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Linear regression and residuals

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5441139
A model uses distance in miles as \(x\) and fuel used in gallons as \(y\). One residual is \(0.7\). State the residual’s units and interpret it in context.

Hints

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

Solution

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.

Answer

The residual is \(0.7\,\text{gallons}\). The observed fuel use was \(0.7\,\text{gallons}\) above the prediction.
5441149
A regression model was built from \(x\)-values between \(10\) and \(40\). Classify each prediction as interpolation or extrapolation: \(x=25\), \(x=8\), and \(x=55\). Which predictions generally deserve more caution?

Hints

- 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.

Solution

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.

Answer

\(x=25\): interpolation \(x=8\): extrapolation \(x=55\): extrapolation The predictions at \(8\) and \(55\) require more caution.
5441289
A regression model gives one observation a residual of \(0\). What does this show, and what does it not show about the model?

Hints

- 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.

Solution

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.

Answer

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.
54414312
A list is reported as the six residuals from an unrounded least-squares line with an intercept: \(2,2,2,2,2,2\). Explain why this list cannot be correct.

Hints

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

Solution

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.

Answer

It is impossible because the residuals would sum to \(12\), not \(0\).
54409912
A least-squares line is \(\hat y=4x+10\). For \(x=1, 2, 3, 4, 5\), the observed \(y\)-values are \(13, 19, 22, 27, 29\). Find all five residuals and their sum. What feature of the result is expected for a least-squares line with an intercept?

Hints

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

Solution

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.

Answer

Residuals: \(-1, 1, 0, 1, -1\) Sum: \(0\)
5441009
A theater uses the model \(\hat y=7.5x+18\), where \(x\) is the number of promotional posts and \(y\) is predicted ticket sales. How many promotional posts correspond to predicted ticket sales of \(93\)?

Hints

- 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.

Solution

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

Answer

\(10\) promotional posts
54410312
For data points \((0, 1), (2, 5), (5, 7), (7, 10)\), a calculator displays the rounded regression equation \(\hat y=1.19x+1.59\). Using this rounded equation, find the four residuals and their sum. Explain why the sum is close to, but not exactly, zero.

Hints

- 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.

Solution

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.

Answer

Residuals: \(-0.59, 1.03, -0.54, 0.08\) Sum: \(-0.02\) The small nonzero sum is caused by rounded regression coefficients.
54410412
Five residuals from an unrounded least-squares line with an intercept are \(1.2, -0.5, r, 0.8, -1.1\). Find the missing residual \(r\).

Hints

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

Solution

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 \(r=-0.4\).

Answer

\(r=-0.4\)
54410512
Two candidate models have these residuals: Model A: \(-2, 0, 0, 2\) Model B: \(-1, -1, 1, 1\) a) Compare their sums of absolute residuals. b) Compare their sums of squared residuals. c) Which model is preferred by the least-squares criterion?

Hints

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

Solution

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.

Answer

a) Both absolute-error sums are \(4\). b) Model A: \(8\); Model B: \(4\) c) Model B is preferred by least squares.
5441119
A regression model is \(\hat y=1.5x+5\). Complete the missing entries. <table><tr><th>\(x\)</th><th>Observed \(y\)</th><th>Predicted \(\hat y\)</th><th>Residual</th></tr><tr><td>\(2\)</td><td>?</td><td>?</td><td>\(-1\)</td></tr><tr><td>\(4\)</td><td>\(12\)</td><td>?</td><td>?</td></tr><tr><td>\(6\)</td><td>?</td><td>?</td><td>\(2.5\)</td></tr></table>

Hints

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

Solution

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 \(y=14+2.5=16.5\).

Answer

<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>
54411712
A least-squares line has slope \(1.8\). The data means are \(\bar x=12\) and \(\bar y=31\). Find the y-intercept and write the regression equation.

Hints

- 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.

Solution

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 \(\hat y=1.8x+9.4\).

Answer

Y-intercept: \(9.4\) Regression equation: \(\hat y=1.8x+9.4\)
54411812
The mean point of a data set is \((5, 12)\). Three candidate regression lines are: A: \(\hat y=2x+2\) B: \(\hat y=x+7\) C: \(\hat y=2x+3\) Use the mean-point property to eliminate any impossible candidate. Can that property identify the actual regression line by itself?

