Effect of Unusual Points on Correlation

How an Unusual Point Can Change Correlation

In “Effect of Unusual Points on Slope,” you used a point’s position relative to the existing mean \(x\) and regression line to predict a slope change. Correlation asks a related but different question: does adding the point make the linear association stronger or weaker, and does it change the association’s direction?

The correlation coefficient \(r\) describes the direction and strength of a linear association. Its sign gives the direction: positive \(r\) means the variables tend to increase together, while negative \(r\) means one tends to decrease as the other increases. The magnitude \(|r|\) describes strength: values closer to 1 or \(-1\) indicate a stronger linear association, and values closer to 0 indicate a weaker one.

Key distinction: “\(r\) increases” means its numerical value moves upward. “The correlation gets stronger” means \(|r|\) increases. For a negative association, \(r\) can increase numerically while the association gets weaker: for example, moving from \(-0.8\) to \(-0.4\) is an increase in \(r\), but a decrease in strength.

A point may strengthen a correlation when it extends the existing pattern: for example, a point far to the right and relatively high in a positive association, or far to the right and relatively low in a negative association. A point that is far from the trend in the response direction can weaken the correlation. These are useful predictions, not rules that every point’s quadrant alone settles. How unusual the point is in both variables matters.

A Formula to Check the Effect

For the existing data, let \(S_{xx}\) be the sum of squared deviations of the \(x\)-values from \(\bar{x}\), \(S_{yy}\) the corresponding sum for the \(y\)-values from \(\bar{y}\), and \(S_{xy}\) the sum of the products of the paired deviations. Then \(r=S_{xy}/\sqrt{S_{xx}S_{yy}}\). Suppose there are \(n\) existing observations and a new point \((x_0,y_0)\) is added. Define its displacements from the existing means as \(d_x=x_0-\bar{x}\) and \(d_y=y_0-\bar{y}\).

The update formulas below show how the point adds to the three sums. The factor \(n/(n+1)\) accounts for the fact that the means change when the point is added. Once the updated sums are found, calculate the new correlation using the same formula for \(r\).

S_{xx,\mathrm{new}}=S_{xx}+\frac{n}{n+1}d_x^2, \qquad S_{yy,\mathrm{new}}=S_{yy}+\frac{n}{n+1}d_y^2, \qquad S_{xy,\mathrm{new}}=S_{xy}+\frac{n}{n+1}d_xd_y

r_{\mathrm{new}} = \frac{S_{xy,\mathrm{new}}} {\sqrt{S_{xx,\mathrm{new}}S_{yy,\mathrm{new}}}}

This makes an important difference from the slope sign rule. The product \(d_xd_y\) tells you whether the point adds positive or negative co-movement, but it does not, on its own, tell you whether \(|r|\) gets larger. The point also changes \(S_{xx}\) and \(S_{yy}\), the amounts of variation in each variable. Those changes affect the denominator of the correlation formula.

For a quick visual prediction, ask whether the new point extends the existing trend or adds substantial variation away from it. If the answer is unclear, the updated sums give a direct check. With technology, entering the data with and without the point into a regression or correlation calculation will also provide both \(r\)-values. Always compare their magnitudes as well as their signs.

Worked Examples: Predicting Changes in \(r\)

Worked Example: A Point Extending a Positive Trend

Original AP-style question. In a fictional greenhouse trial, \(x\) is the number of hours a lamp was used and \(y\) is a seedling’s height in centimeters. Four seedlings have \((2,5)\), \((4,7)\), \((6,8)\), and \((8,11)\). Predict how adding \((14,17)\) affects the correlation.

State. The existing association is positive. We will calculate \(r\) before and after adding the point, then compare \(|r|\) to determine whether the association becomes stronger or weaker.

Plan. Calculate the existing means and deviation sums, then use the point’s displacements from those means to update the sums. No inference procedure is involved: this is a comparison of descriptive correlations for two versions of the data.

Do. The means are \(\bar{x}=5\) and \(\bar{y}=7.75\). The \(x\)-deviations are \(-3,-1,1,3\), so \(S_{xx}=9+1+1+9=20\). The \(y\)-deviations are \(-2.75,-0.75,0.25,3.25\), so \(S_{yy}=7.5625+0.5625+0.0625+10.5625=18.75\). The sum of paired deviation products is \(S_{xy}=8.25+0.75+0.25+9.75=19\). Therefore,

r=\frac{19}{\sqrt{20(18.75)}}=\frac{19}{\sqrt{375}}\approx0.9812

The added point has \(d_x=14-5=9\) and \(d_y=17-7.75=9.25\). Since \(n=4\), the update factor is \(4/5=0.8\). Thus,

S_{xx,\mathrm{new}}=20+0.8(9^2)=84.8, \qquad S_{yy,\mathrm{new}}=18.75+0.8(9.25^2)=87.2, \qquad S_{xy,\mathrm{new}}=19+0.8(9)(9.25)=85.6

The new correlation is

r_{\mathrm{new}}=\frac{85.6}{\sqrt{84.8(87.2)}}= \frac{85.6}{\sqrt{7394.56}}\approx0.9954

Conclude. The correlation increases from about \(0.9812\) to \(0.9954\), and its magnitude increases as well. The added point is far above and to the right of the existing means, in the direction of the positive pattern, so it strengthens the linear association in this fictional data set.

Worked Example: An Unusual Response Weakening the Correlation

Original AP-style question. A fictional school project records \(x\), weekly study hours, and \(y\), a quiz score in points. Four students have \((1,2)\), \((3,4)\), \((5,5)\), and \((7,9)\). A fifth student with \((4,-5)\) is added. What happens to the correlation?

State and plan. The existing association is positive. We will calculate both correlations. In particular, the added point’s \(x\)-value is at the existing mean, while its response is unusually low; the updated sums will show how that affects the strength.

