12.6 Outliers

Identifying Outliers

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. As a rough rule of thumb, we can flag any point that is 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 or errors.

We can do this visually in the scatter plot by drawing an extra pair of lines that are two standard deviations above and below the best-fit line. Any data points that are outside this extra pair of lines are flagged as potential outliers. Or we can do this 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. The graphical procedure is shown first, followed by the numerical calculations. You would generally need to use only one of these methods.

Example 12.12

Problem

In the third exam/final exam example, you can determine if there is an outlier or not. If there is an outlier, as an exercise, delete it and fit the remaining data to a new line. For this example, the new line ought to fit the remaining data better. This means the SSE should be smaller and the correlation coefficient ought to be closer to 1 or –1.

Solution

Graphical Identification of Outliers

With the TI-83, 83+, 84+ graphing calculators, it is easy to identify the outliers graphically and visually. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance were equal to 2s or more, then we would consider the data point to be "too far" from the line of best fit. We need to find and graph the lines that are two standard deviations below and above the regression line. Any points that are outside these two lines are outliers. We will call these lines Y2 and Y3:

As we did with the equation of the regression line and the correlation coefficient, we will use technology to calculate this standard deviation for us. Using the LinRegTTest with this data, scroll down through the output screens to find s = 16.412.

Line Y2 = –173.5 + 4.83x –2(16.4) and line Y3 = –173.5 + 4.83x + 2(16.4)

where ŷ = –173.5 + 4.83x is the line of best fit. Y2 and Y3 have the same slope as the line of best fit.

Graph the scatterplot with the best fit line in equation Y1, then enter the two extra lines as Y2 and Y3 in the "Y="equation editor and press ZOOM 9. You will find that the only data point that is not between lines Y2 and Y3 is the point x = 65, y = 175. On the calculator screen it is just barely outside these lines. The outlier is the student who had a grade of 65 on the third exam and 175 on the final exam; this point is further than two standard deviations away from the best-fit line.

Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult to tell if the point is between or outside the lines. On a computer, enlarging the graph may help; on a small calculator screen, zooming in may make the graph clearer. Note that when the graph does not give a clear enough picture, you can use the numerical comparisons to identify outliers.

Figure 12.18

Try It 12.12

Identify the potential outlier in the scatter plot. The standard deviation of the residuals or errors is approximately 8.6.

Figure 12.19

Numerical Identification of Outliers

In Table 12.5, the first two columns are the third-exam and final-exam data. The third column shows the predicted ŷ values calculated from the line of best fit: ŷ = –173.5 + 4.83x. The residuals, or errors, have been calculated in the fourth column of the table: observed y value−predicted y value = y − ŷ.

s is the standard deviation of all the y − ŷ = ε values where n = the total number of data points. If each residual is calculated and squared, and the results are added, we get the SSE. The standard deviation of the residuals is calculated from the SSE as:

$s=\sqrt{\frac{SSE}{n-2}}$

Rather than calculate the value of s ourselves, we can find s using the computer or calculator. For this example, the calculator function LinRegTTest found s = 16.4 as the standard deviation of the residuals 35; –17; 16; –6; –19; 9; 3; –1; –10; –9; –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. Compare these values to the residuals in column four of the table. 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.

How does the outlier affect the best fit line?

Numerically and graphically, we have identified the point (65, 175) as an outlier. We should re-examine the data for this point to see if there are any problems with the data. 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. For this problem, we will suppose that we examined the data and found that this outlier data was an error. Therefore we will continue on and delete the outlier, so that we can explore how it affects the results, as a learning experience.

Compute a new best-fit line and correlation coefficient using the ten remaining points: On the TI-83, TI-83+, TI-84+ calculators, delete the outlier from L1 and L2. Using the LinRegTTest, the new line of best fit and the correlation coefficient are:

ŷ = –355.19 + 7.39x and r = 0.9121

The new line with r = 0.9121 is a stronger correlation than the original (r = 0.6631) because r = 0.9121 is closer to one. This means that the new line is a better fit to the ten remaining data values. The line can better predict the final exam score given the third exam score.

Numerical Identification of Outliers: Calculating s and Finding Outliers Manually

If you do not have the function LinRegTTest, then you can calculate the outlier in the first example by doing the following.

