Correlation and residual plot practice problems

(a) Read the sign, then the strength. $r = 0.6$ is positive and moderately strong, so schools where students read more minutes at home tend to have higher average reading scores, and the association is roughly linear.

(b) Square the correlation. $r^2 = (0.6)^2 = 0.36$.

(b) Put it in context with the right noun. About 36 percent of the variation in the schools' average reading scores is explained by the linear relationship with average daily home reading minutes. The remaining 64 percent is not.

(c) Name the mix-up. The parent used $r$ where $r^2$ belongs. The value $0.6$ is the correlation, which is not a proportion of anything; the proportion of variation explained is $r^2 = 0.36$, so the figure in that sentence should be 36 percent.

(c) Flag the second problem with the sentence. These are 40 school averages from an observational comparison, so even the corrected 36 percent describes explained variation, not cause. Nothing here shows that adding home reading minutes would raise a school's score.

(d) Ask what rescaling does to a correlation. Dividing every reading time by 60 changes the units of $x$ and nothing else. Correlation is built from standardized values, and standardizing subtracts the mean and divides by the standard deviation, which cancels the units out entirely. So $r$ stays at $0.6$ and $r^2$ stays at $0.36$.

(d) Name what would change. The slope of a fitted line would change, from a per-minute rate to a per-hour rate 60 times larger, and residuals would still be measured in score points. The correlation and the proportion of variation explained do not move, which is why $r$ is safe to quote without saying what the units were.

(a) Find the predicted time at $x = 6$. $\hat{y} = 1.5 + 0.75(6) = 1.5 + 4.5 = 6$ hours.

(a) Take observed minus predicted. $\text{residual} = y - \hat{y} = 7 - 6 = +1$ hour.

(a) Read the sign. The residual is positive, so this job sits above the line: it ran 1 hour longer than the line predicted, and the model underpredicted it.

(b) Find the predicted time at $x = 4$. $\hat{y} = 1.5 + 0.75(4) = 1.5 + 3 = 4.5$ hours.

(b) Rearrange the residual formula to observed equals predicted plus residual. $y = \hat{y} + \text{residual} = 4.5 + (-0.5) = 4$ hours. The negative residual means the model overpredicted this job by half an hour.

(c) Set the predicted value equal to 6.75 and solve for $x$. $1.5 + 0.75x = 6.75$, so $0.75x = 5.25$ and $x = \frac{5.25}{0.75} = 7$ rooms.

(a) Predict at each age from $\hat{y} = 20 - 1.5x$, then subtract predicted from observed.

(a) Check one line of the table by hand. At $x = 4$, $\hat{y} = 20 - 1.5(4) = 20 - 6 = 14$, so the residual is $11 - 14 = -3$ units per hour.

(b) Add them. $1 + (-3) + 2 + 1 + (-1) = 0$. This is forced, not lucky: the least-squares line passes through $(\bar{x}, \bar{y})$, and here $\bar{x} = \frac{30}{5} = 6$ and $\bar{y} = \frac{55}{5} = 11$, with $20 - 1.5(6) = 11$. Any line through the point of averages leaves residuals summing to zero.

(c) Pick out the negative residuals, since negative means predicted above observed. The 4-year-old machine has a residual of $-3$ and the 10-year-old machine a residual of $-1$, so the model overpredicted them by 3 and by 1 unit per hour.

(d) Describe the plot. Against age, the residuals run $+1, -3, +2, +1, -1$: they scatter above and below zero with no curve, no U, and no widening fan, and their sizes do not grow with age. Nothing systematic is left over, so a line is a reasonable model for these machines, though five points is thin evidence for any verdict about form.

(a) Take the square root, then borrow the sign from the slope. $\sqrt{0.5776} = 0.76$, and the slope $-0.45$ is negative, so $r = -0.76$. The square root alone cannot settle the sign, because $r$ and $-r$ have the same square; the slope and $r$ always share a sign.

(b) Interpret $r^2$. About 57.8 percent of the variation in nightly sleep hours is explained by the linear relationship with evening screen time.

(b) Account for the rest. $1 - 0.5776 = 0.4224$, so about 42.2 percent of the variation in sleep is left unexplained by this line and comes from everything else, from school start times to individual differences in how much sleep a person needs.

(c) Interpret the slope with units. For each additional hour of evening screen time, the model predicts about 0.45 fewer hours of sleep per night, which is roughly 27 minutes.

(d) Read the fan. Widening spread means the size of a typical residual grows with $x$, so the line's accuracy is not the same across the range. Near $x = 0$ predictions land close to the truth; at $x = 5$ hours the observed sleep can sit far above or far below the line, so a single predicted value there carries much more uncertainty than $r^2 = 0.5776$ on its own suggests.

(d) Say what the fan does not mean. A fan is not curvature. It does not say the association bends, so a straight line may still be the right shape; it says the scatter about that line is uneven, which is a separate defect from a wrong model form.