Hints

- 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.

Solution

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.

Answer

Line C is impossible. The mean-point property alone cannot determine whether A or B is the actual regression line.
54411912
A data set contains an actual observation at its mean point \((\bar x, \bar y)\). A least-squares line with an intercept is fitted. What is the residual of that observation? Explain.

Hints

- 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.

Solution

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 \(y-\hat y=\bar y-\bar y=0\).

Answer

The residual is \(0\).
5441269
A fitted model has residuals \(-2, 1, 0, 3, -2\). a) Find the sum of the residuals. b) Find the sum of the absolute residuals. c) Find the sum of squared residuals. d) Explain why the three totals measure different features of model error.

Hints

- 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.

Solution

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.

Answer

a) \(0\) b) \(8\) c) \(18\) d) Signed errors can cancel; absolute errors measure total distance; squared errors emphasize large residuals.
5441299
A regression model is \(\hat y=2.5x+7\). An observation has \(y=40\) and residual \(3\). Find its \(x\)-value.

Hints

- 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.

Solution

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 \(x=12\).

Answer

The observation has input \(x=12\).
5441319
A graph has \(x\)-values on the horizontal axis and observed \(y\)-values on the vertical axis, but it is labeled “residual plot.” What should be placed on the vertical axis instead, and why?

Hints

- 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.

Solution

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.

Answer

The vertical axis should show residuals \(y-\hat y\). This makes patterns in the model’s prediction errors visible around the zero line.
5441359
A fitted model is \(\hat y=5.2x-1\). At \(x=6\), an observation currently has \(y=28\). To what \(y\)-value would the observation need to be corrected so that its residual is exactly \(0\)?

Hints

- 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.

Solution

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 \(30.2\).

Answer

\(30.2\)
5441369
A regression model predicts \(47.6\) customer arrivals during an hour, even though an observed customer count must be a whole number. Explain why the decimal prediction can still be useful and state a reasonable way to report it.

Hints

- 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.

Solution

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.

Answer

The decimal is a valid fitted average. Report it as a prediction of about \(48\) customers.
54414112
A data set contains at least two different \(x\)-values, and one line gives residual \(0\) for every observation. Explain why that line is the unique least-squares line.

Hints

- 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.

Solution

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.

Answer

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.
54414212
Model A has residuals \(-3,1,1,1\). Model B has residuals \(-1.8,-1.8,1.8,1.8\). Which model is preferred by least squares? Which model has the smaller maximum absolute residual?

Hints

- 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.

Solution

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.

Answer

Least squares prefers Model A. Model B has the smaller maximum absolute residual.
5440849
A linear regression analysis gives \(r=0.84\). a) Interpret \(r\) in terms of direction and strength. b) Compute \(r^2\) and interpret it as the proportion of variation in the response accounted for by the linear model. c) Explain why neither value means that \(84\%\) of the points lie on one straight line.

Hints

- 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.

Solution

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.

Answer

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.
5440959
A kayaking coach recorded stroke rate and boat speed. <table><tr><th>Stroke rate, \(x\)</th><td>\(30\)</td><td>\(36\)</td><td>\(42\)</td><td>\(48\)</td><td>\(54\)</td></tr><tr><th>Speed in miles per hour, \(y\)</th><td>\(3.2\)</td><td>\(3.8\)</td><td>\(4.4\)</td><td>\(5.1\)</td><td>\(5.5\)</td></tr></table> Use technology to find the least-squares regression line, displaying coefficients to four decimal places. Then use the unrounded regression coefficients to predict the speed at \(32.4\) strokes per minute. Round the prediction to the nearest hundredth.

Hints

- 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.

Solution

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.

Answer

Regression line: \(\hat y=0.0983x+0.2700\) Predicted speed using unrounded coefficients: approximately \(3.46\) miles per hour
5440969
The residual plots compare two linear models. Observations are ordered from smallest to largest \(x\)-value. Which model is more appropriate? Use the residual patterns, not just the average residual.
Figure for problem 544096

Hints

- 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 \(0\).

