Linear regression and residuals

Learning goals

Finds a least-squares regression line from paired data.
Computes predicted values and residuals.
Uses residual patterns to assess linear-model adequacy.
Identifies influential or unusual observations qualitatively.
Avoids extrapolation beyond an appropriate data range.

Difficulty: 1 – 5

Click problems to add them to your worksheet.

Linear regression and residuals 1

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.

Linear regression and residuals 2

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.

Linear regression and residuals 3

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.

Linear regression and residuals 4

A calculator fits the line \(\hat y=3x+2\) to a set of paired observations. What does the symbol \(\hat y\) mean? Does fitting this line mean every observed point must lie exactly on it?

Hints

- Compare a measured response with a value predicted by an equation. - Think about what a line of best fit is supposed to summarize. - Would a line still be useful if some observations are higher or lower?

Solution

1. The symbol \(\hat y\) represents the value of the response predicted from the given x-value by the fitted line. 2. A fitted line summarizes an overall pattern. Observations may lie above or below it, so it need not pass through every data point.

Answer

\(\hat y\) is the predicted response. No, the fitted line does not have to pass through every observation.

Linear regression and residuals 5

A music festival sold advance tickets before its ticket booth opened. A fitted line predicts the total number sold: \(\hat y=50+6x\), where \(x\) is hours since the booth opened. How many tickets does the model predict have been sold after \(3\) hours?

Hints

- Which variable is given in the question? - Substitute that value for the input in the prediction line. - State the response in tickets, not hours.

Solution

1. Substitute \(x=3\) into the fitted equation. 2. The prediction is \(\hat y=50+6\cdot3=68\) tickets.

Answer

\(68\) tickets

Linear regression and residuals 6

A line predicts that a runner will finish a short course in \(11.5\) minutes. The actual time was \(14\) minutes. Find the residual, defined as actual minus predicted. Did the line predict too much or too little time?

Hints

- Which time came from a prediction, and which time came from the stopwatch? - Did the runner finish faster or slower than the line suggested? - What would a residual of zero mean in this situation?

Solution

1. The residual is \(14-11.5=2.5\) minutes. 2. The actual time is longer than the prediction, so the line predicted too little time.

Answer

Residual: \(+2.5\) minutes. The line underpredicted the running time.

Linear regression and residuals 7

The graph shows a fitted line and one measured point marked P. Is P's residual positive, negative or zero? Does the line overpredict or underpredict the observed response?

Hints

- At P's x-value, compare the point's height with the line's height. - Recall that residual means observed minus predicted. - What sign results when the actual value is higher than the prediction?

Solution

1. At \(x=3\), the line predicts \(2\cdot3+1=7\), while the observed point P is at height \(9\). 2. The residual is \(9-7=2\), which is positive. The line underpredicts the observation.

Answer

Positive residual; the line underpredicts P's response.

Linear regression and residuals 8

The graph plots residuals from a fitted linear model. Do the residuals look randomly scattered around \(0\), or do they suggest a pattern the line misses? Explain briefly.

Hints

- Compare the residual heights at the ends with the residual near the center. - Does the pattern change direction in a regular way? - Random residual scatter would not trace a clear shape.

Solution

1. The residuals are positive at low and high x-values but negative near the middle. 2. This curved pattern is systematic, not random scatter around zero. 3. A simple straight-line model is missing a curved part of the relationship.

Answer

The residuals show a curved pattern, so the line does not fully describe the relationship.

Linear regression and residuals 9

A fitted line describes the connection between time spent at a skate park and the number of tricks practiced. The measured times ranged from \(2\) to \(8\) hours. Which prediction uses interpolation: the prediction at \(6\) hours or the prediction at \(12\) hours? Which one would you trust more based only on the measured range?

Hints

- Compare each proposed time with the smallest and largest observed values. - Decide which prediction stays inside the collected data. - A pattern seen in one interval may not continue beyond it.

Solution

1. Six hours is between the smallest and largest measured times, so a prediction there is interpolation. 2. Twelve hours is beyond the measured range, so that prediction is extrapolation. 3. The prediction at six hours is generally safer because the fitted pattern has been observed within that range.

Answer

The prediction at \(6\) hours is interpolation and is generally more trustworthy. The prediction at \(12\) hours is extrapolation.