(a) Find the means. $\bar{x} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6$ weeks and $\bar{y} = \frac{18 + 22 + 30 + 42 + 58}{5} = \frac{170}{5} = 34$ square metres.

(a) Find the deviations and the two sums. The $x$ deviations are $-4, -2, 0, 2, 4$ and the $y$ deviations are $-16, -12, -4, 8, 24$. Then $\sum (x - \bar{x})(y - \bar{y}) = 64 + 24 + 0 + 16 + 96 = 200$ and $\sum (x - \bar{x})^2 = 16 + 4 + 0 + 4 + 16 = 40$.

(a) Compute the slope and intercept. $b = \frac{200}{40} = 5$ square metres per week and $a = \bar{y} - b\bar{x} = 34 - 5(6) = 34 - 30 = 4$, so the line is $\hat{y} = 4 + 5x$.

(b) Predict at each week and subtract.

(b) Check the sum. $4 + (-2) + (-4) + (-2) + 4 = 0$, as least squares requires.

(c) Describe the residual plot. Plotted against week, the residuals go $+4, -2, -4, -2, +4$: positive at both ends, negative through the middle. That is a clear U shape, a systematic pattern rather than random scatter, and it means the line underpredicts at both ends and overpredicts in the middle.

(d) Reconcile the two. $r^2 = 0.947$ says the line removes about 94.7 percent of the variation in area, which sounds excellent, but $r^2$ only measures how much spread a line takes out; it cannot tell you whether some other shape would take out more. The residual plot is the test of form, and its U says the true relationship curves upward. A line is not the right model here: the patch is growing at an accelerating rate, which is what an invasive species usually does.

(d) Confirm $r^2$ from the residuals as a check. $\text{SSE} = 4^2 + (-2)^2 + (-4)^2 + (-2)^2 + 4^2 = 16 + 4 + 16 + 4 + 16 = 56$ and $\text{SST} = \sum (y - \bar{y})^2 = 256 + 144 + 16 + 64 + 576 = 1056$, so $r^2 = 1 - \frac{56}{1056} = 0.947$, matching the value from squaring $r$.

(a) Find the new means. $\bar{x} = \frac{1 + 2 + 3 + 4 + 5 + 3}{6} = \frac{18}{6} = 3$ kg, unchanged, because the new point sits exactly at the old mean of $x$. $\bar{y} = \frac{5 + 8 + 11 + 14 + 17 + 2}{6} = \frac{57}{6} = 9.5$ cm, down from 11.

(a) Find the deviations and sums. The $x$ deviations are $-2, -1, 0, 1, 2, 0$ and the $y$ deviations are $-4.5, -1.5, 1.5, 4.5, 7.5, -7.5$. Their products are $9, 1.5, 0, 4.5, 15, 0$, so $\sum (x - \bar{x})(y - \bar{y}) = 30$, and $\sum (x - \bar{x})^2 = 4 + 1 + 0 + 1 + 4 + 0 = 10$.

(a) Compute the line. $b = \frac{30}{10} = 3$ cm per kg and $a = \bar{y} - b\bar{x} = 9.5 - 3(3) = 9.5 - 9 = 0.5$ cm, so the new line is $\hat{y} = 0.5 + 3x$.

(b) Compare the two lines. The slope is still exactly 3. The bad reading sits at $x = \bar{x}$, so its $x$ deviation is $0$ and its contribution to $\sum (x - \bar{x})(y - \bar{y})$ is $0 \times (-7.5) = 0$, no matter how wrong its $y$ is. Sitting exactly at $\bar{x}$ is the lowest-leverage position a point can occupy, and from there it cannot tilt the line. All it can do is pull the whole line down, which it did: the intercept fell from 2 cm to 0.5 cm.

(c) Compute $r$. $\sum (y - \bar{y})^2 = 20.25 + 2.25 + 2.25 + 20.25 + 56.25 + 56.25 = 157.5$, so $r = \frac{30}{\sqrt{10 \times 157.5}} = \frac{30}{\sqrt{1575}} = \frac{30}{39.686} = 0.756$.

(c) Square it, working from the unrounded sums rather than from $0.756$. $r^2 = \frac{30^2}{10 \times 157.5} = \frac{900}{1575} = 0.571$, down from $1$. One misread ruler took the model from explaining all of the variation in length to explaining about 57 percent of it.

(d) Classify the point. Its residual is $2 - [0.5 + 3(3)] = 2 - 9.5 = -7.5$ cm, easily the largest in the set, so it is an outlier in the $y$ direction. Its $x$ value equals $\bar{x}$, the lowest-leverage position in the set, so it has no pull on the tilt of the line. It is influential for the intercept and for $r$, and not influential for the slope.