Solution

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.

Answer

Model A is more appropriate because its residuals are scattered around \(0\) without a clear pattern. Model B has a systematic increasing residual trend.
54409712
The scatter plot shows four points and an endpoint line that passes through the first and last points. Use technology to find the least-squares line. Then compute the sum of squared residuals for the least-squares line and the endpoint line, and decide which fits better by the least-squares criterion.
Figure for problem 544097

Hints

- Use regression technology on the four plotted observations to obtain one of the lines. - For each line, compare every observed \(y\)-value with the predicted value at the same \(x\)-value. - The least-squares criterion compares sums of squared residuals, not just the signed residual totals.

Solution

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 \(4.20<5.56\).

Answer

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.
5440989
The residual plot comes from a linear model. What shape does the pattern suggest in the original data, and what does it imply about the linear model?
Figure for problem 544098

Hints

- 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.

Solution

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.

Answer

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.
5441019
The scatter plot shows a fitted line, a main data cluster, and labeled points P and Q. Classify each labeled point as primarily a vertical outlier or a high-leverage point. Which point has the larger residual?
Figure for problem 544101

Hints

- 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.

Solution

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.

Answer

Point P is the vertical outlier and has the larger residual. Point Q is the high-leverage point.
5441029
A regression calculator gives \(\hat y=1.2746x+3.918\). Compare the prediction at \(x=85\) from the full coefficients with the prediction from the prematurely rounded equation \(\hat y=1.3x+3.9\). How much error does the early rounding create?

Hints

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

Solution

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 \(114.4-112.259=2.141\).

Answer

Full-coefficient prediction: \(112.259\) Prematurely rounded prediction: \(114.4\) Rounding error: \(2.141\)
5441069
The residual plot comes from a linear model. Describe the pattern and explain what concern it raises about the model.
Figure for problem 544106

Hints

- 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.

Solution

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.

Answer

The residuals form a fan-shaped pattern with increasing variability. The model’s prediction accuracy is not consistent across the response range.
5441079
The residual plot shows residuals against the original \(x\)-values for a candidate line. What does the pattern suggest about the line's slope?
Figure for problem 544107

Hints

- 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.

Solution

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.

Answer

The line's slope is too small. It overpredicts at low \(x\)-values and underpredicts at high \(x\)-values.
5441089
The residual plot shows errors from a linear model of electricity use over two days. Why is a linear time trend inadequate even if the signed residuals average nearly \(0\)?
Figure for problem 544108

Hints

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

Solution

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.

Answer

The linear model is inadequate because the residuals follow a repeating daily cycle. Balanced positive and negative errors do not remove that systematic pattern.
5441099
The residual plot comes from one regression line fitted to delivery times from two warehouses. What does the group pattern suggest, and what model revision should be considered?
Figure for problem 544109

Hints

- 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.

Solution

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.

Answer

The line systematically underpredicts one warehouse and overpredicts the other. Warehouse membership should be incorporated, possibly through separate fitted lines.
5441109
One observation has a residual of \(18.7\), while every other residual is between \(-3\) and \(3\). Give two checks that should be made before deciding whether to remove it, and explain why a large residual alone is not enough.

Hints

- 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.

Solution

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.

Answer

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.
5441129
A linear regression model predicts \(18\) when \(x=2\) and \(42\) when \(x=8\). a) Find the model equation. b) An observation at \(x=5\) has \(y=33\). Find its residual.

Hints

- 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.

Solution

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 \(33-30=3\).

Answer

a) \(\hat y=4x+10\) b) Residual: \(3\)
5441159
A model for material thickness after \(x\) hours of processing is \(\hat y=45-0.8x\), where \(y\) is measured in millimeters. The model was fitted using data from \(5\) to \(30\) hours. At what \(x\)-value does the model first reach \(0\)? Why should the practical use of the model be restricted more than the algebraic equation suggests?

Hints

- 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.

Solution

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.

