Residual plot practice problems with full solutions

(a) Predict first. At $x = 15$, $\hat{y} = 22 + 0.4(15) = 22 + 6 = 28$ minutes.

(a) Subtract predicted from observed. $\text{residual} = y - \hat{y} = 26 - 28 = -2$ minutes, so that crossing finished 2 minutes sooner than the line predicted.

(b) Predict at $x = 30$. $\hat{y} = 22 + 0.4(30) = 22 + 12 = 34$ minutes.

(b) Subtract again. $\text{residual} = 37 - 34 = +3$ minutes, so that crossing ran 3 minutes longer than the line predicted.

(c) A residual plot puts the residual itself on the vertical axis, so a point sits above zero exactly when its residual is positive. The 30 km/h crossing has a residual of $+3$ and plots 3 units above the zero line, while the 15 km/h crossing has a residual of $-2$ and plots 2 units below it.

(a) Predict at $x = 25$. $\hat{y} = 40 + 3(25) = 40 + 75 = 115$ kg.

(a) Take observed minus predicted. $130 - 115 = +15$ kg. The residual is positive, so the observed weight sits above the line and the model underpredicted that week by 15 kg.

(b) Predict at $x = 12$. $\hat{y} = 40 + 3(12) = 40 + 36 = 76$ kg.

(b) Subtract. $70 - 76 = -6$ kg. The residual is negative, so the model overpredicted that week by 6 kg.

(c) Separate sign from size. The sign of a residual reports only which side of the line a case fell on. Accuracy is the size of the miss, and by that measure the 25-household week is the worse of the two fits: the line was off by 15 kg there against 6 kg for the 12-household week.

(c) Say what the positive residual does mean. The program collected 15 kg more than the line predicted for 25 households, which may be welcome news about that week's collection. It is not a statement about how well the model describes the program.

(a) Predict at $x = 8$. $\hat{y} = 6 + 1.5(8) = 6 + 12 = 18$ cm.

(a) Rearrange the residual formula. From $\text{residual} = y - \hat{y}$ you get $y = \hat{y} + \text{residual} = 18 + (-2.5) = 15.5$ cm. The negative residual means the line overpredicted this fish by 2.5 cm, so it is shorter than the hatchery's model expects at 8 months.

(b) Work backwards to the prediction. $\hat{y} = y - \text{residual} = 24 - 1.5 = 22.5$ cm, so the line predicted 22.5 cm for this fish.

(b) Solve for the age that produces that prediction. $6 + 1.5x = 22.5$ gives $1.5x = 16.5$, so $x = \frac{16.5}{1.5} = 11$ months. Check it: $6 + 1.5(11) = 6 + 16.5 = 22.5$ cm.

(c) Read the zero. A residual of $0$ means $y = \hat{y}$, so that fish lies exactly on the regression line and the prediction matched its measured length.

(c) Say what it does not settle. One point on the line is a single lucky hit, not a verdict on the model. Whether a line is the right shape depends on the pattern of all the residuals together, which is what a residual plot shows.

(a) Predict at each guest count from $\hat{y} = 2 + 0.3x$, then subtract predicted from observed.

(a) Check two rows by hand. At $x = 20$, $\hat{y} = 2 + 0.3(20) = 2 + 6 = 8$ kg and the residual is $10 - 8 = +2$ kg. At $x = 40$, $\hat{y} = 2 + 0.3(40) = 2 + 12 = 14$ kg and the residual is $12 - 14 = -2$ kg.

(b) Add them. $(-1) + 2 + 0 + (-2) + 1 = 0$ kg.

(b) Explain the zero. A least-squares line always passes through the point of averages $(\bar{x}, \bar{y})$, read as x-bar and y-bar. Here $\bar{x} = \frac{150}{5} = 30$ guests and $\bar{y} = \frac{55}{5} = 11$ kg, and $2 + 0.3(30) = 11$, so the line does pass through it. That forces the residuals to sum to zero for any data set, which makes the sum an arithmetic check on your subtraction rather than evidence about the fit.