First, square each |y – ŷ|

The squares are 35²; 17²; 16²; 6²; 19²; 9²; 3²; 1²; 10²; 9²; 1²

Then, add (sum) all the |y – ŷ| squared terms using the formula

$\stackrel{11}{\underset{i = 1}{\Sigma }}{\left(|y_i-{\widehat{y}}_i|\right)}^2=\stackrel{11}{\underset{i = 1}{\Sigma }}{\epsilon }_i{}^2$ (Recall that y_i – ŷ_i = ε_i.)

= 35² + 17² + 16² + 6² + 19² + 9² + 3² + 1² + 10² + 9² + 1²

= 2440 = SSE. The result, SSE is the Sum of Squared Errors.

Next, calculate s, the standard deviation of all the y – ŷ = ε values where n = the total number of data points.

The calculation is $s=\sqrt{\frac{\text{SSE}}{n–2}}$.

For the third exam/final exam problem, $s=\sqrt{\frac{2440}{11–2}}=16.47$.

Next, multiply s by 2:
(2)(16.47) = 32.94
32.94 is 2 standard deviations away from the mean of the y – ŷ values.

If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance is at least 2s, then we would consider the data point to be "too far" from the line of best fit. We call that point a potential outlier.

For the example, if any of the |y – ŷ| values are at least 32.94, the corresponding (x, y) data point is a potential outlier.

For the third exam/final exam problem, all the |y – ŷ|'s are less than 31.29 except for the first one which is 35.

35 > 31.29 That is, |y – ŷ| ≥ (2)(s)

The point which corresponds to |y – ŷ| = 35 is (65, 175). Therefore, the data point (65,175) is a potential outlier. For this example, we will delete it. (Remember, we do not always delete an outlier.)

The next step is to compute a new best-fit line using the ten remaining points. The new line of best fit and the correlation coefficient are:

ŷ = –355.19 + 7.39x and r = 0.9121

Example 12.14

The Consumer Price Index (CPI) measures the average change over time in the prices paid by urban consumers for consumer goods and services. The CPI affects nearly all Americans because of the many ways it is used. One of its biggest uses is as a measure of inflation. By providing information about price changes in the Nation's economy to government, business, and labor, the CPI helps them to make economic decisions. The President, Congress, and the Federal Reserve Board use the CPI's trends to formulate monetary and fiscal policies. In the following table, x is the year and y is the CPI.

x	y	x	y
1915	10.1	1969	36.7
1926	17.7	1975	49.3
1935	13.7	1979	72.6
1940	14.7	1980	82.4
1947	24.1	1986	109.6
1952	26.5	1991	130.7
1964	31.0	1999	166.6

Table 12.6 Data

Problem

1. Draw a scatterplot of the data.
2. Calculate the least squares line. Write the equation in the form ŷ = a + bx.
3. Draw the line on the scatterplot.
4. Find the correlation coefficient. Is it significant?
5. What is the average CPI for the year 1990?

Solution

1. See Figure 12.20.
2. ŷ = –3204 + 1.662x is the equation of the line of best fit.
3. r = 0.8694
4. The number of data points is n = 14. Use the 95% Critical Values of the Sample Correlation Coefficient table at the end of Chapter 12. n – 2 = 12. The corresponding critical value is 0.532. Since 0.8694 > 0.532, r is significant.
   ŷ = –3204 + 1.662(1990) = 103.4 CPI
5. Using the calculator LinRegTTest, we find that s = 25.4 ; graphing the lines Y2 = –3204 + 1.662X – 2(25.4) and Y3 = –3204 + 1.662X + 2(25.4) shows that no data values are outside those lines, identifying no outliers. (Note that the year 1999 was very close to the upper line, but still inside it.)

Figure 12.20

Note

In the example, notice the pattern of the points compared to the line. Although the correlation coefficient is significant, the pattern in the scatterplot indicates that a curve would be a more appropriate model to use than a line. In this example, a statistician should prefer to use other methods to fit a curve to this data, rather than model the data with the line we found. In addition to doing the calculations, it is always important to look at the scatterplot when deciding whether a linear model is appropriate.

If you are interested in seeing more years of data, visit the Bureau of Labor Statistics CPI website ftp://ftp.bls.gov/pub/special.requests/cpi/cpiai.txt; our data is taken from the column entitled "Annual Avg." (third column from the right). For example you could add more current years of data. Try adding the more recent years: 2004: CPI = 188.9; 2008: CPI = 215.3; 2011: CPI = 224.9. See how it affects the model. (Check: ŷ = –4436 + 2.295x; r = 0.9018. Is r significant? Is the fit better with the addition of the new points?)

95% Critical Values of the Sample Correlation Coefficient Table