(a) Work the first five. $\bar{x} = \frac{15}{5} = 3$ litres and $\bar{y} = \frac{55}{5} = 11$ cm. The $x$ deviations are $-2, -1, 0, 1, 2$ and the $y$ deviations are $-6, -3, 0, 3, 6$, so $\sum (x - \bar{x})(y - \bar{y}) = 12 + 3 + 0 + 3 + 12 = 30$, $\sum (x - \bar{x})^2 = 10$, and $\sum (y - \bar{y})^2 = 36 + 9 + 0 + 9 + 36 = 90$.

(a) Finish part (a). $b = \frac{30}{10} = 3$ cm per litre and $a = 11 - 3(3) = 2$ cm, so $\hat{y} = 2 + 3x$. The correlation is $r = \frac{30}{\sqrt{10 \times 90}} = \frac{30}{30} = 1$: the five points lie exactly on that line.

(b) Redo the means with all six. $\bar{x} = \frac{1 + 2 + 3 + 4 + 5 + 15}{6} = \frac{30}{6} = 5$ litres and $\bar{y} = \frac{5 + 8 + 11 + 14 + 17 + 8}{6} = \frac{63}{6} = 10.5$ cm.

(b) Recompute the sums. The $x$ deviations are $-4, -3, -2, -1, 0, 10$ and the $y$ deviations are $-5.5, -2.5, 0.5, 3.5, 6.5, -2.5$. Their products are $22, 7.5, -1, -3.5, 0, -25$, which sum to $\sum (x - \bar{x})(y - \bar{y}) = 0$. Also $\sum (x - \bar{x})^2 = 16 + 9 + 4 + 1 + 0 + 100 = 130$.

(b) Compute the line. $b = \frac{0}{130} = 0$ and $a = \bar{y} - 0 \cdot \bar{x} = 10.5$, so the fitted line is $\hat{y} = 10.5$, a flat line at the mean growth.

(c) Find $r$. The numerator of $r$ is the same $\sum (x - \bar{x})(y - \bar{y}) = 0$, so $r = 0$ and $r^2 = 0$, whatever the denominator is.

(d) Explain the mechanism. The sixth seedling sits at $x = 15$ litres while $\bar{x}$ for the six is 5, so its $x$ deviation of $+10$ is as large as all five of the others put together, and squared it supplies 100 of the 130 in $\sum (x - \bar{x})^2$, against 30 from the other five combined. That is extreme leverage. Its growth of 8 cm is far below what the other five would predict there, so it drags the right end of the line down until the line is level. A perfect positive relationship, $r = 1$, now reads as no linear relationship at all, $r = 0$.

(d) Say what you would report. Do not delete a point because it is inconvenient. Report both fits, state that the conclusion rests entirely on one seedling, and give the reason it differs: it was overwatered, so it arguably sits outside the range of watering the study is about. The honest write-up is that growth rises with water from 1 to 5 litres per week and that one overwatered plant at 15 litres flattens the fitted line completely, which is a reason to collect data between 5 and 15 litres rather than to drop a row.

(a) Name the variables, then read the coefficients off the output. Let $x$ be the screen brightness in percent and $\hat{y}$ the predicted battery life in hours. The Constant row is the intercept and the Brightness row is the slope, so $\hat{y} = 12.4 - 0.06x$.

(b) Interpret the slope with units. For each additional percentage point of brightness, the model predicts battery life falls by 0.06 hour. Scaled to something readable, 10 more percentage points of brightness predict $10 \times 0.06 = 0.6$ hour, about 36 minutes, less battery life.

(c) Get $r$ from R-Sq, then fix the sign. $r = \pm\sqrt{0.6889} = \pm 0.83$, and the slope is negative, so $r = -0.83$.

(d) Predict at $x = 50$. $\hat{y} = 12.4 - 0.06(50) = 12.4 - 3 = 9.4$ hours.

(d) Take observed minus predicted and read the sign. $\text{residual} = 8.2 - 9.4 = -1.2$ hours. It is negative, so this machine sits below the line: it ran 1.2 hours, about 1 hour and 12 minutes, shorter than the model predicted for 50 percent brightness, and the model overpredicted it.

(e) Locate the new point first. The tested brightness settings run from 20 to 80 percent, so $x = 100$ is outside the data entirely and far from $\bar{x}$. That makes it a high-leverage point, and any prediction there is an extrapolation.

(e) Compare it with the current line. At 100 percent the line predicts $12.4 - 0.06(100) = 6.4$ hours, but this machine ran 9.0 hours, a residual of $9.0 - 6.4 = +2.6$ hours. The point sits well above the line at the extreme right of the $x$ range.

(e) Read off the effect. Least squares has to chase a high-leverage point, so the right end of the fitted line is pulled up: the slope becomes less steeply negative, moving toward $0$. Because that point lies far from the pattern the other 18 machines follow, the scatter about the fit grows and $r$ moves toward $0$ as well, so $r$ becomes less negative than $-0.83$ and R-Sq drops. One machine tested outside the studied range would weaken a relationship the rest of the data showed clearly.