Answer

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.
5441209
A regression model predicts a temperature response in degrees Celsius: \(\hat C=0.4x+10\). The response is converted to degrees Fahrenheit using \(F=1.8C+32\). a) Write the regression model for predicted Fahrenheit temperature. b) If an observation has a residual of \(2^\circ\text{C}\), what is its residual in degrees Fahrenheit?

Hints

- 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.

Solution

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 \(1.8\cdot 2=3.6^\circ\text{F}\).

Answer

a) \(\hat F=0.72x+50\) b) \(3.6^\circ\text{F}\)
5441219
A data set has regression line \(\hat y=2.4x-7\). Every observed \(y\)-value is increased by \(100\), while the \(x\)-values are unchanged. Write the new regression line and state what happens to every residual.

Hints

- 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.

Solution

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.

Answer

New regression line: \(\hat y=2.4x+93\) Every residual remains unchanged.
5441229
A regression model is \(\hat y=6x+4\), where \(x\) is elapsed time in hours. A new variable \(m\) measures the same elapsed time in minutes. Rewrite the model using \(m\) as the input variable.

Hints

- 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.

Solution

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

Answer

\(\hat y=0.1m+4\)
5441249
The scatter plot shows six paired observations, including repeated \(x\)-values. Use technology to find the least-squares regression line. Explain why repeated \(x\)-values do not prevent fitting a line.
Figure for problem 544124

Hints

- Read each plotted point as a separate observation, even when two points share an \(x\)-coordinate. - Distinguish repeated inputs from every input being identical. - Use all six plotted responses in the regression calculation.

Solution

1. Reading the six plotted observations and using regression 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.

Answer

\(\hat y=2.75x+0.1667\) Repeated \(x\)-values are allowed because \(x\) is not constant across the entire data set.
5441259
The scatter plot shows five paired observations. Use technology to find the correlation coefficient and the least-squares regression line. Explain why a horizontal regression line is valid here.
Figure for problem 544125

Hints

- Read the five plotted observations before entering them into technology. - Look for whether the response rises overall, falls overall, or balances around a horizontal level. - A zero slope does not require all \(y\)-values to be equal.

Solution

1. Reading the five plotted observations and using technology gives \(r=0\). 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 \(x\).

Answer

\(r=0\) \(\hat y=4\)
54412712
A data set has least-squares line \(\hat y=3x+2\). A new observation \((100, 302)\) is added. Determine whether the least-squares line changes, and justify your answer.

Hints

- 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.

Solution

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.

Answer

The least-squares line remains \(\hat y=3x+2\).
5441309
The figure shows a point above a slanted regression line. Two distances from the point to the line are marked: a vertical distance and a shorter perpendicular distance. Which distance represents the residual? Explain why.
Figure for problem 544130

Hints

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

Solution

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.

Answer

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.
5441329
Two analysts make residual plots for the same model. One uses the explanatory variable \(x\) on the horizontal axis; the other uses the fitted value \(\hat y\). Are both choices valid? State what each version helps reveal.

Hints

- 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.

Solution

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.

Answer

Yes. Residuals versus \(x\) emphasize patterns across the input range, while residuals versus \(\hat y\) emphasize patterns across predicted response levels.
5441339
The same five observations are being modeled by two candidate lines. <table><tr><th>\(x\)</th><th>Observed \(y\)</th></tr><tr><td>\(1\)</td><td>\(10\)</td></tr><tr><td>\(2\)</td><td>\(14\)</td></tr><tr><td>\(3\)</td><td>\(13\)</td></tr><tr><td>\(4\)</td><td>\(18\)</td></tr><tr><td>\(5\)</td><td>\(20\)</td></tr></table> Model A: \(\hat y=2x+9\) Model B: \(\hat y=x+10\) a) Calculate the five residuals, in increasing order of \(x\), for each model. b) Describe the residual pattern for each model as \(x\) increases. c) Which model is more reasonable for these observations? What does Model B's residual pattern suggest about its slope?

Hints

- For each candidate line, compare the observed response with that model's prediction at the same \(x\)-value. - Keep the residuals in increasing order of \(x\) so any pattern is visible. - A useful linear model should not leave a clear directional trend in its residuals.

