11.6 Outliers
SLO 2
Identify appropriate graphs and summary statistics for variables and relationships between them and correctly interpret information from graphs and summary statistics.
SLO 4
Demonstrate an understanding of, and ability to use, basic ideas of statistical processes, including hypothesis tests and confidence interval estimation.
Learning Objectives
By the end of this section, you will be able to:
- define an outlier and an influential point, and say how they differ;
- identify outliers graphically using lines two standard deviations from the best-fit line;
- identify outliers numerically by comparing each residual against \(2s\);
- compute \(s\) by hand from the sum of squared errors;
- decide responsibly whether an outlier should be deleted, and describe how removing one changes the line.
Section 11.3 introduced residuals and noted in passing that a large one marks a point the line describes poorly. This section makes that precise. Some points sit far enough from the line to deserve individual attention — and the decision about what to do with them is a judgment about the data, not a calculation. Getting that judgment wrong in either direction is a serious error: deleting a real observation because it is inconvenient, or keeping a transcription mistake that drags the whole line.
11.6.1 Outliers and Influential Points
Outliers are observed data points that are far from the least squares line. They have large "errors", where the "error" or residual is the vertical distance from the line to the point.
Definition 11.6.1 — Outlier: A scatter plot of eleven students' third exam scores (x, 65 to 75) against final exam scores (y, 126 to 198), taken from Table 11.6.1.
Influential points are observed data points that are far from the other observed data points in the horizontal direction. These points may have a big effect on the slope of the regression line.
Picture a tight cluster of points between \(x = 10\) and \(x = 20\), plus a single point at \(x = 90\). That far-right point may sit exactly on the line and have a residual near zero — no outlier at all by the vertical test. Yet it is holding the far end of the line almost single-handedly, and moving it a little swings the slope a lot. Residuals cannot detect this, which is why horizontal position is checked separately.
The two definitions differ in direction, and that is the whole distinction. An outlier is far from the line vertically. An influential point is far from the other points horizontally — it may sit close to the line and still exert enormous leverage on where the line goes.
To begin to identify an influential point, you can remove it from the data set and see if the slope of the regression line is changed significantly. Computers and many calculators can be used to identify outliers from the data, and computer output for regression analysis will often identify both outliers and influential points so that you can examine them.
Definition 11.6.2 — Influential Point: Far from the others horizontally, not vertically — remove it and the slope swings.
For each described point, say whether it is an outlier, an influential point, both, or neither.
a. A point with a residual of 40 when \(2s = 15\), sitting in the middle of the x-range.
b. A point sitting exactly on the line, far to the right of every other point.
c. A point with a residual of 3 when \(2s = 15\), in the middle of the x-range.
Solution — checking vertical and horizontal position separately
Step 1 — Assess a. Its residual of 40 far exceeds \(2s = 15\), so it is far from the line vertically: an outlier. Its x-position is ordinary, so it is not influential.
Step 2 — Assess b. Its residual is essentially zero, so it is not an outlier. But it is far from the other points horizontally, so it is an influential point — and possibly a very strong one.
Step 3 — Assess c. A residual of 3 against \(2s = 15\) is unremarkable, and its x-position is ordinary. Neither.
Step 4 — State the takeaway. The two labels answer different questions, so a point can carry one, both, or neither. Checking only residuals would miss b entirely.
Answer: (a) outlier; (b) influential point; (c) neither.
11.6.2 The Two-Standard-Deviation Rule
As a rough rule of thumb, flag any point located further than two standard deviations above or below the best-fit line as an outlier. The standard deviation used is the standard deviation of the residuals, \(s\).
There are two ways to apply it, and you would generally use only one:
- Graphically, by drawing an extra pair of lines two standard deviations above and below the best-fit line. Any data points outside this extra pair of lines are flagged as potential outliers.
- Numerically, by calculating each residual and comparing it to twice the standard deviation.
On the TI-83, 83+, or 84+, the graphical approach is easier.
