Learning Outcomes
- Identify outliers graphically from a given scatterplot
In some data sets, there are values (observed data points) called outliers. 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.
Outliers need to be examined closely. Sometimes, for some 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 in the data. The key is to examine carefully what causes a data point to be an outlier.
The following video gives an introduction to the idea of an outlier in a set of data.
The IQR can help to determine potential outliers. A value is suspected to be a potential outlier if it is less than (1.5)(IQR) below the first quartile or more than (1.5)(IQR) above the third quartile. Potential outliers always require further investigation.
Besides outliers, a sample may contain one or a few points that are called influential points. 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. 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. Computer output for regression analysis will often identify both outliers and influential points so that you can examine them.
Identifying Outliers
We could guess at outliers by looking at a graph of the scatterplot and best fit-line. However, we would like some guidelines 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 1
In the third exam/final exam example (example 2), 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.
try it 1
Identify the potential outlier in the scatter plot. The standard deviation of the residuals or errors is approximately 8.6.
In the table below, the first two columns are the third exam and final exam data. The third column shows the predicted [latex]\hat{y}[/latex] values calculated from the line of best fit: [latex]\hat{y}[/latex] = –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 − [latex]\hat{y}[/latex].
s is the standard deviation of all the y − [latex]\hat{y}[/latex] = 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:
[latex]s = {\sqrt{\dfrac{SSE}{n - 2}}}[/latex]
Note
We divide by (n – 2) because the regression model involves two estimates.
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.
| x | y | [latex]\hat{y}[/latex] | y – [latex]\hat{y}[/latex] |
| 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. 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.