Linear regression and residuals 10

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>\(\square\)</td><td>\(\square\)</td><td>\(-1\)</td></tr><tr><td>\(4\)</td><td>\(12\)</td><td>\(\square\)</td><td>\(\square\)</td></tr><tr><td>\(6\)</td><td>\(\square\)</td><td>\(\square\)</td><td>\(2.5\)</td></tr></table>

Hints

- Find each prediction from the regression equation first. - Keep residual defined as observed minus predicted. - Work backward from the residual when an observed value is unknown.

Solution

1. At \(x=2\), the prediction is \(1.5\cdot2+5=8\), so the observed value is \(8-1=7\). 2. At \(x=4\), the prediction is \(11\), so the residual is \(12-11=1\). 3. At \(x=6\), the prediction is \(14\), so the observed value is \(14+2.5=16.5\).

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>

Linear regression and residuals 11

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.

Linear regression and residuals 12

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 \(x=12\).

Linear regression and residuals 13

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.

Linear regression and residuals 14

A fitted model is \(\hat y=5.2x-1\). At \(x=6\), the observed response is \(28\). a) Find the predicted response and the residual. b) What observed response at that same \(x\)-value would give a residual of exactly \(0\)?

Hints

- Find the fitted line's value at the given input. - What calculation turns actual and predicted responses into a residual? - Under what condition does that difference equal zero?

Solution

1. At \(x=6\), the prediction is \(5.2\cdot6-1=30.2\). 2. The actual residual is \(28-30.2=-2.2\). 3. A residual of \(0\) would require an observation equal to the prediction, \(30.2\).

Answer

a) Prediction: \(30.2\); residual: \(-2.2\) b) \(30.2\)

Linear regression and residuals 15

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.

Linear regression and residuals 16

A player tracked successful shots across five practice sessions. For each line, find the sum of squared vertical errors at all five points. Which line has the smaller total? Show both calculations.

Hints

- Choose one session and compare the observed height with each line's predicted height. - How can you record the errors so that none of the five observations is skipped? - Does having more points exactly on a line guarantee the smallest total of squared errors?

Solution

1. At sessions \(1\) through \(5\), line a predicts \(3,4,5,6,7\) shots. The observed counts are \(2,5,6,7,8\), so the errors are \(-1,1,1,1,1\). 2. Line a's sum of squared errors is \(1+1+1+1+1=5\). 3. Line b predicts \(2,4,6,8,10\) shots. Its errors are \(0,1,0,-1,-2\), so the sum of squared errors is \(0+1+0+1+4=6\). 4. Line a has the smaller sum of squared errors, even though line b passes through two observed points.

Answer

Line a: \(1+1+1+1+1=5\). Line b: \(0+1+0+1+4=6\). Line a has the smaller sum of squared vertical errors.

Linear regression and residuals 17

A chess club recorded each student's practice time and score on a puzzle quiz. <table><tr><th>Practice hours \(x\)</th><th>Score \(y\)</th></tr><tr><td>\(1\)</td><td>\(2\)</td></tr><tr><td>\(2\)</td><td>\(4\)</td></tr><tr><td>\(3\)</td><td>\(5\)</td></tr><tr><td>\(4\)</td><td>\(7\)</td></tr></table> Use regression technology to find the least-squares line. Give the slope and intercept to the nearest tenth, and state whether the fitted association is positive or negative.

Hints

- Keep the x- and y-values paired when entering the table. - Look for the fitted line's slope and constant in the calculator output. - Use the sign of the slope to describe the fitted direction.

Solution

1. The mean inputs and responses are \(\bar x=2.5\) and \(\bar y=4.5\). 2. The sum of products of deviations is \(8\), and the sum of squared input deviations is \(5\), so the least-squares slope is \(8/5=1.6\). 3. The intercept is \(4.5-1.6\cdot2.5=0.5\). 4. Thus \(\hat y=1.6x+0.5\), with a positive fitted association.

Answer

\(\hat y=1.6x+0.5\); slope \(1.6\), intercept \(0.5\), positive association.

Linear regression and residuals 18

A line fitted to a larger set of observations is \(\hat y=x+3\). The table shows three of those observations. <table><tr><th>\(x\)</th><th>Observed \(y\)</th></tr><tr><td>\(1\)</td><td>\(5\)</td></tr><tr><td>\(2\)</td><td>\(4\)</td></tr><tr><td>\(3\)</td><td>\(8\)</td></tr></table> For each row, find the predicted response and the residual (actual minus predicted).