For the third exam/final exam example, the calculator function LinRegTTest reports \(s = 16.4\) as the standard deviation of the residuals. Table 11.6.1 shows the predicted values and residuals, computed from \(\hat{y} = -173.5 + 4.83x\).
| x | y | \(\hat{y}\) | \(y - \hat{y}\) |
|---|---|---|---|
| 65 | 175 | 140 | 175 − 140 = 35 |
| 67 | 133 | 150 | 133 − 150 = −17 |
| 71 | 185 | 169 | 185 − 169 = 16 |
| 71 | 163 | 169 | 163 − 169 = −6 |
| 66 | 126 | 145 | 126 − 145 = −19 |
| 75 | 198 | 189 | 198 − 189 = 9 |
| 67 | 153 | 150 | 153 − 150 = 3 |
| 70 | 163 | 164 | 163 − 164 = −1 |
| 71 | 159 | 169 | 159 − 169 = −10 |
| 69 | 151 | 160 | 151 − 160 = −9 |
| 69 | 159 | 160 | 159 − 160 = −1 |
We are looking for all data points for which the residual is greater than \(2s = 2(16.4) = 32.8\) or less than \(-32.8\). Comparing these values to the residuals in the fourth column, the only such data point is the student who had a grade of 65 on the third exam and 175 on the final exam; the residual for this student is 35.
Definition 11.6.3 — The Rule of Thumb for Outliers: Table 11.6.1 data with s = 16.47: only (65, 175) falls outside the ±2s = ±32.94 band.
We could guess at outliers by looking at a graph of the scatterplot and best-fit line. However, we would like some guideline as to how far away a point needs to be in order to be considered an outlier.
A regression on 12 points has \(s = 4.5\). The residuals are 2, \(-3\), 11, 1, \(-2\), 4, \(-1\), 0, 3, \(-9\), 2 and \(-8\). Identify which points are flagged as potential outliers.
Solution — comparing each residual against 2s
Step 1 — Compute the threshold.
$$ 2s = 2(4.5) = 9.0 $$Step 2 — Flag any residual at or beyond that, in either direction. Scanning the list, \(11 \geq 9.0\) and \(|-9| = 9 \geq 9.0\).
Step 3 — Check the near misses. The residual of \(-8\) has absolute value 8, which is less than 9.0, so it is not flagged — even though it is close.
Step 4 — Report. Two points are flagged: the one with residual 11 and the one with residual \(-9\).
Step 5 — Note the rule's nature. This is a rough rule of thumb, not a sharp boundary. A residual of \(-8\) against a threshold of 9 is worth a glance even though the rule does not flag it.
Answer: The points with residuals 11 and \(-9\) are flagged as potential outliers; \(-8\) falls just inside the threshold.
In the third exam/final exam example, determine whether there is an outlier using the numerical method, with \(s = 16.4\).
Solution — one residual clears the threshold
Step 1 — Compute the threshold. \(2s = 2(16.4) = 32.8\).
Step 2 — Scan the residuals. They are 35, \(-17\), 16, \(-6\), \(-19\), 9, 3, \(-1\), \(-10\), \(-9\) and \(-1\).
Step 3 — Compare each against 32.8. Every absolute value is below 32.8 except the first, which is 35.
Step 4 — Identify the point. The residual of 35 belongs to the student who scored 65 on the third exam and 175 on the final — the point \((65, 175)\).
Answer: \((65, 175)\) is a potential outlier, with a residual of 35 against a threshold of 32.8.
11.6.3 Computing s by Hand
\(s\) is the standard deviation of all the \(y - \hat{y} = \varepsilon\) values, where \(n\) is the total number of data points. If each residual is calculated and squared, and the results are added, we get the SSE. Then
$$ s = \sqrt{\frac{SSE}{n - 2}} $$We divide by \((n - 2)\) because the regression model involves two estimates — the line has two constants, \(a\) and \(b\), both computed from this same data.