(c) Pick out the negative residuals, since negative means the prediction sat above the observation. The 10-guest event was overpredicted by 1 kg and the 40-guest event by 2 kg.

(d) Describe the plot. Plotted against guests, the residuals run $-1, +2, 0, -2, +1$ kg: they fall on both sides of zero across the whole range, they do not bend into a curve or an arch, and their sizes do not grow as the events get larger.

(d) State the verdict. That apparent randomness is what confirms a linear form, so a line is an appropriate model for these events. Five points is thin evidence for any statement about shape, so treat it as no reason to doubt the line rather than as proof.

(a) Plot A. The residuals scatter randomly around zero with constant spread and no bend, so nothing systematic is left over. Apparent randomness in a residual plot confirms a linear form, so a line is an appropriate model.

(a) Plot B. The sign of a residual here is decided by where you sit in the range: negative at the bottom, positive in the middle, negative at the top. That is curvature, not scatter, and curvature says the linear model is not the appropriate model for these data.

(a) Plot C. Nothing bends, so the plot gives no evidence against a straight-line form. What it does show is that the spread of the residuals grows with order size, so a line is still a defensible description of the form while its predictions are far less precise for large orders.

(b) Read the arch as a set of directions. Negative residuals mean the observations sit below the line, so the model overpredicts at both the low and the high ends of rainfall. Positive residuals in the middle mean it underpredicts there. A model that bends the same way the data do would remove that pattern.

(c) Correct the student. A fan is about spread, not about shape. Curvature is the signal that the form is wrong, and plot C has none: it has unequal spread, which changes how much you trust a prediction rather than whether the line has the right form. The AP course names apparent randomness and curvature as the two things to read off a residual plot, so on the exam the defensible statement about plot C is about the reliability of predictions at large $x$.

(d) All three. Least squares forces the residuals of a fitted line to add to zero no matter what the plot looks like, so the sum cannot tell plot A, plot B, and plot C apart. Only the pattern can.

(a) Compare the sizes of the residuals, ignoring sign, because a miss of $-6$ hours is the same size of error as a miss of $+6$. The sizes are $1, 1, 5, 6, 2, 3$ hours, so Granite has the largest at 6 hours.

(b) Read the sign. Granite's residual is negative, so its observed hours sit below the line: the model overpredicted Granite by 6 hours, and that section took 6 fewer hours than its length would suggest.

(c) Take the largest positive residual, since positive means the model underpredicted. Fern, at $+5$ hours, needed 5 more hours than the line predicted.

(d) Predict for Granite at $x = 8$. $\hat{y} = 4 + 2.5(8) = 4 + 20 = 24$ hours.

(d) Add the residual to the prediction. $y = \hat{y} + \text{residual} = 24 + (-6) = 18$ hours, so Granite actually took 18 hours.

(e) Look for a pattern rather than for a large point. Across lengths of 2 to 12 km the residuals run $-1, +1, +5, -6, -2, +3$: they change sign irregularly, they trace no arch or U, and their sizes do not widen as the sections get longer. Their sum is $(-1) + 1 + 5 + (-6) + (-2) + 3 = 0$, as least squares requires.

(e) Give the verdict. Nothing in the plot says the form is wrong, so there is no reason to abandon the line. The Granite miss of 6 hours is a single point to go and ask about, since something about that section made it cheaper to maintain than its length predicts.

(a) Predict at each speed, then subtract predicted from observed.

(a) Check two rows and the sum. At $x = 30$, $\hat{y} = -18 + 0.7(30) = -18 + 21 = 3$ m, so the residual is $5 - 3 = +2$ m. At $x = 60$, $\hat{y} = -18 + 0.7(60) = -18 + 42 = 24$ m, so the residual is $23 - 24 = -1$ m. The five residuals sum to $2 + (-1) + (-2) + (-1) + 2 = 0$, as least squares requires.