Solution

1. For Model A, the predictions are \(11,13,15,17,19\). Subtracting prediction from observation gives residuals \(-1,1,-2,1,1\). 2. For Model B, the predictions are \(11,12,13,14,15\). The residuals are \(-1,2,0,4,5\). 3. Model A's residuals fluctuate on both sides of zero without a clear increasing or decreasing pattern. 4. Model B's residuals tend to increase as \(x\) increases. The model increasingly underpredicts at larger \(x\)-values, indicating that its slope is too small. 5. Model A is therefore the more reasonable of the two candidate lines for these observations.

Answer

a) Model A: \(-1,1,-2,1,1\) Model B: \(-1,2,0,4,5\) b) Model A's residuals fluctuate around zero without a clear trend. Model B's residuals generally increase as \(x\) increases. c) Model A is more reasonable. Model B increasingly underpredicts at larger \(x\), so its slope is too small.
54413412
The observations are \((-3, 0), (-1, 2), (1, 3), (3, 7)\). Without using a regression calculator, use \(b=\frac{\sum (x-\bar x)(y-\bar y)}{\sum (x-\bar x)^2}\) and \(a=\bar y-b\bar x\) to find the least-squares regression line \(\hat y=bx+a\). Then find all four residuals and the sum of squared residuals.

Hints

- Find \(\bar x\) and \(\bar y\) before forming the deviations in the slope formula. - Keep the numerator and denominator of the slope formula separate until the final division. - After finding the line, compute each residual as observed minus predicted before squaring.

Solution

1. The means are \(\bar x=0\) and \(\bar y=3\). 2. The slope numerator is \(9+1+0+12=22\), and the denominator is \(9+1+1+9=20\). Thus \(b=\frac{22}{20}=1.1\). 3. Since \(\bar x=0\), \(a=\bar y-b\bar x=3\). The least-squares line is \(\hat y=1.1x+3\). 4. The predicted values are \(-0.3, 1.9, 4.1, 6.3\), so the residuals are \(0.3, 0.1, -1.1, 0.7\). 5. The sum of squared residuals is \(0.3^2+0.1^2+(-1.1)^2+0.7^2=1.80\).

Answer

Least-squares line: \(\hat y=1.1x+3\) Residuals: \(0.3, 0.1, -1.1, 0.7\) Sum of squared residuals: \(1.80\)
5441379
The table gives observed values. Use the model \(\hat y=2x+1\) to calculate the predictions. <table><tr><th>\(x\)</th><th>Observed \(y\)</th></tr><tr><td>\(1\)</td><td>\(4\)</td></tr><tr><td>\(2\)</td><td>\(4\)</td></tr><tr><td>\(3\)</td><td>\(8\)</td></tr><tr><td>\(4\)</td><td>\(8\)</td></tr><tr><td>\(5\)</td><td>\(12\)</td></tr><tr><td>\(6\)</td><td>\(12\)</td></tr></table> a) Calculate the residual for each observation. b) Which displayed plot is the residual plot for these data? c) Does the residual pattern support using this linear model? Explain.
Figure for problem 544137

Hints

- 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.

Solution

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.

Answer

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.
5441389
The two residual plots come from candidate models with the same sum of squared residuals. Which model should be preferred? Explain why the tied squared-error total does not settle the decision.
Figure for problem 544138

Hints

- 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.

Solution

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.

Answer

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.
5441399
A regression model was fitted using \(x\)-values \(1, 2, 3, 4, 5, 95, 96, 97, 98, 99\). A prediction is requested at \(x=50\). Is this technically interpolation or extrapolation? Why should the prediction still be treated cautiously?

Hints

- 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.

Solution

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.

Answer

It is technically interpolation, but it should be treated cautiously because \(x=50\) lies in a large gap with no nearby observations.
5441409
In a data set, all observed \(y\)-values are between \(10\) and \(20\). At an \(x\)-value inside the observed \(x\)-range, the least-squares line predicts \(20.8\). Is that prediction automatically an error? Explain.

Hints

- 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.

Solution

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.