For the third exam/final exam problem, first square each \(|y - \hat{y}|\). The squares are \(35^{2}\), \(17^{2}\), \(16^{2}\), \(6^{2}\), \(19^{2}\), \(9^{2}\), \(3^{2}\), \(1^{2}\), \(10^{2}\), \(9^{2}\), \(1^{2}\). Then add them:
$$ \sum_{i=1}^{11}\varepsilon_{i}^{2} = 35^{2} + 17^{2} + 16^{2} + 6^{2} + 19^{2} + 9^{2} + 3^{2} + 1^{2} + 10^{2} + 9^{2} + 1^{2} = 2440 = \textbf{SSE} $$Next, calculate \(s\):
$$ s = \sqrt{\frac{2440}{11 - 2}} = 16.47 $$Then multiply by 2:
$$ (2)(16.47) = 32.94 $$32.94 is two standard deviations away from the mean of the \(y - \hat{y}\) values. If the vertical distance from any data point to the corresponding point on the line of best fit is at least \(2s\), we consider the data point "too far" from the line and call it a potential outlier.
For this example, all the \(|y - \hat{y}|\) values are less than 32.94 except the first, which is 35. Since \(35 > 32.94\), the point corresponding to \(|y - \hat{y}| = 35\) — that is, \((65, 175)\) — is a potential outlier.
The calculator reports \(s = 16.4\) and the hand calculation gives \(16.47\), so the thresholds come out 32.8 and 32.94. The difference is rounding: the hand calculation used residuals already rounded to whole numbers, while the calculator carried full precision throughout. Both identify the same single outlier, which is the point. When a residual sits between two such thresholds, that is a signal to stop trusting the rule of thumb and look at the point directly.
Definition 11.6.4 — Standard Deviation of the Residuals: S = √(SSE / (n − 2)) = √(2440 / 9) ≈ 16.47, so 2s ≈ 32.94; only |y − ŷ| = 35 exceeds the threshold.
If you do not have the function LinRegTTest, you can find the outlier by computing \(s\) directly.
A regression on 8 points produces residuals 4, \(-2\), 7, \(-11\), 1, 3, \(-1\) and 2. Compute the SSE, then \(s\), then \(2s\), and identify any potential outliers.
Solution — the full hand calculation
Step 1 — Square each residual. \(16, 4, 49, 121, 1, 9, 1, 4\).
Step 2 — Add them to get the SSE.
$$ SSE = 16 + 4 + 49 + 121 + 1 + 9 + 1 + 4 = 205 $$Step 3 — Compute \(s\), dividing by \(n - 2 = 6\).
$$ s = \sqrt{\frac{205}{8 - 2}} = \sqrt{34.167} \approx 5.845 $$Step 4 — Double it.
$$ 2s \approx 11.69 $$Step 5 — Compare each residual. The largest absolute residual is 11, and \(11 < 11.69\).
Answer: \(SSE = 205\), \(s \approx 5.85\), \(2s \approx 11.69\). No point is flagged — the residual of \(-11\) falls just below the threshold.
11.6.4 Deciding What to Do with an Outlier
Outliers need to be examined closely. Sometimes, for one reason or another, they should not be included in the analysis of the data. It is possible that an outlier is a result of erroneous data. Other times, an outlier may hold valuable information about the population under study and should remain included. The key is to examine carefully what causes a data point to be an outlier.
Numerically and graphically, we have identified \((65, 175)\) as an outlier. We should re-examine the data for this point to see if there are any problems. If there is an error, we should fix the error if possible, or delete the data. If the data is correct, we would leave it in the data set.
When outliers are deleted, the researcher should either record that data was deleted and why, or provide results both with and without the deleted data. If data is erroneous and the correct values are known — for instance, student one actually scored a 70 instead of a 65 — then this correction can be made to the data. Silently dropping an inconvenient point and reporting only the improved line is not a rounding decision; it is a misrepresentation of what the data showed.
For each case, state whether the outlier should be deleted, corrected, or kept, and why.
a. A weight of 1,500 pounds recorded for an adult human.
b. A student's unusually high exam score, verified as correctly recorded.
c. A temperature entered as 87 that the original log shows as 8.7.
Solution — three different right answers
Step 1 — Assess a. No adult human weighs 1,500 pounds, so this is erroneous data. The correct value is not recoverable from the description, so the point should be deleted — and the deletion recorded.