(b) Describe the shape. Against speed the residuals go $+2, -1, -2, -1, +2$: positive at both ends and negative through the middle, a clear U. That is a systematic pattern, not scatter, and it means the line underpredicts the fastest and slowest stops and overpredicts the ones in the middle.

(c) Say what $r$ actually measures. The correlation $r$ reports how tightly the points cluster around a straight line and the direction of that trend. It cannot report whether a straight line is the right shape, because it has no way to compare a line against a curve.

(c) Confirm the number, then rule against the instructor. Squaring gives $r^2 = 0.986^2 = 0.972$, which you can verify from the residuals: the sum of the squared residuals is $2^2 + (-1)^2 + (-2)^2 + (-1)^2 + 2^2 = 4 + 1 + 4 + 1 + 4 = 14$, the total sum of squares about $\bar{y} = 17$ m is $144 + 64 + 4 + 36 + 256 = 504$, and $1 - \frac{14}{504} = 0.972$. So the line does remove about 97 percent of the variation in braking distance, and the U-shaped residual plot still shows the leftover misses are systematic. The residual plot decides the form, so the instructor's argument does not hold and a curved model would fit better.

(d) Use the line first. $\hat{y} = -18 + 0.7(90) = -18 + 63 = 45$ m.

(d) Now read the data's own pattern. Across equal 10 km/h steps the observed distances rise by $9 - 5 = 4$, $15 - 9 = 6$, $23 - 15 = 8$, and $33 - 23 = 10$ m, so each step adds 2 m more than the step before. Continuing that pattern gives $33 + 12 = 45$ m at 80 km/h and $45 + 14 = 59$ m at 90 km/h.

(d) State the conclusion. The line's 45 m at 90 km/h is far too low, roughly 14 m short, because the true relationship curves upward and the line cannot follow it. The prediction is also an extrapolation past the fastest speed tested, so the honest answer is that this line should not be used at 90 km/h at all.

(a) Predict at each volunteer count, then subtract predicted from observed.

(a) Check two rows and add. At $x = 4$, $\hat{y} = 10 + 3(4) = 22$ kg and the residual is $19 - 22 = -3$ kg. At $x = 12$, $\hat{y} = 10 + 3(12) = 46$ kg and the residual is $47 - 46 = +1$ kg. The sum is $(-3) + 1 + 2 + 2 + 1 + (-3) = 0$ kg.

(b) Answer the coordinator. The sum had to be zero. A least-squares line passes through $(\bar{x}, \bar{y})$, and here $\bar{x} = \frac{54}{6} = 9$ volunteers and $\bar{y} = \frac{222}{6} = 37$ kg, with $10 + 3(9) = 37$. Every least-squares fit produces residuals that sum to zero, including a terrible one, so the sum checks your subtraction and says nothing about the fit.

(c) Describe the plot. Against volunteers, the residuals run $-3, +1, +2, +2, +1, -3$ kg: negative at both ends, positive through the middle, tracing a smooth arch rather than scatter.

(c) Give the verdict and the reason. Curvature in a residual plot means the linear model is not the appropriate model, so a line should not be used here. The data say the same thing directly: each extra pair of volunteers added $10, 7, 6, 5$, and then $2$ kg, so the returns shrink as the crew grows, which is what happens when a fixed amount of litter is being picked up by more and more people.

(d) Predict at $x = 16$. $\hat{y} = 10 + 3(16) = 10 + 48 = 58$ kg, and falling 9 kg below that prediction is a residual of $-9$ kg.

(d) Recover the observation. $y = \hat{y} + \text{residual} = 58 + (-9) = 49$ kg, the same amount the 14-volunteer event collected. That is exactly what the arch predicted: the collection has flattened out, the line keeps climbing at 3 kg per volunteer, and the gap between them grows the further past the data you push it.