Topic 2.9 Notes – Analyzing Departures from Linearity

Influential Points in Regression

The least-squares regression line (LSRL) is built to minimize squared residuals. Because of that, certain points can pull the line around more than others.

An influential point is any observation that, if removed, would noticeably change:

• the slope
• the y-intercept
• the correlation $r$ (and therefore $r^2$)

There are two main types you need to recognize.

Outliers in Regression

An outlier is a point that does not follow the overall linear trend and has a large residual (big vertical distance from the LSRL).

You spot it on a residual plot as a point far above or below 0.

Effects:

• Often reduces $r$
• Increases scatter in residuals
• May shift the y-intercept
• Sometimes changes slope, but not always

Important: A point can be an outlier without being influential if removing it barely changes the line.

High-Leverage Points

A high-leverage point has an unusually large or small x-value compared to the rest.

You see these on the scatterplot, far left or right.

Effects:

• Pulls the regression line toward itself
• Often changes the slope
• Can increase or decrease $r$

Not all high-leverage points are bad. If they follow the pattern, they can actually strengthen the model.

How They Relate

A point can be:

• Outlier only
• High leverage only
• Both

The most dangerous situation is high leverage + large residual. That combination is very likely to be influential.

Look at the scatterplot below. The cluster of points follows a clear positive linear trend, but there is one point far to the right and well above the pattern. That single point has high leverage because its x-value is extreme, and it has a large residual because it sits far above the original line.

High leverage point with large residual changing the LSRL

Notice how the red regression line that includes the extreme point is much steeper than the dashed line that ignores it. That big change in slope is exactly what makes the point influential.

If removing a point causes a big change in slope or correlation, you call it influential.

When you write about it, be specific:

“This point is a high-leverage point because its x-value is much larger than the others. Removing it would decrease the slope substantially, so it is influential.”

Vague answers like “it changes the graph” don’t earn credit.

When the Relationship Is Not Linear

Sometimes the scatterplot curves and the residual plot shows a pattern.

If the residual plot has a curved shape, the linear model is not appropriate.

Exponential data: the straight line misses the curve, the residual plot shows a U shape, and after taking $\ln(y)$ the residuals look random

In panel 1, the LSRL underpredicts at both ends and overpredicts in the middle. That is why the residuals in panel 2 are positive, then negative, then positive again: a clear U shape, so a linear model is not appropriate. Panel 3 shows the residuals after taking $\ln(y)$: no pattern, just random scatter around 0.

A good residual plot should look like random scatter around 0. If you see structure, your model is missing something.

This is where transformations come in.

A transformation applies a function (often natural log) to x and/or y to create a new, more linear dataset.

Evidence a transformation helped:

• Residual plot becomes more random
• $r^2$ moves closer to 1

You’re not chasing a bigger $r^2$ just for fun. You’re trying to justify a better predictive model.

Exponential and Power Models

These are the most common nonlinear models on AP Stats, though other transformations (like squaring x) can also appear.

Exponential Models

Form:

$$\hat{y} = ab^x$$

Used for growth or decay.

To linearize, take the natural log of y:

$$\ln(\hat{y}) = \ln(a) + \ln(b)x$$

Now you run linear regression of $\ln(y)$ on $x$.

From the regression output:

• Intercept = $\ln(a)$
• Slope = $\ln(b)$

Convert back:

• $a = e^{\text{intercept}}$
• $b = e^{\text{slope}}$

For prediction:

1. Plug in x.
2. Compute predicted $\ln(y)$.
3. Exponentiate to return to original units.

Students lose easy points by forgetting step 3.

Power Models

Form:

$$\hat{y} = ax^b$$

Linearize by taking logs of both variables:

$$\ln(\hat{y}) = \ln(a) + b\ln(x)$$

Now run regression of $\ln(y)$ on $\ln(x)$.

From output:

• Intercept = $\ln(a)$
• Slope = $b$

Convert back:

• $a = e^{\text{intercept}}$

Key difference to remember:

• Exponential → log y only
• Power → log both x and y