Step 2 — Assess b. The value has been checked and is correct. An outlier may hold valuable information about the population under study, and an unusually high scorer is a real student. Keep it.
Step 3 — Assess c. The correct value is known from the original log, so this should be corrected to 8.7 rather than deleted. Correcting preserves the observation; deleting throws away a real measurement over a typing error.
Step 4 — State the general order of preference. Correct if you can, keep if the value is real, and delete only when the value is wrong and unrecoverable — recording the deletion either way.
Answer: (a) delete and record; (b) keep; (c) correct to 8.7.
11.6.5 How an Outlier Affects the Line
For the third exam/final exam problem, suppose we examined the data and found that this outlier was an error. We delete it and compute a new best-fit line using the ten remaining points:
$$ \hat{y} = -355.19 + 7.39x \quad \text{and} \quad r = 0.9121 $$The new line with \(r = 0.9121\) is a stronger correlation than the original \(r = 0.6631\) because 0.9121 is closer to one. This means the new line is a better fit to the ten remaining data values, and it can better predict the final exam score given the third exam score.
Notice how much moved. The slope went from 4.83 to 7.39 — more than half again as steep — on the removal of one point out of eleven. A single observation was holding down the whole right-hand end of the relationship.
The data points are \((1,5)\), \((2,7)\), \((2,6)\), \((3,9)\), \((4,12)\), \((4,13)\), \((5,18)\), \((6,19)\), \((7,12)\) and \((7,21)\). The line of best fit is \(\hat{y} = 2.897 + 2.269x\) with \(s \approx 3.06\). Identify the outlier, remove it, recalculate the line, and find \(\hat{y}\) at \(x = 10\).
Solution — finding and removing one point
Step 1 — Compute the threshold. \(2s \approx 2(3.06) = 6.12\).
Step 2 — Find the residuals. Evaluating \(y - \hat{y}\) at each point gives approximately \(-0.17\), \(-0.44\), \(-1.44\), \(-0.70\), \(0.03\), \(1.03\), \(3.76\), \(2.49\), \(-6.78\) and \(2.22\).
Step 3 — Flag any beyond the threshold. Only \(|-6.78| = 6.78\) exceeds 6.12. That residual belongs to \((7, 12)\).
Step 4 — Note why it stands out. There are two points at \(x = 7\), with \(y\) values of 12 and 21. The line predicts about 18.8 there, so one of the pair is far below and the other close.
Step 5 — Recalculate without \((7,12)\). The new line of best fit is
$$ \hat{y} = 1.035 + 2.961x \quad \text{with} \quad r \approx 0.9844 $$a considerably stronger correlation than the original \(r \approx 0.8586\).
Step 6 — Predict at \(x = 10\).
$$ \hat{y} = 1.035 + 2.961(10) \approx 30.65 $$Step 7 — Flag the caution. The observed x-values run only from 1 to 7, so \(x = 10\) is extrapolation and the prediction is not justified by these data.
Answer: The outlier is \((7, 12)\); the new line is \(\hat{y} = 1.035 + 2.961x\) with \(r \approx 0.9844\), giving \(\hat{y} \approx 30.65\) at \(x = 10\) — a value obtained by extrapolation.
Using the new line of best fit based on the remaining ten data points, what would a student who receives a 73 on the third exam expect to receive on the final exam? Is this the same as the prediction made using the original line?
Solution — how much one deleted point moved the answer
Step 1 — Predict with the new line.
$$ \hat{y} = -355.19 + 7.39(73) = 184.28 $$A student who scored 73 points on the third exam would expect to earn about 184 points on the final exam.
Step 2 — Recall the original prediction.
$$ \hat{y} = -173.51 + 4.83(73) = 179.08 $$Step 3 — Compare. The prediction using the new line, with the outlier eliminated, differs from the original prediction by about 5 points.
Step 4 — Draw the lesson. Deleting a single point out of eleven moved every prediction the model makes. That is why the decision to delete must be justified by something about the data, and must be reported.