Do. The existing means are \(\bar{x}=4\) and \(\bar{y}=5\). The deviations give \(S_{xx}=(-3)^2+(-1)^2+1^2+3^2=20\), \(S_{yy}=(-3)^2+(-1)^2+0^2+4^2=26\), and \(S_{xy}=9+1+0+12=22\). Thus,

r=\frac{22}{\sqrt{20(26)}}=\frac{22}{\sqrt{520}}\approx0.9648

For the added point, \(d_x=4-4=0\) and \(d_y=-5-5=-10\). With update factor \(0.8\), the sums become

S_{xx,\mathrm{new}}=20+0.8(0^2)=20, \qquad S_{yy,\mathrm{new}}=26+0.8((-10)^2)=106, \qquad S_{xy,\mathrm{new}}=22+0.8(0)(-10)=22

Therefore,

r_{\mathrm{new}}=\frac{22}{\sqrt{20(106)}}= \frac{22}{\sqrt{2120}}\approx0.4778

Conclude. The correlation remains positive, but its magnitude falls from about \(0.9648\) to \(0.4778\). Since the point’s \(x\)-value equals the existing mean, it adds no \(x\)-variation or cross-product contribution. Its very low response adds substantial \(y\)-variation without adding co-movement, weakening the linear association.

Worked Example: A High-Leverage Point Reversing the Association

Original AP-style question. In a fictional delivery study, \(x\) is a route’s scheduled length in kilometers and \(y\) is the number of parcels delivered on time. Four routes have \((1,2)\), \((2,5)\), \((3,4)\), and \((4,7)\). A route with \((8,-3)\) is added. How does \(r\) change?

State. The original association is positive. We will check whether the added point changes only the strength or also reverses the direction.

Plan. Find the original deviation sums and correlation, then update all three sums. The new point is far to the right and far below the existing trend, so it may add substantial variation in a direction contrary to the original association.

Do. The existing means are \(\bar{x}=2.5\) and \(\bar{y}=4.5\). The deviation sums are \(S_{xx}=5\), \(S_{yy}=13\), and \(S_{xy}=7\). Thus,

r=\frac{7}{\sqrt{5(13)}}=\frac{7}{\sqrt{65}}\approx0.8682

For the added point, \(d_x=8-2.5=5.5\) and \(d_y=-3-4.5=-7.5\). Applying the factor \(0.8\) gives

S_{xx,\mathrm{new}}=5+0.8(5.5^2)=29.2, \qquad S_{yy,\mathrm{new}}=13+0.8((-7.5)^2)=58, \qquad S_{xy,\mathrm{new}}=7+0.8(5.5)(-7.5)=-26

The new correlation is

r_{\mathrm{new}}=\frac{-26}{\sqrt{29.2(58)}}= \frac{-26}{\sqrt{1693.6}}\approx-0.6318

Conclude. The correlation changes from about \(0.8682\) to \(-0.6318\): its direction reverses, and its magnitude is smaller. The added route has a high-leverage \(x\)-value and a response far below the existing positive trend. It contributes a large negative cross-product, changing the sign of \(S_{xy}\) and therefore of \(r\).

What to Check When Removing a Point

To judge a point’s effect on correlation, compare the full data with the data after removing just that point. Calculate or obtain \(r\) for both sets, then compare the signs and magnitudes. A point that strengthens the association when added to the reduced data will, in that same comparison, have strengthened the full-data association. But the numerical change in \(r\) is not a simple slope-style increase or decrease rule: \(r\) depends on all three sums and their ratio.

This is one way to assess whether a point is influential, as discussed in “Influential Points and Their Effect on the Line” and “Testing Influence by Removing a Point.” A large change in \(r\) is evidence that the point affects the measured linear association; whether that change matters depends on the context and the purpose of the analysis.

Common Mistakes and AP Exam Tip

Confusing an increase in \(r\) with a stronger association. For negative \(r\), compare magnitudes. Moving from \(-0.7\) to \(-0.3\) is a numerical increase but a weakening association.
Using only the sign of \(d_xd_y\). That product shows the sign of the added cross-product contribution, not the final strength. The point also changes \(S_{xx}\) and \(S_{yy}\).
Assuming every point in the “right” quadrant strengthens a positive association. Distance matters. A point may be far from the trend, and its contribution to both variation and co-movement must be considered.
Calling a point influential based only on its unusual appearance. Influence is assessed by comparing the results with and without the point. The change in \(r\) may be small or large.
Reporting a number without interpreting it. State the original and new correlations, their directions, and whether the magnitude increased or decreased. In a contextual answer, identify which observations are being compared.

A strong AP response might say: “Adding the route with unusually high scheduled length and unusually low on-time deliveries changes \(r\) from a positive value to a negative value. Its magnitude also decreases, so the linear association becomes weaker and reverses direction.” This describes the data without claiming that route length causes the number of timely deliveries.

Key takeaway: An unusual point can strengthen, weaken, or reverse a correlation. Look for whether it extends the existing linear pattern, but verify uncertain cases by comparing \(r\) and \(|r|\) before and after adding or removing the point. The sign of \(r\) gives direction; its magnitude gives strength.

Check Your Understanding

Answer each question using the difference between correlation direction and strength.

If \(r\) changes from \(0.60\) to \(0.82\), does the association get stronger or weaker? Explain using magnitudes.
If \(r\) changes from \(-0.75\) to \(-0.40\), did \(r\) increase numerically? Did the association get stronger or weaker?
Why can an added point at the existing mean \(x\) weaken a correlation if its response is far from the existing mean \(y\)?
What three deviation sums are updated when a point is added to calculate the new correlation?
When checking a point’s influence on \(r\), what two results should be compared?