Hints

- Begin with all three predictions from the fitted equation. - Keep the order observed minus predicted for each residual. - Check that positive residuals mean the observed point lies above the line.

Solution

1. At \(x=1\), the prediction is \(4\) and the residual is \(5-4=1\). 2. At \(x=2\), the prediction is \(5\) and the residual is \(4-5=-1\). 3. At \(x=3\), the prediction is \(6\) and the residual is \(8-6=2\).

Answer

<table><tr><th>\(x\)</th><th>Predicted \(\hat y\)</th><th>Residual</th></tr><tr><td>\(1\)</td><td>\(4\)</td><td>\(1\)</td></tr><tr><td>\(2\)</td><td>\(5\)</td><td>\(-1\)</td></tr><tr><td>\(3\)</td><td>\(6\)</td><td>\(2\)</td></tr></table>

Linear regression and residuals 19

Four observations from a larger set used to fit a regression model have these residuals: <table><tr><th>\(x\)</th><th>Residual</th></tr><tr><td>\(0\)</td><td>\(-1\)</td></tr><tr><td>\(2\)</td><td>\(1\)</td></tr><tr><td>\(4\)</td><td>\(-2\)</td></tr><tr><td>\(6\)</td><td>\(2\)</td></tr></table> Write the four ordered pairs that belong on a residual plot, with \(x\) on the horizontal axis. What goes on the vertical axis?

Hints

- What would happen if you sorted the two table columns separately before making points? - How should a negative residual appear relative to the zero line? - Does a residual plot show the original response or the model's error?

Solution

1. Combine each explanatory x-value with its residual to form a plotted coordinate. 2. The pairs are \((0,-1)\), \((2,1)\), \((4,-2)\), and \((6,2)\). 3. A residual plot shows residuals on its vertical axis, with zero as the reference line.

Answer

\((0,-1)\), \((2,1)\), \((4,-2)\), \((6,2)\). The y-axis shows the residual.

Linear regression and residuals 20

Plot a) shows observations and their least-squares line. Which residual plot, b), c), or d), matches? Calculate the residuals at \(x=2\) and \(x=5\) to support your choice.

Hints

- At a given x-value, compare the observed point's height with the line's height. - Could two candidate plots agree at one x-value and differ at another? - Check your choice at both requested x-values, not just one.

Solution

1. At \(x=2\), the line predicts \(4\) and the observed value is \(3\), giving residual \(3-4=-1\). 2. At \(x=5\), the line predicts \(7\) and the observed value is \(6\), also giving residual \(6-7=-1\). 3. Both b) and d) show \(-1\) at \(x=2\); both b) and c) show \(-1\) at \(x=5\). Only b) matches both. 4. Plot b) also matches the five remaining observations. Its residuals \(1,-1,-1,2,-1,-1,1\) sum to zero, and their x-weighted sum is zero, as required for the fitted least-squares line.

Answer

Plot b). The residual at \(x=2\) is \(-1\), and the residual at \(x=5\) is \(-1\). Only b) has both values.

Linear regression and residuals 21

Plot a) has four observations that follow an upward pattern. Plot b) includes the same four observations and one extra observation, P. Would adding P make the slope of the least-squares line steeper or flatter? Explain why the location of P matters.

Hints

- First follow the pattern of the original four points. - Compare P's height with the value you would expect if that pattern continued. - Points far from the other x-values can strongly affect a fitted line.

Solution

1. In plot a), the original points lie on \(y=x+1\), with slope \(1\). 2. In plot b), P is far to the right but has a response much lower than the original upward pattern would predict. 3. Because P has an extreme x-value, it strongly affects the fitted slope and pulls it downward, making the line flatter.

Answer

The slope becomes flatter (smaller). P is far to the right and well below the original trend, so it pulls the fitted line downward at large x-values.

Linear regression and residuals 22

Two rules predict the number of hits in three rounds of a ring-toss game. The actual hits and both predictions are shown below. <table><tr><th>Observed</th><th>Model A prediction</th><th>Model B prediction</th></tr><tr><td>\(2\)</td><td>\(2\)</td><td>\(3\)</td></tr><tr><td>\(4\)</td><td>\(5\)</td><td>\(3\)</td></tr><tr><td>\(6\)</td><td>\(6\)</td><td>\(5\)</td></tr></table> Find the sum of squared residuals for each model. Which model fits better by this measure?