Answer: 184.28 with the new line against 179.08 with the original — the predictions differ.
11.6.6 A Significant r That Should Not Be a Line
The Consumer Price Index (CPI) measures the average change over time in the prices paid by urban consumers for goods and services, and is one of the main measures of inflation. Table 11.6.2 gives the year \(x\) and the CPI \(y\) for 14 years. Calculate the least squares line, find the correlation coefficient and test its significance, estimate the average CPI for 1990, and check for outliers.
| x | y | x | y |
|---|---|---|---|
| 1915 | 10.1 | 1975 | 49.3 |
| 1926 | 17.7 | 1979 | 72.6 |
| 1935 | 13.7 | 1980 | 82.4 |
| 1940 | 14.7 | 1986 | 109.6 |
| 1947 | 24.1 | 1991 | 130.7 |
| 1952 | 26.5 | 1999 | 166.6 |
| 1964 | 31.0 | ||
| 1969 | 36.7 |
Solution — every number checks out, and the model is still wrong
Step 1 — Find the least squares line.
$$ \hat{y} = -3204 + 1.662x $$Step 2 — Find the correlation coefficient. \(r = 0.8694\).
Step 3 — Test its significance. The number of data points is \(n = 14\), so \(n - 2 = 12\) and the corresponding critical value is 0.532. Since \(0.8694 > 0.532\), \(r\) is significant.
Step 4 — Estimate the CPI for 1990.
$$ \hat{y} = -3204 + 1.662(1990) = 103.4 $$Step 5 — Check for outliers. Using LinRegTTest, \(s = 25.4\). Graphing \(Y_{2} = -3204 + 1.662X - 2(25.4)\) and \(Y_{3} = -3204 + 1.662X + 2(25.4)\) shows that no data values fall outside those lines, so there are no outliers. The year 1999 comes very close to the upper line but stays inside it.
Step 6 — Look at the scatterplot anyway. Although the correlation coefficient is significant, the pattern in the scatterplot indicates that a curve would be a more appropriate model than a line. A statistician should prefer other methods to fit a curve to this data rather than model it with the line we found.
Answer: \(\hat{y} = -3204 + 1.662x\) with a significant \(r = 0.8694\), a 1990 estimate of 103.4 CPI, and no outliers — but the scatterplot shows a curved pattern, so a line is not the right model despite every number passing.
Every numerical gate passed. \(r\) was significant, no outliers were flagged, the arithmetic was clean — and a line is still the wrong model, because the points curve. In addition to doing the calculations, it is always important to look at the scatterplot when deciding whether a linear model is appropriate. No amount of significance testing substitutes for that look, and this example exists to prove it.
Adding more recent years — 2010 with a CPI of 219.2, 2020 with 258.0, and 2023 with 299.2 — gives \(\hat{y} = -5030 + 2.598x\) with \(r = 0.9067\). The correlation is higher, but the curvature is more pronounced, not less. A better-looking \(r\) on a curved relationship is not evidence that a line has become appropriate.
Problem Set 11.6
Problem 1. Define an outlier and an influential point, and state the direction each is measured in.
Solution
Step 1 — Define an outlier: An outlier is an observed data point that is far from the least squares line; it has a large "error," where the error (residual) is the vertical distance from the line to the point.
Step 2 — State its direction: An outlier is measured in the vertical direction — distance from the point to the line.
Step 3 — Define an influential point: An influential point is an observed data point that is far from the other observed data points in the horizontal direction. Such a point may have a big effect on the slope of the regression line.
Step 4 — State its direction: An influential point is measured in the horizontal direction — distance from the other data points along the x-axis.
Answer: An outlier is a point far from the least squares line (large residual), measured vertically; an influential point is a point far from the other points horizontally, which may strongly affect the slope.
Problem 2. State the rule of thumb for flagging an outlier, and say which standard deviation is used.
Solution
Step 1 — State the rule: As a rough rule of thumb, flag any point located further than two standard deviations above or below the best-fit line as an outlier. Equivalently, flag any point whose residual satisfies \(|y - \hat{y}| \geq 2s\).
