High-Leverage Point vs Outlier

Leverage is a property of the explanatory variable alone. For point $i$ it is $h_i = \frac{1}{n} + \frac{(x_i - \bar{x})^2}{\sum (x_j - \bar{x})^2}$, where $\bar{x}$ is read x-bar, so you can work it out before looking at a single response value. It is at least $\frac{1}{n}$ and at most 1, and a point sitting at the mean of $x$ carries the least leverage available to it.

Five points, $(1, 3)$, $(2, 5)$, $(3, 4)$, $(4, 7)$ and $(5, 8)$, fit $\hat{y} = 1.8 + 1.2x$. Now add a sixth at $x = 20$. Whatever its response, its leverage is 0.967, against values of about 0.17 to 0.26 for the other five. Give it a value on the pattern, $(20, 26)$, and the line barely notices: $\hat{y} = 1.77 + 1.21x$. Give it $(20, 10)$ instead and the slope collapses to $\hat{y} = 4.37 + 0.31x$.

Now the trap. "If a point were distorting the fit, the residual plot would show it." In that second fit, the residual at $x = 20$ is -0.52, the second smallest of the six in size, because the line swung far enough to almost meet the point. The damage shows up on the other five, whose residuals are now -1.68, 0.01, -1.30, 1.40 and 2.09, against 0, 0.80, -1.40, 0.40 and 0.20 under the original fit. A high-leverage point that has already bent the line hides itself behind a small residual.

Leverage is a warning, not a verdict. Both sixth points above have identical leverage, 0.967, and one of them changes almost nothing. Whether a point actually moves the fit is influence, and you settle that by refitting without it and comparing the slope, the intercept and $r$. So keep the three labels apart: an outlier has a large residual, a high-leverage point has an extreme $x$, and an influential point is one whose removal visibly changes the line.

An outlier is a value that stands apart from the rest of the data, and the AP course gives two rules for deciding when apart counts. The first flags any value more than $1.5 \times \text{IQR}$ above the third quartile $Q_3$ or below the first quartile $Q_1$. The second flags any value more than 2 standard deviations from the mean. Neither rule is the definition of an outlier. They are criteria, and they do not always agree.

Take 12, 13, 19, 19, 20, 23, 25, 27, 35, 40. The quartiles are $Q_1 = 19$ and $Q_3 = 27$, so the IQR is 8 and the upper fence sits at $27 + 1.5(8) = 39$. The value 40 clears it, so the IQR rule calls 40 an outlier. Now the other rule: the mean is 23.3 and $s = 8.91$, so two standard deviations reach $23.3 + 2(8.91) = 41.12$, and 40 does not get there. Same data, two rules, opposite answers. Say which rule you used.

"There is an outlier, so it must be a mistake and I should drop it." An outlier is a statement about position inside this one data set and nothing more. A 40 from a broken sensor is an error; a 40 from an unusually productive week is the most informative point you have. Removing a flagged value is a decision about the study that needs a reason you can write down, not an arithmetic step.

The 2 standard deviation rule quietly fails on small samples. No observation among $n$ values can sit more than $\frac{n-1}{\sqrt{n}}$ sample standard deviations from the mean, because the value doing the stretching also inflates $s$. With $n = 5$ that ceiling is 1.79, so the rule flags nothing at all however extreme one of the five values is. The 1.5 IQR rule has no such blind spot.

Outliers sit in topic 1.7 of the Fall 2026 course, which is where both rules are listed, and topic 1.6 names outliers, gaps, and clusters as the unusual features every description of a distribution has to mention in context.