3. Compare \(0.25\,\text{m}^2\) with \(0.36\,\text{m}^2\); Model B has the smaller sum of squared residuals.

Model B has \(0.25\,\text{m}^2\) of squared error, compared with Model A’s \(0.36\,\text{m}^2\). Model B is smaller.

- Use the numerical tie as a starting point, not the final decision. - Inspect whether the residuals show an ordered pattern or irregular scatter. - A useful model should not leave predictable structure in its residuals.

1. Equal squared-error totals mean the models have the same aggregate least-squares error. 2. Plot a) leaves a systematic progression from negative residuals through values near zero to positive residuals, so the model misses predictable structure. 3. Plot b) leaves residuals irregularly scattered above and below zero. 4. Model B should be preferred because its residual plot better supports the model, despite the tied squared-error total.

Prefer Model B, shown in plot b). The totals tie, but plot a) has a systematic residual pattern that shows Model A misses structure in the data.

- Apply the usual range definition first. - Then inspect how the observed \(x\)-values are distributed inside that range. - Distinguish being inside the endpoints from having nearby data support.

1. The value \(50\) lies between the minimum \(x\)-value \(1\) and maximum \(x\)-value \(99\), so it is technically interpolation. 2. However, no observations occur anywhere near \(50\); the data have a large internal gap. 3. The relationship could change inside that gap without being detected, so the prediction has much less local support than ordinary interpolation.

It is technically interpolation, but it should be treated cautiously because \(x=50\) lies in a large gap with no nearby observations.

- Distinguish the range of observed responses from the range of line predictions. - Check whether the requested input is inside the observed \(x\)-range. - Ask whether the predicted value is contextually possible.

1. A least-squares line is not constrained to stay between the smallest and largest observed \(y\)-values. 2. Its prediction follows the fitted slope and intercept, so an endpoint-area prediction can lie slightly outside the observed response range. 3. Because the input is inside the observed \(x\)-range, the prediction is interpolation; it is not automatically invalid, though its residual uncertainty should still be considered.

No. A fitted line may predict slightly outside the observed \(y\)-range even for an input inside the \(x\)-range. The value should be evaluated in context, not rejected solely for exceeding the sample maximum.

- Translate zero residuals into a geometric statement about all points. - Recall how many points with different \(x\)-values determine a line. - Compare zero squared error with any positive squared error.

1. Zero residual for every observation means every point lies on the line. 2. Two points with different \(x\)-values determine exactly one nonvertical line. 3. Any different line would miss at least one of those points and have a positive sum of squared residuals. 4. The zero-error line is therefore the unique least-squares fit.

The line is unique because two observations with different \(x\)-values determine one line, and no other line can retain zero residual for every point.

- Check that each residual list balances around zero. - Compute the squared-error totals and the largest absolute residuals separately. - Do not assume one model must win under both criteria.

1. Both residual lists sum to \(0\), so each could be consistent with a least-squares line that includes an intercept. 2. Model A’s squared-error sum is \(9+1+1+1=12\). 3. Model B’s squared-error sum is \(4\cdot1.8^2=12.96\). 4. Least squares therefore prefers Model A. 5. Model A’s maximum absolute residual is \(3\), while Model B’s is \(1.8\), so Model B has the smaller worst individual error.

Least squares prefers Model A. Model B has the smaller maximum absolute residual.

- Find both lines’ predicted values at every \(x\)-value. - Subtract predictions from observations before squaring. - Compare totals using the criterion named in the problem.

1. The least-squares line is \(\hat y=1.4x+0.9\). 2. Its residuals are \(0.1, 0.7, -1.7, 0.9\), giving a squared-residual sum of \(4.20\). 3. The endpoint line has residuals \(0, \frac{1}{3}, -\frac{7}{3}, 0\), giving a squared-residual sum of \(\frac{50}{9}\approx 5.56\). 4. The least-squares line fits better because \(4.20<5.56\).

Least-squares line: \(\hat y=1.4x+0.9\) Least-squares SSE: \(4.20\) Endpoint-line SSE: approximately \(5.56\) The least-squares line fits better.

- Substitute the new \(x\)-value into the current line. - Determine the new point’s residual. - Consider what a zero-error observation contributes to the fitting criterion.

1. The new point lies exactly on the current line because \(3\cdot 100+2=302\). 2. Its residual from the current line is \(0\), so it adds no squared error and no imbalance to the least-squares conditions. 3. The original line therefore remains the least-squares line after the point is added.

The least-squares line remains \(\hat y=3x+2\).