Step 2 — Name the standard deviation used: The standard deviation used is \(s\), the standard deviation of the residuals, computed as
$$ s = \sqrt{\frac{SSE}{n - 2}} $$where SSE is the sum of squared errors and \(n\) is the number of data points.
Answer: Flag any point more than two standard deviations above or below the best-fit line, using \(s\), the standard deviation of the residuals.
Problem 3. A regression has \(s = 6.2\). State the threshold for flagging an outlier.
Solution
Step 1 — Apply the rule of thumb: The threshold is twice the standard deviation of the residuals:
$$ 2s = 2(6.2) = 12.4 $$Step 2 — Interpret it: Any point whose residual is greater than 12.4 or less than \(-12.4\) is flagged as a potential outlier.
Answer: The threshold is \(2s = 12.4\); flag any residual beyond \(\pm 12.4\).
Problem 4. Residuals are 3, \(-14\), 5, 2, \(-1\) and 8, with \(s = 6.0\). Identify any potential outliers.
Solution
Step 1 — Compute the threshold.
$$ 2s = 2(6.0) = 12.0 $$Step 2 — Compare each residual against \(\pm 12.0\). The residuals are 3, \(-14\), 5, 2, \(-1\) and 8. Their absolute values are 3, 14, 5, 2, 1 and 8.
Step 3 — Flag any that exceed 12.0. Only \(|-14| = 14 > 12.0\). All others fall inside the threshold.
Answer: The point with residual \(-14\) is flagged as a potential outlier; all other residuals are within \(2s = 12.0\).
Problem 5. Write the formula for the standard deviation of the residuals, and explain why the denominator is \(n - 2\).
Solution
Step 1 — Write the formula:
$$ s = \sqrt{\frac{SSE}{n - 2}} $$where SSE is the sum of the squared residuals \(\sum (y - \hat{y})^2\) and \(n\) is the number of data points.
Step 2 — Explain the denominator: We divide by \(n - 2\) rather than \(n\) because the regression model involves two estimates: the line has two constants, the intercept \(a\) and the slope \(b\), both computed from this same data. Each estimated parameter uses up one degree of freedom, leaving \(n - 2\) degrees of freedom for estimating the spread of the residuals.
Answer: \(s = \sqrt{SSE/(n-2)}\); the denominator is \(n-2\) because two parameters (\(a\) and \(b\)) are estimated from the data.
Problem 6. A set of 10 residuals gives an SSE of 512. Compute \(s\) and \(2s\).
Solution
Step 1 — Compute \(s\) using \(s = \sqrt{SSE/(n-2)}\) with \(SSE = 512\) and \(n = 10\):
$$ s = \sqrt{\frac{512}{10 - 2}} = \sqrt{\frac{512}{8}} = \sqrt{64} = 8 $$Step 2 — Double it:
$$ 2s = 2(8) = 16 $$Answer: \(s = 8\) and \(2s = 16\).
Problem 7. An outlier is found to be a data-entry error whose correct value is known. State what should be done.
Solution
Step 1 — Identify the situation: The outlier is erroneous data, but the correct value is known.
Step 2 — State the action: Since the correct value is known, the data should be corrected — replace the erroneous entry with the correct value — rather than deleted. Correcting preserves a real observation that was only mistyped or misrecorded.
Step 3 — Note the reporting requirement: Whatever change is made should be recorded, so the analysis remains transparent about how the data were handled.
Answer: Correct the value to the known correct entry (and record the correction); do not delete the point.
Problem 8. An outlier is verified as a correctly recorded, genuine observation. State what should be done and why.
Solution
Step 1 — Identify the situation: The observation has been verified as correctly recorded and genuine.
Step 2 — State the action: The point should be kept in the data set.
Step 3 — Explain why: An outlier may hold valuable information about the population under study. Deleting a real observation simply because it is inconvenient misrepresents what the data showed, and it can substantially change the slope and predictions of the line.
Answer: Keep the point, because it is a genuine, correctly recorded observation that may carry valuable information about the population.