Answer

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.
5513879
The scatter plot shows quiz scores for six students at different study times. Two candidate lines are shown: line a is \(y=2x+3\), and line b is \(y=x+7\). Without calculating the exact least-squares regression line, which candidate is the more reasonable line of fit? Explain using the vertical distances from the points to the lines. Then use your chosen line to estimate the quiz score for \(4.5\) hours of study.
Figure for problem 551387

Hints

- Compare vertical, not horizontal, distances from the observations to each candidate line. - Look for a line that leaves relatively small gaps with points on both sides rather than a systematic run from one side to the other. - After choosing a line, use its equation for the requested study time.

Solution

1. For line a, the vertical residuals are \(1,-1,1,-1,1,-1\). The points are close to the line and are balanced above and below it. 2. For line b, the residuals are \(-2,-3,0,-1,2,1\). The earlier points tend to lie below the line and the later points tend to lie above it, showing a systematic pattern with larger gaps. 3. Therefore, line a is the more reasonable informal line of fit. 4. At \(x=4.5\), line a gives \(y=2(4.5)+3=12\).

Answer

Line a is the more reasonable fit. It keeps the vertical deviations small and balanced above and below the line. Estimated quiz score at \(4.5\) hours: \(12\) points.
5549569
Beacon Learning Center recorded weekly independent practice time and a unit quiz score for six students. <table><tr><th>Practice time \(x\) (hours)</th><th>Quiz score \(y\)</th></tr><tr><td>\(1\)</td><td>\(52\)</td></tr><tr><td>\(2\)</td><td>\(58\)</td></tr><tr><td>\(3\)</td><td>\(65\)</td></tr><tr><td>\(4\)</td><td>\(67\)</td></tr><tr><td>\(5\)</td><td>\(76\)</td></tr><tr><td>\(6\)</td><td>\(82\)</td></tr></table> a) Describe the direction, form, and strength of the association in the paired data. b) Use technology to find the correlation coefficient \(r\) and the least-squares line. Report each coefficient to the nearest thousandth. c) Interpret the slope and y-intercept in context. Include an appropriate caution about the y-intercept. d) Using unrounded calculator coefficients, predict the quiz score for \(x=5\) and find the residual for the observed score at \(x=5\). Round both to the nearest hundredth. e) Calculate the six residuals to the nearest hundredth. Do they show a clear systematic pattern? Based on the association and residuals, is a linear model reasonable over the observed range?

Hints

- First decide whether the paired values follow a roughly linear trend before relying on a linear summary. - Keep the association measure, fitted equation, and residuals tied to the same paired observations. - Interpret the slope as change in predicted response per one unit of the explanatory variable. - For a residual, compare the observed response with the model's prediction at the same input. - Judge the model using both the overall association and the pattern left in the residuals.

Solution

1. The paired values rise together and stay close to a straight-line trend, so the association is strong, positive, and approximately linear. 2. Technology gives \(r\approx0.993\) and \(\hat y\approx5.886x+46.067\). 3. The slope means the model predicts about \(5.886\) additional quiz-score points for each additional hour of weekly practice. The y-intercept means the model predicts a score of about \(46.067\) at \(0\) hours, but \(0\) hours lies outside the observed range of \(1\) to \(6\) hours, so that interpretation should be treated cautiously. 4. Using the unrounded coefficients, the prediction at \(x=5\) is \(75.50\). The residual is \(76-75.50\approx0.50\). 5. The residuals are approximately \(0.05, 0.16, 1.28, -2.61, 0.50, 0.62\). They fluctuate around zero without a clear curved or directional pattern. Together with the strong approximately linear association, this supports using a linear model over the observed range.

Answer

a) Strong positive, approximately linear association. b) \(r\approx0.993\); \(\hat y\approx5.886x+46.067\) c) Slope: about \(5.886\) predicted quiz-score points per additional hour of weekly practice. Y-intercept: about \(46.067\) predicted points at \(0\) hours, but \(0\) hours is outside the observed data range, so this interpretation is an extrapolation. d) Predicted score: \(75.50\); residual: \(0.50\) e) Residuals: approximately \(0.05, 0.16, 1.28, -2.61, 0.50, 0.62\). No clear systematic pattern is evident, so a linear model is reasonable over the observed range.

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