Hints

- Compare each observed value with each model's prediction. - Square every difference before adding them. - The better fit by this measure has the smaller total squared difference.

Solution

1. Model A has residuals \(0,-1,0\). Their squares sum to \(0+1+0=1\). 2. Model B has residuals \(-1,1,1\). Their squares sum to \(1+1+1=3\). 3. Model A has the smaller sum of squared residuals, so it fits these observations better by that measure.

Answer

Model A: \(1\). Model B: \(3\). Model A fits better by the sum of squared residuals.

Linear regression and residuals 23

Two possible lines have these residuals for four observations: Line A: \(-2, 0, 0, 2\) Line B: \(-1, -1, 1, 1\) a) Add the signed residuals for each line. Can these sums tell you which line fits better? b) Compare the sums of squared residuals. Which line does the least-squares method prefer?

Hints

- Positive and negative differences can cancel when you add them. - Think about why two lines might both have a signed total of zero. - Squaring removes signs and gives greater weight to larger errors.

Solution

1. The signed residuals for A sum to \(-2+0+0+2=0\). For B they sum to \(-1-1+1+1=0\). 2. The equal signed sums do not show which line is closer to the observations overall. 3. Squaring gives \(4+0+0+4=8\) for A and \(1+1+1+1=4\) for B. 4. The least-squares method prefers B because its sum of squared residuals is smaller.

Answer

a) Both signed sums are \(0\); they do not show which line fits better. b) A has squared-residual sum \(8\), and B has \(4\). The least-squares method prefers B.

Linear regression and residuals 24

A fitted model predicts \(\hat y=4+1.5x\). For one observation, \(x=6\) and the residual is \(-2\). What was the actual observed value of \(y\)?

Hints

- Start by finding the prediction at the given x-value. - Use the meaning of residual: actual value minus predicted value. - A negative residual means the observation was lower than the prediction.

Solution

1. At \(x=6\), the predicted value is \(4+1.5\cdot6=13\). 2. A residual is observed minus predicted, so \(y-13=-2\). 3. The observed response was \(y=11\).

Answer

\(y=11\).

Linear regression and residuals 25

Jada timed herself solving the same maze before practicing and after each of four practice sessions. Let \(x\) be the number of practice sessions she has completed. Use regression technology to fit a least-squares line to the five plotted points. Write the equation \(\hat{y}=mx+b\) with both coefficients rounded to the nearest tenth. What does the sign of the slope tell you about Jada's times?

Hints

- Read the coordinates of each point from the plot and keep each pair together. - Enter all five pairs into a linear-regression tool and read its two coefficients. - Relate the sign of the fitted slope to how the times change with practice.

Solution

1. The five points are \((0,10)\), \((1,8)\), \((2,7)\), \((3,4)\), and \((4,3)\). 2. The mean input is \(2\) and the mean time is \(6.4\). The sum of cross-products is \(-18\) and the sum of squared input deviations is \(10\). 3. The least-squares slope is \(-18\div10=-1.8\). The intercept is \(6.4-(-1.8)\cdot2=10.0\). 4. Thus \(\hat{y}=-1.8x+10.0\). The negative slope means the model predicts shorter times as Jada practices.

Answer

\(\hat{y}=-1.8x+10.0\). The negative slope predicts a decrease of \(1.8\) minutes per practice session, not exactly that decrease every time.

Linear regression and residuals 26

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

Linear regression and residuals 27

The residual plots compare two straight-line models of the same measurements. Which model is more appropriate? Use the residual patterns, not just the average residual.

Hints

- Read the residuals from left to right in both panels. - Is each model equally likely to predict too high or too low across the x-values? - Look for structure, not just whether positive and negative residuals balance.

Solution

1. Model A's residuals move above and below \(0\) without a steady trend as \(x\) increases. 2. Model B's residuals generally rise from negative to positive values. 3. The rising pattern suggests that Model B misses a trend. Model A is more appropriate: its errors are smaller overall and have no steady trend.

Answer

Model A. Its residuals have no steady trend and are generally smaller. Model B's residuals tend to increase as \(x\) increases.