Problem 9. A researcher deletes an outlier and reports only the improved line. Explain what is wrong with that.
Solution
Step 1 — Identify the problem: Deleting an outlier and reporting only the improved line hides the fact that the reported results depend on an arbitrary deletion decision.
Step 2 — Explain why this is wrong: A single deleted point can move the slope dramatically (for example, from 4.83 to 7.39 in the exam-score example), so every prediction the model makes changes. Reporting only the improved line misrepresents what the original data showed.
Step 3 — State the correct practice: The researcher should either record that data was deleted and why, or provide results both with and without the deleted data, so readers can judge the effect for themselves.
Answer: It is wrong because silently dropping an inconvenient point misrepresents the data; deletions must be recorded with reasons, or results must be given both with and without the deleted point.
Problem 10. Removing one point changes a slope from 2.1 to 5.8. State what this tells you about that point.
Solution
Step 1 — Interpret the size of the change: Removing one point moved the slope from 2.1 to 5.8 — nearly tripling it. That is a very large change caused by a single observation out of the whole data set.
Step 2 — Name what the point is: This tells you the point is highly influential: it sits far from the other points in the horizontal direction and exerts strong leverage on where the regression line goes.
Step 3 — Draw the practical conclusion: Such a point deserves close examination before any conclusions are drawn — check whether it is an error, and report results with and without it if it is kept.
Answer: The point is highly influential — its removal drastically changed the slope, so it exerts strong leverage on the regression line and must be examined carefully.
Problem 11. A regression has a significant \(r\), no outliers, and a clearly curved scatterplot. State whether a line is appropriate and why.
Solution
Step 1 — Weigh the evidence: The correlation coefficient is significant and no outliers are flagged, so every numerical test passes.
Step 2 — Recall the key lesson: Numerical tests do not substitute for looking at the scatterplot. In the CPI example, \(r\) was significant and there were no outliers, yet the points followed a curve, so a line was still the wrong model.
Step 3 — State the conclusion: Because the scatterplot shows a clearly curved pattern, a linear model is not appropriate, despite the significant \(r\). A curved model should be fit instead.
Answer: No — a line is not appropriate. Even with significant \(r\) and no outliers, a clearly curved scatterplot means a nonlinear model fits the relationship better.
Problem 12. Explain how a point can be influential without being an outlier.
Solution
Step 1 — Recall the two definitions: An outlier is far from the line vertically (large residual); an influential point is far from the other points horizontally (extreme x-value).
Step 2 — Construct the scenario: Picture a tight cluster of points between \(x = 10\) and \(x = 20\), plus a single point at \(x = 90\) that lies exactly on the regression line.
Step 3 — Check each label: That far-right point has a residual near zero, so it fails the vertical test and is not an outlier. But it is far from every other point in the horizontal direction, so it is influential.
Step 4 — Explain the leverage: Because it sits alone at the far end of the x-range, it holds up that end of the line almost single-handedly; moving it even slightly swings the slope considerably. Residual-based tests cannot detect this, which is why horizontal position is checked separately.
Answer: A point can be influential without being an outlier when it has an extreme x-value (far from the other points horizontally) but a small residual because it happens to lie close to the line — such a point strongly affects the slope while never being flagged by the vertical residual test.
Key Terms
outlier — an observed data point far from the least squares line, with a large residual; measured in the VERTICAL direction.
influential point — an observed data point far from the other points in the HORIZONTAL direction; it may strongly affect the slope while having a small residual.
rule of thumb for outliers — flag any point further than two standard deviations above or below the best-fit line, using the standard deviation of the residuals.
standard deviation of the residuals (\(s\)) — \(s = \sqrt{\frac{SSE}{n-2}}\); the denominator is \(n - 2\) because the regression model involves two estimates, \(a\) and \(b\).
potential outlier — a point whose \(|y - \hat{y}|\) is at least \(2s\); "potential" because the rule is a guideline and the point still requires examination.
recording a deletion — the requirement that a deleted outlier be reported, with the reason, or that results be given both with and without it.