Linear regression and residuals 28

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?

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.

Linear regression and residuals 29

A line was fitted to six earlier measurements. Points P and Q are two new measurements that were not used to fit the line. Describe how each labeled point is unusual. Which point has the larger residual? Explain.

Hints

- Compare the horizontal positions of P and Q with the other points. - For each point, compare its actual height with the height of the line at that x-value. - A residual measures the difference in the vertical direction, not horizontal distance.

Solution

1. P is near the group's usual \(x\)-values but far above the line, while Q has an \(x\)-value far from the rest but is close to the line. 2. The line predicts \(5\) at P's \(x=4\), so P's residual is \(10-5=5\). 3. The line predicts \(11\) at Q's \(x=10\), so Q's residual is \(12-11=1\). 4. P has the larger residual, even though Q is farther horizontally from the other points.

Answer

P is unusually far above the line and has the larger residual, \(5\). Q is unusually far to the right but has a smaller residual, \(1\).

Linear regression and residuals 30

The residual plot comes from a linear model. Describe the pattern and explain what concern it raises about the model.

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.

Linear regression and residuals 31

The residual plot shows residuals against the original \(x\)-values for a candidate line. What does the pattern suggest about the line's slope?

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.

Linear regression and residuals 32

Electricity use was measured every \(6\) hours for four days, giving four readings per day. The graph shows residuals from the least-squares line fitted to all \(16\) readings, in time order. Why is a single linear time trend inadequate even if the signed residuals average nearly \(0\)?

Hints

- Compare the first four plotted residuals with the next four, then with the later sets. - What does the same position in each set of four represent? - Does an average near zero rule out an orderly pattern?

Solution

1. Each set of four measurements repeats roughly the same pattern of positive and negative residuals. 2. The signed errors nearly balance, but they do not vary randomly across the measurement times. 3. Because the four positions in each daily cycle correspond to the same times of day, a repeating daily pattern remains unexplained by the linear model.

Answer

The residuals follow a repeated daily pattern rather than random scatter. Their mean being near \(0\) does not make the linear time model adequate.

Linear regression and residuals 33

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?

Hints

- Compare the residual signs for the solid and open dots separately. - Does each warehouse's group look centered at zero? - What could you investigate if the errors differ consistently between warehouses?

Solution

1. All North residuals are positive, so the fitted line predicts delivery times that are too short for North. 2. All South residuals are negative, so the line predicts delivery times that are too long for South. 3. The two groups behave differently. Fitting separate lines for the two warehouses and checking their residuals would be a reasonable next step.

Answer

The single line tends to underpredict North deliveries and overpredict South deliveries. Try fitting separate lines for the two warehouses.

Linear regression and residuals 34

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.

Linear regression and residuals 35

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: \(3\)

Linear regression and residuals 36

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.

Linear regression and residuals 37

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) \(3.6^\circ\text{F}\)

Linear regression and residuals 38

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.

Linear regression and residuals 39

Use technology to find the least-squares regression line for the six plotted observations. Round both coefficients to the nearest hundredth. Do the repeated \(x\)-values prevent fitting a line? Explain.

Hints

- Enter every plotted point as a separate observation, even when two points share one x-coordinate. - Check whether x has different values across the whole set. - Round each fitted coefficient only after the regression calculation.

Solution

1. Enter all six points into regression technology. The slope is \(2.75\), and the intercept is \(\frac{1}{6}\approx0.1667\). 2. Rounding to the nearest hundredth gives \(\hat y=2.75x+0.17\). 3. The \(x\)-values vary over \(1,2,3\). Repeated \(x\)-values simply represent different observed responses at the same input.

Answer

\(\hat y=2.75x+0.17\) Repeated \(x\)-values do not prevent fitting because the explanatory values are not all the same.

Linear regression and residuals 40

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.

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

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

Linear regression and residuals 41

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.

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.

Linear regression and residuals 42

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.

Linear regression and residuals 43

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.

Linear regression and residuals 44

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.

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.

Linear regression and residuals 45

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.

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.

Linear regression and residuals 46

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.

Linear regression and residuals 47

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.

Linear regression and residuals 48

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.

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.

Linear regression and residuals 49

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.

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.

Algebra I math worksheets

Linear regression and residuals

Learning goals

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution